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If $R$ is a domain and $f: R^n \to R^n$ is an $R$-module endomorphism. Suppose $f^m = 0$ for some $m> 0$. Show that $f^n = 0$.

The cases $ m \le n$ is trivial. When $m>n$, I don't have much idea how to start. I tried to apply the Cayley-Hamilton theorem but doesn't seem to help much.

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Do you see why you can WLOG assume that $R$ is a field? – darij grinberg Feb 10 at 3:31
    
@darijgrinberg not really. Why can I do that? – user112564 Feb 10 at 4:26
    
Whether $f^m = 0$ or $f^n = 0$ can be checked equally well over the fraction field of $R$ as over $R$. – Qiaochu Yuan Feb 10 at 5:03
up vote 5 down vote accepted

Let $Q$ be the field of fractions of $R$, and let $F:Q^n\to Q^n$ be the morphism of $Q$-modules which has the same matrix as $f$ (with respect to the standard bases of $R^n$ in the case of $f$ and of $Q^n$ in the case of $F$)

  • Show that $F$ is nilpotent, and that therefore $F^n=0$.

  • If $i:R^n\to Q^n$ is the obvious inclusion, then we have $F\circ i=i\circ f$. Use this, and the fact that $i$ is an injective function, to show that $f^n=0$.

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It's convenient but unnecessary to pass to the fraction field. Your hunch that you should apply Cayley-Hamilton works. By Cayley-Hamilton, we have

$$f^n = a_{n-1} f^{n-1} + \dots + a_i f^i$$

where $a_k \in R$ and $a_i \neq 0$ (of course if all the $a_i$ are zero then we are done). Multiplying by $f^{m-i-1}$ gives

$$a_i f^{m-1} = 0$$

and hence, since $R$ is an integral domain, that

$$f^{m-1} = 0.$$

Now you can multiply by $f^{m-i-2}$, etc. Eventually you conclude that $f^n = 0$ as desired.

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View the matix as lying in a matrix ring of the same size, but having entries in the algebraic closure of the field of factions for $R$.

It therefore has a Jordan Normal Form, and since it is nilpotent, it must have all zeros on the diagonal. The normal form is then a strictly upper triangular.

Finally, it is easy to prove that $M^n=0$ for any strictly upper triangular matrix in the $n\times n$ matrix ring. The same is true for the original matrix, which is a conjugate of this normal form.

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