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Commutative Algebra Background

Commutative Algebra Background

Background

September 18, 2007

1 Spectra and Localization.

1.1 Prime Spectra of rings.

Rings, unless otherwise signaled, will be commutative rings with unit element, and homomorphism of rings will be assumed to preserve the unit element. A denoted k will be understood to be a field, and k[X,Y,...] the (commutative) ring of in the independent variables X,Y,... with coefficients in the field k. The standard list of notation for the rings of rational , rational numbers, reals, and complexes, will also be understood: Z, Q, R, C.

If A is a ring, Spec A, its prime spectrum (for short: its spectrum) is—as a set—the set of prime ideals of A.

Discuss: Spectrum versus Max Spectrum, with examples of Banach algebras.

If f : A → B is a ring-homomorphism, the pullback, i.e., full inverse image, of any prime of B under the homomorphism f is a of A. This gives us a map on spectra Spec B → Spec A that we will denote by Spec(f) (or, sometimes, by just f again) allowing us to consider Spec as a contravariant functor from the of commutative rings to the category of sets. If f : A → B is a surjective ring-homomorphism, then

Spec(f) : Spec B → Spec A is an injection.

1.2 The “Universal solution” to the problem of inverting a given set of elements in a ring.

Let A be a ring, and S ⊂ A any subset. Consider the “problem” of finding ring homomorphisms A → A0 such that the set S gets sent to a set of units in A0. The “universal solution” to this problem (which exists, and is therefore easily seen to be unique up to unique isomorphism) is a ring

1 −1 homomorphism ιS : A → S A which sends S to units, and has the property that any “solution” 0 −1 −1 A → A is obtained by composition of ιS with a unique homomorphism A → S A. The ring S A is called the localization of A with respect to the set S.

Since the problem of inverting a set S is equivalent to the problem of inverting the multiplicative system in A generated by S, we may—when it is convenient for us to do so—suppose that the set S we wish to invert contains the identity element of A and is closed under multiplication.

Exercise 1. If A is a ring and S ⊂ A a subset, let {Xs}s∈S be a collection of independent variables; −1 show that S A can be given as the quotient of the ring A[Xs; s ∈ S] by the ideal generated by the polynomials s · Xs − 1 for s ∈ S. Exercise 2. If A is a ring and S ⊂ A a multiplicative system, show that any element in the of A → S−1A is also in the kernel of A → {f}−1A for some element f ∈ S. Exercise 3. If A is a ring and S ⊂ A a multiplicative system, show that the kernel of A → S−1A consist precisely in the elements of A which are annihilated by some element of S. (Hint: Let a ∈ A be in that kernel. By the previous exercise we may replace S by (the multiplicative system generated by) a single element s and our element a will “still” be in the kernel. But now we’re home since S−1A = A[X]/(Xs − 1) so we get that a = g(X) · (Xs − 1) is a equation in A[X] for some polynomial g(X) ∈ A[X]. Then compute.)

This operation A 7→ S−1A is indeed a localization operation for on the level of spectra we have that the map induced by ιS, i.e., −1 Spec(ιS) : Spec S A → Spec A is injective, and its image is the complement of

{P ∈ Spec A | P ∩ S 6= the empty set}.

Discuss the difficulties of extending this localization procedure to noncommutative rings (or more general set-ups) these difficulties suggesting the scope of geometry.

Here are some important special cases of localization:

• Let A be an integral , and suppose that the multiplicative system S ⊂ A doesn’t contain 0 ∈ A. In this case, if K is the field of fractions of A (K being the localization of A relative to the multiplicative system A − {0}), then

S−1A = {a/s ∈ K | a ∈ A, s ∈ S};

• Suppose the subset S is generated by finitely many elements s1, s2, . . . , sm of A. In this case Q −1 −1 put s = i si and you can take S A to be the ring A[1/s] := A[s]/(st−1) with A → S A the −1 natural homomorphism A → A[1/s]. Denoting X := Spec A and Xf := Spec S A we have that Xs ⊂ X is the open subset consisting of the set of all points of Spec A that correspond to prime ideals P in A that do not contain the element s; visually: “the complement of the locus of zeroes of s.”

2 • For a prime ideal P ⊂ A take S = A − P . Then S is a multiplicative system. We usually −1 denote S A as AP (the localization of A at P ). We have that AP is a , with equal to PAP ⊂ AP , and with residue field AP /P AP = the field of fractions of the A/P . The set of prime ideals of AP pull back to the set of prime ideals of A contained in P .

Define the on spectra as follows: If A is a ring, the open sets on Spec A is given by the complements of the injective mappings Spec(φ) : Spec B,→ Spec A where φ runs through all surjective ring-homomorphisms with domain A. A for the open sets is given by the images −1 2 of the mappings Spec(Sf A) → Spec A for f ∈ A, where Sf = {1, f, f ,...} is the multiplicative set generated by f.

For any homomorphism f of rings, the induced morphism Spec f is a continuous map of spectra.

1.3 Some examples

• Regarding the spectra of local rings:

– If A = k is a field, the zero ideal is the only prime ideal, so Spec A consists of one point. – If A is a local ring, its unique maximal ideal corresponds to the unique closed point in Spec A. – If A is a local ring with nilpotent maximal ideal (e.g., Z/p2Z, or any other artinian local ring) then Spec A consists of one point.

• More generally,

– if A is a ring and N ⊂ A is its nilradical, consider the associated to A, denoted Ared := A/N . The natural projection A −→ Ared induces a homeomorphism on spectra, but wait ...1;

– if A is noetherian and we let P1,P2,...,Pν denote the set of its minimal primes (i.e., those that are connected to a of the zero ideal in A) then Spec A is the union of the closed subsets Spec A/Pj for j = 1, 2, . . . , ν. – if A = B × C is a product then Spec A is the disjoint union of Spec B and Spec C, and conversely;

1.4 Partitions of unity.

The last example of the previous subsection gives us a way of decomposing spectra into a disjoint union of smaller spectra (essentially equivalent to giving orthogonal idempotent decompositions of the unit element). We will now consider something much more general. Let A be a ring. An equation 1 = f1 + f2 + ... + fν

1We will be considering these spectra along with their natural local ring structure.

3 with fi ∈ A will be called a partition of unity attached to which we have a homomorphism of rings, ν Y −1 A −→ {fi} A, i=1 and therefore an open mapping on the level of spectra,

ν ν G −1 Y −1 Spec({fi} A) = Spec ( {fi} A) −→ Spec A. i=1 i=1

−1 The collection of open subsets Ui := Spec {fi} A ⊂ Spec A for i = 1, 2, . . . , ν, form a cover of Spec A. (Proof: If P is a prime ideal of A there is some index i such that fi ∈/ P which strange as it sounds means exactly that the point P of Spec A is in the image of −1 Spec {fi} A.) Exercise 4. If A is an integral domain and K its field of fractions, and we are given a partition of unity for A as above, show that

ν \ −1 A = ({fi} A) ⊂ K. i=1

• Discrete Rings. A ring is a DVR if it is a domain and has precisely one nonzero prime ideal, so Spec A consists of two points, one closed point corresponding to the maximal ideal, the other open point corresponding to the zero ideal. If A is a DVR then its nonzero prime ideal, m(A) ⊂ A is visibly its only maximal ideal; so A is a local ring, and A/m(A) is its residue field. Any element not contained in m(A) is a unit in A; i.e., A∗ = A − m(A). Any generator z of the ideal m(A) is called a uniformizer, the uniformizers being the irreducible elements of A. Fix a uniformizer z. For any nonzero element α ∈ A there is a maximal (≥ 0) n such that α is divisible by zn in A, and for such an n we have that α = uzn for a unit u ∈ A∗ (proof: A is a PID so a UFD and has only one irreducible element up to multiplication by a unit, namely z). Call n the order or the valuation, v(α), of α. If K is the field of fractions of the DVR A, then any nonzero element φ of K can be written uniquely as uzm for m ∈ Z and for u ∈ A∗, where, again, v(φ) ∈ Z is called the valuation of φ. We have an exact sequence

0 → A∗ → K∗ → Z → 0

independent of the choice of uniformizer where the surjective homomorphism K∗ → Z is the valuation.

• Noetherian integral of one. Recall: Krull dimension ... In particular, for A a noetherian inegral domain, and P ⊂ A a prime ideal,

dim AP + dim A/P = dim A.

4 If A is an integral domain finitely generated over a field k then dim A is equal to the transcendence degree of the quotient field of A over k.(Note: dim AP is just the height of P , i.e., the length of a maximal chain of prime ideals all contained in P .) Let A be a noetherian integral domain of Krull dimension one. The closed points of Spec A are the maximal ideals of A, these being all the nonzero prime ideals; the zero ideal is “dense.” Note that the zero ideal corresponds to an open dense point of Spec A if A is semi-local; otherwise it is merely dense. Examples are Dedekind domains (where the prime ideals are all locally principal) and in particular:

– Spec Z, – Spec k[X].

“Draw” these spectra.

• Polynomial rings in n-variables over a field.

Let A = k[X1,X2,...,Xn] be the in n variables. We will call Spec(A) affine n n-space over k and denote it Ak . If n = 1, then A is a PID; in general, A is a UFD. In particular, the minimal nonzero prime ideals are principal ideals (generated by irreducible polynomials). If k is algebraically closed, the Max-Spectrum Max.Spec(A) is in on-one correspondence with kn. “Draw” the rest of the spectrum.

• Rings that are finitely generated over a field. Let fi(X1,X2,...,Xn) ∈ A = k[X1,X2,...,Xn] (i = 1, 2, . . . , m) be polynomials and let I ⊂ A be the ideal that they generate. (Any ideal in A is generated by finitely many polynomial by the finite theorem of Hilbert.) Put B = A/I The natural injection

Spec B = Spec(A/I) ⊂ Spec A

identifies Spec B as a closed subset of Spec A, and the Zariski topology on Spec B is equal to with the inherited topology, this closed subset being “the locus of zeroes of f1, f2, . . . , fm.”

1.5 Affine varieties in n-space over an algebraically closed field.

Compare with Hartshorne’s terminology ([Har] page 3), when k is algebraically closed: Definition 1. Assume k algebraically closed.

n 1. Let T ⊂ A = k[X1,X2,...,Xn] be any set of polynomials. The zero set Z(T ) ⊂ k of T n consist in the set of all points a = (a1, a2, . . . , an) ∈ k such that f(a) = 0 for all f ∈ T . 2. A subset Y ⊂ kn is an algebraic set if there exists a set T ∈ A of polynomials such that Y = Z(T ).

3. Let Y ⊂ kn be any subset. The ideal of Y , denoted I(Y ) ⊂ A, is the ideal consisting of all polynomials that are zero on all points of Y .

5 Of course we have, for any collection of polynomials T ⊂ A:

Z(T ) = Z((T )) where (T ) ⊂ A is the ideal generated by the subset T . Moreover, there is a natural identification, for any collection of polynomials T ,

Z(T ) ↔ Max.Spec(A/(T )) ⊂ Max.Spec(A) = kn.

And we have (Hilbert’s Nullstellensatz) for any ideal J ⊂ A, √ I(Z(J)) = J √ where for any ring A and ideal J ⊂ A the radical of J, denoted J is defined as follows.

√ J := {f ∈ A | f r ∈ J for some r > 0} √ is the radical of J. Note ([A-H] Proposition I.14) that J is the intersection of all prime ideals containing J.

The Zariski topology that we have imposed on Spec A restricts to the Max Spectrum, kn, (also called the Zariski topology on kn) and can be succinctly defined as the topology on kn where the closed sets are the algebraic sets. √ Note that the vocabulary of algebraic sets blurs the distinction between ideals J and J. E.g., 100 Z(f ) = Z(f). You might also think that the vocabulary√ of spectra also blurs this distinction be- cause, after all, the natural homomorphism A/J → A/ J induces a homeomorphism of topological spaces √ ∼ Spec(A/ J) −→= Spec A/J.

If I = P is a prime ideal in A = k[X1,X2,...,Xn] then Z(I) “is” an affine variety defined over k. Discuss: Noetherian induction (Proposition 1.5 of [Har]) (or, more explicitly, the theorem giving primary decompositions in noetherian rings) has as a consequence that any algebraic set is a unique union of (distinct) irreducible algebraic sets (Hartshorne’s terminology: algebraic varieties). Take a look at Hartshorne’s Exercise 1.2 for fun.

I’ll sometimes use the phrase “the locus of zeroes of f1, f2, . . . , fm” to indicate either

Spec k[X1,X2,...,Xn]/I

(where I is the ideal generated by the fi,(i = 1, 2, . . . , m) i.e., to indicate the affine scheme OR to indicate the corresponding algebraic set. I hope that the context will make it perfectly clear which of these concepts I am referring to.

6 1.6 A few examples of affine varieties over algebraically closed fields.

• Plane . Do: Exercise 1. This is Exercise 1.1 of [Har] • The twisted cubic. Exercise 1.2 of [Har]. • General linear Groups. For n > 0 consider the n2 + 1 variables consisting of

{Xi,j for i, j = 1, 2, . . . , n}

and a variable Y . Form A := k[Xi,j; Y ] the polynomial ring over k in those variables and let f(Xi,j; Y ) denote the polynomial

f := det (Xi,j)Y − 1.

Define GLn/k := Spec(A/(f)) ⊂ Spec A.

Discuss. • Hypersurfaces. For more on the following, see page 7 of [Har]. Theorem 1. (Krull’s Hauptidealsatz) Let A be noetherian and f ∈ A an element that is neither a zero-divisor nor a unit. Then every minimal prime P containing f—meaning: every prime ideal P that is minimal among those that contain f—has the property that dim AP = 1 (equivalently: P has height 1). Theorem 2. A noetherian integral domain A is a UFD if and only if every prime ideal of height one is principal. Theorem 3. Any affine variety2 in affine n-space is of codimension one (meaning: of di- mension n − 1, i.e., an affine hypersurface) if and only if it is the zero locus of a single nonconstant irreducible polynomial.

1.7 Recall the notion of Integral Closure.

Integral closure as a local property. Recall with pictures. Discuss “minimal primes” versus minimal non-zero primes in an integral domain, with pictures. Integral closure as a form of “resolution of singularities in codimension one:”

If A is a noetherian integral domain and P a minimal non-zero prime, form AP , a noetherian integral domain of Krull dimension one. If A is integrally closed then AP is then a noetherian, local, integrally closed, integral domain, hence a DVR by Theorem 4 below. When A is equal to 3 k[x1, x2, . . . , xn]/(f1, f2, . . . , fm) so that A can be thought of as the coordinate ring of the (normal ) affine variety V (so, V is “the locus of zeroes of f1, f2, . . . , fm”) then any irreducible subvariety W ⊂ V of codimension one is cut out by a P = PW , and the valuation on the

field of fractions of the DVR APW just records the order of zero or pole of a rational function along W . 2Here we are explicitly using Hartshorne’s definition of variety, meaning irreducible algebraic set. 3Discuss nomenclature: normal affine variety ↔ its coordinate ring is an integrally closed integral domain

7 1.8 Non-examples

• A plane cusp. Consider k a field and put A = k[[x, y]]/(y2 − x3) where k[[x, y]] is the ring of power series in the two independent variables x and y and (y2 − x3) is the ideal you think it is. Here A is a domain; it is a local, as well, and has precisely one nonzero prime ideal; the problem is that it isn’t a PID for (for example) its maximal ideal (x, y) ⊂ A is not generable by a single element. How to remedy this? The ring A is not integrally closed in its fraction field, but we can view its fraction field K as the field of (“finite-tailed”) Laurent y series over k a variable t where we think of t itself as t = x identifying y2 y2 x = x = = t2, x3 x2 and y2 y3 y = y = = t3. x3 x3 Explicitly, then, we have that our A can be identified with the of B := k[[t]] generated by power series in two variables x = t2 and y = t3, and both A and B share the same field of fractions K = k((t)). Since the element t ∈ B ⊂ K satisfies the integral polynomial equation over A, t2 − x, (also t3 −y) we have that B is an integral extension of A and clearly is also the integral closure of A (in its fraction field). The ring B itself is a DVR, and this will set the pattern, as we will shortly see. √ √ • Euler’s “mistake.” Consider the ring A := Z2[ −3] ⊂ K := Q2[ −3] where, again, A is a domain; it is a local, noetherian ring as well, and has precisely√ one nonzero prime ideal. The maximal ideal is generated by two elements (e.g., (2, 1 − −3)) but not by any single

element, and√ its residue field is F√2. The integral closure, though, of A in K is given by 1− −3 1− −3 3 B := Z2[ 2 ] (noting that 2 satisfies the integral relation X − 1 = 0) which is a DVR (it has a unique nonzero prime ideal (2) and is a PID). Its residue field is F4. √ √ • A 2-adic version of a plane cusp. Consider the ring A := Z2[2 2] ⊂ K := Q2[ 2] where, again,√ A is a domain; it is√ a local, noetherian, and has precisely one nonzero prime ideal (2, 2 2). Call x = 2; y = 2 2 and note that y2 − x3 = 0 and you see why I want to call this a 2-adic version of a plane cusp.√ Analogously, the integral closure of A (in its field of fractions√ K) is B := Z2[y/x] = Z2[ 2] which is a PID, its maximal ideal being generated by 2.

2 Discrete Valuation Rings and Integral Closure.

After the above, it should be less of a surprise that we have the following ([A-M] Proposition 9.2)

Theorem 4. If A is a noetherian local integral domain of Krull dimension one ((i.e. having no prime ideals except (0) 6= m), with K its field of fractions, m ⊂ A its maximal ideal, and k = A/m its residue field, the following properties are equivalent.

8 1. There is a valuation v : K → Z ∪ {∞} such that Av, the set of elements in K of valuation ≥ 0, is equal to A ⊂ K.

2. The ring A integrally closed (meaning: in its field of fractions).

3. The ideal m is principal.

4. The k- m/m2 is of dimension one.

5. Every nonzero ideal of A is a power of m.

6. There is a nonzero element x ∈ A such that every nonzero ideal of A is generated by some power of x.

7. The ring A is a DVR.

Proof. We will be using a bit of commutative algebra for this. But first (1) implies (2) because if x ∈ K but not in A is integral over A, then since it has a negative valuation, the ultrametric inequality will rule out that it can satisfy an integral relation over A; i.e., if we have

n n−1 x + an−1x + ... + a0 = 0 the leading term, xn has strictly lower valuation than any of the others on the RHS of this equation, making it impossible for the LHS to be 0.

To see that (2) implies (3) we will go fishing for a generator of the ideal m. Start modestly by choosing any a ∈ m such that a 6= 0. Then the principal ideal generated by a must be sandwiched between m and some (finite) power of m:

mν ⊂ (a) ⊂ m for, m is the only nonzero prime ideal of A so the radical of (a) is m and since A is noetherian a) is contained in a power of its radical. Let mν be the minimal power of m for which this is true, and find some element b ∈ mν−1 (convention: m0 = A) not in mν. We want to show that x = a/b is a generator of m, by noting, first, that x−1 ∈/ A by construction, and therefore x−1 is not integral over A by our hypothesis (2). We also have, by construction, that x−1m ∈ A, giving two possibilities: either

• x−1m = A, or

• x−1m ∈ m ⊂ A

The first of these possibilities will give us that, indeed, x is a generator of m and we would be done; the second of these possibilities would give us an integral relation that x−1 satisfies over A.

2 2 To see that (3) implies (4): (3) already implies that dimk m/m ≤ 1 but if dimk m/m = 0 i.e., m = m2, then an application of Nakayama’s Lemma would give us that m = 0, which it is not.

9 Exercise 5. (4) ⇒ (5) ⇒ (6) ⇒ (7) ⇒ (1).

Remark (John Tate): In a local noetherian domain O of Krull dimension 1, the function f 7→ length(O/fO) on O∗ is the valuation if O is a DVR and otherwise is the sum of the valuations of f at the localizations of the integral closure of O at its maximal ideals, if that integral closure is finite over O as it is in the case of an affine variety V as above.

3 Dedekind Domains.

The basic facts of life are as follows. Theorem 5. Let A be a noetherian integral domain of dimension one. Then every nonzero ideal of A can be expressed uniquely as a product of primary ideals with distinct radicals. Theorem/Definition 1. Let A be a noetherian integral domain of dimension one. Then if A is integrally closed (as we sketched a proof of this above) any localization AP of A at a nonzero prime ideal P is a DVR, and conversely, if every localization AP of A at a nonzero prime ideal P is a DVR, then A is integrally closed. If these conditions apply we say that A is a . Every nonzero ideal of a Dedekind domain A can be written uniquely as a product of powers of nonzero prime ideals.

Let A be a Dedekind domain, and K its field of fractions. A of A is a nontrivial, finitely generated, A-submodule of K, J ⊂ K. Any such fractional ideal may be written uniquely as a product: s Y ei J = Pi i=1 where the Pi are prime ideals in A and the ei ∈ Z. We sometimes refer to the exponent ei as the valuation of J at the prime ideal Pi, and denote it v(Pi) so that we can also write Y J = P v(P ). P

4 Rings of integers in number fields, and function field analogues.

Let K/Q be a number field, which for us will mean an extension field of Q of finite degree. We have, by the primitive element theorem the fact that there exists an element Θ that generates K over Q, so that we may write K = Q[X]/f(X) where f is a monic irreducible polynomial having Θ as a root. There are, of course, many such Θ’s, in fact, and we can (by multiplying Θ by a suitable non-zero rational integer) guarantee that the primitive element Θ is an , i.e., is

10 integral over Z, or equivalently, that the monic polynoimial f alluded to above has rational integer coefficients.

Other than the fact that K is a field rather than merely a finite dimensional vector space over Q, the essential structure that the vector space K carries is a nondegenerate bilinear symmetric (quadratic) form hx, yi := TraceK/Q(xy) (nondegenerate because K/Q is a separable extension). Call this the trace pairing. If Λ ⊂ K is a Z-submodule let Λ? ⊂ K denote the submodule of elements y ∈ K such that hx, yi ∈ Z for all xinΛ. Note that if Λ ⊂ K is actually a in the Q-vector space K generated by basis elements ? λ1, λ2, . . . , λd ∈ K then Λ ⊂ K is also a lattice in the Q-vector space K generated by the dual ? ? ? basis λ1, λ2, . . . , λd ∈ K, where dual of course means with respect to the trace pairing.

Let A ⊂ K denote the ring of elements that are integral over Z ⊂ Q, Note that since h , i : A×A → Z ⊂ Q we have that—viewing A as lattice in K, A ⊂ A?.

Theorem 6. The ring A is a Dedekind domain.

Proof. Choosing a primitive element Θ ∈ K that is integral over Z, we have that Z[Θ] ⊂ A. Comparing with the dual lattices relative to the trace pairing, we have the following string of inclusions

Z[Θ] ⊂ A ⊂ A? ⊂ Z[Θ]? and since Z[Θ] is a lattice, its dual, Z[Θ]? is again a lattice, and therefore A itself is a lattice, is finitely generated as a Z-, hence noetherian. A is of dimension one because if there were a string of prime ideals (0) ⊂ P 0 ⊂ P in A with P 0 nonzero, a simple argument shows that P 0 ∩ Z − P ∩ Z = (p) ⊂ Z for some prime p, and therefore A0 = A/P 0 is an integral domain of finite cardinality, hence a field; from this we see that P 0 = P . The integral domain A, then, is noetherian, dimension one, and integrally closed; hence: a Dedekind domain.

Exercise 6. How would you modify the above discussion to obtain examples of Dedekind domains whose fields of fractions are of transcendence degree one over finite fields, and finitely generated (as fields)?

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