Northcott Multilinear Algebra Universal Property Proof

Solution 1:

The existence of $\lambda$ and $\lambda'$ are furnished by the maps $\phi'$ and $\phi$ via the universal property; that is, we are taking $N = M', M$ respectively. To spell this out further, the map

$$\phi' \colon M_{1} \times \cdots \times M_{p} \to M'$$

is bilinear by assumption, and so the universal property of the pair $(M, \phi)$ says that there is exactly one morphism $\lambda \colon M \to M'$ such that $\lambda \circ \phi = \phi'$. You obtain the morphism $\lambda' \colon M' \to M$ analogously by switching the roles of $M$ and $M'$.

This is a typical situation that arises in universal problems: the data of any two solutions contains information which allows you to relate the two solutions through the universal property. In this case, the crucial piece of information is this bilinear map $\phi \colon M_{1} \times \cdots \times M_{p} \to M$ which allows you to relate any two solutions.

Solution 2:

This answer has the same content as the answer by Alex Wertheim, just explicated in slightly more detail.

In my post I misunderstood the universal problem. I thought the universal problem supplied a $(M, \phi)$ given a particular map $\psi$. But, more powerfully, a solution to the universal problem (for multilinear maps from $M_1\times\ldots\times M_p$) is a pair $(M, \phi)$ with $M$ an R-module and $\phi:M_1\times\ldots\times M_p\rightarrow M$ multilinear, such that for ANY multilinear $\psi:M_1\times\ldots \times M_p\rightarrow N$ there exists a unique R-module homomorphism $h:M\rightarrow N$ such that $h\circ \phi = \psi$.

In this case we have the following. Suppose we have multilinear $\psi:M_1\times\ldots \times M_p \rightarrow N$ and both $(M, \phi)$ and $(M', \phi')$ solve the universal problem. $\phi'$ is a multilinear map from $M_1\times\ldots\times M_p$ into $M'$ so, because $(M, \phi)$ solves the univeral problem from $M_1 \times\ldots \times M_p$, $\phi'$ can be uniquely decomposed as

$$ \lambda \circ \phi = \phi' $$

With $\lambda:M\rightarrow M'$ a R-module homomorphism. Likewise for $\phi$ so that

$$ \lambda'\circ\phi' = \phi $$ with $\lambda':M'\rightarrow M$ also an R-module homomorphism.