# An expression that vanishes over every field

In this question, Jack Schmidt asks to prove a certain identity for $2\times 2$ matrices A and B.

In fact he asks to show that tr(AABABB−AABBAB) = 0. In an answer by user7406, he shows that 3 times this expression must be 0, solving the problem at least when the characteristic of the ground field isn't 3. In a comment by Mariano Suárez-Alvarez he tells us a computercalculation show that it is in fact identically zero.

This made me wonder whether or not this is a surprise. I think the proper way to state the problem is as follows:

Let $\varphi \colon \mathbb Z\langle x_1,\dots x_n\rangle\to \mathbb Q [x_1,\dots,x_n]$ be the morphism from the free non-commutative ring over these variables to the polynomialring. Let $\Phi \colon \mathbb Z\langle x_1,\dots,x_n\rangle \to k[x_1,\dots,x_n]$ be the corresponding morphism by sending every element to 'itself' the ususal way.

Here are my questions:

• Is it true that if $\varphi(x)=0$ then $\Phi(x) = 0$? My guess is it is because one can construct $\psi$ such that $\Phi = \psi\circ \varphi$.
• Does this indeed settle the original problem for characteristic 3 from the solution by user7406, thereby bypassing von Neumann or am I really missing something?

[I first asked this question in a comment to the post by user7406, but I deleted that because I didn't want to hijack the other question.]

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I didn't get the von Neumann reference until you explained it. :) – Grumpy Parsnip Mar 26 '11 at 0:57
What is $k$? I don't see how you can construct a nonzero map $\psi\colon \mathbb{Q}[x_1,\ldots,x_n]\to k[x_1,\ldots,x_n]$ unless $k$ has characteristic zero, which would make it hard to get a $\psi$ if, say, $k=\mathbb{F}_3$ (for which $\Phi$ still makes sense). – Arturo Magidin Mar 26 '11 at 2:32
@Arturo $k$ would be any field. I think you're right, the $\mathbb Q$ should perhaps be $\mathbb Z$ for this to hold. – Myself Mar 26 '11 at 2:39
Yes: the image of $\varphi$ lies inside $\mathbb{Z}[x_1,\ldots,x_n]$ anyway, and $\mathbb{Z}[x_1,\ldots,x_n]$ is initial in the class of polynomial rings in $n$ variables, so you'll want to replace $\mathbb{Q}$ with $\mathbb{Z}$. I'm trying to write up a reply as I type this. – Arturo Magidin Mar 26 '11 at 2:43

Since the image of $\varphi$ lies inside of $\mathbb{Z}[x_1,\ldots,x_n]$, you can replace $\mathbb{Q}[x_1,\ldots,x_n]$ with $\mathbb{Z}[x_1,\ldots,x_n]$ (since the image vanishes in the latter if and only if it vanishes in the former).

Any map from $\mathbb{Z}\langle x_1,\ldots,x_n\rangle$ to a commutative ring factors uniquely through $\mathbb{Z}[x_1,\ldots,x_n]$ by the universal property of the polynomial ring.

Explicitly: given any ring homomorphism $f\colon R\to S$, where $R$ and $S$ are commutative, and given any elements $s_1,\ldots,s_n\in S$, there exists a unique homomorphism $\overline{f}\colon R[x]\to S$ such that $\overline{f}(r)=f(r)$ for all $r\in R$ and $\overline{f}(x_i) = s_i$ for $i=1,\ldots,n$.

In particular, if $\Phi$ is any map from $\mathbb{Z}\langle x_1,\ldots,x_n\rangle$ to a commutative ring $S$, then $\Phi(x_i)$ determines a unique homomorphism $\psi\colon\mathbb{Z}[x_1,\ldots,x_n]\to S$ with $\psi(x_i)=\Phi(x_i)$ and $\psi(1) = \Phi(1)$. This map makes a commuting triangle with the quotient map $\varphi$. Hence we get $\Phi=\psi\circ\varphi$, as you suggest in point (1) (after replacing $\mathbb{Q}$ with $\mathbb{Z}$.

Thus, if $\Phi$ is any map from $\mathbb{Z}\langle x_1,\ldots,x_n\rangle$ to a commutative ring $S$, then necessarily $\mathrm{ker}(\varphi)\subseteq\mathrm{ker}(\Phi)$, which answers your first question.

(Or, you can think about the "multidegree" of a monomial in $\mathbb{Z}\langle x_1,\ldots,x_n\rangle$ which only counts the total exponent of each variable; an element of $\mathbb{Z}\langle x_1,\ldots,x_n\rangle$ lies in the kernel of $\varphi$ if and only if it is a sum of "homogeneous terms", i.e., same multidegree for all monomials, where the coefficients add up to $0$. Any such homogeneous term will also lie in the kernel of any $\Phi$, since the image is commutative).

But I'm not sure why you would need to go through this for your second question: $\mathrm{trace}(AABABB-AABBAB)$ can be written as a (commutative) polynomial in the $8$ variables that form the entries of $A$ and $B$, with integer coefficients; i.e., an element of $\mathbb{Z}[x_1,\ldots,x_n]$ (not of $\mathbb{Z}\langle A,B\rangle$, because we are looking at the trace, not the product itself). If $3$ times this expression is $0$ as an element of $\mathbb{Z}[x_1,\ldots,x_n]$, then this expression is itself $0$ simply because $\mathbb{Z}[x_1,\ldots,x_n]$ is an integral domain. And from there, you get that the corresponding expresssion is also identically zero over any field, because it is zero in the initial polynomial ring $\mathbb{Z}[x_1,\ldots,x_n]$.

That is, the lack of commutativity between the matrices does not matter here, because the expression we are looking for, the trace, is computed in terms of the coefficients of the matrices via commutative (though complicated) operations.

E.g., if you were talking about the trace of $AB-BBA$, you would have \begin{align*} A&=\left(\begin{array}{cc} a_{11} & a_{12}\\ a_{21} & a_{22} \end{array}\right),\\ B&=\left(\begin{array}{cc} b_{11} & b_{12}\\ b_{21} & b_{22} \end{array}\right),\\ \mathrm{trace}(AB - BBA) &= \mathrm{trace}(AB) - \mathrm{trace}(BBA)\\ &= \Bigr(a_{11}b_{11} + a_{12}b_{21} + a_{21}b_{12}+a_{22}b_{22}\Bigr) \\ &\qquad\mathop{+} \Bigl( (b_{11}^2+b_{12}b_{21})a_{11} + (b_{11}b_{12}+b_{12}b_{22})a_{21}\\ &\qquad\quad\mathop{+} (b_{21}b_{11}+b_{22}b_{21})a_{12} + (b_{21}b_{12}+b_{22}^2)a_{22}\Bigr) \end{align*} This is a polynomial in $\mathbb{Z}[a_{11},\ldots,b_{22}]$, with commuting variables; not an element of $\mathbb{Z}\langle a_{11},\ldots,b_{22}\rangle$.

Note that you cannot interpret the trace map as a ring homomorphism from $\mathbb{Z}\langle A,B\rangle$ to something, because the trace is not a ring homomorphism: the trace of the product is not the product of the traces.

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Thanks! I see what you mean that I don't need to go around $\mathbb Z\langle a_{11},\dots,b_{22}\rangle$ to get what I want. I guess I did that because if you state it this way, what it comes down to is that "0 in Z[...] is mapped to 0 in k[...]". Which is, well, so obvious that I wouldn't believe it is just that. – Myself Mar 26 '11 at 3:50
One can look at the trace as a map $\mathbb Z\langle A,B\rangle\to k$ by defining it on monomials as the composition $\mathbb Z\langle A,B\rangle\to M_n(k)\stackrel{\operatorname{tr}}\to k$, with the first map the obvious one. This is what was meant, I guess. This composition is not a ring map, of course. – Mariano Suárez-Alvarez Mar 26 '11 at 6:04
@Mariano: Right: I should change "map" to "ring homomorphism" in that final paragraph. – Arturo Magidin Mar 26 '11 at 18:52

It is certainly true that if you prove a polynomial identity over $\mathbb{Z}$, then you get the same identity over any commutative ring whatsoever for free: there is a well-defined homomorphism $\psi_A : \mathbb{Z}[x_1, \ldots, x_n] \to A[x_1, \ldots, x_n]$, and in particular, $\psi_\mathbb{C}$ is injective, so even if you use complex numbers in your proof, it will still be valid (and the same holds even when you pass to the algebraic completion of $A[x_1, \ldots, x_n]$ itself).

As for the original problem, it is just a matter of proving a certain polynomial identity in 8 variables, so this method is indeed applicable.

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