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 $c,d \in k$
 A: 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.
A: 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.
