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Let $R$ be the ring of complex $n\times n$ matrices, where $n>1$.
Does every nonconstant polynomial in $R[X]$ have a root in $R$?

Note: The "strong" fundamental theorem of algebra for matrices fails, see here. As far as I can see the "weak" fundamental theorem of algebra would not imply the "strong" fundamental theorem of algebra, due to noncommutativity.

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    $\begingroup$ In your linked question, a commenter points out the existence of a $2\times 2$ matrix $N$ which is not a square. The example given is $N = \bigl(\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix}\bigr)$: if $N$ had a square root $\sqrt{N}$, then $N$ and $\sqrt{N}$ would commute; but the commutator of $N$ are the matrices of the form $a + Nb$ for $a,b \in \mathbb{C}$, and $(a + Nb)^2 = a^2 + 2ab N \neq 0 + 1N$. For this matrix $N$, $x^2 - N$ does not have a root. $\endgroup$ Commented yesterday
  • $\begingroup$ @TheoJohnson-Freyd: it's even simpler than that :-), see my answer below. $\endgroup$
    – M.G.
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    $\begingroup$ Perhaps @TheoJohnson-Freyd's example is a little more satisfying because the polynomial is monic. (Although I would say centraliser instead of commutator, which to me has a different meaning.) $\endgroup$
    – LSpice
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    $\begingroup$ @LSpice: I agree! His example does not rely on the cheap commutative algebra characterization of invertible polynomials I used :-) $\endgroup$
    – M.G.
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    $\begingroup$ @M.G. I was also trying to point out that the answer was already available in the OP's source material... $\endgroup$ Commented yesterday

2 Answers 2

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No, for rather trivial reasons. Consider the polynomial $f(X) = \varepsilon X - 1$ with $\varepsilon^2 = 0$, $\varepsilon \neq 0$, in $R = M_2(\mathbb{C})$. Then a root of $f(X)$ would mean that $\varepsilon$ is invertible in $R$, which is of course impossible. Alternatively, just take $\det$ on both sides.

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Although what you are asking for is not true, there is a very interesting related fact that is true. Let $k$ be an algebraically closed field of any characteristic. The theory of matrix factorisations shows that every homogeneous polynomial in $R=k[X_1,\dots,X_n]$ factorises in a suitable matrix ring $\operatorname{Mat}_s(R)$ as a product of linear factors. This follows from Theorem 1.2 of Herzog, Ulrich and Backelin, "Linear maximal Cohen-Macaulay modules over strict complete intersections" (see also Backelin, Herzog and Sanders, "Matrix factorizations of homogeneous polynomials"). To illustrate this theorem, consider the polynomial $f=X_1^2+X_2^2+X_3^2+X_4^2$. Clearly, this does not factor into linear factors in $R$. But in $\operatorname{Mat}_4(R)$ we have $$f.I_4=\begin{pmatrix} X_1&-X_2&X_3&X_4\\X_2&X_1&-X_4&X_3\\-X_3&X_4&X_1&X_2\\-X_4&-X_3&-X_2&X_1\end{pmatrix}\begin{pmatrix} X_1&X_2&-X_3&-X_4\\-X_2&X_1&X_4&-X_3\\X_3&-X_4&X_1&-X_2\\X_4&X_3&X_2&X_1\end{pmatrix}.$$ where $I_4$ is a $4\times 4$ identity matrix.

The theorem is that if $f$ has degree $d$ then it factorises as a product of $d$ linear factors in a suitable size matrix ring over $R$. One consequence of this is that some power of $f$ is the determinant of a matrix whose entries are homogeneous linear polynomials (or zero).

Matrix factorisations come from the theory of maximal Cohen-Macaulay modules over hypersurfaces. I believe the idea originates in the work of David Eisenbud, but there were probably others involved too.

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    $\begingroup$ That sure is a remarkable result! $\endgroup$ Commented yesterday
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    $\begingroup$ And not widely enough known. $\endgroup$ Commented yesterday
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    $\begingroup$ Is there a nice way of interpreting this result geometrically? $\endgroup$ Commented yesterday

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