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Let $K$ denote a commutative ring and let $A$ denote an abelian group.

The tensor product $K\otimes A$ can be looked at as a $K$-module equipped the action determined by: $$r\cdot s\otimes a=\left(rs\right)\otimes a$$

This module is free over abelian group $A$.

If we also have abelian group $B$ then similarly we can construct $K$-modules $K\otimes B$ and $K\otimes\left(A\otimes B\right)$.

My question is:

Are the $K$-modules $\left(K\otimes A\right)\otimes_{K}\left(K\otimes B\right)$ and $K\otimes\left(A\otimes B\right)$ isomorphic in $\mathbf{Mod}\left(K\right)$, and if so how to prove it?


Own effort:

I suspect there is an isomorphism $\left(K\otimes A\right)\otimes_{K}\left(K\otimes B\right)\to K\otimes\left(A\otimes B\right)$ that is determined by: $$\left(r\otimes a\right)\otimes_{K}\left(s\otimes b\right)\mapsto\left(rs\right)\otimes\left(a\otimes b\right)$$

To prove its existence in $\mathbf{Mod}\left(K\right)$ I feel forced to prove the existence of a bilinear function: $$f:\left(K\otimes A\right)\times\left(K\otimes B\right)\to K\otimes\left(A\otimes B\right)$$ and that is the place where I got stuck: if I restrict to stating that $f$ is prescribed by $$\left(r\otimes a,s\otimes b\right)\mapsto\left(rs\right)\otimes\left(a\otimes b\right)$$ then it seems to me that I am not ready yet in the sense that it is not shown that $f$ is well defined. This because I have "defined" $f$ then only on pairs of pure tensors (corresponding bilinear functions are lacking).

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  • $\begingroup$ Yes. $f_!(A)\stackrel{\text{def}}{=}K\otimes A$ constitutes an extension of scalars functor $f:\mathbb{Z}\to K$, $f_!:\mathbf{Mod}\,\mathbb{Z}\to\mathbf{Mod}\,K$, so it's strongly monoidal. $\endgroup$ – K B Dave May 28 at 10:00
  • $\begingroup$ @KBDave Thank you for the link. It is very useful. $\endgroup$ – drhab May 28 at 17:57
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One short way of getting the desired isomorphism is \begin{align*} (K \otimes A) \otimes_K (K \otimes B) &\cong (A \otimes K) \otimes_K (K \otimes B) \\ &\cong A \otimes (K \otimes_K (K \otimes B)) \tag{$\ast$} \\ &\cong A \otimes K \otimes B \\ &\cong K \otimes A \otimes B \\ &\cong K \otimes (A \otimes B) \end{align*} where we use for $(\ast)$ the (generalized) associativity over the tensor product.


To explicitely construct the mutually inverse homomorphisms \begin{align*} F &\colon (K \otimes A) \otimes_K (K \otimes B) \to K \otimes_K (A \otimes B) \,, \\ G &\colon K \otimes_K (A \otimes B) \to (K \otimes A) \otimes_K (K \otimes B) \end{align*} we can proceed like you suspect.

To construct the bilinear map \begin{align*} f \colon (K \otimes A) \times (K \otimes B) &\to K \otimes (A \otimes B) \,, \\ (\lambda \otimes a, \mu \otimes b) &\mapsto (\lambda \mu) \otimes (a \otimes b) \end{align*} we start with the $\mathbb{Z}$-multilinear map \begin{align*} f'' \colon K \times A \times K \times B &\to K \otimes (A \otimes B) \,, \\ (\lambda, a, \mu, b) &\mapsto (\lambda \mu) \otimes (a \otimes b) \,. \end{align*} For fixed $(\lambda, a) \in K \times A$ we get a $\mathbb{Z}$-bilinear map \begin{align*} f''_{(\lambda,a)} \colon K \times B &\to K \otimes (A \otimes B) \,, \\ (\mu, b) &\mapsto (\lambda \mu) \otimes (a \otimes b) \end{align*} and thus an induced $\mathbb{Z}$-linear map \begin{align*} f'_{(\lambda,a)} \colon K \otimes B &\to K \otimes (A \otimes B) \,, \\ \mu \otimes b &\mapsto (\lambda \mu) \otimes (a \otimes b) \,. \end{align*} The mapping $(\lambda,a) \mapsto f''_{(\lambda,a)}$ is itself $\mathbb{Z}$-bilinear by the $\mathbb{Z}$-multilinearity of $f''$, whence the mapping $(\lambda,a) \mapsto f'_{(\lambda,a)}$ is $\mathbb{Z}$-linear. The resulting map \begin{align*} f' \colon K \times A \times (K \otimes B) &\to K \otimes (A \otimes B) \,, \\ (\lambda, a, \mu \otimes b) &\mapsto f'_{(\lambda,a)}(\mu \otimes b) = (\lambda \mu) \otimes (a \otimes b) \,. \end{align*} is therefore $\mathbb{Z}$-trilinear. For fixed $t \in K \otimes B$ we now get again a $\mathbb{Z}$-bilinear map \begin{align*} f'_{t} \colon K \times A &\to K \otimes (A \otimes B) \,, \\ (\lambda, a) &\mapsto f'(\lambda, a, t) \,. \end{align*} that results again in a $\mathbb{Z}$-linear map \begin{align*} f_t \colon K \otimes A &\to K \otimes (A \otimes B) \,, \\ \lambda \otimes a &\mapsto f'(\lambda, a, t) \,. \end{align*} It follows from the $\mathbb{Z}$-trilinearity of $f'$ that the mapping $t \mapsto f'_t$ is $\mathbb{Z}$-linear, whence the mapping $t \mapsto f_t$ is $\mathbb{Z}$-linear. This means that the now resulting map \begin{align*} f \colon (K \otimes A) \times (K \otimes B) &\to K \otimes (A \otimes B) \,, \\ (s, t) &\mapsto f_t(u) \end{align*} is $\mathbb{Z}$-bilinear. For $s = \lambda \otimes a$ and $t = \mu \otimes b$ we have by construction \begin{align*} f(\lambda \otimes a, \mu \otimes b) &= f_{\mu \otimes b}(\lambda \otimes a) \\ &= f'(\lambda, a, \mu \otimes b) \\ &= f'_{(\lambda, a)}(\mu \otimes b) \\ &= f''_{(\lambda, a)}(\mu, b) \\ &= (\lambda \mu) \otimes (a \otimes b) \,. \end{align*} Now we need to check that the $\mathbb{Z}$-bilinear map $f$ is already $K$-bilinear. For this we still need that $$ f(\kappa \cdot s, t) = \kappa \cdot f(s,t) = f(s, \kappa \cdot t) $$ for all $\kappa \in K$ and $s \in K \otimes A$, $t \in K \otimes B$. We may assume that $s$,$t$ are simple tensors $s = \lambda \otimes a$ and $t = \mu \otimes b$ and then \begin{align*} f(\kappa \cdot (\lambda \otimes a), (\mu \otimes b)) &= f((\kappa \lambda) \otimes a, \mu \otimes b) \\ &= (\kappa \lambda \mu) \otimes (a \otimes b) \\ &= \kappa \cdot ((\lambda \mu) \otimes (a \otimes b)) \\ &= \kappa \cdot f(\lambda \otimes a, \mu \otimes b) \,. \end{align*} The other equation can be shown in the same way. It follows that $f$ induces a $K$-linear map \begin{align*} F \colon (K \otimes A) \otimes_K (K \otimes B) &\to K \otimes (A \otimes B) \,, \\ (\lambda \otimes a, \mu \otimes b) &\mapsto (\lambda \mu) \otimes (a \otimes b) \,. \end{align*}

To construct the inverse $G$ of $F$ we can play the same game: We start with a $\mathbb{Z}$-trilinear map \begin{align*} g'' \colon K \times A \times B &\to (K \otimes A) \otimes_K (K \otimes B) \,, \\ (\lambda, a, b) &\mapsto (\lambda \otimes a) \otimes (1 \otimes b) \,. \end{align*} (Note that it dosen’t matter in which tensor factor $K \otimes A$ or $K \otimes B$ we put $\lambda$.) Then we get \begin{align*} g'_{\lambda} &\colon A \times B \to (K \otimes A) \otimes_K (K \otimes B) \,, \\ g_{\lambda} &\colon A \otimes B \to (K \otimes A) \otimes_K (K \otimes B) \,, \\ g &\colon K \times (A \otimes B) \to (K \otimes A) \otimes_K (K \otimes B) \,, \\ G &\colon K \otimes (A \otimes B) \to (K \otimes A) \otimes_K (K \otimes B) \,, \end{align*} It can then be checked on simple tensors that the maps $F$ and $G$ are mutually inverse.


Another way to construct $F$ is by the tensor-$\operatorname{Hom}$-adjunction: The map $$ A \times B \to A \otimes B \,, \quad (a,b) \mapsto ab $$ is $\mathbb{Z}$-bilinear and hence corresponds a $\mathbb{Z}$-linear map $$ A \to \operatorname{Hom}_{\mathbb{Z}}(B, A \otimes B) \,. $$ Every homomorphism $B \to A \otimes B$ can be extended to a $K$-linear map $K \otimes B \to K \otimes (A \otimes B)$, giving a $\mathbb{Z}$-linear map \begin{align*} \operatorname{Hom}_{\mathbb{Z}}(B, A \otimes B) &\to \operatorname{Hom}_K(K \otimes B, K \otimes (A \otimes B)) \,, \\ h &\mapsto \mathrm{id}_K \otimes h \,. \end{align*} By composing the above two $\mathbb{Z}$-linear maps we get a $\mathbb{Z}$-linear map $$ A \to \operatorname{Hom}_K(K \otimes B, K \otimes (A \otimes B)) \,. $$ The right hand side is a $K$-module, so by extension of scalars we get a $K$-linear map $$ K \otimes A \to \operatorname{Hom}_K(K \otimes B, K \otimes (A \otimes B)) \,. $$ This $K$-linear map corresponds to a $K$-bilinear map $$ (K \otimes A) \times (K \otimes B) \to K \otimes (A \otimes B) \,, $$ which is precisely $f$, and hence to a $K$-linear map $$ (K \otimes A) \otimes_K (K \otimes B) \to K \otimes (A \otimes B) \,, $$ which is precisely $F$.

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  • $\begingroup$ This looks okay at first sight. Thank you. I will start with second sight during the day, which will take some time. $\endgroup$ – drhab May 28 at 10:57

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