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Just started studying tensor products. Let $A$ be a commutative ring with unity and let $M$ be an $A$-module. Now let $k$ be a field, I know that a $k$-module is precisely a $k$-vector space.

My question is the following:

Why $k \otimes_{A} M$ is also a $k$-vector space? here $\otimes$ denotes the tensor product with respect the ring $A$.

Is it because we can give $k \otimes_{A} M$ the structure of $k$-module by just taking:

$f: k \times (k \otimes_{A} M) \rightarrow k \otimes_{A}M$ given by:

$f(c,d \otimes m)=(cd) \otimes m$ ? where $c,d \in k$

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    $k$ is playing two roles above: both as an element of the field and the name of the field. That's bad form. You should change the name of the field or of the element.2011-03-23
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    @Arturo Magidin: thanks, just edited it.2011-03-23
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    How does $A$ act on $k$ from the right? This is needed to define the tensor product $k\otimes_A M$.2011-03-23
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    @Rasmus: You need an action, yes, but with commutative ring, any action is a bilateral action.2011-03-23

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