For a positive integer $n$, I have to show that there does not exist a onto ring homomorphism from $M_{n+1 \times n+1}(\mathbb F) \to M_{n \times n}(\mathbb F) $ for any field $\mathbb F.$

If it exists lets say there is $f:M_{n+1 \times n+1}(\mathbb F) \to M_{n \times n}(\mathbb F)$ surjection and since $1$ maps to $1,$ its kernel is trivial and hence an isomorphism. Now if I can show that $f$ is $\mathbb F$- linear then we are done by dimension argument. To show that $f$ is $\mathbb F$-linear enough to show that $f(cI_{n+1})=cI_n$ for each $c \in \mathbb F,$ which I am not able to show. Or may be there is some alternative way to prove it. I need some help to show it.

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    $\begingroup$ How did you conclude that the kernel is trivial? $\endgroup$ – awllower Aug 20 '19 at 8:43
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    $\begingroup$ Well, I think this the statement is false. $\endgroup$ – Wuestenfux Aug 20 '19 at 8:46
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    $\begingroup$ See also mathoverflow.net/questions/126106/… $\endgroup$ – Minseon Shin Aug 20 '19 at 8:51
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    $\begingroup$ We know for any commutative ring $R$ with $1$ the ideals of $M_n(R)$ are precisely $M_n(I)$ where $I$ is an ideal of $R.$ $\endgroup$ – user371231 Aug 20 '19 at 9:06

Let $A\in M_{n+1\times n+1}(\mathbb{F})$ be a nilpotent matrix such that $A^n\neq 0$, but $A^{n+1}=0$.

Since $A^{n+1}=0$ then $0=f(A^{n+1})=f(A)^{n+1}$.

Thus, $f(A)\in M_{n\times n}(\mathbb{F}) $ is also nilpotent. Therefore, $f(A)^{n}=0$. Hence $f(A^n)=0$.

Note that $A^n\neq 0$, so $\ker(f)$ is non trivial.

Since $\ker(f)$ is an ideal of $M_{n+1\times n+1}(\mathbb{F})$ and the only non null ideal of $M_{n+1\times n+1}(\mathbb{F})$ is itself then $f\equiv 0$.

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  • $\begingroup$ I like this answer a lot. For the most part it seems like it is an explanation of where the reasoning of the OP is flawed and then in the last line - surprise twist - it solves the question itself. $\endgroup$ – Vincent Aug 20 '19 at 13:37
  • $\begingroup$ @Daniel: Great Solution ! $\endgroup$ – user371231 Aug 20 '19 at 15:35
  • $\begingroup$ @user371231 You are welcome. $\endgroup$ – Daniel Aug 20 '19 at 19:53

You can also apply an invariant like right composition length, if you know about it.

The two rings being isomorphic would imply that the lattices of right ideals are isomorphic. In particular, the length of a maximal strictly decreasing chain has to be the same in both.

But the maximum length of a strictly descending chain of right ideals in $M_n(F)$ is $n$.

Note that Daniel's answer exploits a different invariant: maxiumum index of nilpotency.

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