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.
 A: $1.$ 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.
A: $2.$ 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.
A: $1.$ 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)
A: $1.$ 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. 
A: For the first question, look at the following chain of ideals:
$$\ker f\subset \ker f^2\subset \ker f^3\ldots$$
A: I am late to the party but just for future reference:
Consider the following chain of submodules: $\ker f \subseteq ker f^2 \subseteq ...$
Since you have a Noetherian Module there exists an $n\in\mathbb{N}: \ker f^n = \ker f^{n+1} = ...$ 
Let $A=\ker f^n$, then $A=\{x\in M: f^k(x)=0,\text{for some } k\in\mathbb{N}\}$. Notice that $f(A)=A$ since $f^i(x)=0$ for some $i\in\mathbb{N} \implies f^{i-1}(f(x))=0$ or just $f(x)=0$ if $i=1$ and in the other direction $f^i(f(x))=0$ for some $i\in\mathbb{N} \implies f^{i+1}(x)=0$
Then we obtain that $f^i(A)=A,\forall i \in\mathbb{N}$ but on the other hand for $i=n$ we have that $f^n(x)=0$ for every $x \in A$. This clearly means $\ker f^n=A=f^n(A)=\{0\} \implies \ker f = 0$.
