why can the kernel be not finitely generated I am very new to modules and couldn't understand this:
Suppose $\phi:R^n\rightarrow M$ is an $R$-linear map. I thought $\ker(\phi)$ is finitely generated since $R^n$ is generated by the canonical basis $\{e_i\}$. But this seems to be wrong as one of the problems sets that as an extra condition.
I really appreciate it if anyone can correct my understanding and hopefully give a basic counterexample.
 A: It's not restrictive to assume that $\phi$ is surjective, in particular that $M$ is finitely generated. The condition that $\ker\phi$ is finitely generated is known to be “$M$ is finitely presented” and, in general, a finitely generated module need not be also finitely presented.
It's not so difficult to prove that if $M$ is finitely presented, that is, there exist $n$ and an $R$-linear map $\phi\colon R^n\to M$ such that $\phi$ is surjective and $\ker\phi$ is finitely generated, then for every surjective $R$-linear map $f\colon L\to M$, if $L$ is finitely generated then $\ker f$ is finitely generated.
On the other hand, a commutative ring $R$ can have nonfinitely generated ideals; if $I$ is such an ideal, then the canonical projection $\pi\colon R\to R/I$ provides an example of a finitely generated module, namely $R/I$, which is not finitely presented. And the classical example is the ring $R=k[x_1,x_2,\dotsc]$ of polynomials in infinitely many variables, with $I$ the ideal generated by the monomials.
In view of the theorem above, for no surjective linear map $\phi\colon R^n\to R/I$, you can have that $\ker\phi$ is finitely generated.
A (commutative) ring $R$ is Noetherian if each of its ideals is finitely generated. This is equivalent to saying that every finitely generated $R$-module is finitely presented.
