# Hilbert's Nullstellensatz without Axiom of Choice

Motivation This question came from my efforts to solve this problem presented by Andre Weil in 1951.

Can we prove the following theorem without Axiom of Choice?

Theorem Let $A$ be a commutative algebra of finite type over a field $k$. Let $I$ be an ideal of $A$. Let $\Omega(A)$ be the set of maximal ideals of $A$. Let $V(I)$ = {$\mathfrak{m} \in \Omega(A)$; $I \subset \mathfrak{m}$}. Let $f$ be any element of $\cap_{\mathfrak{m} \in V(I)} \mathfrak{m}$. Then there exists an integer $n \geq 1$ such that $f^n \in I$.

EDIT So what's the reason for the downvotes?

EDIT What's wrong with trying to prove it without using AC? When you are looking for a computer algorithm for solving a mathematical problem, such a proof may provide a hint. At least, you can be sure that there is a constructive proof.

• Take an arbitrary statement of non-trivial commutative algebra and ask if it holds in ZF - what's the point? Do you know that in ZF there might be nontrivial rings with no maximal ideals at all? So why should one expect that any "spectral" methods take over? Moreover, why do you answer your "question" yourself? – Martin Brandenburg Jul 12 '12 at 13:40
• I can answer that last question, Martin: blog.stackoverflow.com/2011/07/… – Cameron Buie Jul 12 '12 at 13:49
• @MartinBrandenburg Before I answer your question, I'd like you to answer mine: math.stackexchange.com/questions/159699/… – Makoto Kato Jul 13 '12 at 6:34
• @MartinBrandenburg I edited my question. I hope you understand what the point is. – Makoto Kato Jul 15 '12 at 11:31

Lemma 1 Let $A$ be a commutative algebra of finite type over a field $k$. Then there exists a maximal ideal $P$ of $A$ such that $A/P$ is a finite $k$-module.

Proof: If every element of $A$ is algebraic over $k$, then $A$ is a finite $k$-module. Therefore the lemma is trivial. Hence we can assume otherwise. By Noether normalization lemma (this can be proved without AC), there exist algebraically independent elements $x_1,\dots, x_n$ in $A$ such that $A$ is a finitely generated module over the polynomial ring $A' = k[x_1,\dots, x_n]$. Let $\mathfrak{m} = (x_1,\dots, x_n)$ be the ideal of $A'$ generated by $x_1,\dots, x_n$. Clearly $\mathfrak{m}$ is a maximal ideal of $A'$. By the answer by QiL to this question, there exists a prime ideal $P$ of $A$ lying over $\mathfrak{m}$. Since $A/P$ is a finitely generated module over $k = A'/m$, $P$ is a maximal ideal. QED

Lemma 2 Let $A$ be a commutative algebra of finite type over a field k. Let $f$ be a non-nilpotent element of $A$. Then there exists a maximal ideal $P$ of $A$ such that $f \in A - P$.

Proof: Let $S$ = {$f^n; n = 1, 2, \dots$}. Let $A_f$ be the localization with respect to $S$. By Lemma 1, there exists a maximal ideal $\mathfrak{m}$ of $A_f$ such that $A_f/\mathfrak{m}$ is a finite $k$-module. Let $P$ be the inverse image of $\mathfrak{m}$ by the canonical morphism $A \rightarrow A_f$. $A/P$ can be identified with a subalgebra of $A_f/\mathfrak{m}$. Since $A_f/\mathfrak{m}$ is a finite $k$-module, $A/P$ is also a finite $k$-module. Hence $P$ is a maximal ideal. Clearly $f \in A - P$. QED

The title theorem follows immediately from the following lemma by replacing $A$ with $A/I$.

Lemma 3 Let $A$ be a commutative algebra of finite type over a field k. Let $\Omega(A)$ be the set of maximal ideals of $A$. Let $f$ be any element of $\cap_{\mathfrak{m} \in \Omega(A)} \mathfrak{m}$. Then $f$ is nilpotent.

Proof: This follows immediately from Lemma 2.