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.


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.

  • $\begingroup$ 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

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 ...?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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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