I'm working on the problem below. I've put some of my work towards a solution and would welcome hints or comments to clarify whatever I'm not seeing or am misunderstanding.
Let $R$ be an integral domain and $K$ its field of fractions. Let $V$ be a vector space over $K$. Show the map $T: K \otimes_R V \rightarrow V$ by $T(k \otimes v) = kv$ is an $R$-module isomorphism.
Partial solution:
Assuming this map is a homomorphism of $R$-modules, I see why this defines an isomorphism. Given $v \in V$, it has in its pre-image $1 \otimes v$, so the map is surjective.
Given $kv=k'v'$ we have $\frac{k}{k'}v=v'$. Thus, $1 \otimes v' = \frac{k}{k'} \otimes v$, multiplying both sides by $k'$ we get $k' \otimes v' = k'\frac{k}{k'} \otimes v = k \otimes v$. So the map is injective.
Question:
I'm having trouble seeing that the given map is an $R$-module homomorphism. I understand that the $R$-module structure on $K \otimes_R V$ comes from defining multiplication by $R$ as $r\cdot (k \otimes v)=rk \otimes v = k \otimes rv$. So the map respects multiplication by $r$ since we'll have T$r(k \otimes v)) = T(rk \otimes v) = rkv = rT(k\otimes v)$. I'm confused though about how $T$ respects addition in the tensor product, i.e. given two tensors $x,y$ why does $T(x+y)=T(x)+T(y)$?
(In fact, I think I'm really confused about how addition works in the $R$-module $K \otimes_R V$)
I've since realized the following, which is maybe (?) enough to finish the argument:
Since $K$ is a $1$ dimensional vector space over itself, it has basis $\{1\}$ and we can take $\{e_i\}$ as a basis for $V$. Then $K \otimes_R V$ has a basis $\{1 \otimes e_i\}$. We can write two tensors in terms of this basis, say $k \otimes v = \sum_i a_i(1 \otimes e_i)$ and $k' \otimes v' = \sum_i b_i (1 \otimes e_i)$. Then $k \otimes v + k' \otimes v' = \sum_i (a_i+b_i)(1 \otimes e_i)$.
Applying $T$, we get $$T(k \otimes v + k' \otimes v') = T\left( \sum_i (a_i+b_i)(1 \otimes e_i) \right)= \sum_i (a_i+b_i)e_i= \sum a_ie_i + \sum b_i e_i \in V$$ But also, $$ T(k \otimes v) +T( k' \otimes v') = T\left(\sum_i a_i(1 \otimes e_i)\right)+T\left(\sum_i b_i(1 \otimes e_i)\right)= \sum a_ie_i + \sum b_i e_i $$ and so $$T(k \otimes v + k' \otimes v') = T(k \otimes v) +T( k' \otimes v')$$ which together with the above shows $T$ is an $R$-module homomorphism.
Is this now enough?