All Questions
1 question
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: ...