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The questions I'm trying to answer is as follows (all rings are unital):

Let $_{S}B_{T}$ be an $(S,T)$-bimodule, and let $R = S^{op} \otimes_\mathbb{Z} T$. Show that $B$ can be made into a right $R$-module by defining $b(s\otimes t) := sbt$.

I've already shown that three of the four properties required of a right $R$-module hold. However, I'm stuck on the fourth property, namely that $$ b(s_1 \otimes t_1 + s_2 \otimes t_2) = b(s_1 \otimes t_1) + b(s_2 \otimes t_2).$$ Expanding the left hand side gives $$ b(s_1 \otimes t_1 + s_2 \otimes t_2) = b((s_1 + s_2) \otimes (t_1 + t_2) - (s_1 \otimes t_1) - (s_2 \otimes t_2)),$$ whilst expanding the right hand side gives $$ b(s_1 \otimes t_1)+b(s_2 \otimes t_2) = b((s_1 + s_2) \otimes (t_1 + t_2)) - b(s_1 \otimes t_1) - b(s_2 \otimes t_2)).$$ The expressions on the right hand side of the above two equations look very similar, but I don't see how we can go between them by only making use of the given right module action. What am I missing?

Watson
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lokodiz
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1 Answers1

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This is a misconception in the definition of the module structure.

Given $b \in B$, one defines a homomorphism of abelian groups $S \otimes T \to B$ via the universal property of the tensor product, mapping $s \otimes t \mapsto sbt$. Then, one defines the result of $r$ to be $br$. We have $b(r+r')=br+br'$ by construction, namely $r \mapsto br$ really is that homomorphism $S \otimes T \to B$.

General remainder: 1) Tensor products are defined via their universal property. 2) Tensor products don't just consist of pure tensors.

Martin Brandenburg
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  • Just to clarify, should we have, by definition, that $b(\Sigma_i (s_i \otimes t_i)) = \Sigma_i b(s_i \otimes t_i)$ for the definition to even make sense (since an arbitrary tensor is a sum of simple tensors). – lokodiz Sep 23 '14 at 13:33
  • This is not a proper definition - this equations follows from the definition via the universal property. – Martin Brandenburg Sep 23 '14 at 13:51