### The perfect Nullstellensatz just got more perfect

After giving a talk about the perfect Nullstellensatz (the commutative free Nullstellensatz) at the Technion Math department’s pizza and beer seminar, I had a revelation: I think it holds over other fields as well, not just over the complex numbers! (And in particular, contrary to what I thought before, it holds over the reals. It seems to hold over other fields as well).

To explain, I will need some notation.

Let $k$ be a field. We write $A = k[z_, \ldots, z_d]$ – the algebra of all polynomials in $d$ (commuting) variables over the field $k$

Fix $d \in \mathbb{N}$. For $n \in \mathbb{N}$, let $CM_d(n)$ denote the set of all commuting $d$-tuples of $n \times n$ matrices over $k$. We let $CM_d = \sqcup_{n=1}^\infty CM_d(n)$. Now we are looking at all commuting $d$-tuples of commuting matrices of all sizes.

Points in $CM_d$ can be plugged into any polynomial $p \in A$. In fact, points in $CM_d$ can be naturally identified with the space of finite dimensional representations of $A$, by

$X \in CM_d \longleftrightarrow \Big(ev_X : p \mapsto p(X)\Big)$.

(We shall use the word “representation” to mean a linear algebraic homomorphism of an algebra into $M_n(k)$ for some $n$).

Now, given an ideal $J \triangleleft A$, we can consider its zero set in $CM_d$:

$Z(J) = Z_{CM_d}(J) = \{X \in CM_d : p(X) = 0$ for all $p \in J\}$.

(We will omit the subscript $CM_d$ for brevity.) In the other direction, given a subset $S \subseteq CM_d$, we can define the ideal of functions that vanish on it

$I(S) = \{p \in A : p(X) = 0$ for all $X \in S\}$.

Clearly, $I(Z(J)) \supseteq J$ for every $J \triangleleft A$. The interesting statement is the converse.

The following theorem appears as Corollary 11.7 from the paper “Algebras of bounded noncommutative analytic functions on subvarieties of the noncommutative unit ball” by Guy Salomon, Eli Shamovich and myself (though, as I explained already in several previous posts, this result has already been known to algebraists for quite some time).

Theorem (free commutative Nullstellensatz):  For every $J \triangleleft A$,

$J = I(Z(J))$.

Proof: Originally, We intended to prove it only over the complex numbers, I didn’t guess that it could hold over the reals, for example. But an examination of the proof shows that it works when the field $\mathbb{C}$ is replaced by an arbitrary field $k$. For our proof, see this blog post. The only problematic point might be the lemma (there is only one lemma there), where one will need something called “Zariski’s lemma” which says that a field extension $K$ of $k$ which is finitely generated as an algebra, is actually finite dimensional over the base field, that is, it is a finite extension of $k$. This fact is needed for the proof of the lemma, in order to see that the quotient of a commutative unital Noetherian algebra over k by a maximal ideal is a finite dimensional field extension of $k$. Happily, Zariski’s lemma works for every field $k$. Thus the original proof holds and we are done!

Since the statement of the theorem is interesting also for $d = 1$ and $k = \mathbb{R}$, I will concentrate now on explaining why this case is true. In two words, the proof for the case $d = 1$ over the complex numbers can be summarized as: Jordan form (at the end of this older post, why the theorem follows from the Jordan form). Now I will finish by explaining why the real theorem follows form the complex theorem.

So, assume that the theorem is proved over the complex numbers. One can see that the theorem holds over the reals quite easily, by using the fact that complex numbers can be modelled as $2 \times 2$ matrices over the reals, so $n \times n$ complex matrices can be thought of as $2n \times 2n$ matrices over the reals with $2 \times 2$ blocks.

For example, in the one variable case, if $p,q$ are if polynomials with real coefficients, and if $q$ vanishes at every real matrix zero of $p$, then $q$ vanishes at every complex matrix zero of $p$, so by the complex version of the theorem we have $q = rp$ where $r$ is a polynomial with perhaps complex coefficients. But then you can see that $r$ actually has to be a polynomial with real coefficients. (This proves $I(Z(p)) \subseteq \langle p \rangle$).

I still can’t quite believe the theorem holds over any field, so if any readers finds a bug in our proof I will be happy to hear.