Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $0$?

share|improve this question
Are you considering $\text{Frac}(I)$ as a vector space over itself? Or an $I$-module? –  Henning Makholm Oct 21 '11 at 2:54

1 Answer 1

up vote 4 down vote accepted

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 $\bigwedge^2 K = 0$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.