Math 631, Lecture 2

Notes by Bryden Cais

Sept. 9, 2002

Let S, S′ ⊂ k[T_{1}, …, T_{n}] be systems of algebraic equations. Recall that we say that S ≃ S′ if Sol(S; K) = Sol(S′; K) for all k-algebras K.

Definition 1. An affine algebraic variety X over k, or equivalently a k-variety is the set of solutions of any representative of an equivalence class of systems of algebraic equations. Note that this is a good definition since it does not depend on the representative chosen. We denote by X(K) = X ∩ K^{n} = Sol(S; K) for any representative S of the equivalence class.

Example 1. Let S = {0}. then X = 𝔸_{k}^{n} is affine n-space (over k). Clearly, 𝔸_{k}^{n}(K) = K^{n}.

Example 2. Let S = {1}. then X = ∅_{k} and ∅_{k}(K) = ∅.

Definition 2. Let S be an algebraic set of equations and let X(K) = Sol((S); K). Then (S) is the defining ideal of X. Observe that this is also well defined since we proved last time that s ≃ S′ if and only if (S) = (S′).

Notice that we are heavily using the fact that we can take K to be any k-algebra. For example, Sol({X}; K) = Sol({X^{e}}; K) for any field K, but if we take K to be the k-algebra k[X]/x^{e} then we see that X^{e} = 0 but X ≠ 0, so the varieties above are different.

Theorem 1 (Hilbert’s Basis Theorem). Any Ideal I ⊂ k[T_{1}, …, T_{n}] is finitely generated.

Proof. We will induct on n. For n = 1, the Theorem is clear since for any field k, the ring k[T_{1}] is a principal ideal domain. Now suppose that the theorem is true for n − 1 generators and let I ⊂ k[T_{1}, …, T_{n}]. For any F ∈ I, we write F = b_{0}T_{n}^{r} + ⋯ + b_{n}, where b_{i} ∈ k[T_{1}, …, T_{n−1}]. We call b_{0} the leading coefficient of F and say that F is of degree r in T_{n}. Now let J_{r} ⊂ k[T_{1}, …, T_{n−1}] be the set consisting of all leading coefficients of polynomials in I of degree r in T_{n}. It is easy to see that J_{r} is an ideal: since I is an ideal, we can multiply any degree r polynomial (in T_{n}) by any polynomial in k[T_{1}, …, T_{n}] and obtain another polynomial of degree r in T_{n} contained in I; the leading coefficient of this polynomial is therefore in J_{r}. Now observe that we can multiply any degree r in T_{n} polynomial in I by T_{n} to obtain a degree r + 1 in T_{n} polynomial still contained in I (again since I is an ideal). Thus, we see that J_{r} ⊂ J_{r+1}. We therefore define the ideal

J = ⋃_{0≤r} J_{r} ⊂ k[T_{1}, …, T_{n}].

Now by our induction hypothesis, the ideals J_{r} and J are finitely generated. Let (a_{1r}, …, a_{m(r)r}) be a set of generators for J_{r} and (a_{1}, …, a_{s}) a set of generators of J. Moreover, since we certainly have a_{ir} ∈ J_{r} for each 1 ≤ i ≤ m(r), there exist polynomials F_{ir} ∈ I with leading coefficient a_{ir} (this follows from the definition of J_{r}). Similarly, for each 1 ≤ i ≤ s there exists a polynomial F_{i} ∈ I with leading coefficient a_{i}. Let the degree of F_{i} be d(i) and set

N = max{d(1), …, d(s)}.

We claim that I is generated by

S = {F_{ir}, F_{j} : 1 ≤ i ≤ m(r), 0 ≤ r ≤ N, 1 ≤ j ≤ s} .

We will show that any polynomial of degree d > N in T_{n} may be reduced to a polynomial of degree at most N in T_{N}, and then that any such polynomial can be obtained from the J_{ir}. So let G ∈ I be of degree d > N. Certainly, the leading coefficient l_{d} of G is in J (in fact in J_{d}), so that we may write

l_{d} = c_{1}a_{1} + c_{2}a_{2} + … + c_{s}a_{s}

for c_{j} ∈ k[T_{1}, …, T_{n−1}]. Now consider the polynomial

H = G − ((c_{1}T^{d−d(1)})F_{1} + (c_{2}T^{d−d(2)})F_{2} + … + (c_{s}T^{d−d(s)})F_{s}).

Clearly H_{1} ∈ I since F_{i} ∈ I and I is an ideal of k[T_{1}, …, T_{n}]. Moreover, it is clear that H has degree at most n − 1. We can therefore proceed in this manner until we obtain a polynomial H_{t} of degree at most N. It now suffices to show that this H_{t} is in the ideal generated by S. Observe that we can use the F_{ir} to kill the leading coefficient of H_{t}, just as above, since a_{ir} for a fixed r generate all leading coefficients of polynomials of I of degree r in T_{n}. We proceed in this way until we obtain a polynomial of degree 0 in T_{n}, which we can obtain from the F′_{i0}s. This, G is in the ideal of k[T_{1}, …, T_{n}] generated by S. Since G was arbitrary, we are done, and I is generated by S.

Now suppose that K is an algebraically closed field. Observe that in this case we do not have in general that Sol(S; K) = Sol(S′; K) if and only if (S) = (S′). For example, the ideals generated by S = X and S′ = X^{2} are not equal, but over any field, Sol(X; K) = Sol(X^{2}; K) = {0}. However, as the following theorem tells us, this is essentially the only way in which two different ideals can correspond to the same variety.

Theorem 2 (Hilbert’s Nullstellensatz). For any commutative ring A and any ideal I ⊂ A define the radical of I to be the set

rad(I) = {x ∈ A : x^{n} ∈ I for some n ≥ 1} .

It is not difficult to see that rad(I) is an ideal: If x ∈ rad(I) and a ∈ A then (ax)^{n} = a^{n}x^{n} ∈ AI ⊂ I since A is commutative and I is an ideal. Let K be algebraically closed. Then

Sol((S); K) = Sol((S′); K)

if and only if

rad((S)) = rad((S′)).

We first prove two crucial lemmata:

Lemma 1. Let A be a finitely generated algebra over a field k. Then if A is a field, A is an algebraic extension of k.

Proof. Let A be generated by {t_{1}, …, t_{r}}. We induct on r. For r = 1 we have a surjective map

k[T] → A

given by T ↦ t_{1} (this is just what it means for A to be a k algebra). CLearly the map is not injective since k[T] is not a field, so it has some kernel I. Since k[T] is a PID, I = (F) for some f ∈ k[T]. We then have A ≃ k[T]/(F), which realizes A as an algebraic extension of k (by adjoining a root of F). Now suppose that r > 1 and that the lemma is false for A: that is, A is not algebraic over k so that some t_{i}, say WLOG t_{1} is transcendental over k. Now observe that since A is supposed a field, it contains k(t_{1}). Moreover, A is generated over k(t_{1}) by t_{2}, …, t_{r}, so by induction we have that A is an algebraic field extension of k(t_{1}). In particular, each t_{i} for i ≠ 1 is algebraic over k(t_{1}) and so satisfies a polynomial of the form

a_{i}(d(i))t_{i}^{d(i)} + a_{i}(d(i) − 1)t_{i}^{d(i)−1} + …,

where a_{i} ∈ k(t_{1}). For each i we can clear denominators of the a_{i}(m) and assume that t_{i} satisfies a polynomial with coefficients in k[t_{1}], say

b_{i}(d(i))t_{i}^{d(i)} + b_{i}(d(i) − 1)t_{i}^{d(i)−1} + ….

By multiplying this polynomial through by b_{i}(d(i))^{d(i)−1}, we see that b_{i}(d(i))t_{i} satisfies a monic polynomial over k[t_{1}]. Notice that since the t_{i} generate A over k(t_{1}) so do the b_{i}(d(i))t_{i} since each b_{i}(d(i)) ∈ k[t_{1}]. Therefore, A is generated by integral elements over k[t_{1}], which implies that every element of A is integral over k[t_{1}]. However, k(t_{1}) ⊂ A so in particular, every element of k(t_{1}) is integral over k[t_{1}]. But now k[t_{1}] is isomorphic to the polynomial ring k[T] in one variable since t_{1} is transcendental, whence the previous statement is absurd since, for example, 1/T is not integral over k[T]. We have therefore reached a contradiction and A is in fact an algebraic extension of k.

Lemma 2. Suppose that Sol((S); K) = ∅. Then 1 ∈ (S).

Proof. First, observe that given any maximal ideal of k[T_{1}, …, T_{n}]/(S) we can construct a point of V = Sol((S); K). To see this, observe that a maximal ideal M of k[T]/(S) gives us a homomorphism of k[T]/(S) into some field F via the natural projection k[T]/(S) ⟶ (k[T]/(S))/M. By Lemma 1, F is an algebraic extension of k since it is finitely generated, and hence contained in K since K is algebraically closed. But from last lecture, we know that Hom_{k} (K[T/(S)], F) is in bijective correspondence with Sol((S); F). By hypothesis, Sol((S); K) is empty, so Sol((S); F) must be empty also so that there are no maximal ideals of k[T]/(S). On the other hand, maximal ideals of k[T]/(S) are in bijective correspondence with maximal ideals of k[T] containing (S); in short, we have shown that (S) is not contained in any maximal ideal. However, the Hilbert Basis Theorem can be used to show fairly easily that every ideal except for the unit ideal is contained in a maximal ideal. Therefore, (S) must be the unit ideal.

We now prove the main Theorem.

Proof. Given any algebraic k-set V ⊂ K^{n} we define

I(V) = {P ∈ k[T_{1}, …, T_{n}] : V ⊂ Sol(P; K)} ,

that is, I(V) is the set of all polynomials in k[T_{1}, …, T_{n}] that vanish on V. It is not hard to see that I(V) is an ideal. In one direction, suppose that rad((S)) = rad((S′)) and let a ∈ Sol(S; K). Then for any F ∈ S, we have F (a) = 0. Let G ∈ (S′). Then we must show that G(a) = 0. Observe that since G ∈ (S′) we clearly have G ∈ rad((S′)) and hence G ∈ rad((S)); in particular G^{r} ∈ (S) for some r. We therefore see that G^{r}(a) = 0. But since K is a field, this implies G(a) = 0 and hence a ∈ Sol((S′); K), as required. In the other direction, it suffices to show that if V = Sol((S); K) then rad((S)) = I(V). One direction is clear since every polynomial in rad((S)) vanishes on V and is therefore in I(V). By Hilbert’s Basis Theorem, we have (S) = (F_{1}, …, F_{s}) for some F_{i} ∈ k[T_{1}, …, T_{n}]. We must therefore show that if F vanishes on V then we can write

F^{N} = ∑ A_{i}F_{i}

for some A_{i} ∈ k[T_{1}, …, T_{n}]. Since F vanishes on V and F_{i} for 1 ≤ i ≤ s generate (S) we see that the polynomials F_{1}, …, F_{s}, F T_{n+1} −1 in n+1 variables have no common zeroes (since any common zero is a common zero of F_{1}, …, F_{n} and hence in V so that F T_{n+1} − 1 = −1. Therefore, by our lemma we can write

1 = ∑_{i} p_{i}(T_{1}, …, T_{n+1})F_{i}(T_{1}, …, T_{n}) + q(T_{1}, …, T_{n+1})(F (T_{1}, …, T_{n})T_{n+1} − 1).

Now let T_{n+1} = 1/F so that

1 = ∑_{i} p_{i}(T_{1}, …, T_{n}, 1/F)F_{i}(T_{1}, …, T_{n}).

Since each p_{i} is a polynomial, there exists N(i) > 0 such that F^{N(i)}p_{i}(T_{1}, …, T_{n}, 1/F) is a polynomial in k[T_{1}, …, T_{n}]. Set N = max_{i}{N(i)}. then we have

F^{N} = ∑_{i} A_{i}F_{i},

where A_{i} = F^{N}p_{i}(T_{1}, …, T_{n}, 1/F) ∈ k[T]. Hence F ∈ rad((S)), as required.