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 $\mathbb F.$

Question

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.

Answer 1

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 $f\equiv 0$.

Answer 2

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 $n$.

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