Providing non trivial module morphisms I'm starting to study modules, and I would like to get some counterexamples to naive ideas one has in the first approach to the subject.

  
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*Does there exist an ideal I in A and a morphism $f:I \Rightarrow I$ such that $f$ does not extend to $A$  ? In other words, I'm searching for a Morphism in $I$ that is not of the form $f (i)=a*i$.
  If it can be of some aid, this is always true for principal ideals and for ideals generated by irreducible elements.
  
  
  2.Does there exist a module $M$ and an ideal $I$ in $A$ such that $M$ tensor $I$ is not isomorphic to $IM$?  

Thank you for the aid.
 A: For question 1, let $A = k[x, y]/(x^2, xy, y^2)$ and $I = (x, y)$. The $A$-module endomorphisms $\text{End}_A(I)$ can be identified with $2 \times 2$ matrices, and of those, only the scalar matrices show up as multiplication by an element of $A$.
For question 2, this is true for every ideal $I$ if and only if $M$ is flat. This comes from thinking of $IM$ as the kernel of the map $M \to M/I$, and investigating this kernel by tensoring $M$ with the short exact sequence
$$0 \to I \to R \to R/I \to 0$$
to get the Tor long exact sequence
$$0 \to \text{Tor}_1(R/I, M) \to I \otimes M \to M \to M/I \to 0.$$
But you don't need to know that. The simplest examples of modules that are not flat are modules that have torsion. Consider, for example, $M = \mathbb{Z}/2$ as a module over $R = \mathbb{Z}$, and let $I = 2 \mathbb{Z}$. Then $IM = 0$, but $I \otimes M = \mathbb{Z}/2$. 
A: Here is an example for question 2:
Let $J$ be an ideal, and $M=A/J$. The


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*$M\otimes I\simeq I/IJ$.

*$IM=I\cdot A/J=(I+J)/J\simeq I/(I\cap J)$
Now usually, $IJ\neq I\cap J$, unless $I$ and $J$ are coprime, i.e. $I+J=A$.


In particular, if $J=I$, $A/I\otimes I\simeq I/I^2$, while $I\cdot A/I=0$.
