1

Say $A,B,C$ are three $R$-algebras. And given two homomorphisms of $R$-algebras, $f:A\to C$ and $g:B\to C$. We could induce the "product homomorphism" $f\times g$ which is an $R$-bilinear map $A\times B\to C,(a,b)\mapsto f(a)g(b)$ by forgetting the algebra structure.

Is there a natural way to do this merely in the $R$-module category?

CO2
  • 1,373
  • You can take any bilinear map $C\times C\to C$ and feed it with $f$ and $g$. But there is no "natural" bilinear map on $R$-modules. For $R$-algebras, the multiplication is a natural choice for such a bilinear map. – Vercassivelaunos Dec 29 '20 at 12:08
  • Thank you. So, somehow, the tensor product in $R$-algebra and $R$-module are not quite the same as categorial objects? The tensor product in $R$-algebra is the pushout. What is the tensor product in $R$-module? – CO2 Dec 29 '20 at 12:16
  • 1
    The tensor product of modules is a module. The tensor product of algebras is an algebra. Also, viewed as a module, the tensor product of two algebras is the tensor product of their underlying modules. The tensor product of modules is essentially the more basic construction, which can be turned into an algebra if the modules which we tensor have an algebra structure. I'm out of my depth where category theory is concerned, but I think the forgetful functor $F:\textbf{Alg}_R\longrightarrow\textbf{Mod}_R$ satisfies $F(A\otimes B)=F(A)\otimes F(B)$. – Vercassivelaunos Dec 29 '20 at 15:34
  • @CO2: the tensor product in $R$-algebras is not the pushout. That only applies in the commutative case. The tensor product of $R$-modules is the universal object for bilinear maps and that's all. – Qiaochu Yuan Dec 30 '20 at 02:28

1 Answers1

2

No. For modules the best you get is that if $f : M \to N$ is a map of right $R$-modules and $g : P \to Q$ is a map of left $R$-modules then they induce a tensor product map

$$f \otimes g : M \otimes_R P \to N \otimes_R Q.$$

In the algebra case an $R$-algebra $C$ admits an $R$-bilinear map $C \otimes_R C \to C$, namely the multiplication, and you compose by this map to get the map you're looking at. If $C$ is just a module it is of course in no way equipped with such a map.

Qiaochu Yuan
  • 419,620