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I am working on old qualifying problems involving tensor products. I am stuck on a statement about invertible elements in an exterior algebra and was wondering if this was a well known fact in a book somewhere. I think most of this notation is standard from dummite and foote but the notes I have been using are slightly different than what I have seen in textbooks so far.

Let $V$ be a finite dimensional vector space over a field $F$.

Let $T(V) = \oplus_{k=0}^{\infty} T^{k}(V)$ where $T^k(V) = V \otimes V \otimes \ldots \otimes V$ is tensor product of $k$ modules.

Let $\wedge V $ denote the exterior algebra of the $F$-module $V$, that is the quotient of the tensor algebra $T(V)$ by the ideal $A(M)$ generated by all $v \otimes v$ for $v \in V$.

Let $x \in \wedge V$. So that $x = \sum_{k\geq 0} x_k$ where each $x \in \wedge^k V = T^k(V)/A^k(V)$

How do you prove that $x$ is invertible if and only if $x_0 \neq 0$

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Pardon my not knowing anything really about exterior algebras, but what does it mean for an $x$ to be invertible? Anticommutativity precludes an identity element wrt the wedge product, so it must mean something else.. –  anon Oct 19 '11 at 23:20
    
@anon: T^0(V) is just a copy of the field, so F ≤ ⋀V. Maybe the keyword is "graded commutative". (v∧w)∧x = x∧(v∧w) since that is two "swaps", (v∧w)∧x = −v∧x∧w = −−x∧v∧w. –  Jack Schmidt Oct 19 '11 at 23:35

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The set $I = \oplus_{k=1}^\infty T^k(V)$ is an ideal of $T(V)$ containing $A(V)$. I believe $\bigwedge V$ is a finite dimensional local algebra with unique maximal ideal $I/A(V)$. In particular, every element of $I/A(V)$ is nilpotent. In fact if $\dim(V) = n$, then $x^{n+1} = 0$ for every $x \in I$. In particular, the geometric series $\frac{1}{1-x} = \sum_{k=0}^\infty x^k$ converges (is a finite sum plus a bunch of 0s) for every $x \in I$, and shows that such elements are invertible. Multiplying by a nonzero element of the field, gives the result.

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