# Is this tensor product of two modules finitely generated?

Definition. Let $\varphi : R \to S$ be a ring map. Let $R \to R'$ be any ring map. The base change of $\varphi$ by $R \to R'$ is the ring map $\rho: R' \to S \otimes_R R'$.

Now, in the above definition, $S$ and $S\otimes_R R'$ may be considered as an $R$- and $R'$-modules, respectively. Is it true that if $S$ is finitely generated (over $R$), then so is $S \otimes_R R'$ (but over $R'$)? If the answer is affirmative, how does one see this?

Update:

Alright, let me make this question more specific. Let $S$ be generated over $R$ by $m_1,\dots,m_n$. By the definition of the $R$-module structure on $S$, this means that any element $s\in S$ may be written as $s=\sum_{i=1}^nm_i\varphi(r_i)$ for some $r_i\in R$.

Now the claim is that the $n$ elements $m_1\otimes 1,\dots, m_n \otimes 1$ generate the module $S\otimes_R R'$ over $R'$. To establish this, we need to show that every element $\gamma \in S\otimes_R R'$ may be written as $\gamma =\sum_{i=1}^n \rho(r_i') \cdot m_i\otimes 1$ for some $r_i'\in R'$.

My attempt:

As is known, every $\gamma \in S\otimes_R R'$ may be written as $\gamma=\sum_{i=1}^k\rho(\widehat{r_i'})\cdot s_i\otimes r_i'$ for some $s_i\in S$ and $r_i'\in R$. Further, since $S$ is finitely generated over $R$, we have $s_j=\sum_{i=1}^n\varphi(r_{ij})\cdot m_i$ for some $r_{ij}\in R$ and for every $j=1,\dots,k.$ On substituting these $s_j$ in the above expression of $\gamma$, we obtain $$\gamma=\left[\rho(\widehat{r_1'})\sum_{i=1}^n\varphi(r_{i1})\cdot m_i\otimes r_1'\right]+\dots+\left[\rho(\widehat{r_k'})\sum_{i=1}^n\varphi(r_{ik})\cdot m_i\otimes r_k'\right]=\\=\sum_{j=1}^k\rho(\widehat{r_j'})\sum_{i=1}^n\varphi(r_{ij})\cdot m_i\otimes r_j'.$$

How do I proceed?

• Finitely generated over what base? – Hanno Nov 30 '16 at 9:30
• I mean if $S$ is a finitely generated $R$-module, does it follow that $S\otimes_R R'$ is a finitely generated $R'$-module? – Cary Nov 30 '16 at 9:34
• Let $s_1, \dots , s_n$ be generators of $S$. Can you see that $s_1 \otimes 1 , \dots , s_n \otimes 1$ generate $S \otimes_R R'$? – Crostul Nov 30 '16 at 9:40
• I am confused: You give a link to the definition and only a few lines after this definition, there is a proof that the answer is affirmative. How does this fit together? – MooS Nov 30 '16 at 10:17
• @Hanno, Crostul: thanks for your comments, I updated the question. MooS: oops, indeed, I never noticed it. Anyway, it is based on a few lemmas, and the purpose of the updated question is to establish a more direct proof (if possible). – Cary Nov 30 '16 at 10:30

## 1 Answer

Saying that $S$ is finitely generated as $R$-module is the same as saying there is a surjective homomorphism of $R$-modules $f\colon R^n\to S$, for some positive integer $n$.

Then the homomorphism $f\otimes_R\mathit{id}_{R'}\colon R^n\otimes_RR'\to S\otimes_RR'$ is surjective too. Since $R^n\otimes_RR'\cong(R')^n$, you're done.