Finitely Presented Modules Definition

Question

I am a little bit confused with the definition of finitely presented modules. In Lang's Algebra he defines a module $M$ to be finitely presented if and only if there is a exact sequence $F'\to F\to M \to 0$ such that both $F', F$ are free. However the standard definition I have seen elsewhere only demands $F'$ be finitely generated. Are these two definitions equivalent?

Looking at the situation of a non-principal ideal of a ring, say $(x, y)$ of $\mathbb{R}[x, y]$, it appears that this is finitely presented, by the usual definition, but I don't see any way of making it finitely presented by Lang's definition.

Answer 1

If $F' \to F \to M \to 0$ is exact and $F'$ is finitely generated, choose some finitely generated free module $F''$ which maps onto $F'$. Then $F'' \to F \to M \to 0$ is exact.

This shows: A finitely generated module is finitely related iff it is finitely presented.

Of course, this fails for modules which are not finitely generated.

Answer 2

In Lang's Algebra he defines a module $M$ to be finitely presented if and only if there is an exact sequence $F'\to F\to M \to 0$ such that both $F', F$ are free of finite rank, and this is the definition of finitely presented modules. (Note that for each module $M$ there is an exact sequence $F'\to F\to M\to 0$ with $F,F'$ free modules.)

"However the standard definition I have seen elsewhere only demands $F'$ be finitely generated." This is the definition of finitely related modules.

"Are these two definitions equivalent?" In general they aren't: Let $M$ be a finitely presented module, and $L$ a free module which is not finitely generated. Then $M\oplus L$ is finitely related, but not finitely presented. However, if the module is finitely generated the two definitions coincide.

Answer 3

In Lang's definition, both $F$ and $F'$ must be finitely generated (and free). This definition is equivalent to: ‘There exists a short exact sequence: $$0\to K\to F\to M\to 0$$ such the $T$ is a finitely generated free module, and the module of relations $K$ is finitely generated. It is equivalent because there is exists a surjective homomorphism from a finitely generated free module onto $K$.