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Let $A$ be a commutative unital ring with field of fractions $K$. Regarding $K$ as an $A$-module, we have that for each $A$-module $M$, $K\otimes_A M$ is an $A$-module and therefore an abelian group. I need to make sense of $K\otimes_A M$ as a vector space over $K$, a bit like how the complexification of a real vector space is a complex vector space constructed using tensor products.

So it must be necessary to construct a map $K\times(K\otimes_A M)\to K\otimes_A M$. I know that maps of tensor products are usually constructed from bilinear ones on the usual product. We have an $A$-scalar multiplication $A\times(K\otimes_A M)\to K\otimes_A M$ and an $A$-bilinear map $(a,m)\mapsto a\otimes_A m:A\times M\to A\otimes_A M$, but where do I go from here?

user829347
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  • Try $a\cdot (b\otimes m)=(ab)\otimes m$ for $a,b\in K$ and $m\in M$ – Joshua Tilley Jul 18 '22 at 17:00
  • @JoshuaTilley Yes, this is what the multiplication should be, but I need to prove that there is a unique map with this property i.e. construct it from a bilinear one – user829347 Jul 18 '22 at 17:03
  • I don't understand. What is the problem? Surely you just need to show that this is a $K$ vector space under this scaling operation? – Joshua Tilley Jul 18 '22 at 17:07
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    The equivalence you speak of is between linear maps whose domain is a tensor product $A\otimes B$, and bilinear maps whose domain is $A\times B$. In this case we have a bilinear map with domain $K\times (K\otimes_A M)$. If you like, this is equivalent to a linear map with domain $K\otimes_K (K\otimes_A M)$ – Joshua Tilley Jul 18 '22 at 17:19

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I will write simply “ring” for a unital (but not necessarily commutative) ring. See this for the more general notion of tensor product of modules over a ring.

Given an (additive) abelian group $A$, and a ring $S$, a map $S \times A \to A$ that makes $A$ a $S$-module is “equivalent” to a ring homomorphism $S \to \operatorname{End}_\mathbb{Z}(A)$. More precisely, given $m \colon S \times A \to A $, define $S \to \operatorname{End}_\mathbb{Z}(A)$ by $s \mapsto m(s,\_)$; and given $r \colon S \to \operatorname{End}_\mathbb{Z}(A)$, define $S \times A \to A$ by $(s,a) \mapsto r(s)(a)$. It is easy to check that these two correspondences are mutually inverses.

Using the above, I will prove that if $R$ and $S$ are two rings, $M$ a $(S,R)$-bimodule, and $N$ a left $R$-module, then the abelian group $M \otimes_R N$ is a left $S$-module such that $$ \forall s \in S,\, \forall m \in M,\, \forall n \in N \quad s(m \otimes n) = (sm) \otimes n. $$

Indeed, given $s \in S$, the $R$-balanced mapping $$ \mu_s \colon M \times N \to M \otimes_R N, \quad (m,n) \mapsto (sm) \otimes n $$ induces a group homomorphism $\tilde \mu_s \colon M \otimes_R N \to M \otimes_R N$ such that $\tilde \mu_s \circ \otimes = \mu_s$. It is easy to prove that the mapping $s \mapsto \tilde \mu_s$ is a ring homomorphism $S \to \operatorname{End}_\mathbb{Z}(M \otimes_R N)$, as desired.

azif00
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