Let $M$ be a finitely generated module over a ring $R$ and $\phi: M \rightarrow R^n$ be a surjective homomorphism. Show that the $\ker(\phi)$ is finitely generated submodule of $M$.

My effort: Since $M$ is finitely generated, So there exist $x_1,x_2, \ldots, x_k \in M$ such that $M = \langle x_1,x_2, \ldots, x_k \rangle = Rx_1 + Rx_2 + \cdots +Rx_k$. Since $\phi$ is surjective so, $\langle\phi(x_1), \cdots, \phi(x_k) \rangle = R^n$. By first isomorphism theorem, $M/\ker(\phi) \cong R^n$. And $R^n$ is finitely generated so $M/\ker(\phi)$ is finitely generated. We also have the following exact sequence $0 \rightarrow \ker(\phi) \rightarrow M \rightarrow M/\ker(\phi) \rightarrow 0$.

I am stuck after this point. Any help would be appreciated! Thanks in advance!


One way of seeing this is that since $M/\ker(\phi) \simeq R^n$, this is a free module, and thus the projection $M \rightarrow M/\ker(\phi)$ is a retraction, and therefore $\ker(\phi)$ is a direct summand of $M$:

$$ M = \ker(\phi) \oplus S $$

for some submodule $S$. In particular, there exists $y_i \in \ker(\phi), s_i \in S$ for each $i$ such that

$$ x_i = y_i + s_i $$

Now I claim that $\ker(\phi) = \langle y_1 ,\dots,y_k\rangle$. In effect, let $y \in \ker(\phi)$. Now

$$ y = \sum_{1 \leq i \leq k}a_ix_i = \sum_{1 \leq i \leq k}a_i s_i + \sum_{1 \leq i \leq k}a_iy_i $$

but since $y - \sum_{1 \leq i \leq k}a_iy_i = \sum_{1 \leq i \leq k}a_is_i \in \ker(\phi)\cap S = \{0\}$,

$$ y = \sum_{1 \leq i \leq k}a_iy_i $$

  • $\begingroup$ It makes perfect sense! Thanks! $\endgroup$ – Shubham Namdeo Jul 3 '18 at 4:58

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