# Universal property of tensor products of modules

I know the tensor product of $A$-modules $M$ and $N$ is an $A$-module $T$ with a $A$-billinear map $t:M\times N\to T$ such that if $t':M\times N\to T'$ is a $A$-billinear map, there exists a unique $f:T\to T'$ such that $f\circ t=t'$. My question is: is the map $t: (m,n)\mapsto m\otimes_A n$? If that's not the case, $is$ there an explicit function t?

• Yes, if you use the standard explicit construction of tensor products. Note your question does not make sense until the notation for the elements of $T$ is defined. – cjackal Sep 16 '16 at 1:02
• $T$ is some module, so the notation $m\otimes n$ does not make sense. One {\it defines} $m\otimes n=t(m,n)$ and call $T$ by $M\otimes_A N$. – Mohan Sep 16 '16 at 1:02
• The composition goes the other way. – Pedro Tamaroff Sep 16 '16 at 1:02
• basically I'm trying to prove existence of something satisfying this property – the ring leader Sep 16 '16 at 1:03

## 2 Answers

The construction (and thus the existence) of an $A$-module $T=M\otimes_A N$ satisfying the universality property described in the question can be described in various ways:

• In the general case, (no additional structure is supposed in the $A$-modules $M$, $N$), here is an outline: Let $F_{M\times N}$ be the free $A$-module, generated by the elements of the cartesian product $M\times N$. That is, $F_{M\times N}$ is the free $A$-module with basis $M\times N$. Now consider the $A$-submo-dule $\mathcal{J}_{M\times N}$ of $F_{M\times N}$ generated (as a submodule) by all elements of the form: $$(u,av_1+bv_2)-a(u,v_1)-b(u,v_2) \\ (au_1+bu_2,v)-a(u_1,v)-b(u_2,v)$$ where $u,u_1,u_2\in M$, $v,v_1,v_2\in N$, $a,b\in A$. Then we construct: $$M\otimes_A N=F_{M\times N}\Big/\mathcal{J}_{M\times N}$$ as $A$-modules. (For details on the proofs as to why $M\otimes_A N$ constructed like this, satisfies the desired universality property and is unique up to isomorphism, see for example Associative algebras, R.S.Pierce, ch. 9. However, this is standard material and can be found in lots of other books as well).

• On the other hand, if we impose the extra assumptions, that the $A$-modules $M$, $N$ are also vector spaces and that one knows about the tensor product of vector spaces, then the above construction can be seen a little more simplified: Consider the vector space $M\otimes N$ and its subspace $\mathcal{F}$ generated (as a subspace) by all elements of the form: $$ax\otimes y-x\otimes ay$$ where $x,\in M$, $y\in N$, $a\in A$. Then we have: $$M\otimes_A N=\big(M\otimes N\big)\Big/\mathcal{F}$$ as vector spaces, while the $A$=action on $M\otimes_A N$ is given by $$a\cdot (x\otimes y)=ax\otimes y=x\otimes ay$$

Recall that $A$ is assumed commutative and contains 1. Consider the set of functions $\delta_{m.n} : M \times N \to A$ given by $\delta_{m,n}(m,n) = 1$ and $\delta_{m,n}(x,y) = 0$ for $(x,y) \neq (m,n)$.

It's easy to verify that the set of finite linear combinations $\displaystyle \sum_{i=1}^k a_i \delta_{m_i,n_i}$, where $a_i \in A$ and $(m_i,n_i) \in M \times N$ is an $A$-module (under the usual addition and scalar multiplication of functions) with basis $\{\delta_{m,n}\}_{(m,n) \in M \times N}$. This is a concrete construction of the free $A$ module generated by $M \times N$ invoked by @KonKan.

After that you just mod out by the bilinearity relations.