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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!

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  • $\begingroup$ Could you say more about what you find unclear exactly? Is it unclear how the (non)existence of nontrivial alternating bilinear maps would lead to the conclusions? Or is the proof of that (non)existence itself unclear? $\endgroup$ – Eric Wofsey Dec 24 '17 at 1:04
  • $\begingroup$ @EricWofsey I have added some more details about my way of thinking for problems involving exterior algebras and why I don't get their reasoning. Let me know if you need more information. $\endgroup$ – user110320 Dec 24 '17 at 1:18
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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.

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