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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?

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Your confusion arises from the fact that the way to define $T$ normally requires a little more work. One should make use of the universal property of the tensor product in the following way.

Define a map $K\times V \rightarrow V$ which sends the pair $(k,v)$ to $kv$. This map is easily seen to be $R$-bilinear. Hence, there exists a unique homomorphism of $R$-modules $T:K\otimes_R V \rightarrow V$ such that the image by $T$ of an elementary tensor $k\otimes v$ is $kv$, the image of your first bilinear map.

The fact that $T$ is an $R$-homomorphism is then given, so in particular it is additive. Moreover, such a morphism $T$ is unique (again, by universal property). Hence it really is the map given in your exercise.

To my opinion, it is implicit in the exercise that you must, at first, justify in this way that the morphism $T$ is well-defined.

PS: In order to check the injectivity of $T$, it is not enough to show that if an elementary tensor is sent to $0$, then it is $0$, because in general, an element of $K\otimes_R V$ can not be written as an elementary tensor. What you need to verify is the following:

If $x = \sum_{i=1}^n k_i\otimes v_i \in K\otimes_R V$ is such that $T(x)=0$ then $x=0$.

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