The kernel of a surjective homomorphism is finitely generated

Question

Let $M$ be a finitely generated $R$-module and $f:M \rightarrow R^n$ a surjective homomorphism. By letting $e_1,\dots,e_n$ be the standard basis of $R^n$ and choosing $u_i \in M$ with $f(u_i) = e_i$ for $1 \leq i \leq n$,

(i) Show that $M = \ker(f) \bigoplus M_0$, where $M_0 \leq M$ is the submodule generated by the $u_i$.

(ii) Show that $\ker(f)$ is finitely generated.

score 12 · Accepted Answer · answered Feb 08 '13 at 01:53

12

Alternatively to Berci's solution, although definitely more high-tech, is the observation that we have the short exact sequence

$$0\to\ker f\to M\to R^n\to 0$$

which splits because $R^n$ is free. The existence of the back-maps $j:R^n\to M$ and $h:M\to \ker f$ give you i) and ii) respectively.

answered Feb 08 '13 at 01:53

Alex Youcis

54,059

How do you define the back-map $h$? I understand $j$ will be the canonical homomorphism. – winawer Mar 02 '17 at 12:10

score 8 · Answer 2 · edited Feb 08 '13 at 15:13

8

Hints: For (i), assume $f(m)=r_1e_1+\dots+r_ne_n$, then $m-r_1u_1-\dots-r_nu_n\in\ker f$.

For (ii), consider a finite generator set $m_1,\dots,m_k$ of $M$ and write them all in the format according to (i).

edited Feb 08 '13 at 15:13

answered Feb 08 '13 at 00:38

Berci

90,745

How do we prove that (i) is a direct sum? – Ninja May 25 '18 at 22:42
1

Let $m=\sum r_iu_i\in M_0\cap\ker f$. Then $\sum r_ie_i=0$, hence all $r_i=0$. – Berci May 26 '18 at 07:53

The kernel of a surjective homomorphism is finitely generated

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