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

$$\DeclareMathOperator\Ker{Ker}$$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.

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.

$$\DeclareMathOperator\Ker{Ker}$$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.
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 $$\ker\phi\oplus F$$.
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 $$\{j(m_1),\dots,j(m_l)\}$$.