Date: 7/11/17.
Here is a review of the notation: \(\AA ^n\) is affine n-space, \(K\) is an (algebraically closed unless otherwise specified) field, and if \(C \subset K[x_1,\dots ,x_n]\) is an ideal, \(\sqrt C\) is the radical of \(C\) i.e \(\{f \in K[x_1,\dots x_n] | f^n \in C\}\)
Hilbert’s Nullstellensatz is a way of translating geometry (affine algebraic sets) into algebra (radical ideals of polynomial rings) and vice versa. More specifically, given a \(A \subset \AA ^n\), we can consider \(I(A)\) which is the set of polynomials in \(K[x_1,\dots ,x_n]\) vanishing on A. Conversely given \(C \subset K[x_1,\dots ,x_n]\) we can consider the set of common zeroes of all the elements of \(C\). Are these inverses? The answer is yes if we restrict to algebraic sets and radical ideals. This is the Nullstellensatz.
In particular, maximal ideals correspond to points. This is the Weak Nullstellensatz. We will use this to prove the standard form of it. It can be seen in the case of \(K[x]\) easily, as it is a PID so all of the maximal ideals are just irreducibles. K is algebraically closed so indeed we have that the maximal ideals are just \((x-a)\) for \(a \in K\), which is what we want, as its zero set is just the point \(a\).
Now before proving anything let’s develop some facts about integrality that will be used to prove the main lemma, Noether Normalization, which will be used to deduce Nullstellensatz.
Definition 2.1. Let \(R \subset S\) be rings, \(s \in S\). s is integral over \(R\) if it is the root of a monic polynomial in \(R[x]\). We say \(S\) is integral over \(R\) if every element of S is.
Integrality is similar to algebraicity, and reduces to algebraicity in the case that \(R\) is a field. It is a good definition because of the following:
Proposition 2.2. Let \(R \subset S\) be rings, \(s \in S\). The following are equivalent:
Proof.
(1) \(\Rightarrow \) (2) Consider \(R[s]\) with the basis \(1,s,\dots s^{k-1}\) where the degree of the monic polynomial for which s vanishes is k.
(2) \(\Rightarrow \) (1) Suppose \(v_1,\dots v_k\) generate \(T\) as an \(R\)-module. Then consider the endomorphism that takes \(x \in T\) to \(sx\). In particular, for each \(v_i\), with \(a_{ij} \in R\),
\begin{equation} sv_i = \sum _1^k{a_{ij}v_i} \Rightarrow \sum _1^k (\delta _{ij}s-a_{ij})v_i = 0 \end{equation} where \(\delta _{ij}\) is the Kronecker delta. Then by Cramer’s Rule s is a root of the characteristic polynomial of the matrix \(a_{ij}\) which is monic. □
Corollary 2.3. If \(R \subset S\) and \(s,t \in S\) are integral over \(R\), \(s\pm t\) and \(st\) are as well. Integrality is also transitive, ie. If T is integral over S and S is integral over R, T is integral over R.
Proof. If \(s\) and \(t\) are integral over \(R\), let \(v_1,\dots v_k\) be an \(R\)-module generating set of \(R[s]\) and let \(w_1,\dots w_l\) be the ones for \(R[t]\). Then the products \(v_1w_1,\dots v_1w_l,v_2w_1,\dots v_kw_l\) generate \(R[s,t]\) as an \(R\)-module. This proves the first statement. For the second, if \(t \in T\) is integral over \(S\), let \(a_0,\dots a_k\) be the coefficients in a monic polynomial in S for which \(t\) is a root. Then by the same argument, \(R[a_0,\dots a_k]\) is a f.g \(R\)-module, so \(R[a_0,\dots a_k,t]\) is a f.g \(R[a_0,\dots a_k]\)-module hence is also f.g as an \(R\)-module. □
Great, now that we have the basic facts about integrality, let’s prove Noether Normalization!
Lemma 3.1 (Noether Normalization). Let \(B = K[r_1,\dots r_m]\) be a f.g \(K\)-algebra (ie \(B\) is a quotient of a polynomial ring with coefficients in \(K\) in finitely many variables). Then there is an injective map from a polynomial ring over \(K\) such that \(B\) is integral over the image. (This statement doesn’t require K algebraically closed)
Proof. Consider the map \(K[x_1,\dots x_m] \rightarrow B\) sending \(x_i\) to \(r_i\). If the map is injective, it is an isomorphism so B is trivially integral over the image. If not, let \(f\) be a nontrivial element in the kernel, so that \(f(r_1,\dots r_m) = 0\). It would be nice if \(f\) were monic in some variable so that we could say that that variable is integral over the rest and induct on \(m\) by transitivity of integrality, but this is not necessarily the case (say if the polynomial is \(x_1x_2\)). So we will make a change of variables (the ”normalization”) which will force the polynomial to be monic.
Let \(d\) be the degree of \(f\) (the degree of a monomial is the sum of the exponents in each variable), and define \(a_i = (1+d)^i\), \(X_i = x_i-x_m^{a_i}\) for the change of variables. Now let \(g\) be the polynomial: \begin{equation} g(X_1,\dots X_{m-1},x_m) = f(x_1,\dots x_m) \end{equation}
Why is \(g\) (almost) monic in \(x_m\)? Well any monomial in \(f\), say \(cx_1^{k_1}\dots x_m^{k_m}\) corresponds to a \(c(x_1+x_m^{a_1})^{k_1}\dots (x_{m-1}+x_m^{a_{m-1}})^{k_{m-1}}x_m^{k_m}\), so the highest \(x_m\) term it contributes contains only the variable \(x_m\). Now by looking at the exponent of the largest monomials in base \(d+1\), we see that no two monomials contribute the same highest \(x_m\) term so there is no cancelling. Then if the highest degree \(x_m\) term in g has coefficient \(a \in K\), \(\frac g a\) is a monic polynomial in \(K[X_1,\dots X_{m-1},x_m]\) so if we set \(s_i = r_i-r_m^{a_i}\), \(i < m\) then \(g\) shows that \(r_m\) is integral over \(K[s_1,\dots s_{m-1}]\). Now the defining relation of \(s_i\) shows that \(r_i\) is integral over \(K[s_1,\dots s_{m-1},r_m]\), so by transitivity B is integral over \(K[s_1,\dots s_{m-1}]\) so we are done by induction. (The base case is when \(m = 0\), which is trivial). □
We are ready to prove the weak form of the Nullstellensatz. It follows almost immediately from this lemma.
Theorem 3.2 (Weak Nullstellensatz). The maximal ideals of \(K[x_1,\dots x_m]\) are exactly the ideals of the form \((x_1-a_1,\dots x_m-a_m)\) for \(a_i \in K\). In particular, if \(I\) is a proper ideal of \(K[x_1,\dots x_m]\), it has a zero in \(\AA ^n\).
Proof. It is obvious that the ideals of the form above are maximal. For the converse, suppose that I is maximal. \(K[x_1,\dots x_m]/I\) is a field \(K'\). By Noether Normalization, K’ is integral over \(K[x_1,\dots x_q]\). But \(q\) had better be \(0\) as \(\frac 1{x_1}\) is not integral over \(K[x_1,\dots x_q]\) but \(K'\) is a field so would have to contain it. Thus K’ is integral \(\Rightarrow \) algebraic over K but K is algebraically closed so \(K'=K\). Then in the projection map from \(K[x_1,\dots x_m]\) to \(K'\) we have that each \(x_i\) is sent to some element of K so the kernel, \(I\), is of the form we want.
For the second statement, if an ideal \(I\) is proper, it is contained in a maximal ideal \(M\), and so the point \(M\) corresponds to is a zero of \(I\). □
Now we will deduce the stronger form from the weaker.
Theorem 3.3 (Nullstellensatz). Let \(C\) be an ideal of \(K[x_1,\dots x_n]\). Then \(\sqrt C = I(Z(C))\).
Proof. Certainly \(\sqrt C \subset I(Z(C))\) (check this). The hard part is proving the reverse inclusion. Suppose that \(g \in I(Z(C))\). We will ”localize” away from \(g\) and then use the Weak Nullstellensatz to deduce \(g \in \sqrt C\). By the Hilbert Basis Theorem, \(C = (f_1,\dots f_m)\). Now consider the ideal in \(K[x_1,\dots x_{n+1}]\): \((f_1,\dots f_m,x_{n+1}g-1)\) (think of \(x_{n+1}\) as ”\(\frac 1 g\)”). This has no zeroes in \(\AA ^{n+1}\) as if each \(f_i\) is \(0\), \(g\) is too (it’s in \(C\)), so \(x_{n+1}g-1\) is -1. Then By the Weak Nullstellensatz, \((f_1,\dots f_m,x_{n+1}g-1)\) is not a proper ideal, so there are polynomials \(b_i \in K[x_1,\dots x_m+1]\) such that: \begin{equation} 1 = b_1f_1+\dots b_mf_m + b_{m+1}(x_{n+1}g-1) \end{equation} Replacing \(x_{n+1}\) with \(\frac 1 y\) in this equation and multiplying through by a large enough power of \(y\) to clear denominators yields: \begin{equation} y^k = c_1f_1+\dots c_mf_m + c_{m+1}(g-y) \end{equation} where we have changed \(b_i\) to \(c_i\) to absorb the new powers of y introduced. Plugging in g, we get \begin{equation} g^k = c_1f_1+\dots c_mf_m \end{equation} which is exactly \(g \in \sqrt C\). □
In the proof of Noether Normalization, we have done a particularly strange change of variables in order to force our polynomial to be monic, namely we have sent \(x_i \mapsto (x_i+x_n^{a_i})\) where \(a_i\) is quite large. However, most of the time it suffices to make a more reasonable change \(x_i \mapsto (x_i+\alpha x_n)\) instead. This doesn’t always work: for \(\mathbb{F}_q\) we can consider \(x^qy^q-y^{2q}\), for which no change of variables of the form \(y \mapsto y+\alpha x\) will yield a monic polynomial. However, this is the only problem: if we assume our field is infinite, we can do such a change of variables.
To see this, consider the same polynomial \(x^qy^q-y^{2q}\). If we do the change of variables \(y \mapsto y+\alpha x\) we get \(x^q(y+\alpha x)^q-(y^{2q}+\alpha x) = x^qy^q-y^{2q}+(\alpha ^q-\alpha ^{2q})x^{2q}\). The only way this would not be monic in \(x\) is if \(\alpha \) is the root of \(z^q-z^{2q}\), but this polynomial has finitely many roots, so in an infinite field we can always choose a nice enough \(\alpha \).
In general, we can argue inductively as follows: consider the highest total degree monomials of the form \(ax_1^{i_1}\dots x_n^{i_n}\) and choose one such that the first \(i\) so that \(x_i\) is in it is the highest (we will continue to refer to this index as \(i\)). Now if \(i=n\), we are done as the polynomial is already monic in \(x_n\), and if not, we will change the variable \(x_i \mapsto x_i+\alpha x_n\). Any other monomials that might cancel this change of variables give a polynomial in \(1\) variable that \(\alpha \) would need to satisfy for this change of variables to be ineffective, so we then use the fact that the field is infinite, and are left with a polynomial which is closer to being monic, and we can induct.
To illustrate this proof of Noether Normalization, we can consider the hyperbola \(K[x,y]/(xy-1)\) and try to project down to \(K[x]\). Now this won’t give \(a\) nice projection onto the \(x\) axis as the line \(x=0\) is asymptotic, which corresponds to the fact that \(y\) is the inverse of \(x\) and hence is not finite over it. We can tilt the \(y\) axis by making the change of variables \(x\mapsto x+\ee y\), which will make the relation \((x+\ee y)y\), which is monic in \(y\), and will make \(y\) integral by removing the asymptote.