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This is exercise 7.13 from Chapter 6 of Aluffi's Algebra: Chapter 0 textbook. It says that if $R$ is an integral domain, $A \in M_{n}(R)$ is diagonalizable with distinct eigenvalues and $AB = BA$ (where $B$ is a square matrix of the same size as $A$), then $B$ is diagonalizable with respect to a basis of eigenvectors of $A$.

This question has been asked and answered several times in this site for the case of matrices with entries in $\mathbb{R}$ and the usual argument of showing that an eigenvector of $A$ is an eigenvector of $B$ works for matrices with entries in any field, so I could work with the field of fractions $K$ of $R$ and I could certainly show that $B$ is diagonalizable (w.r.t. to a basis of eigenvectors of $A$) as a matrix with entries in $K$. That is, given an eigenvector $v$ of $A$, one can prove that there exists $k \in K$ such that $Bv = kv$, but we'd like to obtain $r \in R$ such that $Bv = rv$ and I don't see how we could conclude this.

My issue basically comes from the fact that the eigenspaces $E_{\lambda_1}, \ldots E_{\lambda_n}$ of $A$ have rank 1, but if $\{ v_1, \ldots , v_n \}$ is a basis of $R^n$ of eigenvectors of $A$, with $v_i \in E_{\lambda_i}$, then I'm not sure whether $\{ v_i \}$ is a basis (over $R$) of the corresponding eigenspace $E_{\lambda_i}$. It's certainly a maximal linearly independent subset, but for free $R$-modules this doesn't imply that this set is a basis and, for example, $\{ 2 \}$ is a maximal linearly independent subset of the free $\mathbb{Z}$-module $\mathbb{Z}$ which is not a basis for $\mathbb{Z}$ as a $\mathbb{Z}$-module. On the other hand, if I start by taking a basis $\{ w_i \}$ for each eigenspace $E_{\lambda_i}$, I'm not sure if $\{ w_1, \ldots , w_n \}$ will form a basis of $R^n$ (again, it would be a maximal linearly independent subset of $R^n$, but I'm not sure if it will generate this $R$-module).

All of this arguments would work fine in a field instead of an integral domain, so, as I said, we could work with the field of fractions of $R$ and I don't know if this is what the author might have meant without saying it or if I'm missing something and B is actually diagonalizable (w.r.t. to a basis of eigenvectors of $A$) as a matrix with entries in $R$?

Any help or insights on this problem would be greatly appreciated.

user594756
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The fact that $\mathcal B = \{v_1,\dots,v_n\}$ is a basis of $R^n$ does ensure that $\{v_i\}$ spans $E_{\lambda_i}$. One argument is as follows.

We know that $\mathcal B$ is a basis of $R^n$. It follows that for any $w \in E_{\lambda_i}\subset R^n$, there exists $r_i \in R$ such that $w = \sum_{j=1}^n r_j v_j$. Using the fact that $w \in E_{\lambda_i}$, conclude that $r_j = 0$ for all $j \neq i$. That is, $w = rv_i$, where $r = r_i \in R$.

Ben Grossmann
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I am not following your "issue". By hypothesis, $R^n=⊕_iE_{λ_i}=⊕_iRv_i$.

Then, the proof in your case works exactly the same way as when $R$ is a field:

$\lambda_iB(v_i)=B(\lambda_iv_i)=BA(v_i)=AB(v_i)$ hence $B(v_i)\in E_{\lambda_i}=Rv_i$.

Anne Bauval
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