Show that $\ker(\phi)$ is finitely generated

Question

$\DeclareMathOperator\Ker{Ker}$Let $\phi : M \to F$ be a surjective homomorphism of a finitely generated module $M$ onto a free module $F$. Show that $\Ker(\phi)$ is finitely generated.

My attempt :

I considered the Short Exact Sequence :$$ 0 \to \Ker(\phi) \to M \to F \to 0.$$ The map from $\Ker(\phi) \to M$ is the inclusion and the map from $M\to F$ is $\phi$. Since, $F$ is free, $\exists \psi : F \to M$ (R-linear map) such that $\phi \circ \psi = id_F$ , in other words , the above Short Exact Sequence splits and hence,$M \cong\Ker(\phi) \oplus F$ . Let $j:M \to \Ker(\phi) \oplus F$ be the isomorphism.

Given any $k \in \Ker(\phi)$ and given $m \in M$ we get a unique, $f \in F$ such that $j(m)=k+f$.

Now since, $M$ is finitely generated $\exists m_1,\dots,m_l,\lambda_1,\dots,\lambda_l$ such that, $m= \lambda_1 m_1 + \dots + \lambda_l m_l$ and also since , $F$ is free, we get a basis $\{f_1,\dots,f_s\}$ and thus $f=c_1f_1 + \dots +c_sf_s$ and hnece, we can write,$$k=\lambda_1j(m_1)+\dots+\lambda_lj(m_l)-c_1f_1-\dots-c_sf_s$$

Since , $k\in \Ker(\phi)$ was arbitrarily chosen we get that $\Ker(\phi)$ is generated by$\{j(m_1),\dots,j(m_l),f_1,\dots,f_s\}$ and hence in particular finitely generated.

Please point out mistakes if there are any in my answer. Thanks in advance for help!

There might be answers else where, But I want to check whether my arguments are correct before looking into others' answers. So please do not tag this as duplicate.

Answer 1

$\DeclareMathOperator\Ker{Ker}$Since $F $ is a free, hence projective, module, the sequence $$\{0\}\to\Ker (\phi)\to M\to F\to\{0\} $$ splits. Consequently, $\Ker (\phi) $ is a direct summand, hence an homomorphic image, of the finitely generated module $M $, thus it is finitely generated as well.

Answer 2

No, $\ker\phi$ is not generated by $\{j(m_1),\dots,j(m_l),f_1,\dots,f_s\}$, but just by $\{j(m_1),\dots,j(m_l)\}$. The former (bigger) set generates $\ker\phi\oplus F$.

You write:

Given any $k\in\ker\phi$ and given $m\in M$, we get a unique $f\in F$ such that $j(m)=k+f$.

You should write: given $k\in\ker\phi$, there exists a unique $m\in M$ such that $k=j(m)$.

Since $M$ is finitely generated, say by $\{m_1,\dots,m_l\}$, we can write $m=\lambda_1m_1+\dots+\lambda_lm_l$. This proves that $\ker\phi$ is generated by $\{j(m_1),\dots,j(m_l)\}$.