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Algebraic Geometry

Sam Macdonald

Section 1.2 Hilbert’s Nullstellensatz

Henceforth, we will assume \(k\) is algebraically closed unless otherwise specified.
Let now \(J\) be any ideal of \(k[x_1, \dots, x_n]\text{.}\) We need to show that \(I(V(J)) \subseteq \sqrt{J}\text{.}\) Pick \(f \in I(V(J))\text{;}\) that is, assume \(f\) is a polynomial such that \(f(a_1, \dots, a_n) = 0\) for every \((a_1, \dots, a_n) \in V(J)\text{.}\) It suffices to prove \(f^m = 0\) in the quotient ring \(k[x_1, \dots, x_n]/J\) for some \(m \geq 1\text{,}\) for this implies \(f \in \sqrt{J}\text{.}\)
Let \(L\) be the ideal of \(k[x_1, \dots, x_n, y]\) generated by (the image of) \(J\) and \(1 - yf(x_1, \dots, x_n)\text{.}\) I claim \(V(L)\) is the empty set: If \((a_1, \dots, a_n, b) \in V(L)\) then \((a_1, \dots, a_n) \in V(J)\) and so \(f(a_1, \dots, a_n) = 0\text{.}\) But we also have \(b f(a_1, \dots, a_n) = 1\text{,}\) which is impossible.
Since \(V(L) = \emptyset\text{,}\) by the Weak Nullstellensatz we must have \(L = \langle 1 \rangle\text{.}\)
Now, by Lemma 1.2.1 there is an isomorphism
\begin{equation*} \frac{(k[x_1, \dots, x_n]/J) [ y]}{(1 - yf)} \xrightarrow{\cong} S^{-1} k[x_1, \dots, x_n]/J \end{equation*}
of commutative rings, where \(S = \{f^m \mid m \geq 1\}\text{.}\) We also have the canonical isomorphism
\begin{equation*} \frac{(k[x_1, \dots, x_n]/J) [ y]}{(1 - yf)} \cong \frac{(k[x_1, \dots, x_n, y]}{\langle J \rangle + (1 - yf)} = k[x_1, \dots, x_n, y]/L. \end{equation*}
Since \(L = \langle 1 \rangle\text{,}\) this proves that \(S^{-1} (k[x_1, \dots, x_n]/J)\) is the trivial ring, and hence \(\frac{1}{1} = \frac{0}{1}\) in this ring. Thus \(f^m = 0\) in \(k[x_1, \dots, x_n]/J\) for some \(m\text{.}\) ◻
Prove: If \(k\) is algebraically closed, then for any set \(S\) of polynomials we have \(I(V(S)) = \sqrt{\langle S \rangle}\text{.}\) In particular, \(I(V(J)) = \sqrt{J}\) for any ideal \(J\text{.}\) Moreover, if \(T\) is any subset of \(\mathbb{A}^n_k\text{,}\) \(V(I(T))\) is the unique smallest affine variety in \(\mathbb{A}^n_k\) that contains \(T\text{.}\)
Prove that \(I(V(J)) = J\) holds for all radical ideals if and only if \(k\) is algebraically closed.

Remark 1.2.5.

The only new/non-elementary assertion in the Strong Nullstellensatz \(I(V(J))\subseteq J\) This can be rephrased as: Suppose \(f\) is a polynomial and \(J\) is an ideal such that \(f \in I(V(J))\text{.}\) Then some power of \(f\) belongs to \(J\text{.}\) Or, equivalently:
Suppose \(f\) and \(g_1, \dots, g_m\) are polynomials such that whenever \(P\) is a zero of each of \(g_1, \dots, g_m\text{,}\) \(P\) is also a zero of \(f\text{.}\) Then \(f^k = \sum_i h_i g_i\) for some \(k \geq 1\) and some \(h_i\)’s.
In other words, if, starting with the system of equations \(g_1 = 0, \dots, g_m = 0\text{,}\) adding on \(f = 0\) has no effect on the solution set, then it is because the new equation is an “algebraic consequence” of the previous — that is, \(f\) is an \(m\)-th root of some \(k[x_1, \dots, x_n]\)-linear combination of the \(g_i\)’s. (Note that the converse is obvious, and holds for any field: if \(f = 0\) is an “algebraic consequence” of the previous equations, then adding it to the list has no effect on the solution set.)

Example 1.2.6.

Recall that for \(J = \langle x^2 + y^2 \rangle\) and \(k = {\mathbb{R}}\text{,}\) we have that \(J\) is a radical ideal and yet \(I(V(J)) = \langle x,y \rangle \supsetneq J\text{.}\)
More concretely, the two systems of equations
\begin{equation*} \begin{aligned} x^2 + y^2 & = 0 \\ \end{aligned} \end{equation*}
and
\begin{equation*} \begin{aligned} x^2 + y^2 & = 0 \\ x & = 0 \\ \end{aligned} \end{equation*}
have the same set of solutions (over \({\mathbb{R}}\)), and yet \(x =0\) is not an algebraic consequence of \(x^2 + y^2 = 0\) (no power of \(x\) is a \({\mathbb{R}}[x,y]\)-multiple of \(x^2 + y^2\)).
The Nullstellensatz tells us that such “bad behaviour” cannot occur if we changed the field to be \(\mathbb{C}\text{.}\) Indeed:
\begin{equation*} V_{\mathbb{C}}(x^2 + y^2) = V_{\mathbb{C}}((x+iy)(x-iy)) = V_\mathbb{C}(x+iy) \cup V_\mathbb{C}(x-iy) \end{equation*}
and thus
\begin{equation*} I(V_\mathbb{C}(x^2+y^2)) =I(V_\mathbb{C}(x+iy)) \cdot I(V_\mathbb{C}(x-iy))=\langle x+iy \rangle \langle x-iy \rangle = \langle x^2+y^2 \rangle = J. \end{equation*}
The Strong Nullstellensatz Theorem is a consequence of the following more “abstract” one:

Remark 1.2.8.

When \(n = 1\text{,}\) the Abstract Nullstellensatz Theorem is clear, since every maximal ideal of \(k[x]\) has the form \(\langle p(x) \rangle\) for an irreducible polynomial \(p\text{.}\)

Remark 1.2.9.

An even more abstract but equivalent form of this theorem is: Given a field extension \(k \subseteq L\text{,}\) if \(L\) is finitely generated as a \(k\)-algebra then \([L:k] < \infty\text{.}\)
We won’t be proving the Abstract Nullstellensatz, but we will use it to draw some consequences.
For a commutative ring \(R\text{,}\) let \(\operatorname{mSpec}R\) denote the set of its maximal ideals.
We will prove this Corollary, taking the Abstract Nullstellensatz on faith.
The function is clearly well-defined (i.e., \((x_1 - a_1, \dots, x_n - a_n)\) is a maximal ideal for any choice of \(a_i\)’s) and one-to-one. (If \(a_i \ne a_i'\) for at least one value of \(i\text{,}\) then \((x_1 - a_1, \dots, x_n - a_n)\) and \((x_1 - a'_1, \dots, x_n - a'_n)\) are distinct ideals). The surjectivity uses the Abstract Nullstellensatz:
Let \({\mathfrak m}\in \operatorname{mSpec}k[x_1, \dots, x_n]\) and consider the composition \(\phi: k \hookrightarrow k[x_1, \dots, x_n]/{\mathfrak m}\) of canonical ring maps \(k \hookrightarrow k[x_1, \dots, x_n] \twoheadrightarrow k[x_1, \dots, x_n]/{\mathfrak m}\text{.}\) (It is injective since \(k\) is a field.) By the Abstract Nullstellensatz, it is a finite field extension. But since \(k\) is algebraically closed, it must be an isomorphism. Let \(a_i = \phi^{-1}(\overline{x_i})\text{.}\) Then it follows that \(\overline{x_i - a_i} = 0\) — i.e., \(x_i - a_i \in {\mathfrak m}\) for all \(i\text{.}\) Thus \({\mathfrak m}\supseteq (x_1 - a_1, \dots, x_n - a_n)\) and since the latter is maximal, we must in fact have an equality. ◻

Remark 1.2.11.

When \(n = 1\text{,}\) the Corollary holds essentially by definition of “algebraically” closed: If \({\mathfrak m}\) is any maximal ideal of \(k[x]\text{,}\) then \({\mathfrak m}= \langle p(x) \rangle\) for a unique irreducible, monic polynomial \(p(x)\text{.}\) If \(k\) is algebraically closed, \(p(x)\) must be linear and thus of the form \((x- a)\) for some \(a \in k\text{.}\)

Remark 1.2.12.

The proof shows, more generally, that for any field \(k\text{,}\) there is a bijection between \(\mathbb{A}^n_k\) and the subset of those maximal ideals \({\mathfrak m}\) in \(k[x_1, \dots, x_n]\) for which the canonical map of fields \(k \to k[x_1, \dots, x_n]/{\mathfrak m}\) is an isomorphism. Such maximal ideals are sometimes called \(k\)-rational. (When \(k\) is algebraically closed, every maximal ideal is \(k\)-rational.)
The function is well defined and one-to-one, since if \((a_1, \dots, a_n) \in V(J)\text{,}\) then
\begin{equation*} \langle x_1 - a_1, \dots, x_n - a_n \rangle= I(\{(a_1, \dots, a_n)\} \supseteq I(V(J)) \supseteq J. \end{equation*}
The Nullstellensatz implies it is onto: If \({\mathfrak m}\in \operatorname{mSpec}(k[x_1, \dots, x_n]/J)\) then \({\mathfrak m}= \tilde{{\mathfrak m}}/J\) for some \(\tilde{{\mathfrak m}} \in \operatorname{mSpec}k[x_1, \dots, x_n]\text{,}\) and by the (Corollary to the) Nullstellensatz \(\tilde{{\mathfrak m}} = (x_1 - a_1, \dots, x_n - a_n)\) for some \(a_1, \dots, a_n \in k\text{.}\)

Remark 1.2.14.

Since \(V(J) = V(\sqrt{J})\) the Corollary implies that \(\operatorname{mSpec}(k[x_1, \dots, x_n]/J) \cong \operatorname{mSpec}(k[x_1, \dots, x_n]/\sqrt{J})\text{.}\) This holds in greater generality.
Since \(J\) is proper it is contained in some maximal ideal \({\mathfrak m}\text{.}\) (As a technical point, this fact does not require Zorn’s Lemma in this setting, since the ring \(k[x_1, \dots, x_n]\) is noetherian.) So the result follows from the previous Corollary.
Let’s return to the question of what can be said about \(I(W \cap W')\text{:}\)
Since both sides are radical ideals, by the Nullstellensatz it suffices to prove
\begin{equation*} V(I(W \cap W')) = V (\sqrt{I(W) + I(W')}). \end{equation*}
This holds since \(V(I(W \cap W')) = W \cap W'\) and \(V (\sqrt{I(W) + I(W')}) = V(I(W) + I(W')) = V(I(W)) \cap V(I(W')) = W \cap W'\text{.}\)

Example 1.2.17.

The Corollary is false over \({\mathbb{R}}\text{:}\) Take \(W = V(y-x^2)\) and \(W' = V(y+1)\) in \(\mathbb{A}^2_{\mathbb{R}}\text{.}\) Then \(I(W) = \langle y-x^2 \rangle\) and \(I(W') = \langle y+1 \rangle\) (why?) and hence \(\sqrt{I(W) + I(W')} \subsetneq k[x,y]\) (since \(k[x,y]/(y-x^2, y+1) \cong k[x]/(x^2+1) \ne 0\)). But \(W \cap W' = \emptyset\) and so \(I(W \cap W') = k[x,y]\text{.}\)

Definition 1.2.18.

Given an affine variety \(W \subseteq \mathbb{A}^n_k\text{,}\) the coordinate ring of \(W\text{,}\) written \(A(W)\) is the \(k\)-algebra
\begin{equation*} A(W) = \frac{k[x_1, \dots, x_n]}{I(W)}. \end{equation*}
Some authors denote this as \(k[W]\) instead.
If \(W \subseteq \mathbb{A}^n_k\) is an affine variety and \(f\) and \(g\) are two polynomials such that \(f(P) = g(P)\) for all \(P \in W\text{,}\) then \(f-g \in I(W)\) and hence \(\overline{f} = \overline{g} \in A(W)\text{.}\) This proves:
Note that \(A(W)\) is a reduced ring since \(I(W)\) is a radical ideal. Moreover, provided \(k\) is algebraically closed, every finitely generated, commutative, reduced \(k\)-algebra occurs, up to isomorphism, as the coordinate ring of some affine variety: Given such an algebra \(A\text{,}\) being finitely generated as a \(k\)-algebra means there is a surjection \(k[x_1, \dots, x_n] \twoheadrightarrow A\) for some \(n\text{,}\) and since \(A\) is reduced, the kernel \(J\) is a radical ideal. Taking \(W = V(J)\) we have \(I(W) =I(V( J)) = J\) by the Nullstellensatz and hence \(A(W) \cong A\text{.}\) The assumption that \(k\) is algebraically closed is needed here, due to examples such that \({\mathbb{R}}[x]/(x^2 + 1)\text{.}\)
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