# Modules $(K\otimes A)\otimes_{K}\left(K\otimes B\right)$ and $K\otimes(A\otimes B)$ isomorphic?

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).

• 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. – K B Dave May 28 at 10:00
• @KBDave Thank you for the link. It is very useful. – drhab May 28 at 17:57

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 $$F$$.

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