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?

  • $\begingroup$ 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 $\mathbb K\times\mathbb K$? $\endgroup$ – Justpassingby Jan 11 '16 at 19:34
  • $\begingroup$ $\mathbb{K} \to \mathbb{K} \times \mathbb{K}$ 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? $\endgroup$ – AnalysisStudent0414 Jan 11 '16 at 20:34
  • $\begingroup$ 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. $\endgroup$ – Justpassingby Jan 12 '16 at 6:33
  • $\begingroup$ 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 $V \times \mathbb{K}$,... $\endgroup$ – AnalysisStudent0414 Jan 12 '16 at 10:48
  • $\begingroup$ ...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? $\endgroup$ – AnalysisStudent0414 Jan 12 '16 at 10:54

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 $w \otimes (c, 0)$.

  • 1
    $\begingroup$ This is exactly the answer i needed. Thank you very much Marc $\endgroup$ – AnalysisStudent0414 Jan 15 '16 at 11:28

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