Finitely generated modules in short exact sequences Let $0\to A\to B\to C\to 0$ be a short exact sequence of $R$-modules where $R$ is a ring.

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*Show that if $B$ is finitely generated, then $A$ and $C$ don't have to be finitely generated.


*If $R$ is a PID and $B$ is finitely generated, then are $A$ and $C$ finitely generated?
I know that if $A$ and $C$ are finitely generated, then $B$ is also finitely generated. It would be nice if this statement would be and iff statement but it isn't, but I suppose that it is true if $R$ is a PID. For the first question I thought of finding a counter example, but I can't come up with one. And for the second question I can't even see why it would be true by just knowing that is a PID (it might not be true though).
Any help is greatly appreciated.

You can take $R$ to be a non-Noetherian ring, for example $\mathbb{R}[X_1,X_2,\dots]$. Then $R$ is obviously a finitely generated module over $R$, but $I=\langle X_1, X_2,\dots\rangle$ is not a finitely generated module over $R$. You get the following exact sequence $0\to I\to R\to R/I\to 0$, with $R$ finitely generated over $R$ but $I$ isn't. Note that as noted in the comments, $R/I$ would necessarily be finitely generated over $R$.
Remark: Every non-Noetherian ring would do, as a ring is Noetherian iff every submodule of finitely generated module is finitely generated. This gives you that in the case of $2$, $A$ and $C$ must be finitely generated, as $A$ can be considered as a submodule of the finitely generated module $B$, and since being PID implies being Noetherian.
