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If $M$ is an artinian module and $f: M\to M$ is an injective homomorphism, then $f$ is surjective.

I somehow found out that if we consider the module $\mathbb Z_{p^{\infty}}$ denoting the submodule of the $\mathbb{Z}$-module $\mathbb{Q/Z}$ consisting of elements which are annihilated by some power of $p$, then it is artinian, but if we have the homomorphism $f(\frac{1}{p^{k}})=\frac{1}{p^{k+1}}$, then we get a $\mathbb{Z}$-module homomorphism, but this map is not surjective, because $\frac{1}{p}$ has no preimage.

I would be very grateful if someone can tell me what is wrong with this counterexample? And how to prove the proposition above if it is correct? Thanks.

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There is no well-defined homomorphism $f: \mathbb Z_{p^\infty} \to \mathbb Z_{p^\infty}$ that satisfies $f\left(\frac{1}{p^k}\right) = \frac{1}{p^{k+1}}$. The existence of such an $f$ would imply $$0 = f(0) = f(1) = f\left(p \frac{1}{p}\right) = p f\left(\frac{1}{p}\right) = p \frac{1}{p^2} = \frac{1}{p}$$ which is a contradiction.


To prove the proposition, consider the descending sequence of submodules $$M \supseteq \operatorname{im} f \supseteq \operatorname{im} f^2 \supseteq \operatorname{im} f^3 \supseteq \ldots.$$ Since $M$ is Artinian, the sequence becomes stationary, say $\operatorname{im} f^k = \operatorname{im} f^{n}$ for all $k \geq n$. Then $$M = \operatorname{ker} f^n + \operatorname{im} f^n.$$ Indeed, for $x \in M$ we have $f^n(x) \in \operatorname{im} f^n = \operatorname{im} f^{2n}$, so there is a $y \in M$ s.t. $f^{2n}(y) = f^n(x)$. Then $x = (x-f^n(y)) + f^n(y) \in \operatorname{ker} f^n + \operatorname{im} f^n$. But $f$ is injective, so $f^n$ is injective as well, i.e. $\operatorname{ker} f^n = 0$. Thus $\operatorname{im} f^n = M$, so $f^n$ and hence $f$ is surjective.

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  • $\begingroup$ Perhaps I'm being dense, but why is $f(0)=f(1)$ necessarily? $\endgroup$ – Alex Becker Jan 9 '13 at 0:49
  • $\begingroup$ Because $0=1$ in $\mathbb Q/\mathbb Z$. $\endgroup$ – marlu Jan 9 '13 at 0:51
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    $\begingroup$ Oh, duh. Silly me. $\endgroup$ – Alex Becker Jan 9 '13 at 0:52
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    $\begingroup$ Dear marlu, for the proof of the proposition it might be a bit simpler to notice that from $M \supsetneq \ f(M)$ we get $f(M) \supsetneq \ f^2(M)$ (because injective morphisms preserve strict inclusions) and similarly the artinianity-contradicting infinite chain $M \supsetneq \ f(M) \supsetneq \ f^2(M)\supsetneq f^3(M)\supsetneq ...$ $\endgroup$ – Georges Elencwajg Jan 9 '13 at 0:53
  • $\begingroup$ You are right, that's better. $\endgroup$ – marlu Jan 9 '13 at 0:57
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HINT: $\phi(M) \subseteq M$ and so on (apply $\phi$ again) is a descending chain, so it must terminate. Then, what happens?

HINT 2: By the descending chain condition, we must have $\phi^{n+1}(M)=\phi^n(M)$ for some $n$. Now let $m \in M$. Then $\phi^n(m)=\phi^{n+1}(k)$ for some $k \in M$ since $\phi^{n+1}(M)=\phi^n(M)$. But $\phi$ is injective, so you can cancel to get ...?

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  • $\begingroup$ Hence there exists an index, say $k_0$ such that for every $j \geq k_0: \phi^{j}(M) = \phi^{k_0}(M)$. $\endgroup$ – gisma Jan 17 '13 at 18:39
  • $\begingroup$ So there is a subset $N$ of $M$ such that $\phi(N) = N$. And now? $\endgroup$ – gisma Jan 17 '13 at 18:49
  • $\begingroup$ @veRSAger: I've added a second, more exhaustive, hint. $\endgroup$ – Fredrik Meyer Jan 17 '13 at 19:29
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Take the descending chain:

$$ M\supseteq\phi(M)\supseteq\phi^2(M)\supseteq\dots\supseteq\phi^n(M)=\phi^{n+1}(M)=\dots $$

where you have found $n$ minimal where the chain stabilizes, using the Artinian hypothesis.

If $M\neq\phi(M)$, there is $m\notin\phi(M)$. What can you say about $\phi(m)$? And then what about $\phi^2(m)$? How does this lead to a contradiction?


You can also dualize this proof to show: a surjective endomorphism of a Noetherian module is injective. Thinking along the same lines, examine this chain:

$$ \ker(\phi)\subseteq\ker(\phi^2)\subseteq\dots\subseteq\ker(\phi^n)=\dots $$

Supposing $\ker(\phi)\neq 0$, you will be able to show there is $y\in \ker(\phi^2)\setminus\ker(\phi)$, $z\in \ker(\phi^3)\setminus\ker(\phi^2)$... etc.

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  • $\begingroup$ Regarding to your assumption with the descending chain: $$M\supseteq\phi(M)\supseteq\phi^2(M)\supseteq\dots\supseteq\phi^n(M)=\phi^{n+1}(M)=\dots$$ Like you said there exists such a $m \not \in \phi(M)$, it follows that $\phi(m) \in \phi(M)$ and $\phi^2(m) \in \phi^2(M)$ till $\phi^n(m) \in \phi^n(M) = \phi^{n+1}(M)$ which is a contradiction to $m \not \in \phi(M)$. Is that right? $\endgroup$ – gisma Jan 17 '13 at 19:12
  • $\begingroup$ I don't see a contradiction in what you wrote, but it is close. The line of thought I had in mind is that $\phi(m)\notin \phi^2(M)$, etc. $\endgroup$ – rschwieb Jan 17 '13 at 19:15
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    $\begingroup$ Ah, yeah. I meant $m \not \in \phi(M) \Rightarrow \phi(m) \not \in \phi^2(M) \Rightarrow \dots \Rightarrow \phi^{n}(m) \not \in \phi^{n+1}(M) = \phi^{n}(M)$ which is a contradiction to the obvious fact $\phi^n(m) \in \phi^n(M)$. So I could deduce that $\phi(M) = M$, hence $\phi$ is surjective. Correct? $\endgroup$ – gisma Jan 17 '13 at 19:29
  • $\begingroup$ Sounds good! :) $\endgroup$ – rschwieb Jan 17 '13 at 19:29
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As a matter of fact, all nonzero endomorphisms of $\mathbb{Z}(p^\infty)$ are surjective. Indeed, the image of an endomorphism must be a divisible subgroup and the only divisible subgroups of $\mathbb{Z}(p^\infty)$ are $\{0\}$ and $\mathbb{Z}(p^\infty)$ itself. Thus you're going the wrong direction into looking for a possible counterexample.

If $\{0\}\ne H\ne \mathbb{Z}(p^\infty)$ is a subgroup, there is a minimum positive integer $n$ such that $p^{-n-1}+\mathbb{Z}\notin H$. Note that if $p^{-k}+\mathbb{Z}\in H$, then $p^{-h}+\mathbb{Z}\in H$, for $h\le k$. It follows quite easily that $H=\langle p^{-n}+\mathbb{Z}\rangle$ is cyclic, so not divisible.


However, the counterexample cannot exist.

Since there exists an injective endomorphism of $M$ (the identity), the set $\mathscr{I}$ of images of injective endomorphisms has a minimal element. Let $f$ be an injective endomorphism with such minimal image. As $f^2$ is an injective endomorphism and $f^2(M)\subseteq f(M)$, we have equality by minimality.

Thus, if $x\in M$, there exists $y\in M$ with $f(x)=f^2(y)$; hence $x-f(y)\in\ker f$. Since $f$ is injective, $x=f(y)$ and so $f$ is surjective. It follows $\mathscr{I}=\{M\}$.

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