Theorem on the Lengths of Modules - Cohn, Theorem 2.5

[Math Amateur]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)[/COLOR]

In Chapter 2: Linear Algebras and Artinian Rings, on Page 61, Cohn presents Theorem 2,5 concerning the lengths of modules.

Cohn indicates that the proof of this theorem is obvious/trivial ... BUT ... I am having trouble even getting started in formulating an explicit and formal proof ... can someone please help ...

The text of Theorem 2.5 is as follows:
View attachment 3325 Help will be appreciated,

Peter

 

[Euge]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)[/COLOR]

In Chapter 2: Linear Algebras and Artinian Rings, on Page 61, Cohn presents Theorem 2,5 concerning the lengths of modules.

Cohn indicates that the proof of this theorem is obvious/trivial ... BUT ... I am having trouble even getting started in formulating an explicit and formal proof ... can someone please help ...

The text of Theorem 2.5 is as follows:
View attachment 3325 Help will be appreciated,

Peter

Hi Peter,

Recall the equation

\(\displaystyle \ell(M) - \ell(\text{ker }f) = \ell(N) - \ell(\text{coker}f) \quad (*)\)

If $f$ is injective, then $\text{ker }f = 0$ and thus $\ell(\text{ker }f) = 0$. So $(*)$ becomes $\ell(M) = \ell(N) - \ell(\text{coker }f)$, which shows that $\ell(M) \le \ell(N)$. If $f$ is surjective, then $\text{coker }f = 0$ and thus $\ell(\text{coker }f) = 0$. The equation $(*)$ reduces to $\ell(M) - \ell(\text{ker }f) = \ell(N)$, which shows that $\ell(M) \ge \ell(N)$.

Finally, suppose $f$ is an isomorphism. Then $f$ is both injective and surjective, so by the previous results, $\ell(M) \le \ell(N)$ and $\ell(M) \ge \ell(N)$. Consequently, $\ell(M) = \ell(N)$. Try proving the converse.