# Is there a simple way to understand the tensor of group rings?

For example, if I had a group $G$ and a subgroup $H$, and the field $F$, then the tensor $FG \otimes FH$ is just $FG$.

Also why do tensors have subscripts? I believe this tensor above should have the subscript $FH$, but I am not sure why. Could someone give a better explanation then "you just have tensor over something is contained by both factors of the tensor."

Thanksa

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Well, you can tensor over whatever you like. If we simply tensor over $F$ then you probably won't have $FG \otimes_F FH \simeq FH$ in the case that you describe. Are you using this to construct an induced representation? – Dylan Moreland Apr 25 '12 at 15:51

As Dylan Moreland points out, if you are doing representation theory then you given an $F[H]$ module $V$ then you can incude this up to an $F[G]$ module by simply taking $F[G]\otimes_{F[H]}V$. Of course, $F[G]$ is naturally an $F[H]$ module via the inclusion $F[H]\hookrightarrow F[G]$.

The reason why this seems likely, from context, is that $V\otimes_{F[H]}F[H]\cong V$ as $F[H]$-modules for any $F[H]$-module $V$. This would explain your equation by taking $V=F[G]$.

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Suppose $A$ and $B$ are resp. right and left $R$-modules and $S$ is a subring of $R$. Tensoring,

$$A\times B\to A\otimes_S B: ~~ (u,v)\mapsto u\otimes v$$

will be an $S$-bilinear map, in the sense that it is additive in both $u$ and $v$ and additionally

$$(us)\otimes v = u\otimes (sv)$$

for all $s\in S$. However, it is not constructed to be $R$-bilinear as well, so for $r\in R,\not\in S$ we could have for instance that $(ur)\otimes v=u\otimes (rv)$ inside $A\otimes_R B$ but not inside $A\otimes_S B$; this is why the subscript, the ring being tensored over, is important to understanding the structure of the tensor product.

Thus as Dylan and Alex note, if $S\subsetneq R$ is a proper subring, there's an inclusion $S\hookrightarrow R$, and for all pure tensors in $R\otimes_R S$ we have $u\otimes v=uv\otimes1_R$ because of $S$-bilinearity and $1_R\in S$, and further that linear combinations of pure tensors will also result in $\Sigma\otimes1_R$ by additivity, which means the tensor product is in fact $\,\cong R$ (all elements and operations are the same as in the latter, there just happens to be a formal "$\otimes1$" at the end of everything in the former). Same for $R\otimes_S S$.

But this reasoning does not obtain in $R\otimes_T S$, for example, which can be bigger than just $R$ where $T\subsetneq S$ is even smaller than $S$. The reason for saying "bigger" and "smaller" here: the pure tensors are like $A\times B$ but with various elements identified (equated with each other artificially), the number of which is proportionate to the tensored-over ring. The less you have in this ring, the less elements you identify, the bigger the product stays.

This is just loose, figurative thinking though. We can be more rigorous. Take $A,B,R,S$ as before (so $S$ is a proper subring of $R$) and define the map, generated by the images on pure tensors,

$$\varphi: ~ A\otimes_S B\to A\otimes_R B: ~~ u\otimes v\mapsto u\otimes v.$$

It extends to all of $A\otimes_SB$ by $S$-linearity. Check that it is indeed well-defined (this amounts to the identification process of $\otimes_S$'s construction being part of that in $\otimes_R$'s.) If $r\in R$ but $\not\in S$ we observe that generally $ur\otimes v=u\otimes rv$ is not necessarily true in $A\otimes_SB$ but is always true in $A\otimes_RB$, hence $\varphi$ is surjective but can have nontrivial kernel. Indeed, $\ker\varphi$ is generated by $\{ ur\otimes v-u\otimes rv\}_{r\in R}$.

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