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Assume that $\mathbb{F}$ is a field, and let $\mathbb{M}_{t}\left( \mathbb{F}\right) $ be the ring of matrices of order $t$ over $\mathbb{F}$.

Does there exist a non-trivial ring homomorphism from $\mathbb{M}_{n+1}\left( \mathbb{F}\right) $ to $\mathbb{M}_{n}\left( \mathbb{F}\right) $?

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No: there would have to be a non-zero kernel ( for example, decompose $1$ as a sum of $n+1$ mutually orthogonal primitive idempotents in $M_{n+1}.$ At least one of these would have to be sent to $0$ in $M_{n}.$ But $M_{n+1}(\mathbb{F})$ is a simple ring, so has no proper non-zero two-sided ideals. – Geoff Robinson Feb 29 '12 at 16:55

$M_{n+1}(\mathbb F)$ is simple, so a nontrivial ring homomorphism from $M_{n+1}(\mathbb F)$ is an isomorphism onto its image. Let $N\in M_{n+1}(\mathbb F)$ be a nilpotent matrix of order $n+1$. Then the image of $N$ under an isomorphism is also nilpotent of order $n+1$, but $M_n(\mathbb F)$ contains no such elements.

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simple and elegant proof. – GA316 Jul 8 '14 at 6:41

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