# Finitely generated modules over noetherian rings

There is a question in here that has a proof which I can’t understand, (maybe they use higher level of algebraic definition and I’m new), could you please explain it?

Let $R$ be a commutative noetherian ring. If $A$ and $B$ are finitely generated $R$-modules, then $\operatorname{Hom}_R(A,B)$ is a finitely generated $R$-module.

The part I have problem is: since $R$ is commutative (or since $R$ is noetherian), a finitely generated $R$-module is a quotient of $R^n$ for some $n$. Then let us write $A \cong \dfrac{R^n}{I}$ and $B \cong \dfrac{R^m}{J}$. Now we've got: $$\operatorname{Hom}_R(A,B) \cong \operatorname{Hom}_R(\dfrac{R^n}{I},\dfrac{R^m}{J}).$$

• You need to clarify what it is you don't understand. Perhaps include the proof you don't understand in your question and indicate what points you don't understand. As it stands, this question is hard to answer. – ÍgjøgnumMeg Jan 26 '18 at 14:04
• Very related (although that post is about proving the result, not asking about details in a specific proof, it is a much better post about the same theorem). – Arthur Jan 26 '18 at 14:09
• I want a new reference to study this, that proof don’t seem useful for me. – Math90 Jan 26 '18 at 14:41

## 1 Answer

If $A$ is finitely generated, then there is a surjective homomorphism $R^n\to A$, so $A\cong R^n/I$, where $I$ is the kernel.

However, using this fact is just confusing.

Since we have a surjective homomorphism $R^n\to A$, we have an embedding $$\operatorname{Hom}_R(A,B)\to\operatorname{Hom}_R(R^n,B)$$ and the codomain is isomorphic, as $R$-modules, to $\operatorname{Hom}_R(R,B)^n$. Thus it suffices to show that $\operatorname{Hom}_R(R,B)$ is finitely generated. This is obvious, because $\operatorname{Hom}_R(R,B)\cong B$.

In a different way, consider a set of generators $\{x_1,\dots,x_n\}$ of $A$ and the map $$\Phi\colon\operatorname{Hom}_R(A,B)\to B^n$$ defined by $$\Phi(f)=(f(x_1),\dots,f(x_n))$$ You just need to prove that $\Phi$ is an injective homomorphism of $R$-modules.

• And why we have that surjective homomorphism from R^n to A ? – Math90 Jan 26 '18 at 16:51
• @Math90 Because $A$ is finitely generated. – egreg Jan 26 '18 at 16:51
• what do you mean by codomain? do you mean $\text{Hom}_R(R^n, B)/ \text{Im}$? – Math90 Jan 26 '18 at 18:35
• @Math90 The codomain is $\mathrm{Hom}_R(R^n,B)$ (domain = what's before the arrow, codomain = what the arrow points to); you're referring to the cokernel. – egreg Jan 26 '18 at 18:40