Are there any nontrivial ring homomorphisms $M_{n+1}(R)\rightarrow M_n(R)$? Let $R$ be a finitely generated ring with identity, $M_n(R)$ the set of $n\times n$ matrices. Are there any nontrivial ring homomorphisms $M_{n+1}(R)\rightarrow M_n(R)$? This should be an elementary question in abstract algebra. But even if $R$ is a field, I couldn't get a quick (negative) proof. Any comments are welcomed.
RMK: If we view the natrual map $M_{n}(R)\rightarrow M_{n+1}(R)$ as a ring homomorphism, we will not require that a ring homomorphism preserves identities.
 A: If $R$ is a local ring, possibly non-commutative, then there is no non-trivial homomorphism.  Let $B_k$ be the semigroup of $k\times k$-matrix units and $0$. It is well known every proper homomorphic image of $B_k$ collapses all elements. So if $M_{n+1}(R)\to M_n(R)$ is nontrivial it must not collapse $B_{n+1}$ (since $B_{n+1}$ spans $M_{n+1}(R)$). But then $B_{n+1}$ embeds in $End(R^n)$ as a semigroup with zero.  So $End(R^n)$ contains $n+1$ orthogonal idempotents. But this implies $R^n$ is a direct sum of at least $n+1$ non-zero projective modules.  But projective is free for local rings and local rings have invariant basis number. This is a contradiction. 
Added. This argument works as long as R has the invariant basis number property for finitely generated free modules and finitely generated projective $R$-modules are free. In particular it applies to free algebras and firs (free ideal rings) if memory serves. 
A: We can also rule out the case of commutative $R$ by appealing to the Artin–Procesi theorem: an Azumaya algebra of constant rank $(n+1)^2$ (e.g. $M_{n+1}(R)$) satisfies all the $\mathbb Z$-multilinear identities of $M_{n+1}(\mathbb Z)$ but no nonzero homomorphic image of it satisfies all the $\mathbb Z$-multilinear identities of $M_n(\mathbb Z)$.
It's perhaps worth noting that if R is a field, then there's a fairly straightforward way of proving that there is no injective ring homomorphism M_{n+1}(R) \to M_n(R). In fact, suppose we have a nonzero ring homomorphism M_{n'}(R) \to M_n(R). Then this allows us to view R^n as a left M_{n'}(R)-module. Now if R is a field, then M_{n'}(R) is simple, and so R^n decomposes into a finite direct sum of irreducible M_{n'}(R)-modules. It's a standard fact (and one that is easy to prove) that each such module is isomorphic to R^{n'}. We thus obtain an isomorphism R^n = R^{n'} \oplus \cdots \oplus R^{n'} of M_{n'}(R)-modules, and hence of R-vector spaces by restricting the action to the subring of scalar matrices. But then linear algebra allows us to conclude that n'|n. Nevermind. :)
Update: It's possible to have a nontrivial ring map $M_{n+1}(R) \to M_n(R)$ with $R$ finitely generated (and necessarily noncommutative). The idea, inspired by my previous mishap and wccanard's comment, is to find a finitely generated ring $R$ for which there is an isomorphism $R^{n+1} \cong R^n$ of left $R$-modules. In this case one obtains ring isomorphisms $$ M_{n+1}(R) \cong \mathrm{End}_R(R^{n+1}) \cong \mathrm{End}_R(R^n) \cong M_n(R). $$
The ring theorists provide us with examples of such rings. In fact, for any positive integers $n < m$, Leavitt gives a finitely generated ring $L_{n,m}$ for which there is a left $L_{n,m}$-module isomorphism $L_{n,m}^n \cong L_{n,m}^m$ and, consequently, a ring isomorphism $M_n(L_{n,m}) \cong M_m(L_{n,m})$.
A: Since it is not clear that $R$ must be commutative (none of the tags eliminate this possibility), the finite generation hypothesis (which I interpret as finitely generated as an algebra over a field) is not  significant, in view of the existence of rings for which all finitely generated projective modules are cyclic and free. 
The standard dense subalgebra of the Cuntz C*-algebra is such an example, and is  finitely generated as an algebra over the complexes, and satisfies $M_n R $ isomorphic to $R$ for all $n$, yielding unital maps (isomorphisms) $M_{n+1}R \to M_n R$. 
A: According to the Amitsur-Levitzki theorem, $n \times n$ matrices over a commutative ring satisfy a polynomial identity of degree $2n$ and none of smaller degree. So there can be no injective ring homomorphism $M_{n+1}(R) \to M_n(R)$, which at least rules out the case when $R$ is a field.
A: Here is a slightly silly example of a non-trivial ring homomorphism $M_{n+1}(R)\rightarrow M_n(R)$ with $R$ noncommutative.
Let $R=\mathbb C \times M_2(\mathbb C)$. Then there is a ring homomorphism $M_2(R)\rightarrow M_1(R)$ sending the ideal $M_2(M_2(\mathbb C)) \subseteq M_2(R)$ to $0$ and $M_2(\mathbb C) \subseteq M_2(R)$ isomorphically to $0 \times M_2(\mathbb C)$.
There might be more interesting examples based on this idea.
