# Tensor product of two f. g. modules vs. algebras

I was wondering if it is different from proving that tensor product of two finitely generated modules is finitely generated versus tensor product of two finitely generated algebras is finitely generated. Namely, Liu gave them in separate problems so what is the difference in proving those?

Let $M,N$ be $A$-modules, and let $B,C$ be $A$-algebras.

Prove that if $M$ and $N$ are finitely generated over $A$, then so is $M\otimes_AN$.

Prove that if $B$ and $C$ are finitely generated over $A$, then so is $B\otimes_AC$.

Taking $A=\mathbb Z$, $M=B=\mathbb Z/2\mathbb Z$, and $N=C=\mathbb Q$, show that the converse of (a) and (b) is false.

Does these needs separate proofs depending whether I use modules or algebras?

• Maybe this depends on the definition of "finitely generated". If $M$ is some module over some ring $R,$ then, to me, the statement "$X\subseteq M$ generates $M$ as an $R$-Module" means that every element of $M$ can be written as an $R$-linear combination of finitely many elements of $X.$ Now, if $A$ is some algebra over some ring $R,$ then the statement "$Y\subseteq A$ generates $A$ as an $R$-algebra" could mean either the same as for modules, or that every element of $A$ is a polynomial of elements of $Y$ with coefficients in $R.$ Jan 11, 2015 at 17:15
• [continued] There's a difference between linear combinations and polynomials. Maybe you should check that in your context. Jan 11, 2015 at 17:16

## 1 Answer

They are different exercises because they are different notions of finitely generated:

1) If $R$ is a ring then an $R$-module $M$ is finitely generated if there exists a surjection $R^n \to M$ from a free module to $M$. Equivalently there's a finite set of elements such that every element of $M$ can be written as a linear combination of these elements with coefficients in $R$.

2) On the other hand if $R$ is a $k$-algebra then $R$ is finitely generated if there exists a surjection $k[x_1, \ldots, x_n] \to R$ from a polynomial ring to $R$. Equivalently there's a finite set of elements such that every element of $R$ can be written as a polynomial in these elements with coefficients in $k$.