How to prove the existence of a bilinear map $B$ extending a given function $f$ on pairs of basis elements

Question

Let $R$ be a ring and $M,N,A$ be free $R$-modules. Let $\mathscr{B}_M,\mathscr{B}_N$ be bases for $M$ and $N$ respectively, and let $f:\mathscr{B}_M\times\mathscr{B}_N\to A$ be an arbitrary map. Then (I am told) there is a unique bilinear transformation $B:M\times N\to A$ which extends $f$. I'm able to show uniqueness, but existence is a little harder. Since $\mathscr{B}_M\times\mathscr{B}_N$ is not a basis of $M\times N$, there is not necessarily a unique representation of each element $m\in M\times N$ as a linear combination of elements of $\mathscr{B}_M\times\mathscr{B}_N$. Thus we can't use the same technique in as the equivalent proof for ordinary linear transformations.

Answer 1

There are various ways of seeing that, with various levels of abstraction, but they all amount to essentially the same thing.

Here's the low-tech version : Take $x\in M, y\in N$ and decompose $x= \sum_i x_i e_i, y = \sum_j y_j f_j$ where $(e_i) $ is the basis $B_M$ of $M$ and $(f_j)$ similarly with $N$.

Then any bilinear map $B$ agreeing with $f$ must satisfy $B(x,y) = \sum_i \sum_j x_i y_j B(e_i,f_j)= \sum_i\sum_j x_iy_j f(e_i,f_j)$.

Why not take this as a definition of $B$ ? $x_i, y_j$ are well defined because $B_M,B_N$ are bases, and $f(e_i,f_j)$ is fixed as well.

So let $p_i$ denote the map defined on $M$ by $x= \sum_k x_k e_k \mapsto x_i$ and $q_j$ defined on $N$ by $y= \sum_k y_kf_k \mapsto y_j$; and define $B := \sum_{i,j} p_iq_j f(e_i,f_j)$.

You then check that this is indeed bilinear, and coincides with $f$ on the bases.

The point is that $B_M\times B_N$ doesn't span $M\times N$ as a module, but we don't care about that, because the definition of a bilinear map $M\times N \to A$ doesn't rely on the module structure on $M\times N$, rather it relies on the structures of $M,N$ independently.

The way to see this more abstractly is to define tensor products which will make clearer how $M,N$ are "independent" somehow (but not completely) in bilinear maps $M\times N \to A$.

Here's a slightly higher tech version, but still without tensor products (because I think you don't know what these are ?) :

a map $f: B_M\times B_N\to A$ gives you, by currying, a map $\tilde{f} : B_M\to F(B_N, A)$ where $F(X,Y)$ is the set of functions $X\to Y$. $\tilde{f}$ is defined by $\tilde{f}(x) = (y\mapsto f(x,y))$

Aha ! But $B_N$ is a basis of $N$ and $A$ is a module, so linear maps $N\to A$ are entirely determined by where they send $B_N$ : in other words there is a bijection $F(B_N,A) \to \hom_R(N,A)$

So in fact, we can see $\tilde{f}$ as a map $f' : B_M\to \hom_R(N,A)$. But now we can play the same trick again : $B_M$ is a basis for $M$, so this map $f'$ yields an $R$-linear morphism $M\to \hom_R(N,A)$.

Now you can "uncurry" this and get $\overline{f} : M\times N\to A$. Now you unravel the definitions to see :

1- this $\overline{f}$ is bilinear (this will come from the fact that $M\to \hom_R(N,A)$ is linear and lands in $\hom_R(N,A)$, not just $F(N,A)$ )

2- it does the correct thing on $B_M\times B_N$ (this will just follow from how everything was defined)

This approach may seem more complicated but in fact you can see that we have done less computations, and relied on a theorem we already knew (we can extend maps on a basis to a linear map on the whole module); so what's happening is conceptually clearer, but also if you're comfortable with this currying process then you can cut down on the number of words that I wrote immensely and in fact get a much shorter, cleaner proof, where you see that essentially what's happening is that we're just using the similar theorem for linear maps