# Tensor product and exterior algebra

I want to show that there is a unique $R$-module isomorphism $M\otimes_{R}N\cong N\otimes_{R}M$, which sends $m\otimes n$ to $n\otimes m$. My idea is to show the map is onto and injective, then how to show its uniqueness?

The second question is that $R$ is an integral domain and $F$ its fraction field, consider $F$ as $R$-module. Show that $\bigwedge^{2}F=0$.

Also, I am confused about the universal properties when I learn the tensor product and exterior algebra, can anyone give me an example of how to calculate the exterior algebra?

Here is another question, the countable direct product of Z is not free. Can anyone give me some hint about how to prove this?

• For your first question define a bilinear map on M x N and use the universal property of the tensor product. This will give you a linear map on the tensor product. You can similarly define a bilinear map on N x M which will give a linear map on the tensor product. You can check they are inverses. To define maps on tensor products, as a matter of course you should define bilinear maps on M x N and then use the UP – Paul Slevin Dec 24 '12 at 22:34
• Just to add to Paul's comment, which is right on, you can't really show that the map $f$ is onto and injective until you've proved it exists. I mean, assuming there exists a linear map $f$ with the desired effect on simple tensors, you can prove that it will be bijective, but ultimately you must resort to the universal property of the tensor product to define it. Uniqueness follows from the fact that every tensor is a sum of simple tensors. – Keenan Kidwell Dec 24 '12 at 23:42

For the first problem: as Paul suggests, you want to use the universal property for tensor products. Define a map $M\times N\to N\otimes M$ by $(m,n)\mapsto n\otimes m$. Since this map is bilinear, there is a unique homomorphism $\varphi\colon M\otimes N\to N\otimes M$ such that $m\otimes n\mapsto n\otimes m$. This is exactly what the universal property of tensor products gives you. Of course, this only gives that the map $\varphi\colon M\otimes N\to N\otimes M$ is a homomorphism, not necessarily an isomorphism. Paul then suggests to do the same thing in the opposite direction: similarly define a map $\psi\colon N\otimes M\to M\otimes N$, again by using the universal property. Now you can check that $\varphi$ and $\psi$ are inverses.
For the second problem: it suffices to check that $x\wedge y = 0$ for all $x,y\in F$. If $x = a/b$ and $y = c/d$ for $a,b,c,d\in R$, then $$x\wedge y = \frac{a}{b}\wedge\frac{c}{d} = \frac{ad}{bd}\wedge\frac{cb}{bd}.$$ Since one can move scalars in $R$ across the wedge, this is equal to $$\frac{cb}{bd}\wedge\frac{ad}{bd} = \frac{c}{d}\wedge\frac{a}{b} = y\wedge x.$$ On the other hand, $x\wedge y = -(y\wedge x)$. The only way $x\wedge y = y\wedge x$ and $x\wedge y = -y\wedge x$ is if $x\wedge y = 0$. (Here I'm assuming the characteristic of $R$ is not $2$).
• $x\wedge y=abcd[(1/bd)\wedge(1/bd)]=0$ – user26857 Dec 24 '12 at 23:53