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6 votes
1 answer
301 views

Is the (super-)symmetric power of the exterior algebra free?

Let $V$ be a vector space over $k$ of dimension $m$. (I'm only interested in the case $k=\mathbb{Q}$.) Let $R:=\Lambda^*V$ be the exterior algebra. It carries the structure of a supercommutative ring: ...
Nicolas Malebranche's user avatar
15 votes
2 answers
2k views

Clifford PBW theorem for quadratic form

$\DeclareMathOperator\Cl{Cl}$Update Feb 3 '12: now with a question 2 which is much more elementary (and should be well-known!). Let $k$ be a commutative ring with $1$. Let $L$ be a $k$-module, and $g:...
darij grinberg's user avatar