# If $x_i$ generate an $A$-module $M$, why do $1 \otimes x_i$ generate the extension of scalars $B \otimes_A M$?

In the following, let "ring" be a synonym for "commutative ring with identity".

For rings $A, B$ and an $A$-module $M$, let $M_B = B \otimes_A M$ be the $B$-module obtained from $M$ by extension of scalars.

In the book by Atiyah and MacDonald on Commutative Algebra, I read:

Proposition 2.17. If $M$ is finitely generated as an $A$-module, then $M_B$ is finitely generated as a $B$-module.

Proof. If $x_1, \ldots, x_m$ generate $M$ over $A$, then the $1 \otimes x_i$ generate $M_B$ over $B$.

I don't understand this. For example, let $M$ be generated by $x_1, x_2$ and let $a_1, a_2 \in A$, $b \in B$. Then we have \begin{align} b \otimes (a_1 x_1 + a_2 x_2) \in M_B. \end{align} Then the proposition tells me that I should be able to find $b_1, b_2 \in B$ such that \begin{align} b_1 (1 \otimes x_1) + b_2 (1 \otimes x_2) = b \otimes (a_1 x_1 + a_2 x_2) \in M_B. \end{align} Naively, I would like to choose $b_1 = a_1 b$, $b_2 = a_2 b$ to get \begin{align} a_1 b (1 \otimes x_1) + a_2 b (1 \otimes x_2) &= a_1 (b \otimes x_1) + a_2 (b \otimes x_2) \\ &= (b \otimes a_1 x_1) + (b \otimes a_2 x_2) \\ &= b \otimes (a_1 x_1 + a_2 x_2). \end{align} However, I would say that this is a manifestation for the fact that $1 \otimes x_1$ and $1 \otimes x_2$ generate $M_B$ as an $(A, B)$-bimodule rather than as a $B$-module, because in one step, I multiply an element of $B$, while in the next step, I multiply by elements of $A$.

How do $1 \otimes x_1$ and $1 \otimes x_2$ generate $M_B$ as a $B$-module?

• $a_1 b$ doesn't make sense, of course. But you can only extend your scalars from $A$ to $B$ if you have a map $f:A\to B$ (this map is unmentioned, but is implicit in your forming the tensor product). Multiplication of an element of $B\otimes_A M$ by $a\in A$ is defined to be multiplication by $f(a)\in B$. So try $f(a_1) b$. Commented Apr 23, 2013 at 4:01
• Recall that B is also an $A$-module via the map $f:A \to B$ given before. Commented Apr 23, 2013 at 4:02

Any element in $B\otimes_{A}M$ can be written as the sum of $b_{i}\otimes_{A}m_{i}$, and since $m_{i}$ can be generated by $x_{j}$ as $m_{i}=\sum a_{ij}x_{j}$, we have $B\otimes_{A}M$ generated by $b_{i}\otimes_{A}\sum a_{ij}x_{j}=\sum b_{i}a_{ij}\otimes x_{j}$. As a module over $B$ it is generated by $1\otimes_{A}x_{j}$ as $b_{i}\phi(a_{ij})$ are scalars in this case. Here $\phi:A\rightarrow B$ is the base changing map.