Exterior product of $\Bbb Z[x,y]$

Question

Let $R$ be the polynomial ring $\Bbb Z[x,y]$ in the variables $x,y$. If $M=R$ (so we are considering the $R$-module $R$) then $\wedge^2 (M)=0$. This was an example in Dummit and Foote's Abstract Algebra, pg. 449. Can someone explain to me why $\wedge^2 (M)=0$? Could you explicitly calculate some exterior products as an example? I'm having a tough time understanding this.

Answer 1

It's easy to see that $\wedge^n R\cong \bigotimes^nR/I$, where $I=\{r_1\otimes...\otimes r_n|r_i=r_j $for $ $some $ i\neq j $}. Now $\bigotimes^2 R=(r_1\otimes r_2|r_i\in R)=(r_1r_21\otimes1|r_i\in R)\subset I$. This means that $\wedge^2 R\cong \bigotimes^2R/I=0$