Sign up ×
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.

I am working some problems that can potentially be on a qualifying exam about tensor algebras and have come across some questions about field of fractions which is something I have not seen for a while and so I have not really been able to work through all the appropriate definitions to arrive at the correct proof. Any help would be greatly appreciated.

Let $I$ be an integral domain and let $Frac(I)$ denote its 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 $T^k(M) = M \otimes M \otimes \ldots \otimes M$ is tensor product of $k$ modules.

Consider an $I$-module $M \subset Frac(I)$.

How do we show $\wedge^k M$ is a torsion module for $k \geq 2$?

I think it is clear that the exterior power should be zero for $k <2$ but I am still not sure this is trivial after putting all the definitions together.

share|cite|improve this question
If M = Frac(I) is a one-dimensional vector space, then the 0th and 1st exterior powers of M are isomorphic to M, I believe. – Jack Schmidt Oct 19 '11 at 23:57

1 Answer 1

up vote 2 down vote accepted

Let K be the field of fractions of the integral domain I, and let IMK. Since tensor product is associative, K ⊗ ⋀(M) = ⋀(KM) = ⋀(K) = KK, which is zero for k ≥ 2.

An I-module N such that KN = 0 is (by definition in some areas) a torsion I-module. If in = 0 and i ≠ 0, then 1 ⊗ n = 1/iin = 0. The other inclusion follows from the flatness of K.

share|cite|improve this answer
Could you explain the first two lines of this answer, please? – yaa09d Nov 17 '11 at 2:11

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.