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If $M$ is an artinian module and $f$ : $M$ $\mapsto$ $M$ 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 $f$: $f(\frac{1}{p^{k}})=\frac{1}{p^{k+1}}$, 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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3 Answers 3

up vote 5 down vote accepted

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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Perhaps I'm being dense, but why is $f(0)=f(1)$ necessarily? –  Alex Becker Jan 9 '13 at 0:49
    
Because $0=1$ in $\mathbb Q/\mathbb Z$. –  marlu Jan 9 '13 at 0:51
    
Oh, duh. Silly me. –  Alex Becker Jan 9 '13 at 0:52
    
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 $M \supsetneq \ f(M) \supsetneq \ f^2(M)\supsetneq f^3(M)\supsetneq ...$ –  Georges Elencwajg Jan 9 '13 at 0:53
    
You are right, that's better. –  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 ...?

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

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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? –  gisma Jan 17 '13 at 19:12
    
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. –  rschwieb Jan 17 '13 at 19:15
1  
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? –  gisma Jan 17 '13 at 19:29
    
Sounds good! :) –  rschwieb Jan 17 '13 at 19:29

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