# 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?

• So $V$ is a $\mathbb K$-vector space, and $\hbox{Ind}(V)$ is a $\mathbb K\times\mathbb K$-module? Then what does $b^{-1}(v\times a)$ mean? Shouldn't the coefficient be an element of $\mathbb K\times\mathbb K$? – Justpassingby Jan 11 '16 at 19:34
• $\mathbb{K} \to \mathbb{K} \times \mathbb{K}$ how? $x \mapsto (x,1)$ or $x \mapsto (x,x)$ ? I have tried both definitions and (unless I made some mistakes) none of them gives an adjunction when I consider Res. My question is exactly that, how are those adjoint functors defined in this case? – AnalysisStudent0414 Jan 11 '16 at 20:34
• If we are in the category of rings with unit then only the second is possible. Perhaps you should add a reference to your question. – Justpassingby Jan 12 '16 at 6:33
• My reasoning: the induction functor is, by definition, tensoring a $B$-module with $A$ to get an $A$-module if $B \subseteq A$ as rings. So given a vector space $V$, $Ind(V) = V \otimes_\mathbb{K} \mathbb{K}^2$, where the action of $\mathbb{K}^2$ is given by left multiplication, so $(c,d). (v \otimes (a,b)) = v \otimes (ca,db)$. Now considering $\mathbb{K}$ linearity of the tensor product we can either get $v \otimes (a,b) = bv \otimes (b^{-1}a, 1)$ (in the diagonal case) or $av \otimes (1, b)$ in the other one. Now, in either case we get a map on $V \times \mathbb{K}$,... – AnalysisStudent0414 Jan 12 '16 at 10:48
• ...the difference is that in the first case we have the action of $\mathbb{K}^2$ defined as $(a,b).(v.c) = (bv, b^{-1}ac)$, and in the second one instead $(a,b).(v,c) = (av, bc)$. Now, you say only the first (diagonal) hypothesis is possible. Why is that? – AnalysisStudent0414 Jan 12 '16 at 10:54

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 $w \otimes (c, 0)$.