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?
 A: 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 $2$). 
