# Endomorphism $\phi$ of free module satisfying $\operatorname{Im}\phi^{n+1}\subsetneq\operatorname{Im}\phi^n$ for every $n$.

Let $R$ be an integral domain that is not a field and $F$ a free $R$-module of finite rank. I'm trying to construct an $R$-module homomorphism $\phi:F\to F$ such that $\operatorname{Im}\phi^{n+1}\subsetneq\operatorname{Im}\phi^{n}$ for all $n\geq 0$.

I know this can be done (intuitively) since $F$ cannot be Artinian but I cannot construct such a map.

I'd appreciate any hints. Thanks in advance.

• Multiplication by a non-unit should be enough. – Bernard Dec 14 '17 at 23:39

Let's just look at the simplest case: where $F$ is free of rank $1$, so $F\cong R$. (Actually, the simplest case would be when $F$ has rank $0$, but that case is not true!) So we may as well assume $F=R$.
We're now looking for an endomorphism $\phi$ of $R$ as an $R$-module with a certain property. But every endomorphism of $R$ is just multiplication by some element of $R$ (namely, $\phi(1)$), so there is some $a\in R$ such that $\phi(x)=ax$ for all $x$. Then $\phi^n(x)=a^nx$. So, the image of $\phi$ is the set of all elements of $R$ divisible by $a^n$.
I'll now leave the rest to you: how can we choose $a\in R$ such that the image of $\phi^{n+1}$ is strictly smaller than the image of $\phi^n$: that is, not every element of $R$ which is divisible by $a^n$ is also divisible by $a^{n+1}$? (For this part, you'll need to use the assumption that $R$ is not a field!) And, having finished that, how can you generalize this idea to the case where $F$ has rank greater than $1$?
• We can choose $0\neq a\in R$ a non-unit and $\phi$ defined by $\phi(x)=ax$. Now $F=R^k$ for $k=rank F$ and if we suppose that $Im\phi^{n+1}=Im\phi^n$ then $a^{n+1}F=a^nF$ and then $(a^n,\dots,a^n)=a^n(1,\dots,1)=a^{n+1}(f_1,\dots,f_k)$ and then $a^n=a^{n+1}f_1$, so $a^n(1-af_1)=0$ and since $R$ is a domain and $a\neq 0$ we have that $af_1=1$ and then $a$ is a unit, which is a contradiction. Then $Im\phi^{n+1}\subsetneq Im\phi^n$. Right? – Wolfgang Fritz Dec 15 '17 at 0:04
• Looks good! ${}{}$ – Eric Wofsey Dec 15 '17 at 0:10