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Let $A$ be a UFD, that is also a $k$-algebra, where $k$ is a field of characteristic $\not=2$ (for instance polynomials over $k$).

Is every involution in $\mathrm{GL}_n(A)$ diagonalisable?

This is of course true over the field of fractions of $A$.

In this question it is explained that $$M=\left(\begin{array}{rr} 3&-1\\-1 & 3\end{array}\right)$$ is diagonalisable over $\mathbb{Q}$ but not over $\mathbb{Z}$. However, it is not an involution.

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    $\begingroup$ @Vladimir Dotsenko: But 2 is invertible in $A$ by hypothesis. $\endgroup$
    – abx
    Commented Dec 3, 2021 at 7:40
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    $\begingroup$ Isn't $A^{n}$ (identified with $n$-long column vectors with entries in $A$) expressible as ${\rm Im}(\frac{1+t}{2}) \oplus {\rm Im}(\frac{1-t}{2})$ as an $A$-module for any such involution $t$ ? $\endgroup$
    – Geoff Robinson
    Commented Dec 3, 2021 at 12:10
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    $\begingroup$ Could you maybe use different letters for the UFD and the matrix? $\endgroup$
    – R. van Dobben de Bruyn
    Commented Dec 3, 2021 at 12:12
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    $\begingroup$ Following @GeoffRobinson's comment, if $A$ is a UFD with $2\in A^\times$ admitting a non-free f.g. projective $A$-module $P$, and $Q$ is another projective module such that $P\oplus Q\cong A^n$, then $\mathrm{id}_P \oplus (-\mathrm{id}_Q)$ should be counterexample. Conversely, if every f.g. projective $A$-module is free (e.g. if $A=F[x_1,\dots,x_r]$ with $F$ a field), then every involution is diagonalizable. $\endgroup$
    – Uriya First
    Commented Dec 3, 2021 at 12:22
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    $\begingroup$ By the way, the question is equivalent to asking whether every idempotent matrix $E \in M_{n}(A)$ is diagonalizable via an element of ${\rm GL}(n,A)$. Note that if $E$ is idempotent, then $2E-I$ is an involution, while if $t$ is an involution, then $t = 2E-I$ for some idempotent $E$ as in my comment above ( here we need $2$ invertible in $A$). $\endgroup$
    – Geoff Robinson
    Commented Dec 4, 2021 at 12:53

1 Answer 1

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Here is an answer based on my comment, and Geoff Robinson's earlier comment.

Let $A$ be a domain with $2\in A^\times$, and let $M$ be an $n\times n$ matrix with $M^2=I$. It is convenient to consider $M$ as an $A$-endomorphism of $A^n$ and $I$ as the identity endomorphism of $A^n$. Put $S_+ =\{x\in A^n\,:\,Mx=x\}$ and $S_-=\{x\in A^n\,:\,Mx=-x\}$. As noted in Geoff Robinson's comment, since $2\in A^\times$, we have $A^n=S_+\oplus S_-$. (This follows readily from the fact that $x=\frac{1}{2}(x+Mx)+\frac{1}{2}(x-Mx)$ for all $x\in A^n$ and $M^2=I$.) In particular, both $S_+$ and $S_-$ are f.g. projective $A$-modules.

Suppose that every f.g. projective $A$-module is free. Then $S_+$ and $S_-$ are free. Let $E_+$ and $E_-$ be bases for these spaces. Then $E=E_+\cup E_-$ is a basis of $A^n$ consisting of eigenvalues of $M$, which means that $M$ is a diagonalizable.

On the other hand, if there exists a f.g. projective $A$-module $P$ which is not free, then we can construct a non-diagonalizable involution $M:A^n\to A^n$ as follows. There exists a f.g. $A$-module $Q$ such that $A^n\cong P\oplus Q$ for some $n\in\mathbb{N}$. Now take $M$ to be the matrix corresponding to $\mathrm{id}_P\oplus (-\mathrm{id}_Q)$. For this $M$, we have $S_+=P$ and $S_-=Q$. If $M$ were diagonalizable, then $A^n$ would have a basis $E$ consisting of vectors $x\in A^n$ with $Mx\in \{\pm x\}$ (here we need $A$ to be a domain). Consequently, $E = (E\cap S_+)\cup (E\cap S_-)$, which means that $E_+:=E\cap S_+$ is a basis of $S_+=P$. This contradicts our choice of $P$, so $M$ is not diagonalizable.

There are examples of UFDs admitting f.g. projective modules which are not free --- see this question, for instance. Quoting from this answer, one can take $A=\mathbb{C}[a,b,c,x,y,z]/(ax+by+cz-1)$, the module $P$ to be the kernel of $(\alpha,\beta,\gamma)\mapsto a\alpha+b\beta+c\gamma:A^3\to A$, and $Q=(x,y,z)A$. One has $P\oplus Q=A^3$, so it should be possible to write the resulting $M\in \mathrm{M}_3(A)$ explicitly.

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    $\begingroup$ Thanks Uriya First for the nice and detailed answer. And thanks to Geoff Robinson too. $\endgroup$
    – Jérémy Blanc
    Commented Dec 6, 2021 at 7:53

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