Exterior algebra of polynomial ring with 2 variables

Question

I am reading Dummit and Foote book on Abstract Algebra, particularly the section of the exterior algebras the example after corollary $37$. They give the example that if $R$ is the polynomial ring $\mathbb{Z}[x,y]$ in the variables $x$ and $y$ and if $M = R$, then $\bigwedge ^2(M) = 0$. But then if we consider that $M = I$ is the ideal generated by $(x,y)$ generated by $x$ and $y$ in $R$, then $I \wedge I\neq 0$. I am struggling a lot to see these conclusions and the explantion they give (that there doesn't exist a nontrivial alternating bilinear map on $R \times R$ and that we can construct a nontrivial alternating bilinear map on $I \times I$, respectively) is not quite clear to me yet. Particularly, I find it unclear why the authors are thinking about alternating bilinear maps on $R \times R$ (or $I \times I$) respectively. For instance, the way I think of some problems on exterior algebras and their existence is by thinking thinking of the following commutative diagram:

$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} M \times M & \ra{\iota} & \bigwedge^2(M) \\ & \searrow{g_1} & \da{g_2} \\ & & A\\ \end{array} $

(Sorry for the awful diagram, I'm not sure how to make them better). Then, just showing that $g_1$ is an alternating bilinear map shows the existence of map $g_2$ by the universal property of exterior algebras. However, in this case I don't see why they conclude tehere are no alternating bilinear maps on $R \times R$ (and the same happens with $I \times I$.

Does anybody have another way of seeing these conclusions? I would really appreciate any explanations or suggestions for further reading! Thanks!

Answer 1

We don't have to consider alternating bilinear maps to see that $\bigwedge^2 M = 0$. Given a simple alternating tensor $f \wedge g \in \bigwedge^2 M$, we have $f \wedge g = fg (1 \wedge 1) = 0$, since $m \wedge m = 0$ for all $m \in M$. (Note that we are allowed to pull out $f$ and $g$ because we are considering $M=R$ as an $R$-module.) Since every element of $\bigwedge^2 M$ is a sum of simple alternating tensors, this shows that $\bigwedge^2 M = 0$.

The universal property of the exterior power (Theorem 36 (2)) implies that there is a bijection between alternating bilinear maps $M \times M \to N$ and $R$-module homomorphisms $\bigwedge^2 M \to N$ where $N$ is any $R$-module. But the only $R$-module homomorphism $\bigwedge^2 M = 0 \to N$ is the zero homomorphism, so by the bijection above the only alternating bilinear map $M \times M \to N$ is the zero map.

This universal property also allows us to show that $\bigwedge^2 I \neq 0$. If $\bigwedge^2 I$ were $0$, then by the same reasoning as above, for any $R$-module $N$ the only alternating bilinear map $I \times I \to N$ would be the zero map. Thus to show that $\bigwedge^2 I \neq 0$, it suffices to exhibit an $R$-module $N$ and a nonzero alternating bilinear map $I \times I \to N$. And that is exactly what the authors do for $N = R/I \cong \mathbb{Z}$.

If you're looking for further reading, I recommend these notes by Keith Conrad.