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: $$\operatorname{Hom}_R(A,B) \cong \operatorname{Hom}_R(\dfrac{R^n}{I},\dfrac{R^m}{J}).$$
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.