# The Unapologetic Mathematician

## Interactions between hom, tensor product, and direct sum

We now have three ways of putting modules together: the abelian group $\hom_{R-{\rm mod}}(A,B)$ of left $R$-module homomorphisms, the tensor product $A\otimes_RB$ of a right $R$-module $A$ and a left $R$-module $B$, and the direct sum $A\oplus B$ of two left $R$-modules. Today we consider their interactions.

First off, the universal property of direct sums tells us that a homomorphism $A\rightarrow B_1\oplus B_2$ is the same as a pair of homomorphisms, $A\rightarrow B_1$ and $A\rightarrow B_2$. Given the first we can compose it with the projections to get the second, and given the second we can use the universal property to get the first. That is, we can make homomorphisms of abelian groups
$\hom(A,B_1)\leftarrow\hom(A,B_1\oplus B_2)\rightarrow\hom(A,B_2)$
$\hom(A,B_1)\rightarrow\hom(A,B_1\oplus B_2)\leftarrow\hom(A,B_2)$
and we can check that these are the injections and projections of a direct sum of abelian groups! That is:
$\hom(A,B_1\oplus B_2)\cong\hom(A,B_1)\oplus\hom(A,B_2)$
where the direct sum on the left is of left $R$-modules, while the one on the right is of abelian groups. Similarly, we can show that
$\hom(A_1\oplus A_2,B)\cong\hom(A_1,B)\oplus\hom(A_2,B)$
and that these both work for all finite direct sums.

For infinite direct sums it’s a little trickier. An infinite direct sum in the second variable works just the same:
$\hom(A,\bigoplus\limits_{i\in\mathcal{I}}B_i)\cong\bigoplus\limits_{i\in\mathcal{I}}\hom(A,B_i)$
but if it shows up in the first variable we have to use the direct product of abelian groups to get the right universal properties to go through. Try following the above argument yourself to see where the difference is.

Tensor products and direct sums are similar, and don’t even have the same difficulties with infinite sums. After playing with universal properties of direct sums like we did above, we find that
$\left(\bigoplus\limits_{i\in\mathcal{I}}A_i\right)\otimes\left(\bigoplus\limits_{j\in\mathcal{J}}B_j\right)\cong\bigoplus\limits_{(i,j)\in\mathcal{I}\times\mathcal{J}}\left(A_i\otimes B_j\right)$

Where things get really fun, though, is with tensor products and homs. Let’s consider a right $R$-module $A$, a left $R$-module $B$, and an abelian group $C$. The universal property of tensor products tells us that a linear function from $A\otimes_RB$ to $C$ is the same thing as a middle-linear function from $A\times B$ to $C$. Let’s consider such a middle linear function $f$.

If we pick an element $a\in A$ and stick it in the first slot of $f$, we get $f(a,_)$. This is a linear function from $B$ to $C$, so $f(a,_)\in\hom(B,C)$. Notice that we’re using $\hom$ for abelian groups, so $B$ has an extra left $R$-module structure. It gets flipped over, turning $\hom(B,C)$ into a right $R$-module. Now, building $f(a,_)$ out of the element $a\in A$ is a homomorphism of right $R$-modules. That is, given $f$ we can build an element of $\hom_{{\rm mod}-R}(A,\hom(B,C))$.

Even better, any such homomorphism gives rise to a middle-linear function from $A\times B$ to $C$. That is, we have an isomorphism:
$\hom(A\otimes_RB,C)\cong\hom_{{\rm mod}-R}(A,\hom(B,C))$

For bonus points, go back through these interactions and try adding extra module structures to each of the modules we used.

May 1, 2007 Posted by | Ring theory | 3 Comments