# Why is this a $k$-module?

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. –  Arturo Magidin Mar 23 '11 at 19:48
@Arturo Magidin: thanks, just edited it. –  user8632 Mar 23 '11 at 19:52
How does $A$ act on $k$ from the right? This is needed to define the tensor product $k\otimes_A M$. –  Rasmus Mar 23 '11 at 19:55
@Rasmus: You need an action, yes, but with commutative ring, any action is a bilateral action. –  Arturo Magidin Mar 23 '11 at 20:03

First: for $k\otimes_A M$ to make sense, you have to have an action of $A$ on $k$; that is, $k$ should be an $A$-module in some way. (It could be the trivial $A$ module, $ar=0$ for all $a\in A$ and $r\in k$, though that would make $k\otimes_AM$ the trivial module).

But: in general, if $S$ and $A$ are commutative rings, and you have an action of $A$ on $S$, then for any $A$-module $M$ you get an $S$-module by taking $S\otimes_A M$: this is called extension of scalars (or extension of the base). The action of $s$ on $S\otimes_A M$ is precisely the one you give: given any $s\in S$ and any generator $t\otimes m$, you define $s(t\otimes m) = (st)\otimes m$ and extend linearly.

Explicitly, we have a $A$-multilinear map $$f\colon S\times S\times M \to S\otimes_A M$$ given by $(s,t,m)\mapsto st\otimes m$. This map is $A$-multilinear: \begin{align*} f(s+s',t,m)&=(s+s')t\otimes m = (st+s't)\otimes m = st\otimes m+s't\otimes m\\ &= f(s,t,m) + f(s',t,m).\\ f(s,t+t',m) &= s(t+t')\otimes m = (st+st')\otimes m = st\otimes m + st'\otimes m\\ &= f(s,t,m) + f(s,t',m).\\ f(s,t,m+m') &= st\otimes(m+m') = st\otimes m + st\otimes m' = f(s,t,m)+f(s,t,m').\\ f(as,t,m) &= (as)t\otimes m = s(at)\otimes m = f(s,at,m)\\ &= a(st)\otimes m = st\otimes am = f(s,t,am)\\ &= a(st\otimes m) = af(s,t,m). \end{align*} Therefore, the universal property of the tensor product (the definition) says that the map $f$ induces a unique map $S\otimes_A(S\otimes_A M)\to S\otimes_A M$. This map makes $S\otimes_A M$ into an $S$-module.

In the particular case where $S$ is a field, as you have, this makes $k\otimes_A M$ into a $k$-module; and $k$-modules are the same thing as $k$-vector spaces.

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great, thanks a lot! –  user8632 Mar 23 '11 at 19:54
ok I saw that somebody had posted an answer :) but I had just finished tipping mine and now I see that you posted almost exactly the same. –  Michalis Mar 23 '11 at 19:59
@Rasmus: Yeah, couldn't make up my mind whether to use $A$ or $R$. I've switched them all to $A$'s. Thanks. –  Arturo Magidin Mar 23 '11 at 20:02

I think you need to assume that k is an A-module (otherwise you can't take the tensor product).

Your statement follows from the following more general lemma: Suppose A,B are rings and N is an A-module and a B-module in a compatible way a(bn) = b(an). Let M be another A-module. Then $M\otimes_A N$ is a B-module.

Just take $B = k$, $N = k$, $M = A$ to get what you want. Your guess for the solution is also correct.

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