2
$\begingroup$

$\DeclareMathOperator\Ker{Ker}$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.

$\endgroup$
0
$\begingroup$

$\DeclareMathOperator\Ker{Ker}$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.

$\endgroup$
0
$\begingroup$

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 $\{j(m_1),\dots,j(m_l)\}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.