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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.

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The kernel of a ring homomorphism is a two-sided ideal, so when $R$ is a field the kernel of $M_{n+1}(R)\rightarrow M_n(R)$ must be $0$ (impossible) or $M_{n+1}(R)$. –  Dag Oskar Madsen Mar 31 '13 at 15:29
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Why is kernel being zero impossible? –  user30035 Mar 31 '13 at 17:00
    
If $R$ is a field, or even a commutative ring, and if you consider $R$-algebra homomorphisms, then the kernel cannot be zero by rank reasons. Ortherwise this seems to be a complicated question. –  Fernando Muro Mar 31 '13 at 17:08
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But even if $R$ is a field, the question does not demand that the ring homomorphism is a vector space homomorphism. There are plenty of fields $K$ and injections $K\to K$ that are far from being surjections. I don't understand your scalar matrices comments. Given $i:K\to K$ an injection which is not $K$-linear, there's an induced ring homomorphism $M_2(K)\to M_2(K)$ which is also not $K$-linear. –  user30035 Mar 31 '13 at 17:39
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I think wccanard is right. There can be many strange ring homomorphisms from a field to itself. Please downvote my previous comment :) –  Dag Oskar Madsen Mar 31 '13 at 17:47
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4 Answers

up vote 24 down vote accepted

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.

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Doesn't the same theorem solve the more general commutative case? Can there be any injective ring homomorphisms $M_{n+1}(R/I) \to M_n(R)$ when $R$ is commutative and $I$ is an ideal? –  Dag Oskar Madsen Mar 31 '13 at 20:53
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@Dag Oskar Madsen: Your comment is on point! If $R$ is commutative then the kernel of any ring hom $f \colon M_{n+1}(R) \to M_n(R)$ is of the form $M_{n+1}(I)$ for some ideal $I$ of $R$ and so we get an injective map $M_{n+1}(R/I) \cong M_{n+1}(R)/\ker f \to M_n(R)$, which forces $f$ to be 0 by A-L. –  Faisal Mar 31 '13 at 21:47
    
Very nice answer and comments. –  yeshengkui Apr 1 '13 at 3:04
    
Very nice, I admire always your leading points. –  Babak Sorouh Apr 6 '13 at 16:58
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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.

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@Dag:Thanks for your answer. But why is $M_2(M_2(R))$ an ideal of $M_2(R)$? Have you assumed that $M_2(R)$ is a direct product of $M_2(M_2(R))$ and $M_2(\bb{C})$ to define the map $M_2(R)\rightarrow M_1(R)$? Why can we have such a product? –  yeshengkui Apr 1 '13 at 7:38
    
@yeshengkui: there is a canonical isomorphism $M_n(A\times B)\cong M_n(A)\times M_n(B)$, and the subspace $M_n(A)$ is an ideal of $M_n(A\times B)$ under this identification. –  Mariano Suárez-Alvarez Apr 1 '13 at 9:32
    
I think I am kind of cheating with this example and not giving you the sort of answer you are after. Maybe there should be more conditions on the ring $R$? –  Dag Oskar Madsen Apr 1 '13 at 10:01
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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.

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If $M_{n+1}\to M_n$ is a ring map, we get $n+1$ idempotents in $M_n$ which are orthogonal and add up to $1$; in particular, they are distinct. This is enough to go, so one does not really need the very nice fact about the set $B_k$ you mention. (Or I am missing something) –  Mariano Suárez-Alvarez Apr 1 '13 at 2:58
    
@Benjamin, what do you mean by $k \times k$-matrix units? –  yeshengkui Apr 1 '13 at 3:05
    
@Mariano, but the ring homomorphism not necessarily preserves identities. Is it easy to classify idemotents in $M_n(R)$? –  yeshengkui Apr 1 '13 at 3:08
    
Ah. Well, orthogonality is enough to show that the $n+1$ idempotents are different, so non-unitality of the map does not hurt. –  Mariano Suárez-Alvarez Apr 1 '13 at 3:10
    
@Yeshegkui, a matrix unit means a matrix of the form $E_{ij}$ which has zeroes in all entries except the ij entry, which is 1. The matrix units are a basis for the matrix ring (in fact it is the contracted semigroup algebra of the matrix unit semigroup). –  Benjamin Steinberg Apr 1 '13 at 11:35
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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 $M_n(L_{n,m}) \cong M_m(L_{n,m})$.

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How do you get in your argument (after "then $M_{n'}(R)$ is simple") that $R^n$ is a finitely generated module over $M_{n'}(R)$? (I guess $n'=n+1$) –  Yves Cornulier Mar 31 '13 at 21:48
    
For the second part I think you are assuming the ring homomorphism preserves identity. –  Dag Oskar Madsen Mar 31 '13 at 21:52
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Finiteness doesn't come for free -- for example take some crazy map $\mathbb{C}\to\mathbb{C}$ making $\mathbb{C}$ into an infinite-dimensional $\mathbb{C}$-vector space, and then let $n=n'=1$. –  user30035 Mar 31 '13 at 21:56
    
Ah, very good point. It seems like the only way to salvage the argument is to assume that the ring map is actually a map of $R$-algebras, but in this case there's a much easier argument... Oh well. –  Faisal Mar 31 '13 at 22:31
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