# Proving that surjective endomorphisms of Noetherian modules are isomorphisms and a semi-simple and noetherian module is artinian.

I am revising for my Rings and Modules exam and am stuck on the following two questions:

$1.$ Let $M$ be a noetherian module and $\ f : M \rightarrow M \$ a surjective homomorphism. Show that $f : M \rightarrow M$ is an isomorphism.

$2$. Show that if a semi-simple module is noetherian then it is artinian.

Both these questions seem like they should be fairly straightforward to prove but I cannot seem to solve them.

• Your two questions are quite unrelated... You could have asked them separately. – Mariano Suárez-Álvarez May 15 '12 at 2:40
• Dear Alex, Regarding your first question, the relevant adjective is Hopfian, which has come up several times on this site, eg. here. Regards, – Matt E May 18 '12 at 0:59
• @user: Dear user, Noetherian modules are Hopfian objects in the category of all $R$-modules. This is discussed in the last paragraph of the answer I linked to above. Regards, – Matt E Nov 18 '13 at 5:26

$$1.$$ Let $$M$$ be a Noetherian $$R$$ - module. We prove that if $$f : M \longrightarrow M$$ is a surjective $$R$$ - module homomorphism from $$M$$ to itself, then $$f$$ is an isomorphism. It is easy to see that it suffices to prove that $$f$$ is injective. Now the kernel of any $$R$$ - module homomorphism is always a submodule, so we can consider the following ascending chain of kernels

$$\ker f \subset \ker f^2 \subset \ker f^3 \subset ....$$

that must eventually become constant by the Noetherian condition so we may suppose that there is $$n \in \Bbb{N}$$ such that

$$\ker f^n = \ker f^{n+1} = \ker f^{n+2} = \ldots = \ker f^{2n} = \ldots .$$

Now we claim that

$$\ker f^n \cap \operatorname{Im} f^n = \{0\}.$$

Clearly $$0$$ is in the left hand side so we just need to show the reverse inclusion. Take $$x \in \ker f^n \cap \operatorname{Im} f^n$$. Then $$f^n(x) = 0$$ and there exists $$y \in M$$ such that $$x = f^n(y)$$. Putting this expression found for $$x$$ into $$f^n(x) = 0$$ we get that

$$f^{2n}(y) = 0$$

and hence that $$y \in \ker f^{2n}$$. But then as noted above $$\ker f^{2n} = \ker f^n$$ and hence that $$f^n(y) = 0$$. But $$x = f^n(y)$$ and so $$x$$ itself is zero proving our claim that $$\ker f^n \cap \operatorname{Im} f^n = \{0\}$$. Now because $$f$$ is surjective it follows that $$\operatorname{Im} f^n = M$$. However $$\ker f^n \subset M$$ so that $$\{0\} = \ker f \cap M = \ker f$$ so that $$f$$ is injective, and hence an isomorphism.

$$\hspace{6in} \square$$

Edit: Supplementary problems:

(1) Prove that any injective $$R$$ - module endomorphism $$\phi : M \rightarrow M$$ for $$M$$ an Artinian module is surjective (and hence an isomorphism). Hint: Consider the descending chain

$$\operatorname{Coker} \phi \supset \operatorname{Coker} \phi^2 \supset \operatorname{Coker} \phi^3 \supset \ldots$$

(2) Considering an Artinian ring $$R$$ as a module over itself, prove that if $$R$$ is an Artinian integral domain then it must be a field (Hint: Consider a suitable $$R$$ - module endomorphism and apply (1) above).

(3) Let $$R$$ be an Artinian local ring with maximal ideal $$\mathfrak{m}$$. Prove that $$\mathfrak{m}$$ of $$R$$ is nilpotent. Hint: By the Artinian condition we have the descending chain $$\mathfrak{m} \supset \mathfrak{m}^2 \supset \ldots \supset \mathfrak{m}^k = \mathfrak{m}^{k+1} = \ldots$$

for some $$k \in \Bbb{N}$$. Suppose that $$\mathfrak{m}^k \neq 0$$ (Yes we are using the same $$k$$). By Zorn's Lemma one can choose an ideal $$I$$ in $$R$$ minimal with respect to the property that $$I\cdot \mathfrak{m}^k \neq 0$$. This is saying that there exists an element $$x \in I$$ such that $$x \mathfrak{m}^k \neq 0$$ and hence by minimality of $$I$$, $$(x) = I$$. Considering $$(x)$$ as an $$R$$ - module and noting it is finitely generated, conclude that

$$(x)\mathfrak{m}^k = (x)$$

and apply Nakayama's Lemma to get a contradiction.

• @AlexKite Please do the supplementary problems that I have given you above. – user38268 May 17 '12 at 23:27
• It's hard to believe that $\operatorname{Coker} \phi \supset \operatorname{Coker} \phi^2 \supset \operatorname{Coker} \phi^3 \supset \ldots$ as all these are factor modules of $M$. – user89712 Nov 9 '13 at 22:34
• I did not understand this: "Now because f is surjective it follows that Im f^n=M". Please explain. Im f^n is a subset of Im f but how to show reverse inclusion? – RagingBull Apr 5 '15 at 11:38

$1.$ Let me give you a proof of the following astonishing result due to Vasconcelos:

Theorem:
Let $M$ be a finitely generated $R$-module, Noetherian or not, and let $\ f : M \rightarrow M \$ be a surjective homomorphism. Then $f : M \rightarrow M$ is injective (hence is an isomorphism).

Proof:
We use the standard trick of converting $M$ into an $R[X]$-module by defining $X\cdot m=f(m)$.
For the ideal $I=XR[X]$ we have $M=IM$ since for any $m\in M$ we can write by surjectivity of $f$ : $m=f(n)=X\cdot n$ and $X\in I$.
Since Nakayama says that $$M=IM\implies m=im$$ there exists $i=P(X)X\in I$ with $m=P(X)X\cdot m=P(f)(f(m))$ for all $m\in M$.
So that finally $f(m)=0\implies m=P(f)(f(m))=P(f)(0)=0 \:$: injectivity of $f$ has been proved.

• I hadn't seen Dan's answer when I posted mine. I'll leave my answer since the write up is a bit different (in particular it emphasises a mnemonic I like for Nakayama). – Georges Elencwajg May 15 '12 at 8:07
• Vasconcelos' result uses commutativity for sure? The proof makes me think so, I'm just checking to make sure. – rschwieb May 15 '12 at 11:03
• Yes, it uses commutativity of $R$. I have the algebraic geometer's habit of implicitly assuming rings to be commutative. This can be dangerous on a generalist site like here (but for my defense the question was tagged "commutative algebra" ...) – Georges Elencwajg May 15 '12 at 11:55
• I had no objection (after all, the OP had commutative-algebra in the tags) I just wanted to be sure not to confuse it for a noncommutative result :) It is definitely a result worth knowing! – rschwieb May 15 '12 at 11:58

$2.$ A semisimple module $M$ is a direct sum of simple modules. If it is noetherian, there are finitely many summands in that decomposition, and then $M$ has finite length. It is therefore artinian.

$1.$ Let $\,f:M\to M\,$ be an epimorphism or $\,R\,$-modules, with $\,M\,$ Noetherian.

i) Show that $\,M\,$ can be made into $\,R[t]\,$-module, defining $\,tm:=f(m)\,,\,\forall m\in M$

ii) Putting $\,I:=\langle t\rangle=tR[t]\,$ , show that $\,MI=M\,$

iii) Apply Nakayama's Lemma to deduce that there exists $\,1+g(t)t\in I\,$ s.t. $\,(1+g(t)t)M=0$

iv) Finally, take $\,y\in \ker f\,$ and show $\,y=0\,$ applying (iii)

• Thank you all for the really helpful responses! I think understand how to do both questions now. – Anon May 15 '12 at 18:11

For the first question, look at the following chain of ideals: $$\ker f\subset \ker f^2\subset \ker f^3\ldots$$

• I'd call them submodules. – user89712 Nov 10 '13 at 20:12