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Let $R$ be an integral domain and $M_n(R)$ the ring of $n\times n$ matrices over $R$.

Question 1: Is there a ring homomorphism $f:R\rightarrow M_n(R)$ such that $f(r)$ is a nondiagonal matrix for some $r\in R$?

Question 2: Can such a homomorphism be chosen injective?

Clearly, both answers are "no" for $R=\mathbb{Z}$ (the only homomorphism is $f: z\mapsto zI$ where $I$ is the identity matrix).

There is an example where both answers are "yes": Let $R=\mathbb{Q}[x]$. Let us write elements of $R$ (uniquely) in the form $p+q$ where $p$ is in the ideal $(x)$ of $R$ and $q\in \mathbb{Q}$. Define $f:p+q\mapsto pX+qI$ where $X\neq I$ is a fixed idempotent matrix over $\mathbb{Q}$ (e.g. $X=\frac{1}{n}J$ where $J$ is the all-ones matrix). Then it is easy to see that $f$ is an injective ring homomorphism and $f(x)=xX$ is nondiagonal.

Can the questions be answered (positively or negatively) for some larger classes of integral domains?

My special interest is in rings $R$ that are models of (Peano or weaker) arithmetic. For this purpose one can assume some additional properties of $R$, e.g. $\pm 1$ are the only invertible elements, prime decompositions are unique when they exist (but they do not have to exist), $R$ is discretely ordered, all elements from the real closure have the unique integer part in $R$, ....

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    $\begingroup$ If the morphism is demanded to be $R$-linear, the answer is obviously no. Otherwise, it's easy to make counterexamples: let $K$ be a field, and $R=L$ a quadratic extension of $K$; say for simplicity $L = K(\sqrt{a})$. Then the map \begin{align*}L&\to M_2(K)\\ x+y\sqrt{a} &\mapsto \begin{pmatrix}x&y \\ ay&x \end{pmatrix} \end{align*} can easily seen to be a ring homomorphism. (This is just saying that every quadratic extension is contained in $M_2(K)$; equivalently that every polynomial $x^2 - a$ occurs as a minimal polynomial of some matrix.) Compose with $M_2(K) \to M_2(L)$ to get the result. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Nov 18, 2017 at 1:21
  • $\begingroup$ @R.vanDobbendeBruyn Thank you for the example. But it is of the same sort as the example $R=\mathbb{Q}[x]$ I gave in my post: my R and your L are both finitely generated over a subring and the homomorphism $f$ is then defined as $f(r)=rI$ on the subring and $f(g)$ a suitable non-diagonal matrix for the generator $g$. I'm more interested about a situation where $R$ isn't of this kind... $\endgroup$
    – PGlivi
    Commented Nov 18, 2017 at 22:48

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