Induction functor from $\mathbb{K}$-mod to ($\mathbb{K}\times\mathbb{K}$)-mod Let $\mathbb{K}$ a field. Given a $\mathbb{K}$-mod (a vector space) there is an induction functor on $\mathbb{K} \times \mathbb{K}$-mod that is, as usual, $- \otimes_\mathbb{K} (\mathbb{K} \times \mathbb{K}) $
Now, I admit I was expecting something like $\operatorname{Ind}(V)=V \times V$. Instead, I get that
$$V \otimes_\mathbb{K} (\mathbb{K} \times \mathbb{K}) \simeq V \times \mathbb{K}$$
because an element is the sum of terms like $v \otimes (a,b)$ and $\mathbb{K}$-linearity tells us that is equivalent to $b^{-1} v \otimes (ab^{-1},1) \mapsto b^{-1}(v \times a)$
Is this correct? What if we do this with a ring instead of a field?
 A: You are asking about vector spaces over fields and modules over rings. Since the latter are more general, let's start with those. 
We know that tensor products are distributive over direct sums. In other words, for every triple $A, B, C$ of $R$-modules, we have a canonical isomorphism between $A\otimes (B \oplus C)$ and $(A\otimes B) \oplus (A \otimes C)$. This map is determined by the rule $a \otimes (b, c) \mapsto (a\otimes b, a \otimes c)$, and its inverse by $(a \otimes b, c \otimes d) \mapsto a \otimes (b, 0) + c \otimes (0, d)$. It is easy to see that these maps are both well-defined, $R$-linear, and each other's inverse. 
In particular, we get $A \otimes (R \oplus R) \cong (A \otimes R) \oplus (A \otimes R) \cong A \oplus A$. So in the case of a vector space $V$ over some field $\mathbb{K}$, we see that $V \otimes (\mathbb{K} \times \mathbb{K}) \cong V \times V$ (notice that the direct sum and the direct product coincide when we are only considering finitely many factors). For completeness, the isomorphisms are given by
$$ V \otimes (\mathbb{K} \times \mathbb{K}) \to V \times V, \quad v \otimes (a, b) \mapsto (av, bv) $$
and 
$$ V \times V \to V \otimes (\mathbb{K} \times \mathbb{K}), \quad (v, w) \mapsto v \otimes (1, 0) + w \otimes (0, 1).$$
So this only leaves the question of why your map $b^{-1}v \otimes (ab^{-1}, 1) \mapsto b^{-1} v \times a$ does not work. I have to admit that I'm not sure about your notation, but after substituting $w = b^{-1}v$ and $c = ab^{-1}$, it seems that your map $V \otimes (\mathbb{K} \times \mathbb{K}) \to V \times \mathbb{K}$ is given by $w \otimes (c, 1) \mapsto (w, c)$. However, this is not well-defined, for example because it does not assign a value to elements of the form $w \otimes (c, 0)$. 
