MAS439 Commutative Algebra and Algebraic Geometry

Dr James Cranch

Office location: G39c Hicks

[email protected]

Course webpage: http://cranch.staff.shef.ac.uk/mas439/

Semester 2, 2014 2

Lecture 1 (2014-02-11) Modules

Fields often make their properties known to us via vector spaces: we care so much about R at least partly because our universe looks like a three- dimensional vector space over it. Modules offer a generalisation for commutative rings of what vector spaces do for fields. The definition is very similar to the definition of a vector space:

Definition 1.1. 1 Let R be a commutative ring. An R-module consists of an abelian group M together with a map

R × M −→ M,

written as (r, m) 7→ rm, satisfying the following axioms:

• for all m ∈ M, we have 1m = m;

• for all r, s ∈ R and m ∈ M, we have (rs)m = r(sm);

• for all r, s ∈ R and m ∈ M, we have (r + s)m = rm + sm;

• for all r ∈ R and m, n ∈ M, we have r(m + n) = rm + rn.

So we can add and subtract elements of M, and multiply them by elements of R. The reader might wonder why there are no axioms explaining that zeroes and subtraction behave nicely. It turns out they’re not necessary:

Proposition 1.2. Let R be a commutative ring and M an R-module. Then for all m ∈ M we have 0m = 0 and (−1)m = −m.

Proof. Firstly, we have

1m + 0m = (1 + 0)m = 1m,

and so cancelling the 1m from both sides, we have 0m = 0. Secondly, we have

0 = 0m = (1 + (−1))m = 1m + (−1)m = m + (−1)m.

By taking the m to the other side, we get that (−1)m = −m.

1In this course I’ll be numbering according to lecture, so that Theorem 15.3 will be the third numbered statement in the fifteenth lecture. 3

In case our ring is a field, we get nothing different: Example 1.3. If K is a field, then a K-module is exactly the same thing as a vector space over K. There is one other ring whose modules are a deeply familiar concept:

Proposition 1.4. A Z-module is the same thing as an abelian group. Proof. A Z-module consists of an abelian group A, plus a multiplication map Z×A → A satisfying the module axioms. But these are uniquely determined by the abelian group structure. For example,

5a = (1 + 1 + 1 + 1 + 1)a = 1a + 1a + 1a + 1a + 1a = a + a + a + a + a, and

−3a = ((−1) + (−1) + (−1))a = (−1)a + (−1)a + (−1)a = −a − a − a.

Hence the scalar multiplication adds nothing new. However, we can often describe a module in terms of linear algebra con- cepts that we know about already: Example 1.5. A C[T ]-module consists of a complex vector space V equipped with a map T : V → V . Indeed, adding and scalar multiplying by constant polynomials give us a C-module, and multiplying by T gives the linear map. Remark 1.6. One nice thing about vector spaces is that they all have a basis: any K-vector space looks like Kn for some (possibly infinite) K. That’s absolutely not true for modules over a general ring, as the example above shows: not every abelian group is of the form n. Z Lecture 2 One occasionally needs to talk about subsets of modules which are mod- (2014-02-13) ules. The definition is what you might expect: Definition 2.1. Let R be a commutative ring. A submodule of an R-module M consists of a subset of M which contains the zero element and is closed under addition and under multiplication by elements of R. Perhaps the most striking use of this is to rephrase something you know about already in this new language: Proposition 2.2. Let R be a commutative ring. Then addition and multi- plication makes R into an R-module, and a submodule of R is exactly the same thing as an ideal of R. 4

Proof. All the axioms of a module (in Definition 1.1) are well-known formulae for multiplication in a commutative ring, so R has the structure of an R- module. A submodule of R is a subset closed under scalar multiplication by ele- ments of R, and under addition and subtraction. That’s exactly the same thing as an ideal of R. Here’s a boring example of a module: Example 2.3. Let R be a commutative ring. Then the trivial module is given by the singleton set {0}, with the only possible addition and scalar multiplication. You can think of that as corresponding to the zero ideal, if you like. Here’s another general way of constructing modules. Example 2.4. Let R and S be a commutative rings, and let f : R → S be a ring homomorphism. Also, let M be an S-module. Then the formula rm = f(r)m makes M into an R-module. Example 2.5. By the above, as S is a module over itself, a ring homomorphism f : R → S makes S into an R-module, with the multiplication given by rs = f(r)s. Conceptually, I sometimes like to think of a ring homomorphism f : R → S as being a way of making S into an R-module, together with a way of multiplying all the elements of S with each other. We can take Cartesian products of modules, though we usually choose to give it another name: Definition 2.6. Let R be a commutative ring, and let M and N be R- modules. We define a new R-module M ⊕ N, the direct sum of M and N to consist of the abelian group M × N together with the following definition of multiplication:

r(m, n) = (rm, rn) (for r ∈ R, m ∈ M and n ∈ N).

If we keep doing this on M, forming M ⊕ · · · ⊕ M, we get something that deserves to be called M n for some n (sometimes written M ⊕n to remind you what the operation you’re repeating is called). A element of M n can be thought of as an n-tuple (m1, . . . , mn) of elements of M. This construction happens particularly frequently when M is R itself (regarded as a module over itself): we say that the resulting module Rn is a (finitely generated) free module over R. There is an infinite version of these ideas, where we direct sum together infinitely many modules, but we won’t need it. 5

The last thing I ought to talk about is the notion of a homomorphism of R-modules. I hope you can guess what this is:

Definition 2.7. Let R be a commutative ring and let M and N be R- modules. A homomorphism f : M → N consists of a homomorphism of abelian groups from M to N with the property that f(rm) = rf(m) for all r ∈ R and m ∈ M.

Example 2.8. If M is an R-module and M 0 a submodule, then the inclusion map M 0 → M is a module homomorphism. Example 2.9. If R is a field, then a R-module homomorphism is exactly the same thing as a vector space homomorphism. Example 2.10. If R = Z, then a Z-module homomorphism is exactly the same thing as a homomorphism of abelian groups. Here’s our last batch of definitions for now:

Definition 2.11. Let R be a commutative ring and let M be an R-module. We say that elements m1, . . . , mk ∈ M generate M if, for all m ∈ M there are r1, . . . , rk ∈ R such that

r1m1 + ··· + rkmk = m.

If there are elements m1, . . . , mk which generate M, we say that it is finitely generated. Lecture 3 We can use this definition to tidy up an embarrassing issue with our (2014-02-18) terminology. Example 3.1. The finitely generated free module Rn is indeed finitely gener- ated (as we’d expect). In fact we can take the generators to be

r1 = (1, 0, 0,..., 0, 0)

r2 = (0, 1, 0,..., 0, 0) . .

rn = (0, 0, 0,..., 0, 1) and then the element r = (r1, . . . , rn) can be written as

r = r1m1 + ··· + rnmn.

Not every module is finitely generated. Here’s an example: 6

Proposition 3.2. Recall that a Z-module is the same thing as an abelian group. The Z-module given by R under addition is not finitely generated.

Proof. Let m1, . . . , mk ∈ R be a finite list of real numbers. Then the set

{r1m1 + ··· + rkmk | r1, . . . , rk ∈ Z} is countable. So it cannot be all of R, as R is uncountable.

Quotient modules Imitating the definitions of quotient abelian groups and quotient vector spaces, we can define quotient modules. In order to do so, let’s first recall definitions and notation for quotients of abelian groups. Given an abelian group A and a subgroup B, we put an equivalence relation on A to define the quotient, setting a ≈ a0 if there is b ∈ B such that a0 ≈ a + b. We write A/B for the quotient abelian group, which is the set of equiv- alence classes under this relation. We write a + B for the equivalence class containing a if we’re being formal, or simply call it a if we’re being informal. That gives us what we need to define quotient modules. Definition 3.3. Let R be a commutative ring, let M be an R-module and let N be a submodule of M. The quotient module M/N has objects consist- ing of the quotient abelian group M/N, with multiplication defined by the multiplication structure on M. Note that the multiplication is well-defined: if we have m ≈ m0 in M, then m0 = m + n for some n ∈ N. Then

rm0 = r(m + n) = rm + rn, and rn ∈ N since N is a submodule of M, so rm0 ≈ rm. The standard examples are that if R is a field, then quotient modules are simply quotient vector spaces, while if R is Z, then quotient modules are simply quotient abelian groups. The following are the standard results explaining the good behaviour of quotient modules: Theorem 3.4 (First isomorphism theorem). Let R be a commutative ring. Let M and N be R-modules, and let φ : M → N be a homomorphism between them. Then: (i) the kernel ker(φ) is a submodule of M, 7

(ii) the image im(φ) is a submodule of N, and

(iii) there is an isomorphism im(φ) ∼= M/ ker(φ).

Proof. The kernel is a subgroup of M because φ is an abelian group homo- morphism. It’s closed under multiplication by R since if f(m) = 0, then φ(rm) = rφ(m) = r0 = 0 for all r, and is hence a submodule. Similarly, the image is a subgroup of M because φ is an abelian group homomorphism. It’s also closed under multiplication by R since rφ(m) = φ(rm). The isomorphism between im(φ) and M/ ker(φ) is defined as follows. Given an element n ∈ im(φ), associate to it an element m ∈ M such that φ(m) = n (which is possible as n is in the image). There may be several possible choices of such m. But given any two such choices m and m0, we have φ(m − m0) = φ(m) − φ(m0) = n − n = 0, and hence m − m0 ∈ ker(φ), and hence equal in M/ker(φ). Conversely to any element m + ker(φ) in M/ ker(φ) we associate the element φ(m) ∈ im(φ), and this is well-defined since ker(φ) is sent to zero by definition. These two can easily be seen to define inverse module maps making an isomorphism.

Theorem 3.5 (Second isomorphism theorem). Let R be a commutative ring. Let M be an R-module, and let S and T be submodules of M. Then:

(i) the set S + T = {s + t | s ∈ S, t ∈ T } is a submodule of M containing S and T as submodules,

(ii) the intersection S ∩ T is a submodule of S, and

(iii) there is an isomorphism (S + T )/T ∼= S/(S ∩ T ).

Proof. The set S + T is a subgroup of M (as proved in the isomorphism theorems for groups, for example). It is closed under multiplication by R as, if s ∈ S and t ∈ T then r(s + t) = rs + rt ∈ S + T . Hence it is indeed a submodule of M, and it contains S (as elements of the form s + 0) and T (as elements of the form 0 + t). The set S ∩ T is clearly a subgroup of S, and it is closed under multipli- cation by R (as if x ∈ S and x ∈ T , then rx ∈ S and rx ∈ T as S and T are submodules of M), and hence rx ∈ S ∩ T . To define an isomorphism between (S + T )/T and S/(S ∩ T ) we must define mutually inverse maps in each direction. 8

The map from (S + T )/T to S/(S ∩ T ) sends an element (s + t) + T to s + (S ∩ T ). This is well-defined. Indeed, suppose we have another way of writing the same element:

(s + t) + T = (s0 + t0) + T.

Then (s + t) − (s0 − t0) ∈ T . But this is equal to (s − s0) + (t − t0). The latter bracket is clearly in T , so we have s − s0 ∈ T . But s − s0 is also in S, so it’s in S ∩ T , and hence s + (S ∩ T ) and s0 + (S ∩ T ) are the same element of S/(S ∩ T ). It can now readily be checked to be a module homomorphism. The map from S/(S ∩ T ) to (S + T )/T sends an element s + (S ∩ T ) to (s + 0) + T . To show that this is well-defined, suppose that we chose another representation s0 + (S ∩ T ). Then s − s0 ∈ S ∩ T , so in particular (s + 0) − (s0 + 0) ∈ T , so their images are the same. This can also be checked to be a module homomorphism. It’s easy to see that these maps are indeed mutually inverse. If we start with s + (S ∩ T ) ∈ S/(S ∩ T ), we get (s + 0) + T ∈ (S + T )/T and then recover s + (S ∩ T ) ∈ S/(S ∩ T ). Conversely if we start with (s + t) + T ∈ (S + T )/T , we get s + S ∩ T ∈ S/(S ∩ T ) and then (s + 0) + T ∈ (S + T )/T . This differs from (s + t) + T by t ∈ T , and is hence equal as an element of (S + T )/T .

Theorem 3.6 (Third isomorphism theorem). Let R be a commutative ring. Let M be an R-module. Let N be a submodule of M, and P a submodule of N. Then:

(i) the quotient N/P is a submodule of the quotient M/P , and

(ii) there is an isomorphism (M/P )/(N/P ) ∼= M/N.

Proof. The first part is reasonably clear: the elements n + P ∈ N/P , where n ∈ N, are a subset of those elements m + P ∈ M/P , where m ∈ M, since N is a submodule of P . For the second part, we define a pair of mutually inverse homomorphisms between (M/P )/(N/P ) and M/N. An element of the (M/P )/(N/P ) is of the form (m + P ) + (N/P ); given such an element we associate to it the element m+N ∈ M/N. We must show that this is well-defined. Suppose (m + P ) + (N/P ) and (m0 + P ) + (N/P ) give the same element in (M/P )/(N/P ); then m + P and m0 + P differ by an element of N/P ; in other words m − m0 ∈ N. But this says exactly that m and m0 are sent to the same coset of N in M, so that this construction is well-defined. 9

Conversely, suppose given an element m + N ∈ M/N. We associate to it the element (m + P ) + (N/P ). Again, it is not clear that it is well-defined. So suppose we have two different representations m + N and m0 + N for the same element, so that m − m0 ∈ N. Then (m + P ) − (m0 + P ) ∈ N/P , so we get a well-defined element of (M/P )/(N/P ). As in the above cases, it’s easy to check that these constructions are module homomorphisms, and it’s obvious that they’re mutually inverse. Lecture 4 (2014-02-20) Chain conditions There are a lot of unpleasant rings out there, and a lot of nasty unpleasant modules over them. It’s good to have notions of “nice”, particularly if we want to do algebraic geometry. Here are some straightforward ones, which come up repeatedly: Definition 4.1. Let R be a commutative ring, and let M be an R-module. • We say that M satisfies the ascending chain condition, or that M is Noetherian if any ascending chain of submodules of M stabilises. That is, if we have R-modules

N1 ⊂ N2 ⊂ N3 ⊂ · · · ⊂ M,

then there is some n such that Nn = Nn+1 = Nn+2 = ··· . • We say that M satisfies the descending chain condition, or that M is Artinian if any descending chain of submodules of M stabilises. That is, if we have R-modules

M ⊃ N1 ⊃ N2 ⊃ N3 ⊃ · · · ,

then there is some n such that Nn = Nn+1 = Nn+2 = ··· . These are actually used more commonly as niceness conditions for rings rather than modules: Definition 4.2. Let R be a commutative ring. • We say that R is a Noetherian ring if R is a Noetherian R-module. Since submodules of R are the same thing as ideals, this means that every increasing chain of ideals

I1 ⊂ I2 ⊂ · · · stabilises. 10

• We say that R is a Artinian ring if R is an Artinian R-module. Simi- larly, this means that every decreasing chain of ideals

I1 ⊃ I2 ⊃ · · ·

stabilises.

Here are some examples of modules which are or are not Noetherian, and which are or are not Artinian. First, there are ones which satisfy both conditions:

Proposition 4.3. Let R be a commutative ring, and let M be a finite R- module (by which I mean, finite as a set). Then M is both Noetherian and Artinian as an R-module.

Proof. Since M is finite, it’s impossible to have an infinite ascending or de- scending chain of subsets of M which does not stabilise, so in particular any chain of submodules stabilises.

There are many modules which are Noetherian but not Artinian:

Proposition 4.4. The abelian group Z is Noetherian but not Artinian as a Z-module. Proof. Decoding the terminology a bit, this says that every ascending chain of ideals in Z stabilises, but not every descending chain. Since Z is a principal ideal domain, any ideal is generated by some single element, and containment is the same as divisibility. Hence an ascending chain of ideals

(n1) ⊆ (n2) ⊆ (n3) ⊆ · · · gives us a chain of nonnegative integers, each a factor of the one before:

· · · | n3 | n2 | n1.

That means that n1 ≥ n2 ≥ n3 ≥ · · · ≥ 0, so certainly the chain stabilises, and hence Z is a Noetherian Z-module. However, it is not Artinian. Consider the following descending chain of ideals: (1) ⊇ (2) ⊇ (4) ⊇ (8) ⊇ (16) ⊇ · · · . This does not stabilise.

There are a few which are Artinian but not Noetherian: 11

Proposition 4.5. The set C× of nonzero complex numbers is an abelian group, and hence a Z-module. Let p be a prime, and consider the subset  × pn U = x ∈ C | x = 1 for some n .

That is a subgroup of C×, and it is Artinian but not Noetherian as a Z- module.

n Proof. Let Un be the subgroup consisting of all p -th roots of unity: that is, n all x ∈ C× such that xp = 1. Then we have an ascending chain

U1 ( U2 ( U3 ( ··· of submodules of U, which does not stabilise as they’re all different. So U is not a Noetherian Z-module. However, it’s not difficult to show (exercise!) that the Ui’s are the only submodules of U. Hence any descending chain stabilises. Lastly, if we take modules which are “infinite-dimensional” in some ap- propriate sense, it’s quite likely that they will be neither:

Proposition 4.6. The set of complex polynomials C[x] is a complex vector space and hence a C-module. It is neither Artinian nor Noetherian as a C-module.

Proof. Let Un be the vector space of polynomials of degree at most n. Being a vector space, it is a submodule of C[x]. Then the chain

U1 ( U2 ( U3 ( ··· is an ascending chain which does not stabilise, showing that C[x] is not a Noetherian C-module. n Similarly, let Vn be the space of polynomials which are a multiple of x (equivalently, those which have a root at 0 of order n). Then the chain

V1 ) V2 ) V3 ) ··· is a descending chain which does not stabilise, showing that C[x] is not an Artinian C-module. The above gives us some examples and nonexamples of Noetherian and Artinian rings: Example 4.7. Proposition 4.4 says that Z is a Noetherian ring but not an Artinian ring. 12

The following may not quite be obvious, and you should let it sink in.

Example 4.8. The reader may think that Proposition 4.6 says that C[x] is neither a Noetherian nor an Artinian ring. This is false. The proof there gives an ascending chain of submodules of C[x] regarded only as a C-module, not a C[x]-module. This means that it does not show that C[x] is not Noetherian. In fact, C[x] is Noetherian, by a theorem of Hilbert. On the other hand, the example there of a descending chain is in fact a descending chain of C[x]-modules (they’re all ideals in C[x]). Hence C[x] is definitely not an Artinian ring. Lecture 5 (2014-02-25) Localisation

We’re going to do some ring theory for the next couple of lectures. I’ll need you to trust me on something, which we won’t prove in this course.

Theorem 5.1. Let R be a commutative ring and I a proper ideal. Then I is contained in some maximal ideal.

Remark 5.2. A proper proof of this takes us well outside this course, into the world of set theory. However, if you know enough set theory, it’s not difficult (assuming the axiom of choice): ask your local set theorist about Zorn’s lemma. One big advantage of Noetherian rings is that it makes this result obvious, and it requires no set theory.

Proposition 5.3. Let R be a Noetherian commutative ring and I a proper ideal. Then I is contained in some maximal ideal.

Proof. Let I1 = I. Now, either I1 is maximal or there is a bigger ideal I2. Then either I2 is maximal or there is a bigger ideal I3. Proceeding in that fashion, unless we reach a maximal ideal at some point we get a ascending chain of ideals

I1 ( I2 ( I3 ( ··· , where each differs from its neighbour, and this contradicts the ascending chain condition. 13

Localisation of rings Recall the definition, last semester, of the field of fractions K(R) of an inte- a gral domain R, as being the set of fractions b , for a, b ∈ R such that b 6= 0. There is some opportunity for generalisation: we could try restricting the things that could occur in the denominator. We want to have 1 as a denominator, and we want to be able to multiply fractions, which means we have to be able to multiply denominators. Accordingly we make the following definition: Definition 5.4. A multiplicative subset U of a ring R is a subset U ⊂ R such that: • 1 ∈ U, and • if a, b ∈ U, then ab ∈ U. In other words, a multiplicative subset is a subset containing the unit, closed under multiplication. Here are some examples of multiplicative subsets, starting with straight- forward ones. Example 5.5. Let x be an element of a ring R. Then the set 1, x, x2, x3,... is a multiplicative subset, since it contains 1, and xmxn = xm+n so it is closed under multiplication. Example 5.6. The set R\{0} is a multiplicative subset of R if and only if R is an integral domain. Indeed, an integral domain has 1 6= 0, and has no elements a, b 6= 0 with ab = 0 so R\{0} is closed under multiplication. Also, if 1 ∈ R\{0}, then 0 6= 1. If \{0} is closed under multiplication, then if a, b 6= 0, then we cannot have ab = 0, so R is an integral domain. Example 5.7. Provided the ring R is nontrivial, the set N of elements which are not zero divisors is a multiplicative subset. Indeed, we have 0 6= 1, and 1 is clearly not a zero divisor. Also, if a and b are not zero divisors, then nor is ab, since if (ab)c = 0, then a(bc) = 0, so a is a zero divisor. Here is our main example: Example 5.8. If P is a prime ideal, then the set R\P is a multiplicative subset. Indeed, 1 ∈/ P so 1 ∈ R\P . Also, if a, b ∈ R\P , then ab ∈ R\P . Indeed, if ab ∈ P , then either a ∈ P or b ∈ P by the definition of a prime ideal. 14

That generalises the example of R\{0} above: the zero ideal is a prime ideal if and only if R is an integral domain. Now we understand multiplicative subsets, we can use them to define rings of fractions. Given a multiplicative subset U of R, we define a new ring U −1R (often pronounced out loud as “R with U inverted”) to have as elements symbols a of the form b , where a ∈ R and b ∈ U, subject to an equivalence relation. The correct equivalence relation which we’ll use might be thought a little a c surprising: we say that b = d if there is t ∈ U such that tad = tbc.

We ought to make the following check:

Proposition 5.9. The relation on fractions defined above is an equivalence relation.

a a Proof. Reflexivity is easy: b = b as we can take t = 1 and get 1ab = 1ab. Symmetry is similarly clear: if tad = tbc then tbc = tad. a c c e Transitivity is the more complex check. Suppose b = d and d = f . This means that there is some t ∈ U such that tad = tbc, and some u ∈ U such that ucf = ude. We need to find v ∈ U such that vaf = vbe. The obvious choice is v = tud (which is in U, as all the factors are), and then

vaf = tuadf = tubcf = tubde = vbe as required.

Now we show that this structure is a ring:

Theorem 5.10. The set U −1R, under the equivalence relation above, equipped with the usual arithmetic operations for fractions, is a commutative ring.

Proof. What we really need to show that the usual operations are well- defined: that is, equivalent inputs give equivalent outputs. a c a e c e First we show that if b = d , then b + f = d + f , or in other words that af+be cf+de bf = df . So we have tad = tbc for some t ∈ U, and we need to show that for some u ∈ U we have u(af + be)df = u(cf + de)bf. If we take u = t we get

t(af + be)df = tadf 2 + tbdef = tbcf 2 + tbdef = t(cf + de)bf as needed. 15

a c a e c e Secondly, we show that if b = d , then b f = d f , or in other words that ae ce bf = df . So, again we have tad = tbc for some t ∈ U, and we need to show that for some u ∈ U we have uadef = ubcef. Again this works with u = t. This having been proved, the rest (the commutative ring axioms) depends simply on standard properties of fraction arithmetic. Lecture 6 (2014-02-27) Proposition 6.1. Let R be a domain, then (R\{0})−1R is exactly the field of fractions K(R) defined last semester.

Proof. The objects look the same, but we have to check that the equivalence relation is the same! a c −1 In K(R) we say that b = d if ad = bc. In (R\{0}) R, we say that a c b = d if there is some t ∈ R\{0} such that tad = tbc. But that means that t(ad − bc) = 0, and since R is a domain that means either that t = 0 (which we assumed not) or that ad − bc = 0, in other words ad = bc. So they really are the same.

−1 a There is a ring homomorphism LU : R → U R sending a 7→ 1 , general- ising the map Z → Q. Note that under this homomorphism, everything in −1 u 1 U becomes invertible in U R, since LU (u) = 1 has inverse u . This observation gives us a way of explaining this construction, which is less explicit but often more helpful:

Theorem 6.2. Let f : R → S be a ring homomorphism which sends elements of U to invertible elements in S. There is a unique ring homomorphism −1 g : U R → S such that f is the composite gLU .

a 1 −1 Proof. We must have g( 1 ) = f(a) for a ∈ R, and we must have g( b ) = f(b) for b ∈ U. Therefore we must have

a a 1 a 1 g = g = g g = f(a)f(b)−1. b 1 b 1 b

As a result of this, we know what g must be: so there is at most one homo- morphism; we merely need to show that this works. a c It sends equal elements to equal elements: if b = d then there is some a −1 t ∈ U with tad = tbc. Under these circumstances, g sends b to f(a)f(b) c −1 and d to f(c)f(d) . 16

These are, however, equal:

f(a)f(b)−1 = f(a)f(d)f(d)−1f(t)f(t)−1f(b)−1 = f(tad)(f(d)f(t)f(b))−1 = f(tbc)(f(d)f(t)f(b))−1 = f(c)f(b)f(b)−1f(t)f(t)−1f(d)−1 = f(c)f(d)−1.

Now we know that it is well-defined, it is easy to check that g is a homomor- phism.

We like to say that U −1R is the universal commutative ring with a ho- momorphism from R which sends elements of U to invertible elements. With all these preliminaries in place, we can define the central thing of interest:

Definition 6.3. Let R be a commutative ring, and let P be a prime ideal in R. The localisation of R at P , written RP , is the commutative ring (R\P )−1R.

a Unpacking this definition, we can consider RP to be a ring of fractions b , where a, b ∈ R and b∈ / P . You may be in need of an example by now, and a good example is given by taking R = Z and P = (p) where p is a prime. For the sake of explicitness, let’s take p = 7. a The localisation Z(7) is the set of fractions b where a, b ∈ Z and 7 - b. It’s a subring of Q: we can add, subtract, and multiply, and we can divide by any prime except 7. The effect of this is to make a ring where 7 is the only prime: all other primes have become units. Number theorists use these rings (for values of p which may or may not be 7) a lot to focus on one prime at a time. They call them the p-local Lecture 7 integers. The lecturer likes to think of the primes as being points in a (2014-03-04) space, and localising at 7 being a way of looking locally at what’s going on near the prime 7 only, ignoring things near all the other primes. This is in fact a point of view with some serious theory backing it up (Grothendieck’s geometry of schemes) but that theory lies outside this course. But in general, localisation is a good way of looking at a ring one prime ideal at a time. We could also ask what the the geometric significance of localisation is. The field of fractions, you’ll remember, gives us the ring of rational functions on a variety: these are much like regular functions, but are permitted to be undefined at a few places. 17

Localisation is a more nuanced operation: it allows us to control the locus on which our functions are permitted to be defined. For example, consider the variety A1, with coordinate ring k[x]. One prime ideal in k[x] is given by (x), corresponding to the subvariety {0}. The localisation at this prime ideal consists of the fractions f/g, where f, g ∈ k[x] and g is not a multiple of x. Geometrically, this is the ring of rational functions which are defined at 0. In general, geometrically, the localisation tells you about rational func- tions which are nonzero along a subvariety. We should ask ourselves: are localisations rings which are easy to work with? It’s not hard to imagine they should be. If we have an integral domain R, then inverting all nonzero elements gives us (R\{0})−1R = K(R), which as a field is a ring with very simple behaviour. Meanwhile, inverting the multiplicative subset {1} gives us {1}−1R = R. It’s not hard to imagine that, given a prime ideal P of R, the localisation −1 RP = (R\P ) R will be not quite as straightforward as K(R), but not far off. We’ll justify that intuition now.

Definition 7.1. We say that a commutative ring is a local ring if it has exactly one maximal ideal.

There are other ways of saying this:

Theorem 7.2. Let R be a commutative ring. Then R is a local ring if and only if sums of noninvertible elements are noninvertible.

Proof. Noninvertible elements are clearly closed under multiplication by any element of the ring. So, if the noninvertible elements are closed under addi- tion, then they form an ideal. Any proper ideal is contained in the set of noninvertible elements (as if u ∈ I is invertible, then I = R). Hence if the noninvertible elements form an ideal, that ideal is maximal and contains all other ideals so is a unique maximal ideal, and hence R is local. Conversely, suppose R is local, and suppose that x and y are noninvertible elements such that x + y is invertible. We’ll produce a contradiction. Note that no proper ideal can contain both x and y, as then it would contain x + y and hence all of R. That means that (using Theorem 5.1) (x) and (y) can be extended to different maximal ideals (the former not containing y and the latter not containing x: this is a contradiction, given that R is local.

The following result may do something to help explain the terminology. 18

Theorem 7.3. Let R be a commutative ring, and P a prime ideal. Then the localisation RP is a local ring. Proof. We claim firstly that the noninvertible elements are exactly those of a the form b where a ∈ P and b∈ / P . a c Firstly, this description actually makes sense: if we have b = d then we have t ∈ R\P with tad = tbc. Also we have b, d ∈ R\P by the definition of localisation. Then if a ∈ P then tad ∈ P so tbc ∈ P . But, by the definition of a prime ideal, that means that t ∈ P , b ∈ P or c ∈ P . Since we know t, b ∈ R\P , that means c ∈ P . Hence if we have two different ways of writting a fraction in RP , then they agree on whether the numerator is in P . a b If we have b with a∈ / P and b∈ / P , then it clearly has inverse a ∈ RP , so such elements are invertible. a c However, if we have b with a ∈ P and b∈ / P . Then suppose we have d a c such that b d = 1. Then we have ac 1 = , bd 1 which amounts to saying that there is some t ∈ R\P such that tac = tbd. However, a ∈ P , so tac ∈ P so tbd ∈ P . By the definition of a prime ideal, this means that t, b or d must be in P , but we’ve assumed none of these are the case. Hence it really is true that the noninvertible elements are those of the form a b with a ∈ P and b∈ / P . We’ll now invoke Theorem 7.2 above. Suppose we a c have two such fractions b and d with a, c ∈ P and b, d∈ / P . Their sum is ad+bc bd , and ad + bc ∈ P since a, c ∈ P . That shows that sums of noninvertible elements are noninvertible, and hence that the localisation RP is local as desired.

Localisation of modules Given a ring R, a multiplicative subset U and a R-module M, we can use −1 −1 a fractions to form a U R-module U M: objects are fractions b with a ∈ M a c and b ∈ U, subject to the equivalence relation that b = d if there is t ∈ U such that tad = tbc. Addition is defined by the usual formula: a c da + bc + = , b d bd (note that this does make sense: we’re never multiplying two elements of M together), and scalar multiplication by elements of U −1R is defined by the 19 usual formula r a ra + = . s b sb It is not hard to check that these definitions work. Lecture 8 It’s also easy to show we have (2014-03-06)

U −1(M ⊕ N) = U −1M ⊕ U −1N.

A family of examples might be helpful. Let U be the multiplicative subset Z\{0} ⊂ Z. Then we have Q = U −1Z. Now suppose we have an abelian group A, and regard it as a Z-module. Hence U −1A is supposed to be a Q-module: a rational vector space. We can explain how this works in the case where A is finitely generated. In this case, A = Zn ⊕ A0, where A0 is finite. For any finite abelian group A0, we have U −1A0 = 0, the trivial vector space! Indeed, any element x ∈ A0 has kx = 0 for some nonzero k ∈ Z, and then x 0 = since kx1 = k01. 1 1 On the other hand, considering the free Z-module Z, we have U −1Z = Q. Indeed, the definition is the same as the standard definition of Q as the fraction field of Z. Hence, for this particular choice of U, we have

−1 −1 n 0 n U A = U (Z ⊕ A ) = Q ; the rule is “throw away the torsion and turn the Zs into Q’s”. One extra feature, which is frequently important, is this:

Proposition 8.1. Let R be a commutative ring, U a multiplicative subset, and M and N both R-modules, and let φ : M → N be an R-module homo- morphism. Then there is a U −1R-module map U −1φ : U −1M → U −1N.

Proof. Define a φ(a) U −1φ = ; b b it is immediate to check that this works.

Professionals call this behaviour functoriality: the phenomenon that many constructions that can be done to objects can also be done to their homo- morphisms. The biggest use of all this is to define localisations of modules: 20

Definition 8.2. Let R be a ring, P a prime ideal of R and M an R-module. −1 We define the localisation MP of M to be the RP -module (R\P ) M. It is not hard to imagine that this should be an important tool: if local- isation lets us look at a commutative ring R “one prime ideal at a time”, then localising modules allows us to look at the effects of each prime ideal of R on their modules, too. By the above, localisation of modules is functorial: if we have R a com- mutative ring, P a prime ideal, and M → N a homomorphism of R-modules, then we get a homomorphism of P -modules MP → NP .

Nakayama’s Lemma

We’re going to do a big theorem in commutative algebra now; the first of several. We’re going to show that, in some sense at least, finitely gener- ated modules over a commutative ring aren’t too badly different from finite- dimensional vector spaces over a field. We’ll start with a bit of a build-up, then prove various versions of our theorem, and showing off some consequences as we go. We’re going to need the following: Definition 8.3. Let R be a ring. The Jacobson radical of R, written J(R), is the intersection of all the maximal ideals of R. Example 8.4. If R is a local ring, then J(R) is its unique maximal ideal. It is normal to think of J(R) as being a collection of “nasty” elements of R. Here’s an alternative characterisation: Theorem 8.5. Let R be a commutative ring. Then for an element x ∈ R, we have x ∈ J(R) if and only if 1 − xy is a unit in R for all y ∈ R. Proof. We’ll show firstly that if x ∈ J(R), then 1 − xy is a unit. Indeed, suppose it isn’t. Then let I = (1 − xy), a proper ideal of R. The ideal I is contained in a maximal ideal M (by Theorem 5.1). Since x ∈ J(R), we have x ∈ M, so xy ∈ M. Also 1 − xy ∈ M since 1 − xy ∈ I. Hence 1 ∈ M, which is a contradiction. Now suppose that 1−xy is a unit for all y, and suppose there is a maximal ideal M with x∈ / M. Then the ideal sum M + (x) (which, you’ll remember, is defined to be {i + j | i ∈ M, j ∈ (x)}) is a bigger ideal than M and is hence the whole of R (as M is maximal). Hence 1 ∈ M + (x), so 1 = u + xy for some u ∈ M and some y ∈ R. Hence 1 − xy ∈ M. But if 1 − xy is a unit, then M = R, a contradiction. 21

Lecture 9 (2014-03-11) We also need some comments about matrices which we’ll use a couple of times. Every square matrix with entries in a commutative ring has a determinant, and your favourite technique for computing the determinant will work fine. However, not every matrix has an inverse: of course, for a general com- mutative ring, not even every 1 × 1 matrix has an inverse! We can form the adjugate matrix, however. (The adjugate matrix is the matrix you have immediately before you divide by the determinant to obtain the inverse: it’s the transpose of the matrix of cofactors). The same proof as usual shows that, for any square matrix A over any commutative ring whatsoever, we have A adj(A) = adj(A)A = det(A) · I where I is the identity matrix and adj(A) is the adjugate of A. For example, if we have a two-by-two matrix a b A = , c d then its adjugate is given by  d −b adj(A) = −c a and we have a b  d −b ad − bc 0  A adj(A) = = c d −c a 0 ad − bc  d −b a b = = adj(A)A, −c a c d and so both are equal to the determinant det(A) = ad−bc times the identity matrix. In particular, whenever we have a matrix with invertible determi- nant, it is invertible. Matrices are useful with modules for the same reason that they’re useful with vector spaces: they define homomorphisms of finitely generated free modules. We’ll also be needing this construction a lot: Definition 9.1. Let R be a commutative ring, I an ideal in R, and M an R-module. We write IM for the set

IM = {i1m1 + ··· + ikmk | ii ∈ I, mi ∈ M} . 22

This is a submodule of M, since it’s a subset of M and is evidently closed under addition and scalar multiplication. Theorem 9.2. Let R be a commutative ring, and I an ideal in R. Let M be a finitely generated R-module and φ : M → M a R-module homomorphism. Suppose that im(φ) ⊂ IM. Then we can produce an equation for φ of the n n−1 form φ + in−1φ + ··· + i1φ + i0 = 0, where all the elements ik are in I. Remark 9.3. This may remind you somewhat of the Cayley-Hamilton theo- rem, if you’ve seen that.

Proof. Let m1, . . . , mk be a set of generators for M (we did say that M was finitely generated). Since im(φ) ⊂ IM, we can write each φ(mi) as a sum of the form j1n1 + ··· + jlnl where ji ∈ I and ni ∈ M. However, all those ni’s can be written as a sum of multiples of the mi’s, since the mi’s generate M. Multiplying out, we can write

φ(mi) = ai1m1 + ··· + aikmk,

for each i = 1, . . . , k and some coefficients aij ∈ I. We can regard that as a matrix equation: if we let δij be the coefficients of the identity matrix (1 if i = j and 0 otherwise), we can write

n X (δijφ − aij)mj = 0. j=1 The determinant of the matrix (with coefficients in R) whose ij-th entry is δijφ − aij is hence zero, as we can see by multiplying by the adjugate matrix. If we expand that determinant out, we get a polynomial in φ with coefficients in I being zero, exactly as we needed. Lecture 10 (2014-03-13) Theorem 10.1 (Nakayama’s Lemma, version 1). Let R be a commutative ring, and let M be a finitely generated R-module. Let I be an ideal of R such that IM = M. Then there exists x ∈ 1 + I such that xM = 0. The notation 1 + I is intended to connote the subset {1 + i | i ∈ I} of R. Also, xM = 0 means “xm = 0 for all m ∈ M”. Proof. Take φ to be the identity in Theorem 9.2 above. This gives us an identity of the form

(1 + in−1 + ··· + i1 + i0)m = 0, valid for all m ∈ M where the coefficients lie in I. The quantity in the brackets is clearly in 1 + I, so is exactly what we need. 23

For one example, consider R = Z and M = Z/2Z and I = (23). We certainly have IM = M, since 23x = x for all x ∈ M. But then 24 ∈ 1 + (23) has the property that 24M = 0. Here’s another version of the same result:

Theorem 10.2 (Nakayama’s Lemma, version 2). Let R be a commutative ring, and M a finitely-generated R-module, such that J(R)M = M. Then M is the trivial module.

Proof. We wish to use the form of Nakayama’s Lemma above (Theorem 10.1). This gives us that xM = 0 for some x ∈ J(R) + 1. By Theorem 8.5 (taking y = −1 in that theorem), x is a unit in R, and hence xM = M, and so M = 0.

I think of this as yet another result saying that J(R) consists of horrid noninvertible elements.

Example 10.3. For a possibly useful example, consider R = Z(7). This ring has Jacobson radical equal to its unique maximal ideal (7). This says then that if (7)M = M for any finitely-generated module M, then M = 0. Note that (7)M = 7M, the multiples of 7. So this tells us that multiplying by 7 cannot be surjective on any nontrivial finitely-generated Z(7)-module. The following result is a handy use of Nakayama’s Lemma; it says that finitely-generated modules are at least in one sense not so very different to finite-dimensional vector spaces.

Theorem 10.4. Let R be a commutative ring and M a finitely generated R-module. Let φ : M → M be a module homomorphism. If φ is surjective, then it is in fact bijective.

Proof. We can make M into an R[t]-module, where multiplication by t cor- responds to taking φ (this is exactly the same trick as in 1.5 in the very first lecture). Then, writing I = (t), we get that M = IM (since tm = φ(m) ∈ IM for all m and φ is surjective). Now, M is a finitely generated R-module and so is also a finitely generated R[t]-module. Now Nakayama’s Lemma (Theorem 10.1) says that there is some element f ∈ R[t] such that (1 + tf)M = 0. Now, suppose that there is an element a ∈ R such that φ(a) = 0. Then we have (1 + tf)a = 0, but (1 + tf)a = a + fφ(a) = a + 0 = a, so a = 0. This proves that φ is injective (and hence bijective). Lecture 11 (2014-03-18) 24

Remark 11.1. Of course, we could ask whether φ being injective implies that it’s bijective, but that’s not true. For example, the homomorphism f : Z → Z of Z-modules defined by f(x) = 2x is an injection but not a surjection of finitely generated modules. Now for our last form of Nakayama’s lemma:

Theorem 11.2 (Nakayama’s Lemma, version 3). Let R be a commutative ring, and let M be a finitely generated R-module. Let N be a submodule of M such that M = J(R)M + N. Then M = N.

Proof. Since M = J(R)M + N, we have nX o J(R)(M/N) = ai(mi + N) | ai ∈ J(R), mi ∈ M nX o = (aimi) + N | ai ∈ J(R), mi ∈ M nX  o = aimi + N | ai ∈ J(R), mi ∈ M = (J(R)M + N)/N = M/N.

Hence Version 2 of Nakayama’s lemma (Theorem 10.2), applied to the module M/N, tells us that M/N is the trivial module, and hence that N = M. The following is a classic use:

Theorem 11.3. Let R be a local ring, with maximal ideal I. Let M be a finitely generated R-module. Let x1, . . . , xn be elements of M such that x1 + IM, . . . , xn + IM generate the quotient M/IM. Then x1, . . . , xn in fact generate M.

Proof. Notice that, since our ring is local, I = J(R) (as mentioned in example 8.4 above). Let N be the submodule of M generated by the elements x1, . . . , xn. Then, since N generates all of M/IM, we have N + IM = M. As a result, we have N = M by Theorem 11.2, so M is generated by x1, . . . , xn.

Integral Extensions

Let R be a commutative ring. Recall from last semester that an R-algebra is a commutative ring S together with a homomorphism f : R → S. In this section we’ll talk about some nice classes of R-algebras. Recall from the first lecture that, when we have an R-algebra S, then S is also an R-module. 25

Suppose we have a chain of homomorphisms of commutative rings

f g R −→ S −→ T.

Then f makes S into an R-algebra, g makes T into an S-algebra, and gf makes T into an R-algebra.

Proposition 11.4. Suppose given a chain of homomorphisms as above. If S is a finitely-generated R-module, and T is a finitely-generated S-module, then T is also a finitely-generated R-module.

Proof. If S is a finitely-generated R-module, then there are elements s1, . . . , sm ∈ S such that any element a ∈ S can be expressed as a sum X a = aisi. 1≤i≤m

If T is a finitely-generated S-module, then there elements t1, . . . , tn ∈ T such that any element b ∈ T can be expressed as a sum

n X b = bjtj. 1≤j≤n

Now, we will show that the elements sitj (for 1 ≤ i ≤ m and 1 ≤ j ≤ n) generate T as an R-module. Indeed, let b be any element of T . We can write X X X X b = bjtj = (aijsi)tj = aij(sitj), 1≤j≤n 1≤j≤n 1≤i≤m 1≤i≤m 1≤j≤n for some elements bj ∈ S and aij ∈ R, exactly as needed.

Now we introduce a definition, extremely useful in algebraic number the- ory and algebraic geometry alike:

Definition 11.5. Let S be an R-algebra. An element x ∈ S is said to be integral over R if it satisfies a monic polynomial equation

n n−1 x + an−1x + ··· + a0 = 0, where an−1, . . . , a0 ∈ R. 26

Formally, what we really mean by that, of course, is that there are ele- ments an−1, . . . , a0 of R such that

n n−1 x + f(an−1)x + ··· + f(a0) = 0, where f : R → S is the structure map of the S-algebra R. But it’s common- place to not mention the structure map, and I’ll usually avoid doing so from now on. Here are some examples and nonexamples. Lecture 12 Example 12.1. Given a commutative ring homomorphism f : R → S, any (2014-03-20) element r ∈ R is integral over R, because it is a root of the equation x−r = 0. Exercise 12.2. The element 1/2 ∈ Q is not integral over Z. √ Example 12.3. The element 17 ∈ R is integral over Z, for example, because it satisfies x2 − 17 = 0. √ 1+ 5 Example 12.4. The golden ratio 2 is integral over Z, because it satisfies x2 − x − 1 = 0. Example 12.5. Any element of C is integral over R. Indeed, if x = a + ib, then (x − a)2 = −b2 and hence x2 − 2ax + a2 + b2 = 0. Now we’ll see what this has to do with things we were talking about earlier:

Theorem 12.6. Let S be an R-algebra. The following are equivalent:

(i) S is a finitely-generated R-module;

(ii) S is generated as an R-algebra by integral elements x1, . . . , xn;

(iii) S is a finitely-generated R-algebra, and every element of S is integral over R.

Proof. The implication (iii) ⇒ (ii) is obvious. We’ll prove (ii) ⇒ (i) and then (i) ⇒ (iii). In order to prove (ii) ⇒ (i), suppose that S is generated as an R-algebra by x1, . . . , xn which are roots of monic polynomials of degrees d1, . . . , dn re- spectively. a1 an Then S is generated as an R-module by the monomials x1 ··· xn . But in fact we don’t need the monomials where ai ≥ di for any i, since we can use integrality to rewrite these. This means we have a finite set of generators. Now we must prove (i) ⇒ (iii). Note firstly that if S is a finitely-generated R-module then it’s certainly a finitely-generated R-algebra: the module gen- erators generate it as an algebra. 27

Let x1, . . . , xn generate S as an R-module, and let s be any element of S; we must show s is a root of a monic polynomial with coefficients in R. We now employ Theorem 9.2, taking I = R so that IM = M, and φ : M → M to be m 7→ sm. This gives us the monic polynomial we want, with coefficients in R.

Definition 12.7. We say that an R-algebra satisfying the conditions of The- orem 12.6 is an integral extension.

Corollary 12.8. Let R be a commutative ring, and S an R-algebra. Then sums and products of elements which are integral over R are integral over R.

Proof. Suppose given two elements x, y ∈ S which are integral over R. Then consider the subalgebra T ⊂ S generated as an R-algebra by x and y. Since it’s generated by integral elements x and y, by Theorem 12.6 all its elements are integral over R, but this includes x + y and xy. Lecture 13 (2014-03-25) Corollary 13.1. Let R be a commutative ring, and S an R-algebra. Then the set of integral elements forms a subalgebra of S.

Proof. This is obvious from Example 12.1 and Corollary 12.8 above.

Here’s a way of describing R-algebras which, by contrast, are definitely not finitely generated as R-modules:

Definition 13.2. Let R be a commutative ring, and S an R-algebra. We say that elements s1, . . . , sn of S are algebraically independent if the R-algebra homomorphism

R[x1, . . . , xn] → S sending xi to si for each i, is an injective homomorphism.

Equivalently, s1, . . . , sn are algebraically independent if there are no nonzero n-variable polynomials with coefficients in R which vanish at (s1, . . . , sn). This allows us to prove a big theorem, describing the structure of finitely- generated algebras:

Theorem 13.3 (Noether Normalisation Theorem). Let K be a field, and S a finitely-generated K-algebra. Then S is an integral extension of a polynomial ring over K. More precisely, there are algebraically independent elements s1, . . . , sn of S such that S is an integral extension of the subring K[s1, . . . , sn]. 28

Proof. Suppose that S is generated over K by elements s1, . . . , sm; we prove it by induction on m. For our base case, if m = 0 then S is generated as an algebra over K by no elements at all, and hence the map K → S is surjective, so 1 generates S as a K-module, and hence K → S is an integral extension. So let’s do the induction step, with m ≥ 1. If the elements s1, . . . , sm are algebraically independent, we’re done (as then S = K[x1, . . . , xm], and certainly S is finitely generated as a module over itself). So suppose instead there is some nonzero polynomial relation f(s1, . . . , sm) = 0. Without loss of generality sm appears in it. We will manipulate to get an integral relation. Let r be an integer, which we’ll fix later in the proof (it will be big, but we can work out how big it needs to be later). We make a change of variables, leaving s1 alone and defining

r t2 = s2 − s1 r2 t3 = s3 − s1 . . rm−1 tm = sm − s1 . Now our polynomial relation is

r rm−1 f(s1, t2 + s1, . . . , tm + s1 ) = 0. The polynomial f is a sum of monomials of the form

α1 αm as1 ··· sm , where a ∈ K, and we can rewrite that as a sum of products of the form

m−1 α1 r α2 r αm as1 (t2 + s1) ··· (tm + s1 ) .

When we multiply out, the term with the highest power of s1 is

m−1 α1+α2r+···+αmr as1 . Now, if we set r bigger than any of the α’s occuring in many of the polynomials, then every monomial gives a different highest power of s1 (we can think of it as encoding the monomials as a number in base r). That means we get a unique highest power, and that’s nonzero. Then, multiplying both sides by a−1 (possible, as we’re in a field), we get a monic equation for s1 over the subalgebra generated by s2, . . . , sn. By induction s2, . . . , sn generate a finite extension of a polynomial alge- bra, and we’re now done by Proposition 11.4. 29

Lecture 14 (2014-03-27) Using some technology we don’t have, it is possible to deduce from Noether Normalisation that every variety admits a quasifinite map to some affine space An (a quasifinite map is one where every point in the codomain has a finite number of preimages in the domain).

Proving the Nullstellensatz There are many proofs of the Nullstellensatz. We give one which does not require much technology beyond the Noether Normalisation theorem. Our method of proof will do it in several steps: hopefully these steps may be of independent interest. We start by gaining a little more understanding of finiteness: Proposition 14.1. Suppose R is a commutative ring, and S an R-algebra which is finitely generated as an R-module. If S is a field, then so is R. Note that this gives us (for example) that Q is not a finitely-generated Z-module, and that C(x) = K(C[x]) is not a finitely-generated C[x]-module. Proof. Suppose x ∈ R. Of course we then have x ∈ S, so x has an inverse x−1 ∈ S. Now, Theorem 12.6 tells us that, since S is a finitely-generated R-algebra, the element x−1 is a root of a monic polynomial with coefficients in R: we have −n 1−n −1 x + an−1x + ··· + a1x + a0 = 0,

for a0, . . . , an−1 ∈ R. Multiplying through by xn−1 we get

−1 n−2 n−1 x = −an−1 − an−2x − · · · − a1x − a0x , and the right-hand side is clearly in R. Hence R is a field. This enables us to prove the following: Theorem 14.2 (Zariski’s Lemma). Let K be a field, and let L be a finitely generated K-algebra. If L is also a field, then L is an integral extension of K. Proof. The Noether Normalisation Theorem (Theorem 13.3) states that L is an integral extension of some polynomial K-algebra K[a1, . . . , an]. Our aim is to show that n = 0. So, in order to get a contradiction, suppose that n ≥ 1. But then we have that L has is integral over the subring K[a1, . . . , an], and Proposition 14.1 then tells us that K[a1, . . . , an] is a field. That’s absurd: for example, a1 is not invertible. 30

Lecture 15 Let us remind ourselves about algebraically closed fields. Recall (from (2014-04-01) Semester 1) that an algebraically closed field is one where every polynomial has a root, and hence where every polynomial can be written as a product of linear factors. The following is another way of thinking about it: Proposition 15.1. If K is an algebraically closed field, then there is no K-algebra L which is a field and which is finitely generated as a K-module. In fact, the result can be strengthened to “if and only if”, but we won’t prove that. Proof. Suppose K is algebraically closed, and L is a K-algebra which is finitely generated as a K-module. Then L is integral over K. So any ele- n n−1 ment x ∈ L satisfies a monic polynomial x + an−1x + ··· + a0 = 0 with coefficients in K. But this polynomial splits as a product of roots:

(x − r1) ··· (x − rn) = 0.

However, L is a field, so x − ri = 0 for some i, so x = ri for some i. Hence any element x ∈ L is in fact an element of K. We now know enough to classify maximal ideals over polynomial rings over an algebraically closed field. Theorem 15.2. Let K be an algebraically closed field. Then the maximal ideals in K[x1, . . . , xn] are exactly the ideals of the form

(x1 − a1, . . . , xn − an) for elements a1, . . . , an ∈ K.

Note this says that maximal ideals in K[x1, . . . , xn] correspond to points in Kn. Proof. We should show that an ideal of this form is indeed maximal. So consider the ideal

(x1 − a1, . . . , xn − an).

By changing coordinates, writing yi for xi −ai we can consider the special case of (y1, . . . , yn) as an ideal in K[y1, . . . , yn]. This ideal consists of all polynomials with zero constant term; the quotient is hence evidently K. Since this is a field, the ideal is maximal. 31

On the other hand, suppose we have a maximal ideal M of K[x1, . . . , xn]. Consider the field K[x1, . . . , xn]/M. This is generated over K by x1, . . . , xn, and so 14.2 says that it is an integral extension of K. But that means it’s isomorphic to K, since K is algebraically closed (Proposition 15.1). Now, let ai be the image of xi under this isomorphism. Now the element xi − ai of K[x1, . . . , xn] is sent to 0 in K[x1, . . . , xn]/M, and so xi − ai ∈ M. Hence (x1 −a1, . . . , xn −an) ⊂ M. But since the left-hand side is maximal, this must be M. Now (at last!) we can do the Nullstellensatz itself. It turns out to be convenient to do one very special case first:

Theorem 15.3 (Weak Nullstellensatz). Let K be an algebraically closed field, and let J be an ideal in R = K[x1, . . . , xn]. If V (J) is the empty set, then J is the unit ideal R.

Proof. We prove the contrapositive: if J is not the unit ideal, then V (J) is nonempty. Suppose J is not the unit ideal, then J ⊂ M for some maximal ideal (by Theorem 5.1). Now, by Theorem 15.2 above, M = (x1 − a1, . . . , xn − an) for some elements a1, . . . , xn ∈ K. That means that (a1, . . . , an) ∈ V (J). Lecture 16 Now we strengthen this to the following, which was stated (but not (2014-04-03) proved) in the notes for Semester 1 as Theorem 14.5.

Theorem 16.1 (Hilbert’s Nullstellensatz). Let K be an algebraically closed field, and let J be an ideal in the polynomial algebra K[x1, . . . , xn]. Then the ideal of elements of K[x1, . . . , xn] which vanish on the zeroes of J is equal to the radical of J: √ I(V (J)) = J. √ √ Proof. It is clear that J ⊂ I(V (J)). Indeed, if f ∈ J, then f N ∈ J for some N. But then f N ∈ I(V (J)) (this just says that f N is a member of the things which vanish wherever the elements of J vanish). But f vanishes N wherever f does, so f ∈ I(V (J)) too. √ We now prove the harder containment: that I(V (J)) ⊂ J. Let J be an ideal of K[x1, . . . , xn], and let f ∈ J. We introduce a new 0 variable y, and consider the ideal J = J + (fy − 1) of K[x1, . . . , xn, y]. Here we use the notation J for the ideal generated by the elements of J in the new ring K[x1, . . . , xn, y]. Now, an element of V (J 0) consists of an element of V (J) where fy−1 also −1 vanishes: in other words, where y = f(x1, . . . , xn) . Given (x1, . . . , xn) ∈ V (J), there is such a y whenever f(x1, . . . , xn) is nonzero. 32

Clearly then if f ∈ I(V (J)), or in other words if f(x1, . . . , xn) = 0 for all 0 (x1, . . . , xn) ∈ V (J), then V (J ) is the empty set. By Theorem 15.3 above, this means that 1 ∈ J 0: in other words, we can write X 1 = figi + g0(fy − 1), i where fi ∈ J and g0, gi ∈ K[x1, . . . , xn, y]. Suppose that N is the degree of the highest power of y that appears in the above. If we multiply though by f N we get

N X f = Gifi + G0(fy − 1), i which is now an equation of polynomials in x1, . . . , xn, fy. If we do long division by (fy − 1), we get an equation

N X f = hifi (mod (fy − 1)), i

N where the h’s are√ polynomials in x1, . . . , xn. Hence f ∈ J, which proves that I(V (J)) ⊂ J: this is the difficult containment done.

Projective varieties

Now, thanks to the last 36 lectures or so, we have a decent understanding of affine space. Unfortunately, affine space is not the right setting for much of modern geometry; I’ll try to explain why. Suppose we have two distinct lines in the affine plane A2. How many times do they intersect? The answer is that they meet at either no points or one point: zero if the lines are parallel and one if not. That’s not a great theorem, so people commonly refer to “points at infinity” in order to make it more exciting: every two distinct lines intersect exactly once, but that intersection point might be a “point at infinity”. This causes us to invent the projective plane. If we take this idea seriously, we can say a bit about the behaviour of these points at infinity. For example, this is because the lines y = 0 and y = 1 are supposed to meet at a point at infinity, and the lines y = 3x + 5 and y = 3x − 7 are also supposed to meet at a point at infinity, and these points have to be different since y = 0 and y = 3x + 5 already meet at 5 (− 3 , 0). Hence there has to be one point at infinity for each direction (that is, for each set of lines under the equivalence relation that says that two lines 33

Lecture 17 are equivalent if they are parallel). Clearly if we are going to go further, (2014-04-29) this description will start to annoy us: we want a way of treating points at infinity like any other point. The standard solution is to go up a dimension, and treat the projective plane as the set of lines through the origin in A3. To see why this is sensible, imagine a lecturer pointing at a screen with a laser pointer. The position of the laser beam forms a line passing through the lecturer, and most such positions determine a unique point on the screen.2 However, if the laser pointer is moved so as to become parallel with the whiteboard, the point on the screen that is being depicted rushes off to infinity. Hence a parallel laser pointer position can be used to represent a point on the screen at infinity. Of course, a line through the origin is determined by any nonzero point on that line. This motivates the following:

Definition 17.1. Let n be a positive integer, and K be an algebraically closed field. We define projective n-space Pn to be the set of lines through the origin in An+1. Equivalently, we define

n n+1 P = (A \{0})/ ∼,

where the equivalence relation identifies (x0, . . . , xn) with (kx0, . . . , kxn), wherever 0 6= k ∈ K and x0, . . . , xn ∈ K. The latter approach is taking the set of points which determine lines through the origin in An+1 and identifying those which determine the same line. We normally use special notation for the coordinates of a point in the latter approach, writing (x0 : ... : xn) for this point. The sole purpose of the colons is to remind us that we are not in affine space: the coordinates (0 : ... : 0) are forbidden, and we are subject to the equivalence relation (x0 : ... : xn) = (kx0 : ... : kxn). These are called homogeneous coordinates for our point. n n As we’d hope, we can regard A as a subset of P : the point (x1, . . . , xn) corresponds to the point (1 : x1 : ... : xn). So the points of the form (0 : x1 : ... : xn) can be regarded as the points at infinity in the affine plane.

2OK, we have to assume that the laser beam goes in both directions. There is a Star Wars film illustrating this concept. 34

(In particular, every point in Pn can be written in exactly one of these two forms). In fact it’s possible to regard An as a subset of Pn in many other ways (imagine the lecturer being told that they had been pointing at completely the wrong wall, and that in fact some other wall was the screen: everything they were doing made sense, but the points at infinity were not the points they thought were at infinity). The easiest is to take the n + 1 different sets with homogeneous coordinates

(x0 : ... : xi−1 : 1 : xi+1 : ... : xn)

for 0 ≤ i ≤ n. These are referred to as the affine pieces of Pn. So every Lecture 18 point of Pn is in at least one affine piece. But, for the minute, let’s keep (2014-05-01) to thinking about Pn as a quotient of An+1\{0}. Our aim was to regard Pn as a home for algebraic varieties together with their points at infinity: that means we need to define “projective varieties”. Since Pn has been defined to be the set of lines through the origin of An+1, the natural thing to do is this:

Definition 18.1. A projective variety X ⊂ Pn is a subvariety of An+1 which is a union of lines through the origin. There’s an obvious way of building these, but we’ll need some algebraic language.

Definition 18.2. A polynomial in K[x0, . . . , xn] is said to be homogeneous if all its terms have equal total degree (that is, the sums of the exponents of the variables are all equal). An ideal J < K[x0, . . . , xn] is said to be homogeneous if it is generated by homogeneous elements. So, for example, x2y + xyz + z3 is homogeneous of degree 3, and x6y − z7 is homogeneous of degree 7, but x2y3 − x2 is not homogeneous as it has one term of degree 5 and one term of degree 2. Why are homogeneous polynomials relevant to us? Well, if a polynomial f is homogeneous of degree k, then

k f(λx0, . . . , λxn) = λ f(x0, . . . , xn).

That means that if J is a homogeneous ideal in K[x0, . . . , xn], then V (J) has the property that if (x0, . . . , xn) ∈ V (J), then (λx0, . . . , λxn) ∈ V (J) for all λ. That almost says that V (J) is a union of lines through the origin in An+1, since if we have any nonzero point we have the whole line through it, 35

but there’s an annoying special case. If J is the ideal (x0, x1, ··· , xn), then V (J) = {0}, which isn’t quite a union of lines. We call this the irrelevant ideal, and frequently have to explicitly exclude it from statements. In fact, we can go the other way as well. But to do that we need to remind ourselves of an old result in algebra, which you may well have seen already:

Proposition 18.3 (Vandermonde’s determinant). Let λ0, . . . , λm be elements of a field K. Then we have the following determinant formula:

2 m 1 λ0 λ ··· λ 0 0 1 λ λ2 ··· λm 1 1 1 1 λ λ2 ··· λm Y 2 2 2 = (λj − λi)...... i

The proof is a fairly straightforward induction, and in fact we really only need that the determinant is nonzero if the λ’s are all different. That’s what we need to complete the story of homogeneous ideals and projective varieties: Theorem 18.4. Let X be a projective variety: a union of lines through the origin in An+1. Then I(X) is a homogeneous ideal.

Proof. Suppose f ∈ I(X). If we let fk be the sum of the degree-k terms of f, then we can write f = f0 +···+fm, and each fk is homogeneous of degree k. Then for any λ, the function f ◦λ obtained by scaling An+1 by λ and then 2 m applying f is also in I(X). This function is f0 + λf1 + λ f2 + ··· + λ fm. But if we take m + 1 different values λ0, . . . , λm ∈ K we get an equation

 2 m     1 λ0 λ0 ··· λ0 f0 f ◦ λ0 1 λ λ2 ··· λm  f   f ◦ λ   1 1 1   1   1  1 λ λ2 ··· λm  f   f ◦ λ   2 2 2   2  =  2  . ......   .   .  . . . . .   .   .  2 m 1 λm λm ··· λm fm f ◦ λm

All the terms f ◦ λi are in I(X), and so, inverting the Vandermonde ma- trix (which we can do, as it has nonzero determinant), we get fk a linear combination of elements of I(X). But as f = f0 + ··· + fm, we have homogeneous generators for I(X). 36

The Nullstellensatz (Theorem 16.1) gives us a bijection between varieties n+1 X ⊂ A and radical ideals J < K[x0, . . . , xn], for an algebraically closed field K. The above result tells us that this restricts to a bijection between projec- n+1 tive varieties X ⊂ P , and homogeneous radical ideals J < K[x0, . . . , xn]. Lecture 19 This is sometimes referred to as the “projective Nullstellensatz”. We some- (2014-05-06) times use extra notation to refer to the V and I constructions in projective geometry: one common choice is V p and Ip. So V p takes a homogeneous ideal and gives a projective variety, and Ip takes a projective variety and gives a homogeneous ideal. Now we understand how to do projective algebraic geometry at least in theory, we should do some examples, so we can see how it works in practice. One thing we haven’t dealt with yet is how to get a projective variety from an affine variety. This is rather important to us: we invented projective varieties so we could talk about points at infinity on affine varieties; all this will have been for nothing unless we can take an affine variety and get a projective variety from it. Let’s take a concrete example: the parabola. This is the affine variety described by the ideal (y − x2) of C[x, y]. In order to produce a projective variety, we need a homogeneous ideal of C[X,Y,Z]. The story is that we make our ideal homogeneous by inserting Z’s wherever necessary. So (y − x2) becomes (YZ − X2). Why is this the right thing to do? Well, when we restrict this projective variety to the standard affine piece Z = 1, the substitution Z = 1 gives us (Y − X2) again, exactly as hoped. So what are the points at infinity on the parabola? This happens when we set Z = 0 instead, which tells us that X2 = 0, so X = 0. So there’s only one point on the parabola at infinity, namely the point (0 : 1 : 0). So this suggests the point of view that a parabola is a kind of degenerate ellipse, whose major axis extends out to infinity (and if you know your conic sections well, this is exactly what it is). A pedant could observe that the ideal we chose, (YZ − X2), is not quite the only way of producing a homogeneous ideal with this property: we could have taken (YZ2 − X2Z) or even (YZ101 − X2Z100). The counterargument to this is that these all vanish whenever Z = 0: they contain the whole line at infinity. So they’re really not what we want: it makes sense to take the most conservative extension, where we make it homogeneous by topping up with Z’s to the largest degree that already exists. Let’s do another example. The affine variety X ⊂ A3 defined by

X = V (x + y + z + 1, x − 2y + 3z − 4) 37

∞ ∞ y y

x x

Figure 1: The projective variety V p((YZ − X2)) (left) and the projective variety V p((YZ2 −X2Z)) (right): they’re the same, except the latter includes the line at infinity. is a line, given as the intersection of two planes. In order to find the corresponding projectivisation, we make these equa- tions homogeneous by inserting powers of a new variable W , and get Xp = V p(X + Y + Z + W, X − 2Y + 3Z − 4W ). We can check what we’ve done by restricting to the affine piece W = 1, where we get the old equation back. What are the points at infinity? These happen, on the other hand, where W = 0, so X + Y + Z = X − 2Y + 3Z = 0. Subtracting these two equations 2 2 5 gives 2Z − 3Y = 0. So Y = 3 Z. Hence X = −Z − 3 Z = − 3 Z. Thus there 5 2 is one point at infinity, given by (− 3 : 3 : 1), or if we prefer by (−5 : 2 : 3).

Tangent spaces

Our next (and final) topic will be the discussion of local behaviour of algebraic varieties. This will bring together many of the things discussed earlier. Here’s our basic definition: n Definition 19.1. Suppose A has coordinates x1, . . . , xn. n Let X ⊂ A be an affine variety, and let a = (a1, . . . , an) be a point on it. The tangent ideal Ia(X) is generated by the polynomials X ∂f (x − a ) (a) i i ∂x i i 38

as f ranges over the defining ideal I(X). The tangent space Ta(X) is defined to be V (Ia(X)).

Each of the generating polynomials of Ia(X) is of degree 1 and vanishes at a. Hence it cuts out a hyperplane of An passing through a. As a result, the tangent space, which is the intersection of these hyper- planes, is an affine linear space passing through a. Proposition 19.2. When calculating tangent spaces, we only need to take f to be the generators of I(X).

Proof. Indeed, suppose f1, . . . , fm are generators of I(X), and suppose f ∈ I(X). Then we can write f as f = g1f1+···+gmfm, where gi ∈ K[x1, . . . , xn]. Then m m   ∂f X ∂gkfk X ∂gk ∂fk (a) = = (a)fk(a) + gk(a) (a) . ∂xi ∂xi ∂xi ∂xi k=1 k=1

But fk(a) = 0 since a ∈ X and fk ∈ I(X). So all we get is a linear combina- tion of the elements of IX (a) arising from the generators. Let’s do an example: let X be the plane curve defined by y2 +x3 −x2 = 0. The generator of its vanishing ideal is hence f = y2 + x3 − x2, and that has partial derivatives ∂f ∂f = 3x2 − 2x, = 2y. ∂x ∂y The tangent space at the point (a, b) is thus given by the equation

(3a2 − 2a)(x − a) + 2b(y − b) = 0.

This is a line, unless 3a2 − 2a = 2b = 0. But that happens only when b = 0 and when a = 0 or 3a − 2 = 0, ie a = 2/3. The point (0, 0) does lie on X but the point (2/3, 0) does not. Hence the dimension of the tangent space of X at a point is 1, except at (0, 0) where it is 2. This result motivates the following definition: Definition 19.3. A point p on a variety X is nonsingular if there is a Zariski- open set p ∈ U ⊂ X with the property that dim Tp(X) ≤ dim Tq(X) for any q ∈ U. The point p is said to be singular if it is not nonsingular (that is, if in any Zariski-open neighbourhood of p there is some point whose tangent space is of strictly smaller dimension than that of p. Lecture 20 (2014-05-08) And, armed with that, we can make a global definition: 39

Definition 20.1. A variety X is smooth if it has no singular points.

It is very useful to have a more high-tech description of the tangent space at a point p. In order to do this, suppose that p is the origin 0 (we can always do this without changing anything, by translating our variety X). Then we have the algebraic advantage that Tp(X) has the structure of a vector space: it’s cut out by linear equations. We give an alternative description of this vector space:

Theorem 20.2. Let X be a variety containing the origin 0. Let M be the maximal ideal of K[X] corresponding to the point 0 ∈ X. 2 Then the vector space T0(X) is naturally dual to the vector space M/M .

Proof. We prove this first for X = An, and then we extend to the case where X is a subvariety of An. n If X = A , the ideal M is given by (x1, . . . , xn), the ideal of polynomials 2 2 2 with zero constant term. Then M = (x1, x1x2, . . . , xn) is the ideal of poly- nomials with zero constant term and zero linear terms, and hence M/M 2 is an n-dimensional vector space generated by the polynomials x1, . . . , xn. n n On the other hand, T0(A ) is K , since there are no generators in the ideal I(An) and hence nothing to restrict the tangent space. Of course Kn is generated as a vector space by the elements

a1 = (1, 0,..., 0), . . . , an = (0,..., 0, 1).

Moreover, these points ai form a dual basis to the polynomials x1, . . . , xn, since xi(ai) = 1 and xi(aj) = 0 for i 6= j. n n Now, what about X ⊂ A ? Then we get an inclusion T0(X) → K , with n ∨ ∨ a dual (K ) → T0(X) . This gives us a surjection

n ∨ ∨ D :(x1, . . . , xn) −→ (K ) −→ T0(X)

(since both maps are surjective). 2 I claim firstly that the kernel of this composite D is (x1, . . . , xn) + I(V ). Indeed, if f ∈ ker D, then X ∂f x = 0 ∂x i i i

on T0(X), by the definition of D. By the definition of T0(X), this is the same as having X ∂f X X ∂gk xi = ak xi, ∂xi ∂xi i k i 40

for some a1, . . . , am ∈ K[x1, . . . , xn] and g1, . . . , gm ∈ I(V ). That in turn is P 2 the same as having f − k akgk ∈ (x1, . . . , xn) , which is the same as having 2 f ∈ (x1, . . . , xn) + I(V ). As a result,

2 2 ∨ M/M = (x1, . . . , xn)/((x1, . . . , xn) + I(V )) = T0(V ) .

Using this description it is possible (and not too difficult, but we don’t have time) to prove the following:

Theorem 20.3. For a projective variety, the tangent space of a point does not depend on the affine piece that we choose to calculate it at.

In fact, much more is true: this description of the tangent space of a point survives localisation (at ideals contained in the maximal ideal corresponding 2 ∨ to that point): we have MI /MI = T0(V ) . These statements unify many of the topics of the course!

What’s next?

This course has consisted of one year and forty lectures. What might we talk about if we had another year, and forty more? Here are some ideas (which are certainly not intended to be a complete list):

(a) One can start studying curves in earnest. It is possible to get a very good understanding of the theory of curves, but classifying higher di- mensional varieties in such a complete fashion is out of reach of modern mathematics. For example, we can ask what the space of all curves looks like. There are infinitely many connected components, one with dimension 0, one with dimension 1, and one with dimension 3n for each n ≥ 1.

(b) At various times, particularly when we’ve been thinking geometrically, we have restricted ourselves to working over a field, and usually an alge- braically closed field. At the expense of more complication here and there, not all of this is absolutely necessary. The resulting theory of schemes is an amazing generalisation of the theory of varieties, which works over an arbitrary ring. This is of particular use to number theorists, who wish to do algebraic geometry over Z and related rings. 41

(c) Varieties are a great source of nice spaces. It is perfectly reasonable to ask if we can get information about their topology (such as their homology) directly from the algebra. Most well-known and easy approaches to algebraic topology work fairly badly in algebraic geometry (for example because the Zariski topology is a terrible space). But the theory of sheaves does an excellent job, and gives us a good construction of topological invariants of varieties. Index

+ (of submodules), 7 image ∩ (of submodules), 7 (of module homomorphism), 7 ⊕ (of modules), 4 integral, 25 integral extension, 27 adjugate matrix, 21 inverting elements, 14 affine pieces, 34 irrelevant ideal, 35 algebraically closed, 30 isomorphism theorem Artinian module, 9 first —, 6 Artinian ring, 10 second —, 7 ascending chain condition, 9 third —, 8 axiom of choice, 12 Jacobson radical, 20 cartesian products, 4 Cayley-Hamilton theorem, 22 kernel cotangent space, 40 (of module homomorphism), 6 descending chain condition, 9 line at infinity, 36 determinant, 21 local integers, 16 direct sum, 4 local ring, 17 localisation field — of a module, 20 — of fractions, 13 — of a ring, 16 finitely generated, 5, 20 counterexample, 6 module, 2 free module, 4 multiplicative subset, 13 functoriality, 19 Nakayama’s Lemma, 22–24 generate, 5 Noether Normalisation Theorem, 27, 29 Hilbert, 31 Noetherian module, 9 homogeneous, 34 Noetherian ring, 9 homogeneous coordinates, 33 nonsingular point, 38 homomorphism, 5 Nullstellensatz, 31 weak —, 31 ideal, 3 prime —, 13 projective Nullstellensatz, 36

42 INDEX 43 projective plane, 32 projective space, 33 projective variety, 34 projectivisation, 36 quotient module, 6 rings of fractions, 14 singular point, 38 singular variety, 39 submodule, 3 tangent space, 38

Vandermonde’s determinant, 35 vector space, 3

Zariski’s Lemma, 29 Zorn’s lemma, 12