Commutative Algebra

Home , Integrally closed domain, Monoid ring, Radical of an ideal

Commutative Algebra

D. Zack Garza

Sunday 10th May, 2020

Contents

1 Preface 7

2 Wednesday January 87

3 Monday January 139 3.1 Logistics ...... 9 3.2 Rings of Functions ...... 9 3.3 Rings ...... 11

4 Wednesday January 15th 11 4.1 Ideals and Quotients ...... 12

5 Friday January 17th 13

6 Wednesday January 22nd 15 6.1 Pushing / Pulling ...... 16

7 Friday January 24th 18 7.1 Ideals and Products ...... 18 7.2 Modules ...... 20

8 Monday January 27th 21 8.1 Localization ...... 21 8.2 Modules ...... 22

9 Wednesday January 29th 23 9.1 Universal Mapping Properties ...... 24 9.2 Free Modules ...... 24

10 Friday January 31st 26 10.1 Tensor and Hom ...... 26 10.2 Free and Torsion Modules ...... 27

11 Monday February 3rd 29 11.1 Noetherian and Artinian Modules ...... 29 11.2 Tensor Products ...... 30

1 Contents

12 Wednesday February 5th 32

13 Friday February 7th 34 13.1 Projective Modules ...... 34

14 Wednesday February 12th 37 14.1 Projective Modules and Ideals ...... 37 14.2 The Picard Group ...... 39

15 Friday February 14th 40 15.1 Flat Modules ...... 42

16 Monday February 17th 45 16.1 Injective Modules ...... 46

17 Monday February 24th 49 17.1 Divisible Modules ...... 49 17.2 Toward Localization ...... 50 17.3 Radicals ...... 51

18 Wednesday February 26th 52 18.1 Radicals ...... 52

19 Friday February 28th 55 19.1 Radicals: The Jacobson Radical ...... 55 19.2 Proposition (Commutative Algebra Analog of Euclid IX.20: Inﬁnitely Many Primes) 55 19.3 Monoid Rings ...... 57

20 Monday March 2nd 58 20.1 Semigroup and Monoid Rings ...... 58 20.2 Localization ...... 59

21 Wednesday March 4th 60 21.1 Pushing and Pulling ...... 61 21.2 Localization for Modules ...... 63

22 Monday March 30th 63

23 Wednesday April 1st 65 23.1 Chapter 8: Noetherian Rings ...... 67

24 Friday April 3rd 68

25 Monday April 6th 69

26 Wednesday April 8th 71

27 Friday April 10th 74 27.1 Boolean Rings ...... 76

Contents 2 List of Deﬁnitions

27.2 Stone Duality ...... 76 27.2.1 Statement of Stone Duality ...... 77

28 Monday April 13th 77 28.1 Nullstellansatz ...... 77

29 Wednesday April 15th 79

30 Monday April 20th 81

31 Wednesday April 22nd 84 31.1 Integral Extensions ...... 85

32 Monday April 27th 87 32.1 Normalization ...... 87 32.2 Invariant Theory ...... 87 32.3 Normalization Theorems ...... 88

List of Deﬁnitions

2.1.1 Definition – Boolean Spaces ...... 8 2.1.2 Definition – Boolean Rings ...... 8 4.1.1 Definition – Ring Morphisms ...... 12 5.0.1 Definition – Proper Ideals ...... 13 5.0.2 Definition – Lattice Structure of Ideals ...... 13 5.1.1 Definition – Product of Ideals ...... 14 6.0.1 Definition – Reduced Monoids ...... 15 6.0.2 Definition – Zero Elements in Monoids ...... 16 6.0.3 Definition – Prime Ideals ...... 17 6.0.4 Definition – Maximal Ideals ...... 17 6.0.5 Definition – Max Spectrum ...... 18 7.3.1 Definition – Connected Rings ...... 20 7.3.2 Definition – Faithful Modules ...... 21 9.0.1 Definition – Direct Product of Modules ...... 24 9.0.2 Definition – Direct Sum of Modules ...... 24 9.0.3 Definition – Spanning, Linear Independence, and Basis ...... 24 10.0.1 Definition – Reflexive Modules ...... 27 10.0.2 Definition – Torsion Submodule ...... 27 10.0.3 Definition – Torsion and Torsionfree Modules ...... 27 10.2.1 Definition – Divisible and Uniquely Divisible Modules ...... 28 11.0.1 Definition – Noetherian Posets ...... 29 11.0.2 Definition – Artinian Posets ...... 29 11.0.3 Definition – Order Dual ...... 29 11.5.1 Definition – Tensor Product ...... 31 12.0.1 Definition – Invariant Basis Number ...... 33 12.0.2 Definition – Rank Condition ...... 33 12.0.3 Definition – Strong Rank Condition ...... 33 12.1.1 Definition – Noetherian and Artinian Modules ...... 33

List of Deﬁnitions 3 List of Theorems

13.1.1 Definition – Splitting an Exact Sequence ...... 35 13.1.2 Definition – Projective Module ...... 35 14.2.1 Definition – Picard Group ...... 39 14.2.2 Definition – Dedekind Domain and Class Group ...... 40 15.0.1 Definition – Reduced Group ...... 41 15.1.1 Definition – Local Ring ...... 42 15.4.1 Definition – Flat Modules ...... 43 15.9.1 Definition – Non-generator ...... 45 16.3.1 Definition – Simple Modules ...... 46 16.3.2 Definition – Injective Modules ...... 46 16.4.1 Definition – Semisimple Rings ...... 47 16.7.1 Definition – Injective Modules ...... 47 16.8.1 Definition – Self-Injective Modules ...... 48 17.2.1 Definition – Nilpotent Elements ...... 51 17.4.1 Definition – Nil ...... 51 17.4.2 Definition – Nilradical ...... 51 17.5.1 Definition – Radical Ideals ...... 52 18.0.1 Definition – Radical of an Ideal ...... 52 18.0.2 Definition – Radical Ideals ...... 52 18.0.3 Definition – Closure Operators ...... 53 18.1.1 Definition – Jacobson Radical ...... 54 19.0.1 Definition – The Jacobson Radical ...... 55 19.0.2 Definition – Semiprimitive ...... 55 21.0.1 Definition – Saturated Multiplicatively Closed Sets ...... 61 22.2.1 Definition – Fractional Ideal ...... 64 22.4.1 Definition – Rank Function ...... 65 24.1.1 Definition – Length ...... 68 27.4.1 Definition – Boolean ...... 76 28.0.1 Definition – Incidence Relation ...... 78 29.0.1 Definition – Jacobson ...... 80 30.6.1 Definition – Irreducible ...... 83 31.0.1 Definition – Noetherian Space ...... 84 31.6.1 Definition – Integral Elements of Rings ...... 85 32.1.1 Definition – Regular Prime Ideals ...... 87 32.3.1 Definition – Ring of Invariants ...... 87

List of Theorems

2.1 Theorem – Swan ...... 8 3.1 Theorem – Function Spaces Can Have Large Unbounded Chains ...... 10 4.1 Theorem – Cayley ...... 11 4.2 Theorem – Every Ideal Determines a Quotient Ring ...... 13 5.1 Theorem – Lattice Isomorphism Theorem for Rings ...... 14 7.2 Proposition – Ideals of Products are Products of Ideals ...... 18 7.3 Theorem – Chinese Remainder ...... 19 8.1 Theorem – Negata ...... 22 9.1 Theorem – Free Modules are Quotients ...... 25

List of Theorems 4 List of Theorems

9.2 Proposition – Characterization of Freeness in terms of Rings ...... 25 10.1 Proposition – Torsionfree implies submodule of f.g free ...... 28 10.2 Proposition – Implication Chain ...... 28 11.1 Proposition – Noetherian/Artinian Duality ...... 29 11.4 Proposition – 2 out of 3 Property for Noetherian/Artinian ...... 30 11.6 Proposition – Existence of Base Change ...... 31 12.1 Proposition – Commutative Rings have Invariant Basis Number ...... 33 13.1 Theorem – Swan ...... 34 13.2 Theorem – Projective is Direct Summand of Free ...... 36 13.3 Theorem – Lifting Property of Projectives ...... 36 14.2 Theorem – First Isomorphism Theorem for Rings ...... 38 14.3 Theorem – Bass, 1962 ...... 40 15.1 Theorem – Bott-Milnor: Trivial Tangent Bundles of Spheres ...... 41 15.3 Theorem – Projective Modules Over Local Rings are Free ...... 42 15.4 Theorem – Corollary of Nakayama’s Lemma ...... 42 15.5 Proposition – Flat Implies Torsionfree in Domains ...... 43 15.6 Proposition – Projective Implies Flat ...... 43 15.7 Theorem – Existence of Minimal Polynomials ...... 44 15.8 Theorem – NAK, a.k.a Nakayama-Azumaya-Krull ...... 44 15.11Theorem – 3.44, Generalized NAK ...... 45 16.1 Theorem – Nakayama ...... 46 16.2 Theorem – Generalized Nakayama ...... 46 16.3 Theorem – Characterization of Semisimplicity ...... 46 16.4 Proposition – Characterization of Semisimple Modules ...... 46 16.6 Theorem – Semisimple Rings are Products of Fields ...... 47 16.7 Theorem – Wedderburn-Artin ...... 47 16.8 Theorem – Characterization of Injective Modules ...... 47 16.10Theorem – Baer’s Criterion ...... 49 17.1 Proposition – Multiplicative Avoidance ...... 50 17.2 Proposition – Prime Ideals Behave Like Primes ...... 50 17.3 Proposition – Nilpotent Implies Nil ...... 51 17.4 Proposition – Nil and FG implies nilpotent ...... 51 17.5 Proposition – Universal Property of Nil ...... 51 17.6 Proposition – Prime Implies Radical ...... 52 18.1 Proposition – Algebraic Properties of Radicals ...... 53 18.2 Proposition – Characterization of Jacobson Radical in Terms of Units ...... 54 19.1 Proposition – Characterization of Jacobson Radical ...... 55 19.2 Theorem – Euclidean Criterion ...... 55 19.3 Theorem – Chinese Remainder ...... 56 20.1 Proposition – When Monoid Rings are Domains ...... 59 21.2 Proposition – Push-Pull Equality for Ideals in Localizations ...... 61 21.4 Proposition – Properties of Spec for Localization ...... 62 21.7 Proposition – Complements of Prime Ideals are Local? Extremely Important! . . . . 62 22.2 Proposition – 7.11: Expressing Colon Ideal as Annihilator ...... 63 22.3 Theorem – Local-Global Principle for Lattices ...... 64 22.4 Theorem – Kaplanksy, Very Important: Projective Over Local Implies Free . . . . . 64 22.5 Proposition – Determining if a Projective is Free ...... 65 23.1 Proposition – FG Projective of Rank N Generated by N Elements ...... 66

List of Theorems 5 List of Theorems

23.4 Proposition – Flatness is Local ...... 67 23.5 Theorem – 7.2, Extremely Important: FG Projective iff FG by Localizations . . . . . 67 23.7 Theorem – Artinian iff Noetherian and Primes are Maximal ...... 67 24.1 Theorem – Jordan Holder ...... 68 24.2 Proposition – Length is Additive over SESs ...... 68 24.3 Proposition – Quotient/Localization )f Noetherian is Noetherian ...... 68 24.4 Proposition – Ideal Poset of Quotient Order-Injects into Ideal Poset of Full Ring . . 68 24.5 Proposition – Artinian Domains are Fields ...... 68 24.6 Theorem – Properties of Artinian Rings ...... 69 24.7 Theorem – Akizuki-Hopkins ...... 69 25.1 Theorem – Akizuki-Hopkins ...... 69 25.2 Proposition – Powers of Maximal Ideals in Local Rings ...... 69 25.3 Theorem – Primary Decomposition ...... 70 25.4 Theorem – Hilbert’s Basis ...... 70 25.5 Theorem – Formal Power Series over Noetherian is Noetherian ...... 70 26.1 Theorem – Krull Intersection, 8.39 ...... 71 26.2 Theorem – Principal Ideal Theorem / Krull’s Hauptfdealsatz ...... 72 27.1 Theorem – Ideal Intervals in Noetherian Rings are Empty or Infinite ...... 74 27.2 Proposition – Generalized Hauptfidealsatz ...... 75 27.5 Proposition – Characterization of Finite Boolean Rings ...... 76 28.1 Theorem – Hilbert’s Nullstellansatz ...... 79 29.2 Proposition – Rabinovitch? Trick ...... 81 30.1 Proposition – Spec Is Quasicompact ...... 81 30.2 Theorem – Idempotents Corresponds to Clopen Subsets of Spec ...... 82 30.3 Theorem – Topological Properties of Zero Dimensional Rings ...... 82 30.4 Theorem – 13.10: Characterization of Gelfand Rings ...... 82 30.5 Theorem – 5.9,11: Maximal Ideals in Rings of Continuous Real-Valued Functions . . 82 30.6 Theorem – Hocharter: When Spaces Are the Spec of a Ring ...... 82 30.7 Proposition – Properties of Irreducible Spaces ...... 83 30.8 Proposition – Important: Variety is Irreducible iff Radical is Prime ...... 83 31.1 Proposition – 13.14: In Noetherian Spaces, Every Subset is Quasicompact ...... 84 31.2 Proposition – Noetherian Spaces are Finite Direct Sum )f Connected Components . 84 31.3 Theorem – Spec R is Noetherian iff ACC Holds for Radical Ideals ...... 84 31.4 Proposition – Noetherian Spaces Have Finitely Many Irreducible Components . . . . 85 31.7 Theorem – Characterization of Integral Elements ...... 86 31.8 Proposition – Integral Properties Push to Quotients and Localizations ...... 86 32.1 Theorem – 14.19: Integral Closure is Local ...... 87 32.2 Theorem – 14.21: Going Down ...... 87 32.3 Theorem – 14.22, Noether Normalization ...... 87 32.4 Theorem – Invariant Rings are Integral ...... 87 32.5 Proposition – Ring of Invariants Over Integrally Closed is Integrally Closed . . . . . 88 32.6 Theorem – 14.32, Noether 1928 ...... 88 32.7 Theorem – 18.1, First Normalization Theorem ...... 89

List of Theorems 6 1 Preface

These are notes live-tex’d from a graduate course in Commutative Algebra course taught by Pete Clark at the University of Georgia in Spring 2020 As such, any errors or inaccuracies are almost certainly my own.

D. Zack Garza, Sunday 10th May, 2020 15:36

2 Wednesday January 8

Course text: http://math.uga.edu/~pete/integral2015.pdf Summary: The study of commutative rings, ideals, and modules over them. The chapters we’ll cover: • 1 (Intro), • 2 (Modules, partial), • 3 (Ideals, CRT), • 7 (Localization), • 8 (Noetherian Rings), • 11 (Nullstellensatz), • 12 (Hilbert-Jacobson rings), • 13 (Spectrum), • 14 (Integral extensions), • 17 (Valuation rings), • 18 (Normalization), • 19 (Picard groups), • 20 (Dedekind domains), • 22 (1-dim Noetherian domains)

In number theory, arises in the study of Zk, the ring of integers over a number ﬁeld k, along with localizations and orders (both preserve the fraction ﬁeld?).

In algebraic geometry, consider R = k[t1, ··· , tn]/I where k is a ﬁeld and I is an ideal. Some preliminary results:

1. In Zk, ideals factor uniquely into primes (i.e. it is a Dedekind domain). ∼ [k:Q] 2. Zk has an integral basis (i.e. as a Z-modules, Zk = Z ). 3. The Nullstellansatz: there is a bijective correspondence

n {Irreducible Zariski closed subsets of C } ⇐⇒ {Prime ideals in C[t1,··· ,tn]} .

4. Noether normalization (a structure theorem for rings of the form R above).

All of these results concern particularly “nice” rings, e.g. Zk, C[t1, ··· , tn]. These rings are • Domains • Noetherian

2 WEDNESDAY JANUARY 8 7 • Finitely generated over other rings • Finite Krull dimension (supremum of length of chains of prime ideals) – In particular, dim Zk = 1 since nonzero prime ideals are maximal in a Dedekind domain • Regular (nonsingularity condition, can be interpreted in scheme-theoretic language)

Note: schemes will have “local charts” given by commutative rings, analogous to building a manifold from Euclidean n-space. General philosophy (Grothendieck): Every commutative ring is the ring of functions on some space, so we should study the category of commutative rings as a whole (i.e. let the rings be arbitrary). This does not hold for non-commutative rings! I.e. we can’t necessarily associate a geometric space to every non-commutative ring. A common interesting example: k[G], the group ring of an arbitrary group. Good references: Lam, ‘Lectures on Modules and Rings’.

Example: Let X be a topological space and C(X) be the continuous functions f : X −→ R. This is a ring under pointwise addition/multiplication. (This generally holds for the hom set into any commutative ring.) Example: Take X = [0, 1] and C(X) as a ring. Exercise 1. Show that C(X) is a not a domain.

Hint: ﬁnd two nonzero functions whose product is identically zero, e.g. bump functions. Note that they are not analytic/holomorphic. 2. Show that it is not Noetherian (i.e. there is an ideal that is not ﬁnitely generated). n o

3. Fix a point x ∈ [0, 1] and show that the ideal mx = f f(x) = 0 is maximal. 4. Are all maximal ideals of this form? Hint: See textbook chapter 5, or Gilman and Jerison ‘Rings of Continuous Functions’. The following is a theorem about topological vector bundles over C([0, 1]), see textbook.

Theorem 2.1(Swan). There is a categorical equivalence between vector bundles on a compact space and f.g. projective modules over this ring.

So commutative algebra has something to say about other branches of Mathematics!

Definition 2.1.1 (Boolean Spaces). A topological space is called boolean (or a Stone space) iff it is compact, hausdorff, and totally disconnected.

Example 2.1. A projective variety over p-adics with Qp points plugged in.

Deﬁnition 2.1.2 (Boolean Rings). A ring is boolean if every element is idempotent, i.e. x ∈ R =⇒ x2 = x.

2 WEDNESDAY JANUARY 8 8 Exercise (Boolean Domains are F2): If R is a boolean domain, then it is isomorphic to the ﬁeld with 2 elements.

Lemma 2.2. There is a categorical equivalence between Boolean spaces, Boolean rings, and so-called “Boolean algebras”.

3 Monday January 13

3.1 Logistics Some topics for final projects • The cardinal Krull dimension of Hol(X). • Galois connections • Ordinal filtrations • Lam-Reyes prime ideal principal • C(X) • Hol(X) • Semigroup rings • Swan’s Theorem – Vector bundles on a compact space • Boolean rings and Stone duality • More Nullstellansatz – Beyond Hilbert’s usual one • Hochster’s Theorem – Characterizes Spec R as a topological space, i.e. when is a topological space homeomorphic to the spectrum of some commutative ring. • Invariant theory (quotients of rings under finite group actions, i.e. RG for |G| < ∞) – For R = k a field, this is Galois theory – Easy case of geometric invariant theory, when G is infinite • UFDs – What conditions does a ring need to have to ensure unique factorization? • Euclidean rings • Claborn (Leedham-Green-Clark): Every commutative group is (up to isomorphism) the class group of some Dedekind domain. – A type of inverse problem, class group measures deviation from being a UFD – Uses ordinal filtrations, transfinite induction – See proof in elliptic curves course

3.2 Rings of Functions Let k be a ﬁeld, X a set of cardinality |X| ≥ 2, and deﬁne kX := Maps(X.k) = {f : X −→ K} is a ring under pointwise addition and multiplication. As a ring, this is a (big!) cartesian product. Some facts: • kX is not a domain (exercise), and there are nontrivial idempotents (e2 = e)

3 MONDAY JANUARY 13 9 3.2 Rings of Functions

Note: it could be worse and have nilpotents.

• kX is reduced, i.e. it has no nonzero nilpotents, where z ∈ R is nilpotent iﬀ ∃n ≥ 1 such that zn = 0. – Note: domains are reduced, cartesian products of reduced rings are reduced. • Every subring of kX is reduced. Moral: should be viewing every ring as functions on some space, but this can’t literally be true because of the above restrictions. Nilpotent elements are “hard to view as functions”. X • For X a topological space, C(X) the ring of continuous functionals to R, then C(X) ⊂ R . Exercise When is C(X) a domain? (Note that we can have products of nonzero functions being identically zero.) n o

Example: Let R be the ring of holomorphic functions C , i.e. Hol(C, C) := f : C −→ C f is holomorphic . The set of zeros of such an f must be discrete, the example of bump functions doesn’t go through holomorphically. This is a domain, not Noetherian, not a PID, but every f.g. ideal is principal (thus this is a Bezout domain, a non-Noetherian analog of a PID). It has infinite Krull dimension: recall that ideals are prime iff xy ∈ p =⇒ x ∈ p or y ∈ p iff R/p is a domain, and the Krull dimension is the supremum S of lengths of chains of prime ideals (only when S is finite). If C ⊂ (X, ≤) is a finite-length chain in a totally ordered set, then the length `(C) = |C| − 1 (1 less than the number of elements appearing). The cardinal Krull dimension of a ring R is the actual supremum. Note: in Noetherian rings, there can still be finite but unbounded length chains. n Letting X be a complex manifold (i.e. covered by subsets of C with holomorphic transition functions) and let Hol(X) be the holomorphic functionals f : X −→ C. Then Hol(X) is a domain iff X is connected. Note that if X is disconnected, we can take a function that is constant on one component and zero on another, then switch, then multiply to get a zero function. If X is a compact connected projective variety, then Hol(X) is just constant functions by the open mapping functions. So Hol(X) = C, and carddimC = 0 because for any field there are only two ideals, and here (0) is prime. Moreover, carddimHol(C) ≥ α0. Note that for complex manifolds, X is either compact or supports many holomorphic functions.

Theorem 3.1(Function Spaces Can Have Large Unbounded Chains). If X is a connected complex manifold which has a nontrivial holomorphic function, i.e. Hol(X) ⊃ C, then there exists a chain of prime ideals in Hol(X) of length |R| > ℵ0, i.e. it has at least the cardinality of the continuum.

Note: the cardinality could be even bigger!

3 MONDAY JANUARY 13 10 3.3 Rings

Maximals are prime: equivalent to ﬁelds are integral domains.

3.3 Rings Take all rings to be unital, i.e. containing 1. A ring without identity is referred to as an rng. In this course, all rings are commutative. Example: This is a fairly special restriction. Take (A, +) a commutative group and deﬁne End(A) = {f : A −→ A} the ring of group homomorphisms under pointwise addition and composition. This is generally not commutative, i.e. End(Z/(2) ⊕ Z/(2)) = M2(Z/(2)) the ring of matrices with Z/(2) entries, which is not commutative. n M Exercise Given (A, +), show that End( A) = Mn(End(A)).

Generally, if R is a ring and M is as R-module, then EndR(M) = {f : M −→ M} of R-module homomorphisms is always a ring under pointwise addition and composition, and is (probably) non-commutative.

4 Wednesday January 15th

Theorem 4.1(Cayley). ∼ For G a group, then there is a canonical injective group homomorphism Φ: G,→ Sym(G) = Sn for n = |G|. The map is given by g 7→ g · , i.e. multiplying on the left.

Is there an analog for rings? Take a similar map:

R −→ EndZ(R, +) r 7→ (x 7→ rx).

Unfortunately there is no specialization for commutative groups/rings – Sym(G) for example is noncommutative when |G| ≥ 2. Similarly, even if R is commutative, End(R, +) is probably not. As per the Grothendieck philosophy, we ﬁnd that all rings are a ring of functions on something – namely themselves, since this map is injective. n o × × All rings are commutative here, so take R = x ∈ R ∃ys.t.xy = 1 . For R a group, R is a commutative group, so this is an interesting invariant. Another interesting invariant: the class group. Notation: Let R• = R \ 0. An element x ∈ R is a zero divisor iﬀ there exists y ∈ R• such that

xy = 0. For x, y ∈ R we write x y iff ∃z ∈ R such that xz = y. R is a domain iff 0 is the only zero divisors, i.e. xy = 0 =⇒ x = 0 or y = 0. (R•, ·) is a commutative monoid (group without inverses) iff R is a domain. Observe that R is a field iff R• = R×.

4 WEDNESDAY JANUARY 15TH 11 4.1 Ideals and Quotients

For rings R,S we have the usual deﬁnition of ring homomorphism, additionally requiring f(1) = 1. Note that f(0) = 0 follows from f(x + y) = f(x) + f(y), but f(1) = 1 does not. Rings have products R1 × R2 which is again a ring under coordinate-wise operations. Note that there are canonical projections πiR1 × R2 −→ Ri. There is a dual inclusion ι1 : R1 −→ R1 × R2 given by x 7→ (x, 0), but these are not ring homomorphisms (although everything is a group homomorphism). This is because ι1(1) = (1, 0) 6= (1, 1), the identity of R1 × R2. Note that 1 always has to map to an idempotent element, i.e. e2 = e, and idempotents are always zero divisors. Also note that the map x 7→ 0 is not a ring homomorphism unless S = 0.

Deﬁnition 4.1.1 (Ring Morphisms). A ring homomorphism is a map f : R −→ S is an isomorphism iﬀ it has a two-sided inverse, i.e. there exists a morphism g : S −→ R with g ◦ f = idR and f ◦ g = idS.

Exercise Check that this is equivalent to f being a bijection.

Exercise Check that the zero ring is the ﬁnal object in the category of rings. Show that Z is the initial object in this category? R is a subring of S iﬀ R ⊂ S and the inclusion R,→ S is a morphism. Adjoining elements: Suppose R ≤ S is a subring and X ⊂ S is just a subset. Then there exists a ring R[X] such that • Top-down description: R[X] ≤ S is a subring containing R and X, and is minimal with respect to this property (obtained by intersecting all such subrings) X • Bottom-up description: things resembling rixi Exercise (1.6 in Notes) Take R = Z,S = Q, P a arbitrary set of prime numbers. Let 1 = p ∈ P . ZP Z p ∼ a. When do we have ZP1 = ZP2 ?

Hint: take P1 = {3, 7, 11} , P2 = {5}. Need P1 = P2!

b. Show that every subring T such that Z ≤ T ≤ Q is of the form ZP for some unique set of primes P. Note that if T is any intermediate ring between R and S, then R[T ] = T .

4.1 Ideals and Quotients For f : R −→ S a ring homomorphism, deﬁne I = ker f = f −1({0}). Then I is a subgroup of (R, +), and for all i ∈ I and all r ∈ R we have ri ∈ I, since

f(ri) = f(r)f(i) = f(r)0 = 0.

In other words, RI ⊆ I. By deﬁnition, an ideal I of R is an additive subgroup of R that satisﬁes these properties. Is every ideal the kernel of a ring homomorphism? The answer is yes, namely the quotient π : R −→ R/I.

4 WEDNESDAY JANUARY 15TH 12 Theorem 4.2(Every Ideal Determines a Quotient Ring). Let I ⊂ (R, +), then TFAE: a. I is an ideal of R, written I E R. b. There exists a ring structure on the quotient group R/I such that the projection r 7→ r +I is a ring morphism. When these conditions hold, the ring structure on R/I is unique and we refer to this as the quotient ring.

5 Friday January 17th

For a R ⊂ T a subring of a ring, the set of intermediate rings is a large/interesting class of rings. Recall: uncountably many rings between Z and Q! Taking R a PID and T its fraction ﬁeld, a similar result will hold.

Define I E R as the kernel of a ring morphism. This implies that I ⊂ (R, +) with the absorption property RI ⊂ I. Conversely, any I satisfying these two properties is the kernel of a ring morphism: namely R −→ R/I. This makes sense as a group morphism. Exercise Define xy + I = (x + I)(y + I), need to check well-definedness. Write out

(x + i1)(y + i2) = ··· Need to check that

i1y + i2x + i1i2 ∈ I, but the absorption property does precisely this. Note that if we were in a non-commutative setting, this would deﬁne a left ideal. These don’t have to coincide with right ideals – there are rings where the former satisfy properties that the latter does not.

Example: The subrings of R = Z are of the form nZ for n ≥ 0, with the usual quotient. Deﬁnition 5.0.1 (Proper Ideals). An ideal I E R is proper iﬀ I ( R.

Exercise An ideal I is not proper iff I contains a unit. Exercise R is a field iff the only ideals are 0,R.

Deﬁnition 5.0.2 (Lattice Structure of Ideals). Let I(R) be the set of all ideals in R. What structure does it have? It is partially ordered under inclusion. It is a complete lattice, i.e. every element has an inﬁmum (GLB) and a supremum (LUB). Namely, for a family of D E

ideals {Ij}, the inﬁmum is the intersection and supremum is deﬁned as Ij j ∈ J , the smallest ideal containing all of the Ij, i.e. ( ) n X hyi = riyi n ∈ N>0, ri ∈ R, yi ∈ y .

5 FRIDAY JANUARY 17TH 13 Exercise For I1,I2 E R, it is the case that I1 + I2 := {i1 + i2} = hI1,I2i. Theorem 5.1(Lattice Isomorphism Theorem for Rings). Let I E R and φ : R −→ R/I, and deﬁne `(I) = {I ⊂ J E R}. Then we can deﬁne maps

Φ: `(R) −→ `(R/I) I + J J 7→ , J and

Ψ: `(R/I) −→ `(R) −1 J E R/I 7→ φ (J). We can check that Ψ ◦ Φ)(J) = I + J, and Φ ◦ Ψ(J) = J (= J/I?) So Ψ has a left inverse and is thus injective. Its image is the collection of ideals that contain J, and Ψ: `(R/I) −→ `I (R) is a bijection and is in fact a lattice isomorphism with `I (R) ⊂ `(R).

Note that this gives us everything above a given ideal in the ideal lattice (blue); the dual notion will come from localization (red):

Spec (R/I) Spec (Rp)

p1 p2 p3 p1 p p2

I q

(0)

Here we take p to be a maximal ideal. Remarks The ideal lattice `(R) is • A complete lattice under subset inclusion, • A commutative monoid under addition • A commutative monoid under multiplication, which we’ll deﬁne.

Deﬁnition 5.1.1 (Product of Ideals). For I,J E R, we deﬁne D E

IJ = ij i ∈ I, j ∈ J .

5 FRIDAY JANUARY 17TH 14 Note that we have to take the ideal generated by products here. For hxi = (x) a principal ideal and hyi principal, we do have (x)(y) = (xy). Note that IJ ⊂ I \ J, whereas the sum was larger than I,J. Exercise Note that (`(R), ·) has an absorbing element, namely (0)I = (0). For (M, +) a commutative monoid and M,→ G a group, then multiplication by x is injective and so for all y ∈ M, xz = yz =⇒ x = y, so M is cancellative. Question: what if we consider I•(R) the set of nonzero ideals of R. Does this help? We will see next time.

6 Wednesday January 22nd

Let R be a ring and let I(R) be the set of ideals I E R. This algebraic structure is • Partially ordered under inclusion • Forms a complete lattice with sup the ideal generated by a family and inf the intersection. • Forms a commutative monoid under I + J • Forms a commutative monoid under IJ For any commutative monoid (M, +), there exists a group completion G(M) such that • G(M) is a commutative group • g : M −→ G(M) is a monoid homomorphism • For any map φ :(M, +) −→ (G, +) into a commutative group, we have the following diagram

∀φ M G

g ∃!Φ

M(G)

So φ factors through M(G). If this exists, it is unique up to unique isomorphism (as are all objects defined by universal properties). It remains to construct it. Exercise For (M, +) a commutative monoid, show that TFAE 1. There exists an injective ι : M,→ G monoid homomorphism for G some commutative group. 2. The map g : M −→ G(M) is an injection. 3. M is cancellative, i.e. ∀x, y, z ∈ M we have x + z = y + z =⇒ x = y, i.e. the map pz(x) = x + z is injective. The content here is in 3 =⇒ 1. Definition 6.0.1 (Reduced Monoids). A commutative monoid is reduced iff M × = (0), i.e. if “∀m ∈ M∃n such that m + n = 0” =⇒ m = 0

6 WEDNESDAY JANUARY 22ND 15 6.1 Pushing / Pulling

Example 6.1. + (N, +) and (Z , ·) are cancellative and reduced.

Deﬁnition 6.0.2 (Zero Elements in Monoids). z ∈ M is a zero element iﬀ z + x = z for all x ∈ M.

Remark If M has a zero element, then G(M) = {0}. (0) is a zero element of (I(R), ·), so this is not cancellative. If we take I• the set of nonzero ideals with multiplication, then this is a submonoid of I(R) iﬀ R is a domain. • × For R a domain, let I1(R) be the set of nonzero principal ideals of R, then I1(R) = R /R , so this is reduced and cancellative. What is the group completion? In this case, it will consist of fractional ideals. • • If R is a PID, then I1 (R) = I (R) is reduced and cancellative. Example 6.2. • ∼ + I = (Z , ·).

4! Warning: If R is not a PID, then I•(R) need not be cancellative. √ D √ √ E Exercise Take R = Z[ −3] and p2 := 1 + −3, 1 − −3 . Show that |R/p2| = 2, |R/(2)| = 4,

2 2 • and p2 = p2(2) and R/p2 = 8. Conclude that I (R) is not cancellative. √ √ 1 + −3 What went wrong here? Take K = [ −3], then [ ] is the integral closure of in K. Q Zk 2 Z Zk is a Dedekind domain, and there are inclusions

√ √ 1 + −3 ⊂ [ −3] [ ] ⊆ K. Z Z (Z 2

√ • Here the problem is that Z[ −3] is not a Dedekind domain. If R is a Dedekind domain, then I (R) is cancellative. Exercise Does the converse hold? Things that are too small to be the full rings of integers, and things tend to go wrong (??).

6.1 Pushing / Pulling Let f : R −→ S be a ring homomorphism. We can deﬁne a pushforward on the set of ideals I(R):

6 WEDNESDAY JANUARY 22ND 16 6.1 Pushing / Pulling

f∗ : IR −→ I(S) I 7→ hf(I)i . and a pullback

f ∗ : I(S) −→ I(R) J 7→ f −1(J).

−1 Exercise: Show that f (J) E R. For I E R and J E S, then

∗ f f∗(I) ⊇ I ∗ f∗f (J) ⊆ J.

Exercise: These are not equal in general, and give examples where equality does and does not hold. ∗ If f is surjective, f∗f J = J. Will also hold for localization, which is dual to taking a quotient.

∗ ◦ ∗ Deﬁne I := f f∗(I) and J := f∗f (J), the closure and interior respectively. Show that these operations are idempotent.

Deﬁnition 6.0.3 (Prime Ideals). An ideal p is prime iﬀ ab ∈ p =⇒ a ∈ p or b ∈ p.

Exercise: I is prime iﬀ R/I is a domain.

Deﬁnition (Prime Spectrum) Spec (R) = {p E R} the collection of prime ideals is the spectrum. Exercise Show that for I E R, if we deﬁne n o

V (I) := p ∈ Spec (R) p ⊇ I ⊆ Spec (R), n o

then V (I) I ∈ I(R) are the closed sets for a topology on Spec (R) (the Zariski topology). Exercise If f : R −→ S and J ∈ Spec (S) then f ∗(J) ∈ Spec (R). Show that f ∗ : Spec (S) −→ Spec (R) is a continuous map. Conclude that Spec ( · ) is a functor.

Deﬁnition 6.0.4 (Maximal Ideals). I E R is maximal iﬀ I is proper and is not contained in any other proper ideal.

Exercise I is maximal iﬀ R/I is a ﬁeld. Exercise Show that maximal ideals are prime.

6 WEDNESDAY JANUARY 22ND 17 Deﬁnition 6.0.5 (Max Spectrum). n o

Let Spec max(R) be the set of maximal ideals and deﬁne V (I) = m ∈ Spec max(R) m ⊇ I .

Exercise Show that these form the closed sets for a topology, and that this is the subspace topology for the Zariski topology. ∗ Exercise Show that if f : R −→ S and m ∈ Spec max(S) that f (m) is prime but need not be maximal. Exercise Show that if f is an integral extension, then maximal ideals pull back to maximal ideals.

7 Friday January 24th

7.1 Ideals and Products Recall: Prime and maximal ideals.

Fact: If I E R then there exists a maximal ideal I ⊂ m E R. Proof . Use Zorn’s lemma.

Corollary 7.1. maxSpec R 6= ∅ ⇐⇒ R 6= 0.

\ Later: Multiplicative avoidance, if S ⊂ R is nonempty with SS ⊂ S, let I E R with I S = ∅, then a. There exists an ideal J ⊇ I with J \ S = ∅ which is maximal with respect to being disjoint from S. b. Any such ideal J is prime. Taking S = {1} recovers the previous fact.

Exercise Let f : R −→ S be a ring homomorphism and p ∈ Spec (R). Show that f∗(p) need not be prime in S.

We can consider products of rings, and correspondingly I(R1 × R2). Exercise Show that if φ is surjective, φ(I) is an ideal.

Proposition 7.2(Ideals of Products are Products of Ideals). Let I ∈ I(R1 × R2). Take πi −→ Ri the projections, and let Ii be the corresponding images of I. Then I = I1 × I2.

Note: a suspiciously strong result! Not every group is the cartesian product of some subgroups.

It’s clear that I ⊂ I1 × I2.

7 FRIDAY JANUARY 24TH 18 7.1 Ideals and Products

Proof . Showing I1 × I2 E R1 × R2 is an ideal, since it equals hI1 × {0} , {0} × I2i. To show I1 × I2 ⊆ I, show that I1 × 0, 0 × I2 ⊆ I. E.g. I1 × 0 ⊆ I: take (x, 0) ∈ I1 × 0 such that there exists a y ∈ R2 with (x, y) ∈ I. Then (x, y) · (1, 0) = (x, 0) ∈ I, then similarly 0 × Is ⊆ I.

Exercise Use I(R1 × R2) = I(R1) × I(R2) to describe Spec (R1 × R2) in terms of Spec (R1) and Spec (R2). ∼ Question: For a ring R, when is R = R1 × R2 for some nonzero R1,R2? Exercise Show that comaximal ideals correspond with coprime ideals when R = Z. Theorem 7.3(Chinese Remainder). If I1,I2 are comaximal, so I1 + I2 = R, then the map

7 FRIDAY JANUARY 24TH 19 7.2 Modules

Φ: R −→ R/I1 × R/I2

x 7→ (x + I1, x + I2).

\ CRT Then ker Φ = I1 I2 = I1I2 and Φ is surjective, and

\ ∼ R/(I1 I2) = R/I1I2 = R/I1 × R/I2.

Proof . \ ∼ Case 1: Let I1 + I2 = R and I1 I2 = 0 (equivalently I1I2 = (0)), then R = R/I1 × R/I2.

Conversely, let R = R1 × R2 with R1,R2 nonzero. Let e1 = (1, 0) and e2 = (0, 1). Then 2 e1e2 = 0 and e2 = (1 − e1), so 0 = e1(1 − e1) = e1 − e1 and e1 is idempotent.

So e1, e2 are complementary nontrivial idempotents. Then I1I2 = e1e2 = (0), I1 + I2 = he1, e2i = R, and thus R = R/e2R × R/e1R. Note that e2R = 0 × R2 and e1R = R1 × 0, thus

R1 × R2 R/e2R = = R1 0 × R2 R1 × R2 R/e1R = = R2. R1 × 0

We thus have a correspondence

{Nontrivial product decompositions R=R1×R2} ⇐⇒ {I1,I2 E R such that I1I2=0 and I1+I2=R} ⇐⇒ {Idempotents e6=0,1} .

Thus a ring can be decomposed as a product iﬀ it contains nontrivial idempotents.

Definition 7.3.1 (Connected Rings). ∼ R is connected iff there do not exists nonzero R1,R2 such that R = R1 × R2 iff R does not contain an idempotents e 6= 0, 1.

Exercise Show that R is connected iﬀ Spec (R) is connected as a topological space. Note: Not every ring is a ﬁnite product of connected rings.

7.2 Modules

For (M, +) a commutative group, we want an action R y M for R a ring. Recall that End(M) for a group is a (potentially noncommutative) ring. An R-module structure is a ring homomorphism

7 FRIDAY JANUARY 24TH 20 R −→ End(M). Equivalently, it is a function R × M −→ M with rs(x) = r(sx), r(x + y) = rx + ry, and 1 · x = x for all x ∈ M. Note that this deﬁnes a left R-module, but right/left modules coincide for commutative rings. Exercise Let M be an R-module and for m ∈ M deﬁne n o

Ann(m) = r ∈ R xm = 0 E R,

show this is in fact an ideal. Note: skipped chapter on Galois connections, i.e. some binary relation on a pair of sets. This is an instance of such a connection, where x ∼ m ⇐⇒ xm = 0. Exercise For any subset S ⊂ M, deﬁne n o

Ann(S) := x ∈ R xm = 0 ∀m ∈ S .

Show that Ann(S) = \ Ann(m) and m∈S n o

Ann(M) = x ∈ R xM = 0 = ker(R −→ End(M)).

Definition 7.3.2 (Faithful Modules). M is faithful iff Ann(M) = 0 iff R,→ End(M) is an injection.

Exercise Any M is naturally a faithful R/Ann(M)-module.

8 Monday January 27th

8.1 Localization

Consider rings T such that Z ⊆ T ⊆ Q, and let P be a set of prime numbers. We’ve shown that if ∼ P,Q are two sets of prime numbers, then Zp = Zq ⇐⇒ Zp = Zq ⇐⇒ P = Q. Let R be a domain with fraction field K. Let P be a set of mutually nonassociate prime elements. Note that p ∈ R is a prime element iff (p) is a prime ideal. We say x, y are associates iff there exists a u ∈ R× such that y = ux. Since we’re in a domain, (exercise) this is equivalent to (x) = (y). Fact: We can then consider 1 R definedasR p ∈ P , P p and the fact is that the previous statement still holds.

But if R = Z, we also have (exercise) if Z ⊂ T ⊂ Q then T = Zp for a unique P . a Exercise: How do we ﬁnd such a P ? This comes down to looking at ∈ T with gcd(a, b) = 1, b 1 then ∈ T . b

8 MONDAY JANUARY 27TH 21 8.2 Modules

Hint: In a PID, gcd(a, b) exists and is a Z-linear combination of a and b. The solution should work for an arbitrary PID. Let R be a domain and S multiplicatively closed (so (S, ·) ≤ (R, ·) is a submonoid). Then S is primal if S is generated as a monoid by its prime elements. Suppose that S is saturated, i.e. if s ∈ S

and r ∈ R with r s, then r ∈ S.

Can always add in all divisors. We can then deﬁne the localization of R at S,

a R := a ∈ R, s ∈ S . s s

This satisﬁes R ⊂ RS ⊂ K, and is a multiplicative partial group completion. If we took nonzero elements, this would yield exactly the fraction ﬁeld.

Theorem 8.1(Negata). Let R be a Noetherian domain with S ⊂ R primal as above. If RS is a UFD, then R is a UFD.

Exercise Show that the converse holds. Fraction ﬁelds are always UFDs? Localizing makes it easier for irreducibles to be prime. This helps prove that some interesting rings are UFDs.

8.2 Modules

If M is an R-module, then an R-submodule N ≤ M is a subgroup of (M, +) such that R y N ⊂ N. Every ring R is an R-module over itself, and the R-submodules of R are precisely the ideals of R. Can express certain concepts about rings/commutative algebra in the language of modules. A morphism of R-modules f : M −→ N is a homomorphism (M, +) −→ (N, +) such that f(r y m) = r y f(m). Exercise: Any module morphism that is a bijection is an isomorphism. (Usually true in algebraic settings.) M We can form quotient modules which is an R-module with r (m + N) = (r m) + N, and N y y M M −→ is a surjective morphism. N If I E R is an R-submodule of R, then R/I is an R-module. We have Ann(R/I) = I. Fact: Every ideal in R is the annihilator of some R-module. Fact: Suppose R is a ring such that every every nonzero R-module is faithful, then R is a ﬁeld. The converse also holds. General idea: we study rings by looking at modules over them.

8 MONDAY JANUARY 27TH 22 For an R-module M and S ⊂ M, then we can consider hSi the R-submodule generated by S. We can write this as ( )

\ X N = risi ri ∈ R, si ∈ S . N s.t. S⊂N⊆RM i=1n

We say R is finitely generated iff there exists a finite generating set S ⊂ M. We say M is cyclic iff it is generated by a single element, i.e. M = hsi. Y Let {Mi}i∈I be a family of R-modules. Let Mi be the cartesian product with a coordinate-wise i∈I R-action be the direct product. Let n o M Y Mi = (xi) ∈ Mi xi 6= 0 for finitely many i , i∈I Y which is a submodule of Mi. When I is finite, these are equal. Recall: If R is a PID and M is a finitely generated R-module, then there exist finitely many cyclic ∼ M R- modules {C1, ··· ,Cn}, then M = Ci. Exercise: Let R be a ring and C a cyclic R-module, then show that C =∼ R/Ann(C) as R-modules. We’ll later see that the class of rings R such that every R-module is free are exactly fields.

Remark: Let I E R, then I is cyclic as an R-module iﬀ I is principal. Exercise:

a. Let I E R for R a domain, then I is indecomposable, i.e. I 6= M1 ⊕ M2 for any nonzero M1,M2 R-modules.

b. If R is additionally Noetherian and not a PID, then there exists an I E R where I is ﬁnitely generated, not principal, and so I is not a cyclic R-module. Converse to structure theorem! Mild assumptions negate cyclic direct sum decomposition.

9 Wednesday January 29th M Coming up: the modules Z, homR(M,N),M ⊗R N, as well as various properties: • Torsion • Torsionfree • Free • Projective • Flat • Injective • Divisible We have a series of implication

free =⇒ projective =⇒ ﬂat =⇒ torsionfree

9 WEDNESDAY JANUARY 29TH 23 9.1 Universal Mapping Properties

9.1 Universal Mapping Properties

Deﬁnition 9.0.1 (Direct Product of Modules). For a collection {Mi} of modules, the direct product is characterized by ϕj Mj πj

∃! Y N Mi

πk

Mk ϕk

Here we deﬁne the canonical projection by πj(m1, ··· , mj, ··· ) = mj.

Fact Y Y homR(N, Mi) = homR(N,Mi)

Deﬁnition 9.0.2 (Direct Sum of Modules). For a collection {Mi} of modules, the direct sum is characterized by ϕj Mj ιi

M ∃! Mi N

ιk

Mk ϕk

Here we deﬁne the canonical injection by ιj(m) = (0, 0, ··· , m, 0, ··· ).

X In this case, we can deﬁne φ(m1, m2, ··· , mi, ··· ) = φi(mi), which makes sense because coﬁnitely many of the terms in this sum are zero. Fact ! M Y S homR R,N = homR(R,N) = N s∈S s∈S ∼ Fact homR(R,N) = N via the map f ∈ hom(R,N) 7→ f(1).

9.2 Free Modules Deﬁnition 9.0.3 (Spanning, Linear Independence, and Basis). For M an R-module and S ⊂ M, 1. S spans M if hSi = M, where hSi is the set of all ﬁnite linear combinations of elements in S.

9 WEDNESDAY JANUARY 29TH 24 9.2 Free Modules

X 2. S is R-linearly independent iﬀ rimi = 0 =⇒ ri = 0 for all i. 3. S is a basis for M iﬀ S is a spanning R-linearly independent subset of M. If M admits a basis, M is said to be free.

Theorem 9.1(Free Modules are Quotients).

a. If S = {si} is a basis for M, then there is a surjection

M R −→ M s∈S X ri ← risi. [ b. For any set S, the module M R has a canonical basis s∈S

es = (0, 0, ··· , 0, 1, 0, ··· , 0) M c. If φ : R −→ M is an isomorphism, then {φ(es)}s∈S is a basis for M. s∈S

Fact Let F be a free R-module, then Ann(F ) = R if F = (0) and 0 otherwise. Moreover, M \ • Ann( Mi) = Ann(Mi) • Ann(R) = {0}

Proposition 9.2(Characterization of Freeness in terms of Rings). For a ring R 6= 0, TFAE: a. Every R-module is free b. R is a ﬁeld

Proof . a =⇒ b: If R is not a ﬁeld, then 0 < I E R is proper, and since Ann(R/I) = I, we have 0 < Ann(R/I) < R. So Ann(R/I) is proper, and R/I is thus not a ﬁeld. The reverse implication is linear algebra. Every vector space has a basis by AOC (note that this is equivalent to Zorn’s Lemma).

Fact Every R-module N is the quotient of a free module. M This follows by taking the generating set S = N, then R N using a previous fact. n∈N Fact N is quotient of a finitely generated free module iff N is finitely generated. Exercise Show that for 0 −→ A −→ B −→ C −→ 0 a SES of R-modules, a. If A, C are finitely generated, then so is B. b. If B is finitely generated, then so is C. Example 9.1.

9 WEDNESDAY JANUARY 29TH 25 It is possible for B to be finitely generated with A < B and A not finitely generated. Let R be non-Noetherian. Equivalently, there exists I E R that is not finitely generated. So take B = R and A = I. For example, take M = C([0, 1],R) the module of continuous functionals, which is non-Noetherian.

Examples of non-Noetherian rings: Y 1. {Ri} where each Ri is inﬁnite and ???; then Ri is non-Noetherian. n o

2. For k a field, T = tn 1 ≤ n < ∞ , take R = k[T ]. Then I = hT i is not finitely-generated. Fact If R is a Noetherian ring, then every finitely generated R-module is a Noetherian module. Example 9.2. Take R,M = Z, which are free modules, and S = {2}. Note that S is R-linearly independent in M, 0 but can not be extended to a basis, and hSi = 2Z 6= Z. Similarly, S = {2, 3} can not be reduced to 0 a basis, while S = Z.

Question: can M have basis sets of diﬀerent cardinalities? Answer: sometimes, commutative rings have the invariant basis property.

10 Friday January 31st

10.1 Tensor and Hom Let M,N be R-modules, then we deﬁne

n o

homR(M,N) := f : M −→ N f is an R-module map .

Recall that R-module maps satisfy • f :(M, +) −→ (N, +) a morphism of abelian groups • For all r ∈ R, for all m ∈ M, f(rm) = rf(m).

Note that homR is a commutative group, and is in fact an R-module with structure given by (r · f) · m 7→ rf(m) = f(rm). Note that the proof of this fact uses commutativity in a key way. Facts:

homR(R,N) = N ! M S homR Rs,N = N s∈S ∨ homR(M,R) := M .

10 FRIDAY JANUARY 31ST 26 10.2 Free and Torsion Modules

Note: Inﬁnite dimensional vector spaces over ﬁelds are never isomorphic to its dual. Exercise: Think about M ∨ and (M ∨)∨. Recall the map

∨ ∨ ι : M −→ (M ) = homR(homR(M,R),R) x 7→ (` : M −→ R 7→ `(x) ∈ R).

Exercise: If R = k is a field, then show that ι is injective iff dim M is finite. ∨ Is this always injective? No! Counterexample: Take R = Z and M = Z/pZ, then M = homZ(Z/pZ, Z) = 0. It can also fail to be surjective in the infinite dimensional case – the space M ∨ is strictly larger than M. Definition 10.0.1 (Reflexive Modules). M is reflexive if ι : M −→∼ (M ∨)∨ is an isomorphism.

Exercise: Show the following: • If M is free and finitely generated, then M is reflexive. • If R = k is a field, then M is reflexive iff M is finitely generated. • There exists a ring R and a reflexive R-module M that is not finitely generated.

10.2 Free and Torsion Modules • Let R be a domain, and for all a ∈ R the map [a]: R −→ R is injective, and [a] ∈ homR(R,R) = R.

Deﬁnition 10.0.2 (Torsion Submodule).

n o

M[tors] := m ∈ M Ann(m) 6= (0) ≤ M

is the torsion submodule of M.

Definition 10.0.3 (Torsion and Torsionfree Modules). M is torsion iff M = M[tors], and M is torsionfree iff M[tors] = (0).

Exercise Show that if 0 −→ A −→ B −→ C −→ 0, then • Show that if B is torsion then A, C are torsion. • If A, C are torsion, must B be torsion? • Show that if B is torsion-free then A is torsion-free but C need not be torsion-free. • If A, C are torsion-free, must B be torsion-free?

Note: 0 −→ Z/2 −→ Z/4 −→ Z/2 −→ 0 is an extension that isn’t a semidirect product! Fact Free modules are torsion-free.

10 FRIDAY JANUARY 31ST 27 10.2 Free and Torsion Modules

Note that we need to be in a domain to even talk about torsion.

Proposition 10.1(Torsionfree implies submodule of f.g free). Let R be a domain and M an R-module. Then a. M/M[tors] is torsion-free. b. If M is finitely generated, then M is torsion free iff M is isomorphic to a submodule of a finitely-generated free module.

Proposition 10.2(Implication Chain). Free =⇒ projective =⇒ ﬂat =⇒ R a domain torsion-free.

Proof (of (a)). Let x ∈ M/M[tors] such that ∃r ∈ R• such that rx = 0. Lift x to x˜ ∈ M, then rx˜ ∈ M[tors]. Then ∃r0 ∈ R• such that

0 0 0 = r (rx˜) = (r r)˜x := r2x˜

for some r2 6= 0. But then x˜ ∈ M[tors], and so x = 0 in M/M[tors].

Proof (of (b)). Let M = hx1, ··· , xri with r ≥ 1 and xi =6 0. After reordering, there exists some s with [ 1 ≤ s ≤ r such that x1, ··· , xs are R-linearly independent, and for all i > s, {xj}j≤s xi is linearly dependent. Then deﬁne F := hx1, ··· , xsi; this is a ﬁnitely generated free module. If r = s, we done. • Otherwise, r < s, then ∀i > r there exists an ai ∈ R such that aixi ∈ F . So we can take a := as+1 ··· ar 6= 0; then aM ⊂ F . Since M is torsion-free, the multiplication maps are ∼ injective, so [a]: M −→= M ⊂ F , so M,→ F embeds M into a free module.

Does this work with M not finitely generated? No, we can’t take an infinite product for a. Is every torsion-free module a submodule of a free module? No. Remark This fails without finite generation, see Theorem 3.56 on ordinal filtration. If R is a PID and F is a free R-module and M ≤ F as an R-submodule, then M is free.

Thus if R is a PID, “subfree” ⇐⇒ free. Does torsion-free imply free? No, take R = Z and M = (Q, +), this is not ﬁnitely generated and torsion-free but not a free Z-module.

Deﬁnition 10.2.1 (Divisible and Uniquely Divisible Modules). For R a domain, M is divisible if ∀a ∈ M • iﬀ

[a]: M M

∼ is a surjection. M is uniquely divisible if for all a ∈ M •, [a]: M −→= M is an isomorphism, i.e. M is torsion-free and divisible.

Exercise Show that (Q, +) is a uniquely divisible Z-module.

10 FRIDAY JANUARY 31ST 28 Exercise Let R be a domain with fraction ﬁeld K, with R =6 K. Show that a nonzero free R-module is not divisible but (K, +) is a divisible torsion-free R-module. Thus (K, +) is a torsion-free module R-module that is not free. Remark Finitely generated torsion free modules are embedded in free modules. Note that in the spectrum of properties earlier (projective, free, etc), the two extremes are equal for f.g. PIDs.

Exercise Let R be a Noetherian domain which is not a PID. Then an ideal I E R with I f.g., not principal, and a torsion-free R-module. Show that since I is not principal, I is not free as an R-module. So ideals can’t contain linearly independent elements, so they have to be free of rank 1 and thus principal. Thus ﬁnitely generated torsion-free is strictly weaker then free in this setting.

11 Monday February 3rd

Some module topics from Chapter 8.

11.1 Noetherian and Artinian Modules

Definition 11.0.1 (Noetherian Posets). A poset (X, ≤) is said to satisfy the ACC or to be Noetherian iff there does not exist an infinite sequence (a chain) {xn} with strict inequalities x1 < x2 < ···. Equivalently, every weakly ascending chain x1 ≤ x2 ≤ · · · eventually stabilizes, i.e. there exists an N such that xN = xN+1 = ···.

Definition 11.0.2 (Artinian Posets). Similarly, a poset satisfies the DCC or is Artinian iff there does not exist an infinite decreasing sequence x1 > x2 > ···.

Deﬁnition 11.0.3 (Order Dual). For (X, ≤), deﬁne the order dual (X∨, ≤) where x ≤ y ∈ X∨ ⇐⇒ y ≤ x ∈ X.

Proposition 11.1(Noetherian/Artinian Duality). X is Noetherian iﬀ X∨ is Artinian.

Lemma 11.2. The ACC holds iﬀ every nonempty subset has a maximum (and similarly the DCC with minimums).

Proof . Otherwise use AOC to pick elements xi; if xi isn’t the maximum then there is some xi+1 > xi, and this yields an inﬁnite ascending chain iﬀ no maximum.

Let M be an R-module, and deﬁne SubRM = {(R-submodules of M, ≤)}.

11 MONDAY FEBRUARY 3RD 29 11.2 Tensor Products

Lemma 11.3. M is Noetherian ⇐⇒ every submodule N ≤ M is ﬁnitely generated.

Proof . Apply the DCC.

0 0 Exercise Let M ⊂ M and q : M −→ M/M and N1 ⊂ N2 ⊂ M such that \ 0 0 • N1 M = N2 = M , and • q(N1) = q(N2).

Then N1 = N2.

Proposition 11.4(2 out of 3 Property for Noetherian/Artinian). If 0 −→ M 0 −→ M −→ M 00 −→ 0 is exact and M is Noetherian (resp. Artinian) then M 0,M 00 are both Noetherian (resp. Artinian).

Proof . 0 00 Note that SubRM , SubRM ,→ SubRM in an order-preserving manner. If we then have N + M 0 N ⊂ N ⊂ · · · with N ≤ M submodules of M, we can consider N = n , which is 1 2 i n M 0 weakly increasing in M 0.

Note: this is how we push forward into quotients.

0 0 \ 0 \ 0 Thus this chain stabilizes, so for i, j 0 we have Ni + M = Nj + M . So then Ni M = Nj M , and by the exercise, Ni = Nj for all i, j 0.

Corollary 11.5. R is Noetherian (resp. Artinian) iﬀ every ﬁnitely-generated R-module is Noetherian (resp. Artinian)

Proof . =⇒ : Suppose R is Noetherian. Note that 0 −→ R −→ R2 −→ R −→ 0 since R2 is an extension of R by R. Thus R2 is Noetherian, and inductively Rn is a Noetherian R-module.

If M is a ﬁnitely-generated R-module, it is a quotient of a ﬁnitely-generated free R-module, and in particular 0 −→ K −→ Rn −→ M −→ 0 is exact. So M is Noetherian, by the previous proposition (middle of a SES Noetherian =⇒ ends are Noetherian).

11.2 Tensor Products Motivation from Representation Theory: For G ﬁnite, H ≤ G, and ρ : G −→ V a ﬁnite-dimensional C-representation, this data is equivalent to a C[G]-module structure on V . If W is a representation G G on H, then IndH W is a representation of G given by V = IndH W = W ⊗C[H] C[G].

11 MONDAY FEBRUARY 3RD 30 11.2 Tensor Products

Deﬁnition 11.5.1 (Tensor Product). Let M,N be R-modules, then the tensor product M ⊗R N is an object characterized up to canonical isomorphism by the following universal property: If P is an R-module and Φ: M × N −→ P is any bilinear map, then there exists a unique lift such that the following diagram commutes: M ⊗R N

∃!ψ ι

M × N Φ P where ι : M × N −→ M ⊗R N is R-bilinear and for all (m, n) ∈ M × N, we denote m ⊗ n := ι(m, n).

By dimension counting in the ﬁnite-dimensional case of vector spaces, it’s clear that ι need not be surjective. In general, elements in M ⊗R N are ﬁnite sums of simple tensors, not just simple tensors, i.e. M ⊗R N = hι(m, n)i.

Proof (existence). n o

Let F be the free R-module on M × N with basis (m, n) m ∈ M, n ∈ N . Mod out by the following relations: for all m, m1,M2 ∈ M and for all n, n1, n2 ∈ N and all r ∈ R,

• (m1 + m2) ⊗ n − m1 ⊗ n − m2 ⊗ n • m ⊗ (n1 + n2) − m ⊗ n1 − m ⊗ n2 • r(m ⊗ n) − (rm) ⊗ n • r(m ⊗ n) − m ⊗ (rn)

Let R be the ideal generated by these relations, the define M ⊗R N = F/R by (m, n) 7→ (m, n) + R. Then (straightforward check) the universal mapping property holds. How do we work with tensor products? Namely, how do we even know whether an arbitrary element is zero or not in this complicated quotient. • To show m ⊗ n = 0, use bilinear relations (reduce to relations above) • To show m ⊗ n 6= 0, find an R-module and a bilinear map ψ : M ⊗R N −→ P such that im (m ⊗ n) 6= 0. ∼ • To show M ⊗R N = X, show that X satisfies the universal property.

Exercise R ⊗R M ≡ M by (r, m) 7→ r · m, with · the R-module action on M. Let P be arbitrary, let φ : R × M −→ P be arbitrary, and deﬁne ψ : M −→ P by m 7→ φ(1, m).

Exercise Z/mZ⊗Z Z/nZ ≡ Z/ gcd(n, m)Z. Show that every element is both n-torsion and m-torsion.

Proposition 11.6(Existence of Base Change). For M and R-module and f : R −→ S, we can create an S-module S ⊗R M by base change.

Deﬁnitely the most important concept thus far!

11 MONDAY FEBRUARY 3RD 31 12 Wednesday February 5th

Recall that if M,N are R-modules then there is a map

Φ M × N −→ M ⊗R N (r, m) 7→ r ⊗ m which is universal wrt the property that any bilinear map φ : M × N −→ A factors through Φ uniquely. We have a notion of pullback, where if i : R −→ S and N is an S-module then i∗N is an R module i · with action given by composition R −→ S −→ EndZ(N).

Dually, we have a notion of base change, where for M an R-module we can form i∗M := S ⊗R M X X an S-module where s( si ⊗ mi) = ssi ⊗ mi.

An R-algebra is i : R −→ S a ring morphism, where algebra morphisms f : S1 −→ S2 are given by commutative diagrams

i1 R S1

i2 f

For S, T R-algebras, the tensor product S⊗RT is an R-algebra with (s1⊗m1)·(s2⊗m2) = s1s2⊗m1m2. Note that the tensor product satisﬁes the universal property of the direct sum or coproduct:

t ψ

T 1 ⊗ t

∃! R S ⊗R T W

S s ⊗ 1 φ s

Exercise (Important) Verify the following identities ∗ One: Let M be an R-module and N an S-module with ι : R −→ S. homR(M, ι N) = homS(ι∗M,N) = homS(S ⊗R M,N). What’s the map? s ⊗ m 7→ sf(m). ∗ ∗ Two: For P and R-module and M,NS-modules, we have M ⊗X (i N ⊗RP ) = i (M ⊗S N)⊗RP . So for N = S, then M ⊗S (S ⊗R P ) = M ⊗R P .

Three (Good Exercise! Very important!): For M an R-module and I E R, we have IM ⊂R M. Show that we can identify the base change as R/I ⊗R M = M/I.

12 WEDNESDAY FEBRUARY 5TH 32 Hint: Show that the RHS satisﬁes the appropriate universal property. Four:

• (⊕Mi) ⊗R N = ⊕(Mi ⊗R N). • The tensor product of free modules is free. • If F is a free R-module and we base change with ι : R −→ S then S ⊗R F is a free S-module. Definition 12.0.1 (Invariant Basis Number). Let R be a ring, then R satisfies the invariant basis number property (IBN) iff any two bases for a free left R-module have the same cardinality.

Definition 12.0.2 (Rank Condition). m n R satisfies the rank condition iff whenever there exists a q : R R , n ≤ m.

Definition 12.0.3 (Strong Rank Condition). R satisfies the strong rank condition iff whenever q : Rm ,→ Rn then n ≤ m.

Facts If R is commutative or (left)-Noetherian, then strong rank condition =⇒ rank condition =⇒ IBN. Note: this is not obvious, since if R is not Noetherian there are submodules that aren’t finitely generated but can still have bounded rank. Exercise (Non-Commutative) Let k be a field and V an infinite dimensional k-vector space, i.e. V =∼ V ⊕ V . Let R := Endk(V ); then R does not satisfy the IBN.

Proposition 12.1(Commutative Rings have Invariant Basis Number). If R is nonzero and commutative then R satisﬁes IBN.

Proof . ∼ Suppose there exist I,J such that ⊕i∈I R =R ⊕j∈J R. We want to show that |I| = |J|. Since R 6= 0, there is a maximal ideal m ∈ maxSpec (R). Since R/m is a ﬁeld, we base change to it M M M to obtain R/m ⊗R ( R) = R/m. We know this equals R/m ⊗R ( R) = ⊕j∈J R/m. So i∈I i∈I j∈J I,J are bases of isomorphic vector spaces and thus |I| = |J| by linear algebra.

Definition 12.1.1 (Noetherian and Artinian Modules). A module M is Noetherian iff ACC on submodules, and Artinian iff DCC on submodules.

Exercise If R = k is a field and V is a k-vector space, then V is Noetherian iff Artinian iff infinite-dimensional.

Exercise If R = Z, R is Noetherian but not Artinian. Find a Z-module that is Artinian but not Noetherian.

Try all 2n possibilities for adjectives!

12 WEDNESDAY FEBRUARY 5TH 33 Exercise If R is finite, it is both Artinian and Noetherian, and moreover has only finitely many ideals. Artinian is much stronger, and implies Noetherian? Converse iff every ideal is maximal. The only Artinian integral domains are fields. Very small class of rings. It’s not true that Artinian alone implies finitely many ideals. 2 2 2 Exercise (8.29 in Notes) Let I = (x , xy, y ) = (xy) E C[x, y] and take R = C[x, y]/I. a. Show that a C-basis for R is given by {1 + I, x + I, y + I}. b. Deduce that R is Noetherian and Artinian.

c. Show proper ideals of R are precisely the C-subspaces of hx, yi + I. d. Deduce that R has continuum many ideals.

13 Friday February 7th

13.1 Projective Modules For X a topological space and π : E −→ X a real vector bundle on X. Then n o

Γ(E,X) = σ : X −→ E π ◦ σ = idX is naturally a module over the ring C(X, R) of continuous real-valued functions. For p ∈ X, the −1 ﬁbers σ(p) ∈ π (p) are vector spaces, and we can consider f(p)σ(p) for any f ∈ C(X, R). For n π trivial bundles R × X −→ X with a global section

n σ : X −→ R × X p 7→ (˜σ(p), p).

n Then σ˜ : X −→ R , or equivalently a collection of n continuous functions σ˜j −→ R. Thus ∼ n Γ(X,E) = C(X, R) .

Theorem 13.1(Swan). Suppose X is compact. Then a. Γ(X,E) is a ﬁnitely generated projective C(X, R)-module, i.e. π is a direct summand of a trivial vector bundle on X, and b. There is an equivalence of categories between vector bundles on X and ﬁnitely generated projective C(X, R)-modules.

Example 13.1. Let X be the two points space {1, 2}. Take a 0-dimensional vector space over 1 and a 1-dimensional vector space over 2.

13 FRIDAY FEBRUARY 7TH 34 13.1 Projective Modules

Remark Such cheap examples exist on X iﬀ X is disconnected.

Deﬁnition 13.1.1 (Splitting an Exact Sequence). f Recall that it 0 −→ A −→ B −→ C −→ 0 is exact, then a splitting is a map σ : C −→ B such ∼ that f ◦ σ = idC . Then B = A ⊕ σ(C) = A ⊕ C.

∼ Exercise Take R = Z and find a SES such that B =Z A ⊕ B but the sequence is not split. Definition 13.1.2 (Projective Module). A module P is projective iff 0 −→ M −→ N −→ P −→ 0 is split.

Exercise Show that free implies projective. Hint: Lift basis and use universal property.

13 FRIDAY FEBRUARY 7TH 35 13.1 Projective Modules

Theorem 13.2(Projective is Direct Summand of Free). If P is projective, then there exists a K such that P ⊕ K is free.

Idea: summands can be both a submodule and a quotient module.

Proof . Choose a free F and an R-module surjection q : F P with K = ker q to obtain 0 −→ K −→ F −→ P −→ 0. Since P is projective, this sequence splits and thus F =∼ K ⊕ P is free.

Comment: If P is finitely generated, then we can take K (and hence F ) to be finitely generated module. A quotient of a finitely-generated module is also finitely generated, and F =∼ K ⊕ P .

Theorem 13.3(Lifting Property of Projectives). If there exists a K such that P ⊕ K is free, then P satisﬁes this lifting property: P

f ∃f˜ M N 0

Proof .

Choose K such that P ⊕ K is free, and let {fi}i∈I be a basis for F . Then write F = P ⊕ K and fi = pi + ki where pi ∈ P, ki ∈ K. Then we can construct a unique g : F −→ M by sending fi to mi: {fi},

F = P ⊕ K

π ι(p)=(p,0) ∃!g P

q M N 0

{mi} {ni} Then q ◦ g ◦ ι = (q ◦ g) ◦ ι = (f ◦ π) ◦ ι = f ◦ (π ◦ ι) = f since ι is a section.

Todo: Revisit! This P is projective iﬀ • Every length 2 resolution of P splits.

13 FRIDAY FEBRUARY 7TH 36 • P is a direct summand of a free module. • P satisﬁes this lifting property. If P satisﬁes this lifting property, we have: P

∃σ idP

0 M N P 0 Exercise Show free implies projective in as many ways as you can (using any of these properties).

Remark An easy consequence of the lifting property implies that the functor M 7→ homR(P,M) is covariant and exact. Natural question: for any new property of modules, is there a class of rings for which this coincides with known properties? Question: How diﬀerent is projective from free?

Free =⇒ projective =⇒ subfree =⇒ R a domain torsion-free. For R a PID and M ﬁnitely generated, these are all equivalent (hence the diminished importance of projectivity when studying the structure theorem). Recall (Theorem 3.56) that if R is PID, then subfree =⇒ free and projective ⇐⇒ free, but (Q, +) is torsion-free but not free. a Recall Spec (R1 × R2) = Spec R1 Spec R2

Example 13.2 (Projective does not imply free):*). Let R1,R2 be rings and consider R = R1 × R2. Then recall that I E R implies I = I1 × I2 for Ii E Ri. Take M1 := R1 × 0 E R, and M2 := 0 × R2 E R.

Then M1 ⊕ M2 = M1 + M2 = R, so both Ri are projective. They are not free though, since AnnM1 = M2 and vice-versa.

Example 13.3. Let R = C × C, so Spec R = {1, 2}, then M1 = C × 0 −→ Spec R, and we can construct “cheap” bundles in analogy to the disconnected topological case.

Next question: What is an example of a nonfree projective module over a domain.

14 Wednesday February 12th

14.1 Projective Modules and Ideals

Summary: Free =⇒ projective =⇒ ﬂat =⇒ R a domain torsion free. Moreover, projective =⇒ reﬂexive.

14 WEDNESDAY FEBRUARY 12TH 37 14.1 Projective Modules and Ideals

If M,N are cyclic R-modules, then Ann(M ⊗R N) = AnnM + AnnN. Does this hold for every M,N? The answer is no; we have

Ann(M ⊗R N) ⊇ AnnM + AnnN.

See MSE post: let I E R and M an R-module, we have M ⊗R R/I = M/IM. Is there an equality Ann(M/IM) = Ann(M) + I? No, take R = C[x, y]. Recall that an R-module is reflexive iff ι : M −→ (M ∨)∨ is an isomorphism, where M ∨ = homR(M,R). This is injective for R a field, and then surjective iff R is finite-dimensional. As shown in the problem sessions, finitely generated free modules are reflexive.

Exercise: Show that direct summands of reflexive modules are reflexive, and M1 ⊕ M2 is reflexive iff Mi are reflexive. Conclude that finitely generated projective modules are reflexive. Example 14.1. 2 2 To get a projective module that is not free, take C = (C × 0) ⊕ (0 × C) = C , which is free, so the summands are projective, but not free.

Note: this corresponds to taking a vector bundle over a disconnected space, and letting the ﬁbers just be diﬀerent dimensions. Letting the summands above be I,J, note that I + J = R and IJ = 0, which is a comaximality condition.

Lemma 14.1(3.17). Let I, J, K1, ··· ,Kn E R. Then a. (I + J)(I \ J) ⊆ IJ b. If I + J = R (so I,J are comaximal), then I \ J = IJ. c. If I + ki = R for all 1 ≤ i ≤ n then I + K1 ··· Kn = R.

Proof . Omitted.

Theorem 14.2(First Isomorphism Theorem for Rings). Let R be a domain, IJ E R such that I + J = R. We can form a map: I

q I ⊕ J R

J where

14 WEDNESDAY FEBRUARY 12TH 38 14.2 The Picard Group

q : I ⊕ J −→ R (i, j) 7→ i + j.

Then a. q is surjective n o \ ∼ \ b. ker q = (x, −x) x ∈ I J =R I J = IJ. q ∼ c. There is a SES 0 −→ IJ −→ I ⊕J −→ R −→ 0, and since R is projective, I ⊕J =R IJ ⊕R. ∼ d. If IJ is principal (so IJ =R R) then I is projective.

See “monogenic”. This gives a criterion for determining if ideals are projective. √ D √ E D √ E Exercise Let R = Z[ −5] with p1 = 3, 1 + −5 and p2 = 3, 1 − −5 . ∼ ∼ a. Show that R/p1 = R/p2 = Z/3Z.

b. Show p1 + p2 = R.

c. Show p1p2 = h3i.

d. Show neither p1, p2 are not principal. ∼ e. Conclude p1 =R p2 is a ﬁnitely generated projective but not free R-module.

14.2 The Picard Group

Definition 14.2.1 (Picard Group). Let R be a domain with fraction field K, we’ll define picard group Pic (R) in the following way: For I E R with I 6= 0. we say I is invertible iff there exists a J E R such that IJ is principal. Then Pic (R) is the set of invertible ideals modulo the equivalence I ∼ J iff there exist a, b ∈ R• such that hai I = hbi J. This is a group under [I] + [J] = [IJ] (check that this is well-defined).

Note that if I is principal, then [I] = 1 is the identity, and if IJ = hxi, then [I][J] = [hxi] = 1. Fact If I is invertible, then I is a projective R-module.

Fact (Stronger) If I E R in a domain, then • I is invertible iff I is a projective R-module. • [I] = 1 in Pic R iff I is principal iff I is a free R-module.

Proof . Later!

Every nontrivial element gives a projective but not free R-module! Note that Pic R = 0 for R a PID.

14 WEDNESDAY FEBRUARY 12TH 39 Deﬁnition 14.2.2 (Dedekind Domain and Class Group). R is a Dedekind domain iﬀ every nonzero I E R is invertible, and Pic R is referred to as the class group of R.

In this case, Pic R = 0 iff every ideal is principal iff R is a PID. So the class group measures how far R is from a PID. Any Dedekind domain that is not a PID yields projectives that aren’t free. Rings of integers over number fields are Dedekind domains. Embarrassingly open problem: are there are infinitely many number fields K such that the ring of integers ZK is a PID, or equivalently Pic ZK = 0? Example 14.2 (Important). + Let k be a field and n ∈ Z , and define R := k[t1, ··· , tn]. Since k is a PID, R is a PID, and every finitely generated module over a PID is free.

Theorem 14.3(Bass, 1962). Let R be connected (recall: rules out silly case!) and noetherian. Then every inﬁnitely generated (i.e. not ﬁnitely generated) projective module is free.

So we can restrict our attention to the ﬁnitely generated case.

n Analogy: is every topological vector bundle trivial? E.g. for C , yes. Are all holomorphic bundles trivial? In general, no, see Stein manifolds. Question (Serre, 1950s): Is every projective R-module free? Answer: Yes, showed by Quillen, Suslin 1976. See book about this by T.Y. Lam.

Upcoming: Algebraic K-theory, built from f.g. projective R-modules. Trivial in K0 doesn’t quite imply free, usually weaker. Tries to analyze distinction between projective and free. Also some results about ﬂat modules.

15 Friday February 14th

Let R be a ring and consider K0(R). Measures diﬀerence between f.g. projective and free modules over R. Deﬁne (M(R), +) := the commutative monoid of isomorphism classes of f.g. projective R-modules with addition given by direct sum, i.e. [P ] + [Q] = [P ⊕ Q] with identity the zero modules, and K0(R) = G((M(R), +)) is the group completion, which any map M(R) −→ G a group factors through.

Concretely, any element of K0(R) is of the form [P ] − [Q], where

[P1] − [Q1] ∼ [P2] − [Q2] ⇐⇒ [M] + [P1] + [Q2] = [M] + [P2] + [Q2] for every ﬁnitely generated projective R- module M. Note that excluding the R here fails transitivity and thus doesn’t yield an equivalence relation.

15 FRIDAY FEBRUARY 14TH 40 If P,Q are finitely generated projective R-modules, then [P ] = [Q] iff ∃M such that P ⊕M =∼ Q⊕M iff there exists N a finitely generated projective such that M ⊕N =∼ Rn for some n, i.e. P ⊕Rn =∼ Q⊕Rn. In such a case, we say P,Q are stably isomorphic. Note that [P ] = 0 iff [P ] has rank zero, or [P ] ⊕ Rn =∼ Rn. Also note that [P ] =∼ [Q] can occur without necessarily having P =∼ Q as modules.

We can actually make K0(R) into a ring with [P ] · [Q] := [P ⊗R Q]. Note that the tensor product of two finitely-generated R-modules is still finitely generated as an R-module. Example: Let R be a PID, then M(R) is a commutative semiring (no additive inverses) and is equal to (N, +, ·) (occurs whenever )very finitely generated projective is free). Similarly G(R) = (N, +, ·). Since R has invariant basis number, there is always an injective group morphism

(Z, +) 7→ (K0(R), +) n 7→ [Rn].

Yields no cancellation among free modules. We want to essentially ignore this case, so we’ll mod out.

Deﬁnition 15.0.1 (Reduced Group). ˜ The reduced K group is given by K0(R) := (K0(R), +)/(Z, +).

˜ m ∼ n Note that [P ] = [G] in K0(R) iff there exist m, n such that P ⊕ R = Q ⊕ R . Moreover [P ] = 0 iff ∃m, n such that P ⊕ Rn =∼ Rn. In this case we say P is stably free. Exercise If P is a projective module (possibly not finitely generated) then there exists a free module F such that P ⊕ F is free. Example 15.1. DX 2 E For n ∈ Z, define Rn := R[t0, ··· , tn]/ ti − 1 . This is the ring of polynomial functions on the n-sphere. To construct a stably free module that is not free, take TSn for any n for which it’s trivial.

By Poincare Hopf, need euler characteristic zero, which happens when n is odd. Tangent bundle also trivial for lie groups.

Theorem 15.1(Bott-Milnor: Trivial Tangent Bundles of Spheres). The tangent bundle of Sn is trivial iﬀ n ∈ {1, 3, 7}.

If every module is free, they are stably free, yielding K0 = 0. ˜ Fact: If R is a dedekind domain, K0(R) = Pic (R), the ideal class group. So f.g. projectives need not be free, since ideals need not be principal. Theorem of Clayborn: Pic (R) can be any commutative group!

15 FRIDAY FEBRUARY 14TH 41 15.1 Flat Modules

Analogy: bundles are locally trivial, are projective modules “locally free”? We’ll need localization to make sense of this, but such a theorem turns out to be true.

Deﬁnition 15.1.1 (Local Ring). A local ring is a ring R with a unique maximal ideal, usually written (R, m).

× Exercise R is local iﬀ R \ R E R is an ideal. Localizing in the right way will yield local rings.

Lemma 15.2. Let q : R −→ R/m and x ∈ R, then x ∈ R× ⇐⇒ q(x) ∈ (R/m)×.

Proof . The forward implication holds for any ring. The converse doesn’t usually (think Z −→ Z/pZ). But if q(x) ∈ (R/m)× = R/m \ 0, then x ∈ R \ m = R×.

Theorem 15.3(Projective Modules Over Local Rings are Free). A f.g. projective module over a local ring is free.

This turns out to be true with “f.g.” dropped, but that is a harder theorem. To prove this, we’ll need the following:

Theorem 15.4(Corollary of Nakayama’s Lemma). For (R, m) a local ring and M a ﬁnitely generated R-module. Take a ﬁnite collection {mi} such that {mi} ∈ M/mM are generators as an R/m module. Then M is generated by {xi}.

Usually identiﬁed as Nakayama’s Lemma.

Proof (of ﬁrst theorem). Let P be a f.g. projective R-module for R a local ring. Choose Q such that P ⊕ Q = Rn. By base change, P/mP ⊕ Q/mQ = (R/m)n. a b So choose R/m bases {pi} of P/mP and {qj} of Q/mQ. Choose any lifts pi ∈ P, qj ∈ Q. Let A ∈ Mn,m(R) be the matrix formed by setting the ﬁrst columns to pi and the remaining to qj. Then det(A) mod m ∈ (R/m)×, and by the lemma, det(A) ∈ R× and thus A is invertible. So {pi, qj} are R-linearly independent, so {pi} span P by Nakayama’s lemma. Thus P is a free R-module.

15.1 Flat Modules Why are projective modules called such? See notes, characterization in terms of linear algebra and projection operators. Suppose we have a SES of R-modules and we tensor with some R-module M:

15 FRIDAY FEBRUARY 14TH 42 15.1 Flat Modules

0 N1 N2 N3 0

· ⊗RM

··· ··· N1 ⊗R M N2 ⊗R M N3 ⊗R M 0 Note that the induced map of the injection need not remain an injection. Example 15.2. ×2 Take Z 7→ ×2Z −→ Z/2Z, then taking · ⊗ Z/2Z yields Z/2Z −−→ Z/2Z, which is the zero map.

Definition 15.4.1 (Flat Modules). A module M is flat iff M ⊗R · is exact, i.e. if N1 ,→ N2 =⇒ M ⊗R N1 ,→ M ⊗R N2.

Proposition 15.5(Flat Implies Torsionfree in Domains). If R is a domain, then ﬂat =⇒ torsionfree.

Proof . For the contrapositive, suppose M is not torsionfree, then there exists some nonzero r ∈ R• and 0 6= m ∈ M such that rm = 0. Then take R −−→×r R, which is injective since R is a domain. Then tensoring with M yields M −−→×r M, which has nonzero kernel by assumption. M Exercise (Important) Let Mi be a family of R-modules, then Mi is flat iff Mi is flat for all i. i Use the fact that tensor commutes with direct sum, use functoriality of direct sum to sum maps.

Proposition 15.6(Projective Implies Flat). Projective =⇒ ﬂat.

We now have the chain:

Free =⇒ projective =⇒ ﬂat =⇒ R a domain torsionfree.

Proof (easy). By the exercise, P projective implies existence of a Q where P ⊕ Q is free, so it’s enough to ∼ show that free =⇒ flat. If F is free, F = ⊕iR, so F is flat iff R is flat. But R ⊗R R = R, which does not change a SES at all.

So ﬂat is somewhere between projective and torsionfree. The next theorem is related to Cayley-Hamilton.

15 FRIDAY FEBRUARY 14TH 43 15.1 Flat Modules

Theorem 15.7(Existence of Minimal Polynomials). Let M be a ﬁnitely generated R-module with generators {xi} and I E R, and take φ ∈ n EndR(M) such that φ(M) ⊆ IM. Then there exist a set of coeﬃcients {ai} such that n n−1 φ + an−1φ + ··· + a1φ + a0 = 0 ∈ EndR(M).

Proof (Sneaky). n n X For all i ≤ n, there exists a set {aij}j=1 ⊂ I such that φ(xi) = aijxj. Equivalently, for all i, j=1

n X (δijφ − aij) = 0. j=1

Let P be a matrix with entry i, j equal to δijφ − aij ∈ Mn×n(R[φ]) where R[φ] ≤ EndR(M) is the subalgebra generated by φ. Note that this makes the base ring commutative, so this t matrix makes sense. We can rewrite this as P · [x1, x2, ··· , xn] = 0. Claim: If S is a ring and P ∈ Mn×n(S), then there is an identity

P adj(P ) = adj(P )P = det(P )In.

Note that expanding this in the 2 × 2 case yields a collection of polynomial identities, which tend to remain true in arbitrary rings (see “permanence of polynomial identities”). Then det(P )Inx = adj(P )Mx = 0 (often called the determinant trick). Thus det(P )xi = 0 for all i. But then det(P )M = 0, and since M is a faithful R[φ]-module, we have det(P ) = 0. Then thinking of φ as a variable, expanding the determinant yields a monic polynomial in φ with coeﬃcients that are products of aij, which are in I.

Note the analogy to det(Iλ − A), so this yields the usual characteristic polynomial in the case of ﬁelds. Theorem 15.8(NAK, a.k.a Nakayama-Azumaya-Krull). Let J E R be an ideal and M ∈ R − mod ﬁnitely generated such that JM = M. Then a. ∃x ∈ R such that x =∼ 1 mod J and xM = 0. b. Suppose J ∈ J (the Jacobson radical), i.e. J is in every maximal ideal; then M = 0.

Note that if R were local, this reduces to a simple case.

Proof (of (a)). Apply the previous proposition to φ = idM and I = J; then the polynomial relation reduces to the existence of some x = 1 + an−1 + ··· + a0 with ai ∈ J, and this is equal to the zero endomorphism and thus xM = 0 and x = 1 mod J as desired.

Proof (of (b)). If J ∈ m for all m ∈ maxSpec (R), then if x = 1 mod J and x = 1 mod m, this forces x 6∈ m and so x ∈ R×. So if yx = 1 and $xM = 0 $, then 0 = yxM = M.

15 FRIDAY FEBRUARY 14TH 44

Corollary 15.9. Suppose J ∈ J and M ∈ R − mod is f.g. with N ≤R M a submodule such that JM + N = M, then N = M.

Proof . JM + N Apply part (b) above to M/N. If M is f.g. then so is M/N, and J(M/N) = = M/N N (just from pushing into quotients).

Deﬁnition 15.9.1 (Non-generator). An element x ∈ M is a non-generator if whenever S is a generating set for M, then S \ x is still a generating set.

Thus if you’re trying to ﬁnd generators for a module, it never helps to add elements of J.

Corollary 15.10. Let J ∈ J , M ∈ R − mod f.g., x1, ··· , xn such that {xi} ∈ M/JM are generators. Then M is generated by {xi}.

Proof . N + JM M Take N = h{x }i ≤ M. Then im (N) ⊂ M/JM is given by im (N) = = since i JM JM im (N) was assumed a generating set. But then N = M by the previous corollary.

Theorem 15.11(3.44, Generalized NAK). Let J E R, M ∈ R − mod f.g., then JM = M ⇐⇒ J + AnnM = R.

Proof .

⇐= JM = M ⇐⇒ M/JM = 0 ⇐⇒ AnnM/JM = R, and AnnM/JM = Ann(M ⊗ R/J) ⊇ J + AnnM = R. =⇒ : Exercise.

Exercise Why does this imply part (b) in NAK? Use the assumption that J, AnnM are comaximal, and J ∈ Z, which forces AnnM = R and thus M = 0.

16 Monday February 17th

Last time: R is a ring, M a ﬁnitely-generated R-modules, J E R.

16 MONDAY FEBRUARY 17TH 45 16.1 Injective Modules

Theorem 16.1(Nakayama). If M = JM, then there exists an x ∈ R with x = 1 mod (J) such that xM = (0).

Theorem 16.2(Generalized Nakayama). M = JM ⇐⇒ J + AnnM = R.

Proof . The reverse implication is immediate, the forward is by Nakayama.

16.1 Injective Modules

Exercise Show that for R = R1 × R2 and M = M1 × M2, Mi is an Ri-module. Recall that every R-module is free ⇐⇒ R is a ﬁeld. Question: What is the analogous condition for every R-module to be projective? Answer: Every SES

0 −→ M1 −→ M2 −→ M3 −→ 0 of R-modules splits.

Focusing on M2: every submodule M1 of M2 is a direct summand.

Theorem 16.3(Characterization of Semisimplicity). For an R-module M, TFAE 1. Every submodule of M is a direct summand. 2. M is a direct sum of simple modules (semisimple). 3. M is generated by its simple submodules.

Deﬁnition 16.3.1 (Simple Modules). M is simple iﬀ ∃0 ( N (R M.

In this case, M =∼ R/AnnM (i.e. cyclic, monogenic) and the AnnM is maximal.

Proof . Omitted, see chapter 8 of notes.

Thus every R-module is projective iﬀ every R-module is semisimple.

Deﬁnition 16.3.2 (Injective Modules). Dually, now focusing on M1, every SES starting with M1 is split iﬀ whenever M1 ≤ M2, M1 is a direct summand. In this case we say M1 is injective.

Proposition 16.4(Characterization of Semisimple Modules). For R a ring, TFAE

16 MONDAY FEBRUARY 17TH 46 16.1 Injective Modules

1. Every SES of R-modules splits 2. Every R-module is projective 3. Every R-module is semisimple 4. Every R-module is injective 5. (Claim) R is itself a semisimple R-module.

Proof . 3 =⇒ 5 is clear, and we’ll prove 5 =⇒ 3 shortly using Baer’s Criterion.

Deﬁnition 16.4.1 (Semisimple Rings). R is semisimple iﬀ for all I E R, there exists a J E R such that I ⊕ J = R. Moreover, Ann(I) = J and Ann(J) = I.

Exercise (easy) If Ri are semisimple, R1 × R2 is semisimple.

Corollary 16.5. Fields are semisimple, so any ﬁnite product of ﬁelds is semisimple.

In fact, the converse is true:

Theorem 16.6(Semisimple Rings are Products of Fields). If R is semisimple, then R is a product of ﬁelds.

Note that everything works here for left modules over non-commutative rings. Let R be a ring.

Theorem 16.7(Wedderburn-Artin). r a ∼ Y A ring R is semisimple iﬀ R = Mni (Di) a product of matrix rings over division rings. i=1

aPotentially non-commutative, but reduces to previous theorem in commutative case.

ι Let 0 −→ M1 −→ M2 −→ M3 −→ 0 be a SES. ∼ Note that splitting is slightly stronger than M2 = M1 ⊕ M3.

This sequence is split iﬀ there exists a retraction π : M2 −→ M1 such that ι ◦ π = idM2 . In this case, ∼ M2 = ι(M1) ⊕ ker π.

Deﬁnition 16.7.1 (Injective Modules). An R-module E is injective iﬀ every SES 0 −→ E −→ M2 −→ M3 −→ 0 admits a retraction π : M2 −→ E.

Theorem 16.8(Characterization of Injective Modules). For an R-module E, TFAE 1. E is injective

16 MONDAY FEBRUARY 17TH 47 16.1 Injective Modules

2. Reversing arrows of projective condition, there exists a lift of the following form: E

ϕ ∃Φ

0 M N 3. If M,→ N, then hom(N,E) hom(M,E). 4. The contravariant functor hom( · ,E) is exact.

Note big diﬀerence: no analog of being a direct summand of a free module! Free modules are usually not injective. Example 16.1. ×2 Z is a free but not injective Z-module. Take 0 −→ Z −−→ Z −→ Z/2Z −→ 0. If this splits, we would ∼ have Z = Z ⊕ Z/2Z as Z-modules. Why isn’t this true? Z is a domain, the LHS is torsionfree, and the RHS has torsion.

Example 16.2. Suppose now R is a domain and not a ﬁeld, then let a be a non-unit and run the same argument with multiplication by a. This would yield R =∼ R ⊕ R/aR, where the LHS is torsionfree and the RHS has torsion. So R itself need not be an injective R-module.

Deﬁnition 16.8.1 (Self-Injective Modules). A ring R is self-injective iﬀ R is injective as an R-module.

Example 16.3. A ﬁeld or a semisimple ring.

Claim 16.9. + n Let R˜ be a PID, π a prime element, n ∈ Z , then take R := R/˜ (π ). Then R is self-injective.

Example 16.4. n Let R = Z/p Z, and let M be a ﬁnite p-primary commutative group (i.e. a p-group). Then n n n exp M = p ⇐⇒ AnnM = (p ). M is a faithful Z/p Z-module, so there exists an element x ∈ M n n such that # hxi = p . There is a SES of Z/p Z-modules

0 −→ hxi −→ M −→ M/ hxi −→ 0.

∼ n ∼ Since hxi = Z/p Z, which is self-injective, so there exists a module N such that M = hxi ⊕ N = n Z/p Z ⊕ N. Continuing on N yields a decomposition of M into a sum of cyclic submodules.

16 MONDAY FEBRUARY 17TH 48 Conclusion: a ﬁnitely generated torsion module over a PID is a direct sum of cyclic modules. In general, to show a module is injective, we need to consider lifts over all pairs of modules M,→ N. How to do this in practice?

Theorem 16.10(Baer’s Criterion). If suﬃces to check the lifting condition for N = R and M = I E R. I.e. if there is a lift of the following form: E

ϕ ∃Φ

0 I R then E is injective.

Proof . Omitted for time.

Application Let R be a semisimple R-module and let E be any R-module. Let I E R, and f ∈ hom(I,E). If R is semisimple, then there exists a J E R such that R = I ⊕ J. So extend f to f ⊕ 0, which yields a lift: E

f (f,0)

0 I R = I ⊕ J

Exercise Prove the claim that R is self-injective for R = R/˜ (πn) above.

17 Monday February 24th

17.1 Divisible Modules We know that injective implies divisible, and uniquely divisible implies injective. Fact: quotients of divisible modules are divisible Exercise If R is a domain that is not a ﬁeld and M is a ﬁnitely-generated divisible R-module, then M = 0.

Proof (of exercise). Claim: for any ring R, any nonzero f.g. R-module M has a nonzero cyclic (monogenic) quotient given by modding out by all but one of the generators. Thus if M admits a ﬁnitely generated divisible R-module, it admits a cyclic module. Then M =∼ R/AnnM, and there are two cases: • AnnM = 0, in which case M =∼ R. Then choosing r ∈ R• \ R×, then [r]: R −→ R is not a surjection.

17 MONDAY FEBRUARY 24TH 49 17.2 Toward Localization

• Otherwise, choose x ∈ Ann(M) \{0}. Then ×x : R −→ R is the same map as ×0 : R −→ R, so it is not surjective.

Fact: there is a classification of divisible (iff injective) Z-modules: • (Q, +), since the fraction field of any domain is divisible. M n • (Q/Z, +) = Qp/Zp, where Qp/Zp = lim Z/p Z. This is isomorphic to the group of p primes power roots of unity. On the other hand, Q/Z is the group of all roots of unity Fact (Classification of Divisible Z-Modules: Any divisible Z-module is isomorphic to a direct sum of copies of

• (Q, +) • (Qp/Zp, +) Note that any direct sum of divisible groups is still divisible. Moreover, this decomposition is unique.

17.2 Toward Localization

Proposition 17.1(Multiplicative Avoidance). n o \ Let S ⊂ R with SS ⊂ S, 1 ∈ S, 0 6∈ S. Deﬁne I(S) = I E R I S = ∅ . Then 1. I(S) 6= ∅ 2. Every element of I(S) is contained in a maximal element of I(S). 3. Every maximal element of I(S) is prime.

Proof . In parts: a. (0) ∈ I(S) by construction. b. Standard Zorn’s lemma argument. c. Let I ∈ I(S) be a maximal element, and let x, y ∈ R such that xy ∈ I with x 6∈ I. Then \ hx, Ii ) I, so S hx, Ii= 6 ∅ by maximality. I.e., there exists s1 ∈ S such that s1 = i1 +ax for some a ∈ R. Continuing this way, if y 6∈ I, produce an s2 = i2 + by1 for some b ∈ R. Since S is multiplicatively closed, s1s2 ∈ S. But we also have s1s2 = (i1 +ax)(i2 +by) ∈ I, a contradiction.

See Kaplansky’s Commutative Algebra book.

Proposition 17.2(Prime Ideals Behave Like Primes). Y Let p ∈ Spec (R) and I1, ··· ,In E R, then if p ⊃ Ii, then p ⊃ Ii for some i.

Proof . Y Y Suppose p ) Ii for any i, and let xi ∈ Ii \ p. Consider x := xi ∈ Ii ⊂ p; then since p is prime, some xi ∈ p.

17 MONDAY FEBRUARY 24TH 50 17.3 Radicals

Corollary: If p ⊃ In, then p ⊃ I. I.e. prime ideals are radical.

17.3 Radicals

Definition 17.2.1 (Nilpotent Elements). n n An element x ∈ R is nilpotent iff x = 0 for some n ∈ Z. An ideal is nilpotent iff I = (0) for some n, and is nil iff every element x ∈ I is nilpotent.

Proposition 17.3(Nilpotent Implies Nil). Nilpotent =⇒ nil.

Proof . If In = (0), then for any x ∈ I, xn ∈ In = (0) so xn = 0.

Proposition 17.4(Nil and FG implies nilpotent). If I is ﬁnitely generated and nil, then I is nilpotent.

Proof . P ei ei Let I = hx1, ··· , xni. Then for each i, choose ei ∈ Z such that xi = 0. The (check) I = (0).

Deﬁnition 17.4.1 (Nil). An ideal is nil iﬀ all generators are nilpotent.

Corollary: If R is Noetherian, I is nilpotent iﬀ I is nil. Exercise Exhibit a ring with an ideal that is nil but not nilpotent. Need to choose a non-Noetherian ring, e.g. a polynomial ring in inﬁnitely many indeter- D E n minates {ti}, and consider tn n ∈ N .

Deﬁnition 17.4.2 (Nilradical). The nilradical of R, nil(R), is the set of all nilpotent elements.

Proposition 17.5(Universal Property of Nil).

n n 2n a. nil(R) E R, since a = b = 0 =⇒ (xa + yb) = 0. b. R/nil(R) is reduced, and this quotient map is universal wrt morphism into a reduced ring. I.e., if R −→ S with S reduced, there is commutative diagram

17 MONDAY FEBRUARY 24TH 51 f R S

π ∃f˜

R/nil(R) c. nil(R) = \ p. prime ideal

Proof (of c). ⊆: If x ∈ nil(R), then xn = 0 for some n, so xn ∈ p and since p is prime, x ∈ p. ⊆: We’ll show that if x is not nilpotent, then it avoids some prime ideal. Deﬁne S := n o n x n ∈ N ; since x is not nilpotent, S is multiplicatively closed and does not contain zero, so by a previous result, there is some p ∈ Spec (R) such that S \ p = ∅.

Deﬁnition 17.5.1 (Radical Ideals). n An ideal I E R is radical iﬀ for all x ∈ R there exists an n such that x ∈ I =⇒ x ∈ I.

Proposition 17.6(Prime Implies Radical). Prime ideals are radical.

Idea: the set of radical ideals is much easier to work with than the set of prime ideals.

18 Wednesday February 26th

18.1 Radicals n o n For R a ring, we deﬁned nil(R) := x ∈ R ∃n ∈ N, x = 0 E R. We had a theorem:

nil(R) = \ p p∈Spec (R)

Deﬁnition 18.0.1 (Radical of an Ideal). (Flat Implies Torsionfree in Domains) n o n For I E R, we deﬁne rad (I) = x ∈ R ∃n, x ∈ I ⊇ I.

Fact I E R. To see this, note that for any I E R, then nil(R/I) E R = rad (I). Deﬁnition 18.0.2 (Radical Ideals). I is a radical ideal iﬀ I = rad (I).

Example 18.1. Prime ideals are radical.

18 WEDNESDAY FEBRUARY 26TH 52 18.1 Radicals

Deﬁnition 18.0.3 (Closure Operators). Deﬁne a closure operator ` : I 7→ rad (I). In general, if (X, ≤) is a poset, then a Moore closure operator is a map c : X −→ X satisfying 1. c(c(x)) = c(x) 2. x ≤ c(x) 3. x ≤ y =⇒ c(x) ≤ c(y).

This is most often applied to X the family of subsets of a set A and ≤ subset inclusion. Note that this doesn’t completely correspond to a topological closure, since this would also require preservation of intersections. Related to Galois connections, not covering in this class but good for a ﬁnal topic. We can produce a nice characterization:

rad (I) = nil(R/I) = \ p = \ p p∈R/I p⊇I

\ Exercise (easy) If {Ii} is any family of radical ideals, then Ii is radical. i 2 Exercise Let R = Z. What are the radical ideals? (0), (p), but (p ) is not radical – i.e. (0), (n) for n squarefree.

∼ Y ai Fact I is radical iﬀ R/I is reduced. Noting that by the CRT, Z/nZ = Z/pi Z, which is reduced iﬀ ai = 1 for all i. If R is a PID, π1 ··· πr radical ideals, then (π1 ··· πr) nonassociate prime elements ??

Exercise Let R be a ring, p1 6= p2 prime ideals.

1. Must p1p2 be radical? \ 2. If p1 + p2 + R, then p1p2 = p1 p2, and is thus radical.

Product may be smaller than intersection in general.

Proposition 18.1(Algebraic Properties of Radicals). Let I,J E R. a. I ⊂ rad (I), rad (rad (I)) ⊂ rad (I), and I ⊂ J =⇒ rad (I) ⊂ rad (J). b. rad (IJ) = rad (I \ J) = rad (I) \ rad (J) c. rad (I + J) = rad (rad (I) + rad (J)) d. rad (I) = R ⇐⇒ I = R e. rad (In) = rad (I) for n ≥ 1 f. If R is Noetherian and J ⊂ rad (I), then J n ⊂ I for some n ≥ 1.

So for Noetherian rings, two radicals are equal iﬀ powers of each ideal land in the other.

Proof (of (b)). IJ ⊆ I \ J and thus rad (IJ) ⊂ rad (I \ J). If x ∈ rad (I \ J), there exists an n such that xn ∈ I \ J. Then x2n = xnxN ∈ IJ =⇒ x ∈ rad (IJ).

18 WEDNESDAY FEBRUARY 26TH 53 18.1 Radicals

Proof (of b). Since I \ J ⊂ I,J, we have rad (I \ J) ⊂ rad (I) \ rad (J). If x ∈ rad (I) \ rad (J), then xn ∈ I, xm ∈ J for some n, m, so xm+n ∈ I \ J ⊂ rad (I \ J).

Proof (of (f)). By replacing R with R/I, assume I = (0), then J ∈ nil(R) and since R is Noetherian, J is nilpotent and J n = (0) for some n.

So we simplify things by passing from I to rad (I). There is a class of rings for which it’s feasible to understand all radical ideals, and hopeless to understand all ideals. Example 18.2. n n Take R = C[x], a PID. Suppose I E R and rad (I) = x , then I = (x ). So this is no big deal, it’s just an extra integer parameter.

Now instead take R = C[x, y], and let I = hx, yi. Note that applying (f) above to J = rad (I), we ﬁnd that I ⊃ hx, yin for some n. But note that D E hx, yin = xn, xn−1y, ··· , xyn−1, yn .

2 Exercise Suppose I ⊃ hx, yi . For such I, dimC R/I < ∞. So for each d, try to ﬁnd all ideals I such that rad (I) = hx, yi and dimC R/I = d. Note that these correspond to “fat points” in algebraic geometry. The idea hx, yi corresponds to a fat point at zero. When doing AG, we hope to restrict attention entirely to radical ideals.

Deﬁnition 18.1.1 (Jacobson Radical). The Jacobson radical is deﬁned by

J (R) = \ m. m∈maxSpec (R)

Fact J (R) ⊃ nil(R), since not every prime ideal is maximal. Example 18.3. If (R, m) is a local domain, then nil(R) = 0 and J (R) = m.

Exercise Let R be a domain, show that J (R[t]) = (0).

Proposition 18.2(Characterization of Jacobson Radical in Terms of Units). x ∈ J (R) ⇐⇒ 1 ± xR ⊂ R×.

Exercise Show directly that xn = 0 =⇒ ∀y, 1 − xy ∈ R×.

18 WEDNESDAY FEBRUARY 26TH 54 19 Friday February 28th

19.1 Radicals: The Jacobson Radical

Deﬁnition 19.0.1 (The Jacobson Radical). J (R) = \ m ∈ maxSpec (R)m.

For a noncommutative ring, instead of intersecting just two-sided ideals, need to intersect either left ideals or right ideals (the intersections turn out to be equivalent). Fact If R is finite dimensional over a field, then J (R) = 0 ⇐⇒ R is semisimple. By Wedderburn, Y this happens iff R = Mni (Di).

Deﬁnition 19.0.2 (Semiprimitive). A ring is semiprimitive (or J -semisimple or Jacobson-semisimple) iﬀ J (R) = 0.

Proposition 19.1(Characterization of Jacobson Radical). x ∈ J (R) ⇐⇒ 1 − xR ⊂ R×.

Proof . Let x ∈ J (R) and suppose 1 − xy 6∈ R×, so 1 − xy ∈ m for some maximal ideal. But then x ∈ m, so xy ∈ m, so 1 = m + xy ∈ m, a contradiction.

Suppose instead that x 6∈ J (R), so there exists some maximal such that hm, xi = R. Thus for y ∈ R, m ∈ m, we have 1 = m + xy so 1 − xy = m ∈ m and thus 1 − xy 6∈ R×. In other words, R× + J (R) ⊂ R×, and is the largest ideal with this property. Thus the elements are “close to zero” in the sense that it doesn’t take you outside of the unit group.

19.2 Proposition (Commutative Algebra Analog of Euclid IX.20: Infinitely Many Primes) Let R be a domain, then recall that p ∈ R• is irreducible iff p 6∈ R× and p = xy =⇒ x ∈ R× or y ∈ R×. If p is irreducible and u ∈ R×, then up is irreducible and associate to p, and (up) = (p). Define an atom to be the principal ideal generated by an irreducible element. Define a Fursentenberg domain to be a domain such that x ∈ R• \ R× has an irreducible divisor. Note that we have a chain of implication, UFD =⇒ Noetherian = ACC =⇒ ACC on principal ideals =⇒ nonzero nonunits factor into irreducibles (atomic domain) =⇒ Fursentenberg. So this is a weak factorization condition.

Exercise Let R = Hol(C) be the ring of holomorphic functions, which is a domain by the identity theorem. Show that R is semiprimitive, Furstenberg but not atomic.

Theorem 19.2(Euclidean Criterion). Let R be a domain, not a ﬁeld, and semiprimitive. ∞ a. There exists a sequence of pairwise comaximal elements {an}n such that ham, aN i = R

19 FRIDAY FEBRUARY 28TH 55 19.2 Proposition (Commutative Algebra Analog of Euclid IX.20: Inﬁnitely Many Primes)

for m 6= n. b. If R is Forstenburg, then there is a sequence of primitive pairwise comaximal irreducible elements, and thus inﬁnitely many atoms.

Note that applying this to R = Z, the only unit ideals are generated by ±1, and the result follows immediately.

Proof . Exercise.

For what class of rings does this criterion apply? Application For R a Noetherian domain, then by Hilbert’s basis theorem R[t] is Noetherian and semiprimitive. So by the above result, R[t] is has inﬁnitely many elements. Most interesting for R = Fq, since for e.g. R = R we can consider ideals (x − r). Fact a. If I,J R and r(I) + r(J) = R, then I + J = R, and r(r(I)) + r(J) = r(I + J). E Y b. If I,J1, ··· ,Jn E R and I + Ji = R for each i, then I + Ii = R. Y \ c. Suppose I1, ··· ,In are pairwise comaximal, then Ii = Ii (note: could be smaller and general).

Theorem 19.3(Chinese Remainder). Suppose R is arbitrary with I1, ··· ,In E R pairwise comaximal. Then there is a natural map

n Y Φ: R −→ R/Ii i=1

r 7→ (r + I1, ··· , r + Ii). \ 1. Φ is surjective, and ker Φ = Ii. Y ∼ Y 2. By pairwise primality, R/ Ii = R/Ii.

Note that as modules, both sides are cyclic.

Proof . By induction on n, with trivial base case. n−1 0 Y 0 0 Let R := R/Ii and assume by induction that Φ : R −→ R is surjective by induction. Let i=1 n−1 0 0 0 Y 0 0 (r , s) ∈ R × R/In. By hypothesis, ker Φ = . So there exists an r ∈ R such that Φ (r) = r . i=1 Lifting to s ∈ R such that s + In = s + In. By assumption, n−1 ! 0 Y I + I := Ii + In = R. i=1 0 0 0 So there exist x ∈ I , y ∈ In such that s − r = x + y. Note that Φ (r + x) = r since x ∈ ker Φ,

19 FRIDAY FEBRUARY 28TH 56 19.3 Monoid Rings

rx = r + x + y = x mod In.

But then Φ(r + x) = (r0, s). Y Exercise (Converse to CRT (Good for Problem Sessions)) Let I1, ··· ,In E R. If R/Ii is a cyclic R-module, then the Ii are pairwise comaximal. Immediately reduce to n = 2 case. Also a nice proof using tensor products, use characterization of R/I ⊗ R/J.

19.3 Monoid Rings Here let R be a ring∗ (potentially noncommutative) and (M, ·) a monoid (i.e. a group without requiring inverses). Goal: we want to define a monoid ring R[M]. If M is finite, the definition is unambiguous, but for infinite M we require an extra condition. In this case we define the big monoid ring R[[M]]. Example 19.1. For R a nonzero ring and M = (N, +), R[M] = R[t], and R[[M]] = R[[t]].

Step 1: suppose M is ﬁnite, then

= M R[M] R = {f : M −→ R} , R−mod the set of all functions. Note that (f + g)(m) = f(m) + g(m), and deﬁne a new multiplication

(f ∗ g)(m) := X f(x)g(y), (x,y)∈M 2, xy=m the convolution product. One must check that this actually satisfies the axiom of a ring, since we are building this by hand. This is a ring iff R is a ring and (M, ∗) is commutative. There is an identity, namely 1 7→ 1 and x 7→ 0 for x 6= 1. Distributivity isn’t difficult, but we need to check that ∗ is associative. This follows from

((f ∗ g) ∗ h)(m) = X f(x)g(y)h(z) = (f ∗ (g ∗ h)). x,y,z∈M 3, xyz=m

! ! X X Deﬁne [m] · [n] = [mn], then check that rm[m] sm[m] =?. m∈M m∈M

19 FRIDAY FEBRUARY 28TH 57 20 Monday March 2nd

20.1 Semigroup and Monoid Rings For today, let R be a ring∗, so not necessarily commutative, and (M, ·) be a nonzero monoid, we then deﬁne the monoid ring R[M] by the following condition:

2 If M is infinite and divisor-finite, i.e. for all m ∈ M, the set {(x, y) ∈ M xy = m} is finite. Note that M finite implies divisor-finite, and M a group and divisor-finite implies finite. M For S a set, the free commutative monoid on S is given by (N, +). s∈S Example 20.1. ∞ >0 M (Z , ·) = (N, +) by the fundamental theorem of arithmetic. The map is given by n=1

n Y ti M = pi 7→ (ti). i=1

We deﬁne the big monoid ring R[[M]]. Note that R[M] and R[[M]] are commutative iﬀ R,M are commutative. For R[M], we try R[M] = RM = {f : M −→ R} with pointwise addition and instead of point wise multiplication, we take the convolution product

(f ∗ g)(m) := X f(x)g(y). xy=m

Note that this is a finite sum ⇐⇒ M is divisor-finite. Moreover, if M is divisor-finite, then this defines R[[M]].

For any M, we deﬁne R[M] as above not RM but rather M R ⊂ RM , i.e. those f : M −→ R with m∈M

ﬁnite support, so for all f, {m ∈ M f(m) 6= 0} < ∞.

Note that this makes the convolution product again a finite sum. For M divisor-finite, R[M] ,→ R[[M]], but in general the latter is larger. X Define [m] = δm, and expand f = rm[m]. Forming the product fg comes down to defining m∈M what [m1] ∗ [m2] := [m1m2] should be. We saw that this yields an associative product, since both ways of associating parentheses yield a sum that ranges over triples. X n For M = (N, +), we find R[M] = rn[n] where [n] ∗ [m] := [n + m], so we can define [n] := t , so n this is R[t]. M More generally, take M = (N, +) for S an arbitrary set, then s∈S

R[M] := R[{ts s ∈ S}]

20 MONDAY MARCH 2ND 58 20.2 Localization is a multivariate polynomial ring.

Consider also M = (Z, +). Since this construction should be functorial, there should be a containment ∼ −1 R[(Z, +)] ⊃ R[(N, +)] = R[t]. In this case, M = R[t, t ], the ring of Lauren polynomials. We can also identify R[[(N, +)]] = R[[t]] is the ring of formal power series over R, since we’re dropping the ﬁniteness condition.

Note that R = Z is not divisor ﬁnite, so we can’t necessarily take R[[M]].

Proposition 20.1(When Monoid Rings are Domains). For R a ring and (G, +) a commutative group, R[G] and R[[G]] are domains iﬀ R is a domain and G is torsionfree.

Note that R being a domain is necessary because it occurs as a subring via r 7→ r[1].

Proof (Idea). See notes. If g ∈ G[tors] \{0}, then [g] − [0] is a zero divisor.

See Kaplansky’s Group Ring Conjecture. Exercise Let R be a field, identify the fraction field of R[[t]]. Should be formal, finite-tailed Lauren series – but what does this mean? There is a universal property of monoid rings: Let R be a ring, (M, ·) a commutative monoid. Let B be an R-algebra. Then homR−alg(R[M],B) = hommonoid(M, (B, ·)) given by restriction. Thus if f : M −→ B is a monoid morphism, there exists a unique map

F : R[M] −→ B X X rm[m] 7→ rmf(m).

Note that this is the only possible map that could extend f. Exercise Check that this gives an R-algebra morphism. Note that the monoid ring is thus adjoint to the forgetful functor R − alg −→ Monoids. M Note that if M = (N, +), then s∈S

homMonoid(M,T ) = homSet(S, T ), i.e. it is fully determined by where it sends basis elements.

This yields a universal mapping property for polynomial rings, i.e. homR−alg(R[T ],B) = homSet(T,B).

20.2 Localization Let S ⊂ R with SS ⊂ S, with 1 ∈ S, then there exists a ring S−1R and a ring morphism ι : R −→ S−1R such that 1. For all s ∈ S, ι(s) ∈ (S−1R)×

20 MONDAY MARCH 2ND 59 2. ι is universal for property 1, i.e.

S−1R

∼ ∃!F

f R T Remark: when R is a domain, this reduces to the fraction ﬁeld construction, i.e. (R•)R = K = ff(R). For S any multiplicatively closed subset,

1 S−1R = R[ s ∈ S]. s

Make sense of partial group completion of a monoid with respect to a submonoid.

Construction: We’ll deﬁne S−1R = (R × S)/ ∼, where

(r1, s1) ∼ (r2, s2) ⇐⇒ ∃s ∈ S such that sr1s2 = sr2s1.

Recall that this stabilization is needed, and becomes clear if you try to carry out the construction without it. If R is a domain, the s appearing can just be canceled. Deﬁne maps

(r1, s1) + (r2, s2) := (r1s2 + r2s1, s1s2)

(r1, s1) · (r2, s2) := (r1r2, s1s2)

Need to check that this is well-deﬁned. Exercise Check that localization satisﬁes the universal mapping property. Question: what is ker(R −→ S−1R) where R 7→ (r, 1)? This has to do with the s that appears in the stabilization.

21 Wednesday March 4th

For R a ring and S ⊂ R such that S2 ⊂ S and 1 ∈ S, there exists a ring S−1R and a ring morphism ι : R −→ S−1R such that 1. ι(S) ⊂ (S−1R)×, and f f˜ 2. ι is universal for such morphisms, i.e every R −→ T with f(S) ⊂ T × lifts to S−1R −→ T .

Last time we constructed it as R × S/ ∼ where (r1, s1) ∼ (r2, s2) iﬀ there exists an s ∈ S such that sr1s2 = sr2s1 (needed to obtain transitivity). We then have ι(r) = [(r, 1)]; what is its kernel? If (r, 1) ∼ (0, 1) then there exists s ∈ S such that s · r · 1 = s · 1 · 0 = 0, so Ann(r) \ ker ι 6= ∅. Note that if 0 ∈ S then ker ι = R and thus S−1R = 0. Conversely, if 0 6∈ S, then Ann(1) \ S = ∅, so S−1R 6= 0. Thus S−1R = 0 iﬀ 0 ∈ S.

21 WEDNESDAY MARCH 4TH 60 21.1 Pushing and Pulling

Example 21.1. −1 n 2 o For f ∈ R, Rf := Sf R where Sf = 1, f, f , ··· , then Rf = 0 ⇐⇒ f ∈ nil(R).

Deﬁnition 21.0.1 (Saturated Multiplicatively Closed Sets). A multiplicatively closed set S is saturated iﬀ for s ∈ S, f ∈ R with f dividing S, then f ∈ S. Denote the S the saturation of S obtained by adding all divisors, then S−1R = S−1R.

Recall link to early problem of characterizing rings between Z and Q. There are more localizations than such rings, since localizing at n is as good as localizing at kn. If R is a domain, then for any S with 0 6∈ S, there is a diagram

R S−1R

K where K = ff(R). In any ring, take S to be the nonzero divisors, then there is a maximal injective localization.

ιR S−1R

Total fraction ﬁeld

Can generalize results from domains to arbitrary rings this way.

−1 Exercise Take R1,R2 nonzero rings, R = R1 × R2, and take S = R1 × {1}. What is S R? (First ﬁgure out the kernel of the localization.)

21.1 Pushing and Pulling

Note that we can push/pull for quotients and get back what we started with – want something similar for localization. Consider the map ι : R −→ S−1R.

Lemma 21.1. x I R implies that ι (I) = x ∈ I, s ∈ S . E ∗ s

Proof . Easy.

Proposition 21.2(Push-Pull Equality for Ideals in Localizations). −1 For all J E S R,

∗ ι∗ι J = J.

21 WEDNESDAY MARCH 4TH 61 21.1 Pushing and Pulling

Proof . Note that we always have containment, just need to show reverse containment.

Lemma 21.3. For I E R,

−1 \ i∗(I) = S R ⇐⇒ I S 6= ∅.

Proposition 21.4(Properties of Spec for Localization).

a. For p ∈ Spec (R), TFAE: −1 • ι∗p ∈ Spec (S R) −1 • ι∗ ( S R • p \ S = ∅ \ ∗ b. If p S = ∅, then ι ι∗p = p.

Corollary 21.5. ∗ i and i∗ are mutually inverse, order-preserving bijections

i∗ n o −1 \ Spec (S R) p ∈ Spec (R) p S = ∅ . i∗

Lemma 21.6. For I E R, S a multiplicatively closed set, let f : R R/I be the quotient map and S := q(S). Then

S−1R ∼ −→= S−1(R/I) IS−1R a a + I + IS−1R 7→ . s s + I

Thus localizing commutes with taking quotients.

Let p ∈ Spec (R), then Sp := RSetminusp is multiplicatively closed. (Note that localizing at any −1 non-prime ideal gives you the zero ring.) Let Rp := Sp R.

Proposition 21.7(Complements of Prime Ideals are Local? Extremely Impor- tant!). Rp is a local ring with a unique maximal ideal pRp,

Proof . n o \ The poset Spec (Rp) = q ∈ Spec (R) q (R \ p) = ∅ ⇐⇒ q ≤ p .

21 WEDNESDAY MARCH 4TH 62 21.2 Localization for Modules

This gives us a way to construct a local ring from any maximal ideal. Perhaps the most important construction thus far.

Exercise Let (R, m) be local and S ⊂ R be multiplicatively closed. Show that S−1R need not be local.

21.2 Localization for Modules Let M be an R-module and S ⊂ R multiplicatively closed. We want S−1M to satisfy: • S−1M is an R-module [s] • There is a morphism M −→ S−1M such that for all s ∈ S, the map S−1M −→ S−1M is an −1 −1 isomorphism, i.e. S −→ EndR(S M) with i(S) ⊂ EndR(S M) • This is universal wrt the above property. There are two potential constructions. Construction 1: Adapt the S−1R construction, defining S−1M = M × S/ ∼. −1 −1 −1 Construction 2: Define S M := S R ⊗R M, where ι : M −→ S M where m 7→ 1 ⊗ m. It can be checked that these both satisfy the appropriate Universal mapping property. Exercise If M is an R-module, then M has an S−1R-module structure iff S acts invertibly (so [s]: M −→ M is invertible), and if so the structure is unique.

22 Monday March 30th

We’ll cover localization and Noetherian rings. We’ll need some local to global results.

Corollary 22.1. If R is a domain with fraction ﬁeld K, then \ Rm = R m∈maxSpec R

It follows from this that the analogous statement for prime ideals holds.

Proposition 22.2(7.11: Expressing Colon Ideal as Annihilator). Let M1,M2

n o M + M 1 2 (M1 :M M2) = x ∈ R xM2 ⊆ M1 = Ann . M1 So the colon is an ideal. Suppose that S ⊂ R be a multiplicatively closed subset satisfying a. If M is ﬁnitely generated, S−1AnnM = AnnS−1M. −1 −1 −1 b. If M2 is ﬁnitely generated, then S (M1 : M2) = (S M1 : S M2).

22 MONDAY MARCH 30TH 63 Proof (of Proposition). Omitted, see notes.

Proof (of Corollary). n o \ We want to show that R,→ Rm. If x ∈ K \ R and I := (R : Rx) := r ∈ R rx ∈ R , note that 1 6∈ I. Thus I is a proper ideal, so let m ∈ maxSpec R with m ⊃ I. Then (Rm : Rmx) = Im =? ⊂ mRm which is a proper ideal in Rm. So 1 is not in the colon ideal.

These colon ideals aren’t the obvious thing to look at, but come up in applications to algebraic geometry and number theory. n o

Remark For I,J E R, we have (I :R J) = x ∈ R xJ ⊂ I . Thus this construction formalizes the idea of a “quotient I/J”. This works for ideals in a domain, but also for fractional ideals.

Deﬁnition 22.2.1 (Fractional Ideal). A fractional R-ideal is a nonzero R-submodule I of K such that there exists an x ∈ R• such that xI ⊂ R.

Any ideal is a fractional ideal by taking x = 1. Note that some books define fractional ideals as finitely generated R-submodules, but this isn’t a great definition. Exercise (Easy) If I ⊂ K is finitely generated, then I is a fractional ideal. Idea: scale all generators. n o

Note that I is a fractional R-ideal iﬀ (R :K I) = x ∈ K xI ⊂ R is nonzero. Next up: local-global theory for lattices.

Theorem 22.3(Local-Global Principle for Lattices). Let R be a domain with fraction field K and V a finite dimensional K-vector space. Let Λ ⊂ V \ be a finitely generated R-submodule. Then Λm = Λ. m∈maxSpec R

Note that if V = K and Λ = R, this recovers the previous theorem. Thus a global lattice over an integral domain can be recovered in terms of its localizations. Next up: rounding out some theorems about projective and free modules.

Theorem 22.4(Kaplanksy, Very Important: Projective Over Local Implies Free). A projective module over a local ring is free.

We proved this for ﬁnitely generated modules, which is the most important case. Note that projective modules are locally free, i.e. if M is a projective R-module then for all p ∈ Spec R, Mp is a free Rp-module. We’ll now deﬁne a notion of “the least number of generators” locally.

22 MONDAY MARCH 30TH 64 Definition 22.4.1 (Rank Function). Suppose M is a finitely generated R-module. For p ∈ Spec R, denote k(p) = ff(R/p) the residue field at p. We have R R/p ,→ k(p), so we define the rank function as

rankM : Spec R −→ N p 7→ dimk(p) M ⊗R k(p).

where the RHS is base-changing to k(p) to get a ﬁnite dimensional vector space over k(p).

Exercise Show the following properties of the rank function for M,N ﬁnitely generated R-modules:

a. rankM⊕N = rankM + rankN and rankM⊗RN = rankM · rankN . b. Compute the rank function on Z/nZ for R = Z. c. For R a PID, compute rankM . • Taking p = pZ in Z yields a delta function at p. • If M is ﬁnitely generated and free, then

n n M n rankM (p) = dimk(p) R ⊗R k(p) = dimk(p) R ⊗R k(p) = dimk(p) k(p) = n. i=1

∼ rankM (p) • If M is locally free, then for all p ∈ Spec R, we have Mp = Rp . R Rp

k(p) – Thus the rank can be thought of as the fiberwise dimension for bundles. m ∼ n • If M is stably free, i.e. there exists an m, n ∈ N such that M ⊕ R = R , then rankM = n − m. Note that projective implies locally free. In order for a finitely generated projective module to be free, it must have constant rank function. The geometric analog here would be that the fibers having constant dimension is necessary for a bundle to be trivial.

Proposition 22.5(Determining if a Projective is Free). Suppose M is ﬁnitely generated projective of constant rank n. Then M is free iﬀ M can be generated by n elements.

23 Wednesday April 1st

Let M be a ﬁnitely generated R-module, then we deﬁne a rank function

rankM : Spec R −→ N p 7→ dimk(p) M ⊗R k(p) where k(p) = ff(R/p).

Question: If p1 ⊂ p2, how do the ranks compare?

23 WEDNESDAY APRIL 1ST 65 Example: Take R = Z and M = Z⊕Z/(10), then take (0) ⊆ (p). Then rankM ((0)) = dimQ M ⊗ZQ = 1. However,

(2 p = 2, 5 rankM ((p)) = . 1 else

If M is locally free, e.g. projective, and p ⊆ p then M = M ⊗ R is a canonical isomorphism. 1 2 p1 p2 Rp2 p1 Then

Rrankp1 = M = M ⊗ = Rrankp2 ⊗ R =∼ Rrankp2 . p1 p1 p2 Rp2 p2 p1 p1

Thus rank(p1) = rank(p2). In other words, tensor to the fraction ﬁeld and take the dimension.

Proposition 23.1(FG Projective of Rank N Generated by N Elements). If M is ﬁnitely generated projective of constant rank n then M is free ⇐⇒ M can be generated by n elements.

Lemma 23.2. If M is ﬁnitely generated projective and I ∈ J (R), then M/IM free implies M is free.

Nakayama: “ﬁnite generators in quotient” lifts to ﬁnite generators in total module.

Proof . M/IM =∼ (R/I)n since it’s ﬁnitely generated and free, so this ﬁts the hypothesis of Nakayama’s lemma. So the last number of generators for M is n. Then for any m ∈ maxSpec R, then after base change we get a diagram m M/IM

M/mM Since M/mM is free of ﬁnite rank n, ???????.

Corollary 23.3(FG Projective Over Semilocal Implies Free iﬀ Constant Rank). If R is semilocal and M is a ﬁnitely generated projective module, then M is free ⇐⇒ rankM is constant.

Proof . =⇒ : We know free modules have constant rank function. n n n \ Y Y ⇐= : maxSpec R = {m1, ··· , mn}, and J (R) = mi = mi. Thus R/F(R) = R/mi, i=1 i=1 i=1 n ∼ Y and M/J (R)M = M/miM. So we just need to show that the dimension is independent of i=1 i. But this follows from the rank function being constant, since the rank equals the dimension for each factor.

23 WEDNESDAY APRIL 1ST 66 23.1 Chapter 8: Noetherian Rings

Proposition 23.4(Flatness is Local). For M an R-module, then M is flat ⇐⇒ Mp is flat for all p ∈ Spec R ⇐⇒ Mm is flat for all m ∈ maxSpec R.

Noetherian: f.g. projective iﬀ locally free.

Theorem 23.5(7.2, Extremely Important: FG Projective iff FG by Localizations). For an R-module M, TFAE • M is finitely generated and projective • M is finitely presented and locally free

• There exist f1, ··· , fn ∈ R such that hf1, ··· , fni = R and for all 1 ≤ i ≤ n, Mfi is a free

Rfi -module.

Here Mfi means localize at the powers of fi, i.e. M ⊗ Rfi .

Corollary 23.6.

• Finitely generated and flat implies projective. • For M finitely generated and R Noetherian, projective ⇐⇒ locally free ⇐⇒ flat (important!)

A module M is Z-locally free (Zariski) iff there exists elements fi ∈ R such that hfii = R and Mfi is free for all i. Note that Z-locally free implies locally free. ∞ Y Example: R = F2 and let I E R not be finitely generated. Note that R is not Noetherian since i=1 Z it’s an infinite product of nonzero rings – just identify as functions F2 and take the maximal ideal where the first coordinate is zero (?). Then R/I is an R-module is finitely generated and flat (even though R isn’t a domain) but not projective, locally free but not Z-locally free. Thus the conditions in the hypotheses of the corollary are necessary, particularly being Noetherian.

23.1 Chapter 8: Noetherian Rings For (X, ≤) a poset, then X is Noetherian iff it satisfies the ACC, i.e. there does not exist an + order-embedding Z ,→ X, and X is Artinian iff it satisfies the DCC, i.e. there’s no embedding − Z ,→ X. Note that X is Noetherian iff X∨ is Artinian, where X∨ is given by x ≤∨ y ⇐⇒ y ≤ x. We’ll generally be interested in the poset of submodules of a given module and set inclusion. Recall that M is Noetherian ⇐⇒ every submodule is finitely generated, which is easy to show.

Exercise Show that for Z-modules, Noetherian and Artinian are two diﬀerent conditions by exhibiting the 4 possibilities.

Theorem 23.7(Artinian iﬀ Noetherian and Primes are Maximal). R is Artinian iﬀ R is Noetherian and dim R = 0, i.e. prime ideals are maximal, maxSpec R = Spec R.

23 WEDNESDAY APRIL 1ST 67 So Artinian is much much stronger than Noetherian.

24 Friday April 3rd

Recall that the deﬁnition of a normal series for G a group.

Theorem 24.1(Jordan Holder). Any two composition series for the same group G are equivalent (same isomorphism classes of quotients and multiplicities).

There is an analog of this for modules, even over a noncommutative ring: this is just a sequence of submodules inclusions, since normality is automatic. There is similarly a notion of Schreir refinement. For p groups, the composition factors have to be cyclic of order p. On one hand, we could fix the series and ask for what modules have a compatible composition series – this is the extension problem, and is difficult in general. Here we will fix the module and see what the possible composition series are. Question: When does an R- module admit a finite composition series? Answer: When R is both Noetherian and Artinian.

Suppose that M satisﬁes the ACC and DCC. Then there exists a minimal simple module M1 < M, an M2 properly containing M1 such that M2/M1 is simple, and so on. This sequence of inclusions terminates due to the ACC, so this yields a ﬁnite composition series.

Deﬁnition 24.1.1 (Length). ?

Proposition 24.2(Length is Additive over SESs). For 0 −→ M 0 −→ M −→ M 00 −→ 0, M has ﬁnite length iﬀ M 0,M 00 do, and `(M) = `(M 0) + `(M 00).

Dream of commutative algebra: every theorem at the level of generality of “Let M be a module over a Noetherian ring”.

Proposition 24.3(Quotient/Localization )f Noetherian is Noetherian). Quotients and localizations of Noetherian rings are Noetherian.

Proposition 24.4(Ideal Poset of Quotient Order-Injects into Ideal Poset of Full Ring). For I E R, I(R/I) ,→ I(R) is an isotone inclusion of posets.

Thus Noetherian-Artinian properties in the RHS imply the same properties in the LHS. For localizations, we also have I(S−1R) ,→ I(R) by push-pull properties.

Proposition 24.5(Artinian Domains are Fields). If R is an Artinian domain, then R is a ﬁeld.

24 FRIDAY APRIL 3RD 68 Proof . • × 2 For the contrapositive, let a ∈ R \ R , then (a) ) (a ) ) ··· is an inﬁnite descending chain.

Theorem 24.6(Properties of Artinian Rings). For R Artinian, a. dimKrull R = 0. b. J (R) = nil(R). n c. maxSpec R = {mi}i=1 is ﬁnite. d. nil(R) is a nilpotent ideal.

Proof .

a. If p ∈ Spec k, R/p is an Artinian domain and thus a ﬁeld, so p is maximal. n n+1 Y Y b. Produce a descending chain m1 ⊃ m1m2 ⊃ · · · and suppose that mi = I, then n Y mi ⊂ mn+1 and thus mn+1 ⊃ mi for some i, which is a contradiction. c.? d.?

Theorem 24.7(Akizuki-Hopkins). R is Artinian ⇐⇒ R is Noetherian and dim R = 0.

25 Monday April 6th

Last time: a characterization of Artinian rings, the Akizuki-Hopkins theorem

Theorem 25.1(Akizuki-Hopkins). R is Artinian iﬀ R is Noetherian with Krull dimension 1, i.e. all primes are maximal.

Proposition 25.2(Powers of Maximal Ideals in Local Rings). Suppose (R, m) is Noetherian and local. Then either n n+1 1. m m for all n, or 2. mn = (0) for some n. Moreover, (2) holds iﬀ R is Artinian.

Proof . If mn = mn+1 for some n, then by Nakayama mn = (0). If R is Artinian, then (1) can not hold, so (2) must hold. Conversely, if (2) holds, then m ∈ nilR = \ p, Spec R but this can only happen if Spec R = {m} is precisely one ideal. But then every prime ideal is maximal.

25 MONDAY APRIL 6TH 69

Note that Artinian rings have ﬁnitely many maximal ideals.

Theorem 25.3(Primary Decomposition). n n Y Let R be a nonzero Artinian ring and suppose maxSpec R = {mi} . Then R = ri where i=1 ri := Rmi .

n n Y ai Y ai Note that for R = Z and N = pi , this recovers Z/NZ = Z/pi Z. i=1 i=1 Punchline: Artinian rings split as a product of local rings. Modules over local rings are free, and thus simple, so this gives a local-to-global principle. Moreover, every localization is a projection onto one of the factors. Exercise For R Artinian, show that every x ∈ R is either a unit or a zero divisor. Thus R is its own total fraction ring. Exercise For a ring R, TFAE 1. R is semilocal (ﬁnitely many maximal ideals). 2. R/J (R) is Artinian. 3. R/J (R) has ﬁnitely many ideals. Most important theorem in the course: Hilbert’s basis theorem!

Theorem 25.4(Hilbert’s Basis). If R is Noetherian, then R[t] is Noetherian.

Proof . Toward a contradiction, let J E R[t] that is not finitely generated. Construct a sequence {fn} and take Jn := hf1, ··· , fni ⊆ J. This can be done by taking f0 = 0 and fn+1 any element of J/Jn of minimal degree. Note that this ensures that deg fn ≤ deg fn−1. Set an to be the leading coefficient of fn, and let I = h{ai}i. Since R is Noetherian, I is finitely generated, so there is some N ∈ Z such that I = ha1, ··· , aN i. N N X X deg fN+1−deg fi Thus an+1 = uiai for some ui ∈ R. So set g := uifit . Then g ∈ JN and i=1 i=1 fN ∈ J/Jn and fN+1 − g ∈ J \ JN . X Now the leading term of g is uiai = aN+1, and since deg g = deg fN+1 where aN+1 is also the leading term of fN+1. Thus deg(fN+1 − g) < deg fN+1, contradicting minimality.

Theorem 25.5(Formal Power Series over Noetherian is Noetherian). R Noetherian implies that R[[t]] is Noetherian.

Exercise If R is a ring with R[t],R[[t]] both Noetherian, then R is Noetherian. Idea: use the fact that quotients of Noetherian rings are again Noetherian.

25 MONDAY APRIL 6TH 70 Corollary 25.6(Single Most Important Result!). If R is Noetherian then every ﬁnitely generated R-algebra is Noetherian.

If such an algebra is ﬁnitely generated by n generators, it’s a quotient of R[x1, ··· , xn]. Some historical notes on the Hilbert Basis Theorem: Given G a group and T a ring, we can consider actions G −→ Aut(R) and the ring of invariants n o G T = t ∈ T gt = t ∀g ∈ G ⊂ T.

Note that Galois theory fits into this framework. For a classical example, consider T = C[t1, ··· , tn]. Then G ⊂ GL(n, C) y T by linear automorphisms, i.e. it acts on each ti by taking it to some linear combination of tj. Note that G can be chosen to be “nice”, i.e. a linear algebraic group. G Question: is T finitely generated as a C-algebra? Hilbert proved that this is true when G is a linear algebraic group, and the main step was the basis theorem. Previously, people were proving these kinds of theorems for a single group at a time, whereas this encompassed all of them simultaneously! Quote by Gordon: “This is not Mathematics, this is theology.” This is an early triumph of abstraction in algebra, as opposed to writing out lines upon lines of equations for single proofs. How effective is this proof? This doesn’t necessarily lead to a good algorithm for finding a finite generating set, see computational commutative algebra. Exercise For k a field, k[x, y] is Noetherian. Show that k[y, xy, x2y, ··· ] is not Noetherian. Note that things work out very nicely for Noetherian rings of dimension 0 and 1, but many theorems fail in higher dimensions. Exercise (Possibly difficult to prove) Show that every k-subalgebra of k[x] is Noetherian.

26 Wednesday April 8th

Theorem 26.1(Krull Intersection, 8.39). For R Noetherian and I E R a proper ideal, ∞ a. If there exists an x ∈ \ In, then x ∈ xI. i=1 b. Suppose either R is a domain or I ⊂ J (R), then \ In = (0).

Proof .

a. Omitted. b. If x ∈ \ In, then (a) implies that there exists an a ∈ I such that x = xa. Then x(1 − a) = 0. Since R is a domain and a 6= 0, then 1 − a is not a zero divisor.

\ n Exercise Exhibit a proper ideal I E R Noetherian such that I 6= (0). n Note: there is a very small ring that will work.

26 WEDNESDAY APRIL 8TH 71 ∞ Exercise (Recommended, Diﬃcult Calculus) Consider R = {f : R −→ R, f ∈ C } ⊂ RR and n o

m := f ∈ R f(0) = 0 .

1. Show m E R and R/m = R and thus m is maximal. 2. Show m = (x). n o n n 0 n+1 3. Show that for all n ∈ Z, we have m = (x ) = f f(0) = f (0) = · · · − f = 0 . n o \ n 4. Show that m = f T0(f) ≡ 0 6= (0) 5. Show that f 6∈ fm (so f 6= xg for some other smooth g) therefore R is not Noetherian.

Start with a smooth function vanishing at zero, divide by x, deﬁne value at zero to make it 2 continuous. Example: f(x) = e−1/x for x 6= 0 and f(0) = 0.

Theorem 26.2(Principal Ideal Theorem / Krull’s Hauptfdealsatz). For R Noetherian, x ∈ R \ R×, p ∈ Spec (R) minimal over (x) (to be explained). Then ht(p) ≤ p, where we recall that that height is given by the number of ideals below it.

Corollary 26.3. Every nonzero ring has a minimal prime.

In particular if p = (x) is a prime ideal in a Noetherian ring, ht(p) ≤ 1.

Corollary 26.4. Under these conditions, if x is not a zero divisor then ht(p) = 1.

Proof . We can reduce R to Rp (using how ideals are pushed into localizations). Then Rp is a Noetherian local ring of krull dimension 0, hence is Artinian by Akazuki. By a previous theorem, pRp is thus nilpotent.

n So there exists an n such that x = 0 in Rp which is thus in the kernel of the localization map, and there exists an a ∈ R/p such that axn = 0 in R. Since x is not a zero divisor, so multiplication by x (and all compositions) is injective, xn is not a zero divisor and thus a = 0. But since a ∈ R/p, so a 6= 0, a contradiction.

Thus the number of generators of a prime ideal in a Noetherian ring is bounded below by its height.

Lemma 26.5(Page 44). [ For p, {qi} ∈ Spec (R), if p ⊂ qi then p ⊂ qi for some i. i

Proof omitted.

Exercise In R = C[x, y], m := hx, yi, show that ht(m) ≥ 2 and m is the union of all of the principal ideals it contains. Thus the ﬁniteness in the previous lemma is necessary. n o

For (X, ≤) a poset and x ≤ y, we deﬁne the interval (x, y) := z ∈ X x ≤ z ≤ y :

26 WEDNESDAY APRIL 8TH 72 Corollary 26.6. For p ⊂ q a proper containment in a Noetherian ring, (p, q) is either empty or injective.

Note that this implies that for possible line order of length n, there is a ring with Spec (R) having that structure.

26 WEDNESDAY APRIL 8TH 73 27 Friday April 10th [ Last time: prime avoidance, for p, {qi} ∈ Spec R, if p ⊂ qi, then p ⊂ qi for some qi.

Theorem 27.1(Ideal Intervals in Noetherian Rings are Empty or Inﬁnite). For R Noetherian and p ( q ∈ Spec R, then (p, q) := {n ∈ Spec R s.t. p ( n ( q} is either empty or inﬁnite.

Proof . Wlog by passing from R to R/p, assume p = (0), and toward a contradiction assume that the [ theorem doesn’t hold. Then (∅, q) = {pi}, and by prime avoidance, there exists an x ∈ q \ pi and thus q is a minimal prime over x. By Hauptﬁdealsatz, ht(q) ≤ 1, but this is a contradiction because ht(q) ≥ 2.

27 FRIDAY APRIL 10TH 74 Proposition 27.2(Generalized Hauptﬁdealsatz). For R Noetherian, I = hx , ··· , x i . R, if p ∈ Spec R is minimal over I, then ht(p) ≤ n. 1 n .

Corollary 27.3. If p ∈ Spec (R) for R Noetherian, then R needs at least ht(p) generators.

Complete intersection: minimal number of deﬁning equations. Last item on Noetherian rings, used to prove Nullstellansatz:

Lemma 27.4(Artin-Tate Lemma). If R is Noetherian and R ⊂ T ⊂ S with S finitely generated as a T -module and finitely- generated as an R-algebra, then T is finitely generated as an R-algebra.

27 FRIDAY APRIL 10TH 75 27.1 Boolean Rings

27.1 Boolean Rings

Deﬁnition 27.4.1 (Boolean). A ring R is Boolean if every element is idempotent, i.e. x2 = x for all x ∈ R.

Exercise If R is Boolean, then 1. R = R×, 2. Every quotient is Boolean, 3. Every subring is Boolean, 4. (More interesting) every ideal is radical. Exercise Suppose R is Boolean. ∼ 1. If R is a domain then R = Z/2Z. 2. dim R = 0 ∼ Exercise If R is Boolean and local, then R = Z/2Z.

Proposition 27.5(Characterization of Finite Boolean Rings). For R Boolean, TFAE: 1. R is ﬁnite. 2. R is Noetherian. 3. Spec R is ﬁnite. n In any of these cases, (Z/2Z) .

Proof .

• 1 =⇒ 2 is clear because any finite ring is Noetherian. • 2 =⇒ 3: using Akazuki-Kopkins, R is Artinian and thus Spec R = maxSpec R is finite. • 3 =⇒ 1: By a previous exercise, since maxSpec R is finite, R/J (R) is a finite product of fields (by CRT essentially). Then J (R) = nil(R) = {0} therefore R is a finite product ∼ Y of fields, forcing R = Z/2Z.

27.2 Stone Duality This leads to Stone duality, an early example of categorical equivalence. Let R be a Boolean ring, Spec R ⊂ maxSpec R equipped with the Zariski topology generated by the basis

{U(f) s.t. m ∈ Spec R, f 6∈ m}f∈R . Then V (f) := Spec R \ U(f) = {m ∈ Spec R s.t. f ∈ m} .

For any R, V (f) is closed and V (f) ⊆ U(1 − f). But R is boolean iff f(1 − f) = 0, and if 1 − f 6∈ m, then f ∈ m and thus V (f) is open. So Spec R has a base of clopen sets, which is referred to as a zero-dimensional space. Moreover Spec R is also Hausdorff, and we’ll later see that Spec R is Hausdorff iff dim R = 0.

27 FRIDAY APRIL 10TH 76 Exercise For any R, Spec R is quasicompact (where compact is quasicompact and Hausdorff). Spec R is zero-dimensional and compact. Fact: • A zero-dimensional Hausdorff space is totally disconnected, • A totally disconnected locally compact space is ? X is a Boolean space if it is ? and compact, iff compact and totally disconnected. There is a notion of Stone space, equivalently a profinite space is an inverse limit of finite discrete spaces, which happens iff compact and totally disconnected. To any Boolean topological space X we attached its characteristic ring C(X). The elements are clopen subsets of X with U + V := U∆V = U \ V [ V \ U and U · V := U \ V . Note that any algebra of sets can be made into a ring in this way, and we note that U · U = U \ U = U, so C(X) is a Boolean ring.

27.2.1 Statement of Stone Duality For notation, set M(R) := Spec R, and deﬁne functors

M {Boolean rings} {Boolean spaces} . C

This gives a mutually inverse pair of functors which are naturally isomorphic to the identity, i.e. there is a canonical isomorphisms C(M(R)) =∼ R and M(C(X)) =∼ X. Exercise There exists a ring R and x, y ∈ R such that (x) = (y) but there exists a u ∈ R× such that x = uy

28 Monday April 13th

28.1 Nullstellansatz n Let k be a ring (later a ﬁeld) and Rn := k[t1, ··· , tn]. If x ∈ k and f ∈ Rn, we can evaluate f at x since f(x) ∈ k, and we thus get an evaluation map

kn E : Rn −→ k f 7→ (x 7→ f(x)) which is a ring homomorphism, where the RHS is regarded as a large direct product of rings. When can we identify polynomials (abstract elements of a ring) with the corresponding polynomial function? Exercise Suppose k is a domain. a. If k is inﬁnite, then E is injective but not surjective. b. If k is ﬁnite, the E is surjective but not injective.

28 MONDAY APRIL 13TH 77 28.1 Nullstellansatz

Deﬁnition 28.0.1 (Incidence Relation). n Put a relation on R × k where f ∼ x ⇐⇒ f(x) = 0R.

We’ll deﬁne an antitone Galois connection

V kn I(R) 2 I where the RHS is the powerset. We deﬁne V which takes an ideal and yields a subset of kn, n o n V (J) = x ∈ k ∀f ∈ J, f(x) = 0 n o

I(S) = f ∈ R ∀x ∈ S, f(x) = 0 .

That this forms an ideal is immediate. The pair (I,V ) are antitone (order-reversing) and ﬁts into the (extremely general formal) framework of Galois connections. This can be seen by writing V (J) = \ V (f) and I(S) = \ I({x}). f∈J x∈S Therefore (by the formalism) there are associated closure operators: kn 1. 2 ,S 7→ S := V (I(S)) 2. I(R) ,J 7→ J := I(V (J)). These are both isotone maps (order-preserving) and are closure operators on posets, i.e. they satisfy • (X, ≤) 7→ (X, ≤), x ≤ y =⇒ C(x) ≤ C(y) (isotonicity) • x ≤ C(x) • C(C(x)) = C() (idempotence) n o n o n n By construction, S = x ∈ k f(s) = 0 =⇒ f(x) = 0 and J = f ∈ k g(x) = 0∀g ∈ S =⇒ f(x) = 0 . Example: Take k a ﬁeld, S ⊂ k. Then

(SS is ﬁnite S = . k else

[ [ Exercise For k a ﬁeld, show that for all S1,S2 ⊂ k, we have S1 S2 = S1 S2.

This mirrors what closures in a topological space would do, so S −→ S is in fact a Kuratowski closure operator and therefore is the closure operator for a unique topology (here, the Zariski topology, where the closed subsets are given by V (I)). For n = 1, the Zariski topology on k is the cofinite topology. If k is finite, this is discrete, and is very coarse and non-Hausdorff when k is infinite since any two open sets intersect. These maps satisfy a tridempotence relation, i.e. VIV = V and IVI = I, and are antitone bijections when restricted to closed sets. Exercise The closure operator on ideals never satisfies the Kuratowski property. So we don’t have a topology on the ring/ideal side, since even unions of ideals may fail to be ideals. We take the description sending S −→ S to be fairly explicit, but what is the corresponding description for J −→ J?

28 MONDAY APRIL 13TH 78 n n For all S ∈ k , I(S) k[t1, ··· , tn], we use the fact that f (x) = 0 =⇒ f(x) = 0 and thus \E I(S) = rad I(S) = p, and V (J) = V (rad J). Moreover, for all J E R, we have rad J ⊂ J. p⊃I(S)

Passing from an ideal to its radical is a closure operator, so does J = rad J for all J E R? Not for D 2 E all k, e.g. take k = R, n = 1,J = t + 1 . Then V (J) = ∅, but J = I(∅) = R[t] vacuously. Note ∼ that R/J = C, so J is maximal and hence radical, so J ) rad J = J. What went wrong? Potentially k 6= k. + Exercise Let k be a ﬁeld, ﬁx n ∈ Z , and show that if all m ∈ maxSpec (R) satisfy m = rad m, then L is algebraically closed.

Exercise Suppose k = k is algebraically closed and n = 1, show that for all J E k(t), we have J = rad J. Theorem 28.1(Hilbert’s Nullstellansatz). + If k = k and for all n ∈ Z and J E R = k[t1, ··· , tn], J = I(V (J)) = rad J Therefore I,V induce a galois correspondence

nRadical idealso Zariski closed subets n of k[t1,··· ,tn] ⇐⇒ of k .

Exercise If k = k, show that the above theorem implies the weak Nullstellansatz, i.e. the map

n =∼ I : k −→ maxSpec k[t1, ··· , tn]

x 7→ mx = I({x}) = ht1 − x1, ··· , tn − xni

is a bijection. Try without using the formalism of Galois connections. Proof sketch: we’ll spend most of our time proving the weak Nullstellansatz, and use a trick to bootstrap up to the full theorem. The closure operator is only the radical for algebraically closed – what is the closure operator for other ﬁelds.

29 Wednesday April 15th

Recall that we had mutually inverse maps V,I and for J E R, we had a closure J = I(V (J)) and S = V (I(S)). Hilbert’s Nullstellensatz says that if k = k, then for all J, we have J = rad J and there is a Galois n correspondence between radical ideals of k[x1, ··· , xn] and Zariski closed subset of Ak . The weak Nullstellensatz gives a correspondence between maximal ideals and points of kn. For S = S = kn, we have S = [ {x} and by the Galois correspondence, we ﬁnd that if J is radical x∈S then J = \ m ⊃ Jm.

29 WEDNESDAY APRIL 15TH 79 Deﬁnition 29.0.1 (Jacobson). A ring R is Jacobson if every radical ideal is the intersection of the maximal ideals containing \ it, i.e. for all I E R we have rad I = m m⊃I

Exercise For x ∈ kn, deﬁne n o

mx := I({x}) = f ∈ R f(x) = 0 .

1. Show that m = ht1 − x1, ··· , tn − xni and R/mx = k. 2. Show that the following map is injective:

I : kn −→ maxSpec R

x 7→ mx. n o n 3. Show that I(k ) = m ∈ maxSpec R R/m = k .

Hint: consider the images of t1, ··· , tn in R/m. The following result will imply the weak nullstellensatz:

Lemma 29.1(Zariski, 1947). Let k be a field, R a finitely generated k-algebra, m ∈ maxSpec R. Then [R/m : k] < ∞. Equivalently, if K/k is a field extension that is finitely generated as a k-algebra, then K is finitely generated as a k-vector space.

Proof .

Case 1: K = k(α1, ··· , αn) is algebraic.

In this case you get the obvious tower of extensions where each extension is ﬁnite, so [K : k] is ﬁnite.

Case 2: K/k is not algebraic.

We can choose a transcendence basis t1, ··· , tn, and since K is finitely generated as a field extension over k, the transcendence degree is finite. Then k ⊂ k(t1, ··· , tn) ⊂ K and K is algebraic over k(t1, ··· , tn), so [K : k(t1, ··· , tn)] < ∞. By the Artin-Tate lemma, k(t1, ··· , tn)/k is a finitely generated k-algebra. But then k(t1, ··· , tn) is finitely generated over k(t1, ··· , tn−1), so it suffices to consider the case of one variable. In other words, we need to show that for all k, k(t) is not a finitely generated k-algebra.

Supposing otherwise, let {ri(t)} be a finite set of rational functions that are generators. Factor each ri as fi/gi Since k[t] is a PID with infinitely many primes, i.e. infinitely many nonassociate irreducible polynomials.

So choose q monic, irreducible, not equal to any of the f ; then 1 , a contradiction. i q6∈k[f1,··· ,fn]

29 WEDNESDAY APRIL 15TH 80

So rational function fields are not finitely generated over their base fields, or even finitely generated over their polynomial rings. Exercise Show that a PID is Jacobson iff Spec R is infinite. Exercise Show that R is Jacobson iff for every p ∈ Spec R, we have p = \ m. p⊂m∈maxSpec R

Proposition 29.2(Rabinovitch? Trick). + For k a ﬁeld and n ∈ Z , k[t1, ··· , tn] is a Jacobson ring.

To see why the weak Nullstellansatz and this trick imply the Nullstellansatz, we show that for J E R = k[t1, ··· , tn], rad J = J = I(V (J)) = I(V (rad J)) = rad J, so wlog we can assume J is radical and show that J = J. Taking J a radical ideal,

\ RT \ WN \ J = p = m = ma. n p⊂J m⊃J a∈k ,ma⊃J

Then J ⊂ ma iﬀ for all f ∈ J, f(a) = 0, i.e.

a ∈ V (J) = \ I({a}) = I(V (J)) = J. a∈V (J)

Next time: proof of RT.

30 Monday April 20th

Exercise If f : R −→ S is a ring morphism, then f ∗ : Spec S −→ sec R is a continuous map of topological spaces. Exercise Deﬁning Rred := R/mR, show that q∗ : Spec Rred −→ Spec R is a homeomorphism. Y a Exercise The bijection Spec Ri = Spec Ri is a canonical homeomorphism for ﬁnite (co)products.

Proposition 30.1(Spec Is Quasicompact). For any ring, Spec R is quasi-compact, i.e. every open cover has a ﬁnite subcover.

Here, “compact” will denote quasi-compact and Hausdorﬀ.

Proof (Sketch, Important to Check). The open sets U(f) := Spec R \ V (f) are a base for the Zariski topology. It suﬃces to check

30 MONDAY APRIL 20TH 81 that every open cover by basic open sets has a ﬁnite subcover. Can check that that the unit ideal is generated by ﬁnitely many elements.

Theorem 30.2(Idempotents Corresponds to Clopen Subsets of Spec). For any ring R, there is a bijection

{Idempotents of R} ⇐⇒ {clopen subsets of Spec R} e 7→ U(e)

where on the RHS we have e(1 − e) = 0 =⇒ V (e) = V (1 − e).

Note that a space is connected iff the only clopen sets are trivial, thus Spec R is connected iff the only idempotents of R are 0, 1. Thus a ring is connected, i.e. it’s not a nontrivial product of rings, iff its spectrum is a connected space.

Theorem 30.3(Topological Properties of Zero Dimensional Rings). For a ring R, TFAE • dim R = 0, so every prime ideal is maximal. • Spec R is separated, i.e. T 1, points are closed, i.e. for disjoint points, disjoint neighborhoods exist (note that maxSpec R is always separated and quasicompact) • Spec R is Hausdorﬀ • Spec R is a Boolean space (compact and totally disconnected)

Theorem 30.4(13.10: Characterization of Gelfand Rings). A ring R is Gelfand iﬀ every prime ideal is contained in a maximal ideal (e.g. a local ring). TFAE: 1. R/J (R) is Gelfand. 2. maxSpec R is compact. 3. maxSpec R is Hausdorﬀ.

Theorem 30.5(5.9,11: Maximal Ideals in Rings of Continuous Real-Valued Func- tions). For a topological (Hausdorﬀ?) space X, deﬁne C(X) to be the ring of R-valued functions f on X. If X is compact, then there is a bijection

∼ X −→= maxSpec C(X) n o

c 7→ mc = f ∈ C(X) f(c) = 0 .

Recall Kolmogorov spaces are deﬁned as those for which distinct points never have precisely the same set of neighborhoods.

Theorem 30.6(Hocharter: When Spaces Are the Spec of a Ring). Let X be a topological space. ∼ a. X =Top maxSpec R for some R ⇐⇒ X is quasi-compact and separated. b. X ∼ Spec R ⇐⇒ X = lim Y where Y is a ﬁnite Kolmogorov space. =Top ←− i i

30 MONDAY APRIL 20TH 82 For fun: ﬁnd where this theorem is applied in the literature!

Deﬁnition 30.6.1 (Irreducible). A topological space is irreducible if it is nonempty and not the union of two proper closed subspaces.

Exercise Show that X is Hausdorﬀ and irreducible iﬀ #X = 1. Exercise For a topological space X, TFAR: 1. X is irreducible. 2. Finite intersections of nonempty open sets are nonempty. 3. Every open subset is dense. 4. Every open subset is connected.

Proposition 30.7(Properties of Irreducible Spaces). For X a nonempty topological space, a. If X is irreducible, then every nonempty open subset is irreducible. b. If Y ⊂ X is irreducible, then its closure Y is irreducible. \ c. If X is is covered by {Ui} and for all i, j we have Ui Uj 6= ∅, then X is irreducible. d. The continuous image of an irreducible space is irreducible.

Recall that we have a correspondence

I {Radical ideals of R} {Closed subsets of Spec R} V .

Question: What is V (p) for p prime? Note that the radical of an ideal could be prime without the ideal itself being prime.

Proposition 30.8(Important: Variety is Irreducible iﬀ Radical is Prime). For I E R, then V (I) is irreducible iﬀ rad I is prime.

Proof . WLOG, we can take I = rad I. If p ∈ Spec R is prime and V (p) is reducible, then we will have a contradiction. In this case, there exist I,J E R such that V (I),V (J) ( V (p) with V (p) = V (I) [ V (J) = V (IJ). So p = rad IJ ⊃ IJ, so p ⊃ I or p ⊃ J. WLOG p ⊃ J, then V (p) ⊂ V (I), which is a contradiction.

Conversely, suppose V (I) is irreducible and I is radical but not empty? Then there exist a, b ∈ R such that ab ∈ I but a, b 6∈ I. So rad a, rad b 6⊂ I, so V (I) 6⊂ V (a) or V (b). But then V (a) \ V (I) and V (b) \ V (I) are proper closed subsets with intersection (?) V (I). Then (ab) ⊂ I implies that V (I) ⊂ V (b) = V (ab) = V (a)V (b) (?).

30 MONDAY APRIL 20TH 83 Under what hypotheses does Spec R have only ﬁnitely many closed subsets? Uses this correspondence, and the next proof will show that this is true when the ring is Noetherian.

31 Wednesday April 22nd

Recall that we had an antitone bijection

prime ideals of R ⇐⇒ irreducible closed subsets of sec R maximal ideals ⇐⇒ Singletons {p} , p closed minimal primes ⇐⇒ maximal irreducible closed subsets.

Note that the last item on the RHS corresponds to an irreducible component.

Deﬁnition 31.0.1 (Noetherian Space). A topological space X is Noetherian iﬀ the ascending chain conditions holds for open sets.

Proposition 31.1(13.14: In Noetherian Spaces, Every Subset is Quasicompact). X is a Noetherian space iﬀ every open subset U ⊂ X is quasicompact. iﬀ all subsets are quasicompact.

Note that this is a strange class of spaces! Exercise If X is Noetherian and Hausdorff, then X is finite. Conversely every finite space is Noetherian, regardless of whether or not it is Hausdorff.

Proposition 31.2(Noetherian Spaces are Finite Direct Sum )f Connected Compo- nents). A Noetherian space has ﬁnitely many connected components and is the direct sum (disjoint union) of its connected components.

This holds e.g. for manifolds, but not every topological space. Thus if we’re considering the topology of such spaces, we may as well assume they’re connected. Recall Spec R is connected when R is connected (no nontrivial idempotents), and a similar condition holds here: Theorem 31.3(Spec R is Noetherian iﬀ ACC Holds for Radical Ideals). For a ring R, TFAE 1. Spec R is Noetherian. 2. The ACC holds on radical ideals of R.

Note that if R is Noetherian, (2) holds, so this is a weaker condition.

Proof . This follows from the Galois correspondence.

31 WEDNESDAY APRIL 22ND 84 31.1 Integral Extensions

Thus if R is Noetherian, Spec R is Noetherian. Next up: any Noetherian ring has ﬁnitely many minimal primes.

Proposition 31.4(Noetherian Spaces Have Finitely Many Irreducible Compo- nents). A Noetherian space has ﬁnitely many irreducible components,

Step 1: Show that X is a finite union of irreducible closed subsets. Let f be the set of counterexamples: closed subsets of X that are not a finite union of irreducible closed subsets. Since X is Noetherian, if f 6= ∅ it has a minimal element Y . Since Y is not [ irreducible, we have Y = Y1 Y2 which are proper closed subsets. By minimality, Y1,Y2 6∈ f, so they must be finite unions of irreducible closed subsets, and thus so is Y . a a So f must be empty, and X = Y1 ··· Yn with the Yi irreducible. If there exists an i, j such that n [ Yi ( Yj, so remove Yi and similarly all redundancies. This yields a cover X = Yi where Yi 6⊂ Yj i=1 for any i, j.

Claim 31.5. The irreducible components of X are precisely the Yi.

Proof . n [ \ \ Let Z be irreducible and closed, Z = Z Yi . Since Z is irreducible, Z Yi = Z for i=1 some i, so Z ⊂ Yi.

Corollary 31.6. R Noetherian implies MinSpecR is ﬁnite.

This wraps up everything we’ll say about the Zariski topology.

31.1 Integral Extensions Leading toward normalization theorems, potentially Krull-Akuzuki. We’ll ﬁrst generalize the notion of “algebraic” for ﬁelds

Deﬁnition 31.6.1 (Integral Elements of Rings). For R ⊂ S rings, α ∈ S is integral over R if there exists a monic P ∈ R[t] such that P (α) = 0.

Note that we don’t require P to be monic. We say R ⊂ S is integral iﬀ every α ∈ S is integral over R.

31 WEDNESDAY APRIL 22ND 85 31.1 Integral Extensions

Theorem 31.7(Characterization of Integral Elements). For R ⊂ T 3 α, TFAE: 1. α is integral over R 2. R[α] is ﬁnitely generated as an R-module. 3. There exists an S such that R ⊂ S ⊂ T such that α ∈ S and S is ﬁnitely generated as an R-module.

This says that an integral extension is “locally finite”. Example 31.1. For L/K an algebraic field of extension of infinite degree (e.g. Q/Q), then K ⊂ L is integral but not finitely generated as a K-module.

Note that we’ve used this to prove Zariski’s lemma. n o

Exercise Take Z to be the set of algebraic integers, i.e. Z = α ∈ C α is integral over Z , then Z ⊂ Z is integral but Z is not ﬁnitely generated as a Z-module.

E.g. show that Z is not Noetherian. If R ⊂ S is integral, then S is finitely generated as an R-module iff S is finitely generated as an R-algebra.

Proposition 31.8(Integral Properties Push to Quotients and Localizations). Let R ⊂ S be integral, then \ a. For J E S, there is an injection R/ J R ,→ S/J and this is integral. • b. For T ⊂ R multiplicatively closed, localization induces RT ,→ ST which is integral.

Corollary 31.9(Towers of Successive Integral Containments are Integral). For R ⊂ S ⊂ T , if R ⊂ S and S ⊂ T are integral then R ⊂ T is integral.

Proof . Take α ∈ T , then there exist b0, ··· , bn−1 ∈ S such that

n n−1 α + bn−1α + ··· b1α + b0 = 0.

Then R[{bi} , α]$ is finitely generated as an R[{bi}]-module. Since R ⊂ S is integral, R[{bi}] is finitely as an R-module. Therefore R[{bi} , α] is finitely generated as an R-module as well. By condition (3) in the integrality condition theorem, α is thus integral over R.

31 WEDNESDAY APRIL 22ND 86 32 Monday April 27th

32.1 Normalization

Theorem 32.1(14.19: Integral Closure is Local). A domain R is integrally closed iﬀ Rp is integrally closed for all p ∈ Spec R iﬀ Rp is integrally closed for all m ∈ maxSpec R.

Recall that the height of a prime ideal is the supremum of lengths of chains. Height one corresponds to principal.

Deﬁnition 32.1.1 (Regular Prime Ideals). pRp For R a Noetherian ring, p ∈ Spec R is regular iﬀ dimRp/pRp 2 = ht(p). (pRp)

This corresponds to a mild nonsingularity property for varieties.

Theorem 32.2(14.21: Going Down). For R ⊂ S an integral extension of domains, R is integrally closed

∃p1 p2 Spec S

⊂ p1 p2 Spec R

Theorem 32.3(14.22, Noether Normalization). Let k be an arbitrary field, R a domain and a finitely generated k-algebra. Then a. There exists a d ∈ N and algebraically independent elements y1, ··· , yd such that k[y1, ··· , yd] ⊂ R is a finitely generated module. b. dim R = d equals the transcendence degree of ff(R)/k.

Application: Noether normalization implies Zariski’s lemma. Remark (Separable Noether Normalization): In the above setup, if k is perfect that one can choose the yi such that ff(R)/k(y1, ··· , yd) is ﬁnite and separable.

32.2 Invariant Theory For G a ﬁnite group and R an arbitrary object of a category C, the notion of a group action by makes sense: namely a group morphism G −→ AutC(R).

Deﬁnition 32.3.1 (Ring of Invariants). For a ring R and a group acting by ring automorphisms, the ring of invariants is given by n o G R := x ∈ R gx = x∀g ∈ G .

Theorem 32.4(Invariant Rings are Integral). RG ⊂ R is integral.

32 MONDAY APRIL 27TH 87 32.3 Normalization Theorems

Proof . Y For x ∈ R, deﬁne Φx(t) := (t − gx). We can extend the action of G by acting on coeﬃcients g∈G G G and acting on t trivially, yielding Φx ∈ (R[t]) = R [t] and thus Φx(x) = 0.

Proposition 32.5(Ring of Invariants Over Integrally Closed is Integrally Closed). If R is integrally closed, so is RG.

Note that this doesn’t hold with regularity! Here RG corresponds to “quotient varieties” in AG, which have mild “quotient singularities”. Important point: in dimension 1, normal coincides with regular. Example 32.1 (Important in Other Areas). If G = {±1}, R = C[x, y] and let −1 act by C-algebraic automorphisms acting trivially on C and sending x 7→ −x, y 7→ −y. Then

G 2 2 D 2E R = C[x , xy, y ] = C[X,Y,Z]/ XZ − Y which is a conic, and in particular an aﬃne quadric cone. Since we started out with a UFD, which is integrally closed (and in fact regular), this is integrally closed but the maximal ideal hX,Y,Zi + (XZ − Y 2) is singular.

Note that here the Picard group is trivial but the divisor class group has order 2.

Theorem 32.6(14.32, Noether 1928). Suppose R is a finitely generated k-algebra and G acts on R by k-algebra automorphisms. Then a. R is finitely generated as an RG-module. b. RG is finitely generated as k-algebra.

This says one can take a quotient of an affine coordinate ring by an affine variety again yields an affine coordinate ring.

Proof .

a. R is finitely generated as an RG-algebra and RG ⊂ R is integral, which implies that R is finitely generated as an RG-module. b. Apply the Artin-Tate lemma for k ⊂ RG ⊂ R where k ⊂ R is a finitely generated algebra and RG ⊂ R is a finitely generated module.

32.3 Normalization Theorems Normalization is a geometric synonym for integral closure. Next up: theorems giving good properties of the integral closure.

32 MONDAY APRIL 27TH 88 32.3 Normalization Theorems

Let R be an integrally closed domain with fraction ﬁeld K, L/K a ﬁeld extension, and set S = IL(R) the integral closure.

S ⊂ L

R ⊂ K We know that S is integrally closed, ff(S) = L, and R ⊂ S is integral, we can conclude that dim R = dim S. Questions: 1. Is S finitely generated as an R-module? 2. Is S Noetherian? WLOG, we can assume L/K is algebraic since we can pass to fraction fields, and integral over a field implies algebraic. Also note that if [L : K] is infinite, then we shouldn’t expect either answer to be “yes”. Example 32.2. Take R = Z, k = Q,L = Q,S = Z. Then Z is not Noetherian and not finitely generated as a Z-module.

Is S ﬁnitely generated over R or even Noetherian? Not necessarily, we need to assume R Noetherian integrally closed and [L : K] ﬁnite.

Theorem 32.7(18.1, First Normalization Theorem). Supposing that R is an integrally closed domain, L/K ﬁnite and separable, then a. If R is Noetherian, so is S. ∼ [L:K] b. If R is a PID, S =R R (free of ﬁnite rank).

Proof uses discriminant and some linear algebra. There are (complicated) counterexamples: R a DVR in characteristic 2, L/K a quadratic extension. Upcoming: two more normalization theorems.

32 MONDAY APRIL 27TH 89