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The following is based on a follow up to some of the comments from this post Exterior powers of a module contained in a field of fractions

Let $I$ be an integral domain and let $\operatorname{Frac}(I)$ denote its field of fractions http://en.wikipedia.org/wiki/Field_of_fractions.

let $\wedge^k M$ be the $k$-th exterior power of $M$ that is $T^k(V)/A^k(V)$ where $A(M)$ is the ideal generated by all $m \otimes m$ for $m \in M$ and $T^k(M) = M \otimes M \otimes \cdots \otimes M$ is tensor product of $k$ modules.

How do we show $\wedge^2 \operatorname{Frac}(I)$ is $0$?

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    $\begingroup$ Are you considering $\text{Frac}(I)$ as a vector space over itself? Or an $I$-module? $\endgroup$ – Henning Makholm Oct 21 '11 at 2:54
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I guess you mean as an $I$-module because as a vector space over itself it would be trivial: in general, since $r \wedge s = rs(1 \wedge 1) = 0$ we have that $\bigwedge^2 R = 0$ when seen as a module over itself for any ring $R$ (with unit 1).

Now, let $K$ be the fraction field of $I$. If $a,b,c,d$ are all in $I$, each non-zero. Then $a/b \wedge c/d = da/bd \wedge cb/bd = dbac(1/bd \wedge 1/bd)=0$. This shows that $\bigwedge^2 K = 0$.

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