Let $S$ be extension of $R$ and $M$ is an $R$-module. If $M$ is finitely generated of $R$-module then $S\otimes_RM$ is finitely generated of $S$-module.
Proof We can construct a ring homomorphism $f:R\to S$ such that $f(r)=r$ for all $r\in R$. Let $\{m_1,m_2,\dots,m_n\}$ be a set of generator of M. Then, forall $m\in M,m\displaystyle\sum_{i=1}^{k}r_im_i$. We knew that, every element of $S\otimes_R M$ is sum of elements as the form $s\otimes m$ with $s\in S, m\in M$. So $$s\otimes m=s\otimes\left(\displaystyle \sum_{i=1}^{k}r_im_i \right)=\displaystyle \sum_{i=1}^{k}r_i(s\otimes m_i)=\displaystyle \sum_{i=1}^{k}(r_i s)\otimes m_i=\displaystyle \sum_{i=1}^{k}(f(r_i)s)\otimes m_i=\displaystyle \sum_{i=1}^{k}[f(r_i)s](1\otimes m_i) $$ Hence, $\{1\otimes m_1,1\otimes m_2,\dots,1\otimes m_n\}$ be a set of generator of $S\otimes_RM$ so $S\otimes_RM$ is finitely generated.
Does have any problem in my proof? Please show me. Thankyou.