# Exterior powers of a module contained in a field of fractions

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 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 $T^k(M) = M \otimes M \otimes \ldots \otimes M$ 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.

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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

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.

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