Ascending chain - Isomorphism theorems So that what I did so far : 
Interesting results : 


*

*$f$ is injective $\iff$ $\ker f = \{0\}$

*$\ker f$ is a submodule of $X$

*As $X$ is noetherian, then each ascending chain $M_1 \subset M_2 \subset \dots$ has the property that there exists $k \in \mathbb{N}$ such that $M_k = M_{k+1}= M_{k+2}= \dots$


So in considering those results, suppose that $f$ is non injective. The objective is to construct an ascending chain doesn't having the property $3$.
Let $M_1= \ker f$. Then we have $M_1 \subset X \subset Y$. According to the first isomorphism theorem $X/M_1 \cong f(X)=Y$. Furthermore, I have the intuition we could use the correspondance theorem, i.e. we have that the submodules of $X$ containing $M_1$ are in correspondance with the submodules of $X/M_1$ (i.e. $K \subset K'$, where $K, K' \supset N_1 \iff K'/M_1 \subset K / M_1$). 
I would like to continue the construction of this chain, but I am blocked. Is anyone could give me a hint how could I continue this problem? I think I have to consider the composition of the function $f$ with itself, but it is unclear.
Thanks!
P.S. Please don't give me more than a good hint to continue.
 A: $\require{AMSmath}$
You are on the right track in wanting to consider the composition of $f$ with itself.
Knowing $f: X\to Y$ is surjective, consider $f^{-1}(X) \subset X$. Notice that ker$(f) \subset f^{-1}(X)$. Applying $f$ again (to the $X$ in the image of the first application of $f$) gives a map from $f^{-1}(X)$. 
$$X \xrightarrow{f} Y\supset X\xrightarrow{f}Y$$
The important part: the kernel of this new map contains the kernel of $f$. To build an ascending chain, we take a similar repeated application of $f$:
$$X \xrightarrow{f} Y\supset X\xrightarrow{f}Y\supset X\xrightarrow{f}Y\supset X\xrightarrow{f}...$$
Composing these together, we get a series of functions $f$, $f^2$, $f^3$... from the first $X$ in the sequence, and we can show that the kernels form an ascending chain in $X$. The kernel of the second application of $f$, for example, contains the 0 element in $Y$, and the preimage of that element is exactly ker$(f)$. We know that $f$ fails to be injective in all these applications, so that ker$(f^n)\subset $ ker$(f^{n+1})$ $\forall n\ge 1$. Accordingly, we have the ascending chain ker$(f)\subset$ ker$(f^2)\subset$ ker$(f^3)$...
