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).

  • $\begingroup$ 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

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 $F$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.