Suppose \(A\) is a commutative ring and \(M\) is a finitely presented module. Given any surjection \(\phi:A^n\rightarrow M\) from a finite free \(A\)-module, show that \(\ker\phi\) is finitely generated.
Solution
1. Since \(M\) is finitely presented, there is an exact sequence \(A^m \xrightarrow{\psi} A^r \xrightarrow{\pi} M \to 0\).
definition of finitely presented.
2. We have a surjection \(\phi : A^n \to M\), and we want to show \(\ker \phi\) is finitely generated.
setup.
3. By Schanuel’s lemma (applied to the two surjections \(\phi : A^n \to M\) and \(\pi : A^r \to M\)), we get\[\ker \phi \oplus A^r \cong \ker \pi \oplus A^n.\]
Schanuel’s lemma.
4. \(\ker \pi = \operatorname{im} \psi\) is finitely generated (it is the image of the finitely generated module \(A^m\)).
1.
5. Hence \(\ker \phi \oplus A^r \cong \operatorname{im}\psi \oplus A^n\) is finitely generated.
3 and 4.
6. A direct summand of a finitely generated module is finitely generated, so \(\ker \phi\) is finitely generated.
5 (since \(\ker \phi\) is a direct summand of \(\ker \phi \oplus A^r\)).
7. Q.E.D.
6.