Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This precise question grew out from the question whether a smooth commutative $k$-algebra (char($k$)=$0$) is always cofibrant as a non-positively graded commutative differential graded co-chain $k$-algebra. I think the answer is no (while the converse is true). For this I'd need at least one example of the following situation: a smooth commutative $k$-algebra $R$, a commutative $k$-algebra $S$, together with a NON square-zero ideal $I\subseteq S$, and a $k$-algebra morphism $R\rightarrow S/I$ that does NOT admit a lift to a $k$-algebra morphism $R\rightarrow S$. (Of course, any such example should have $R$ not a polynomial $k$-algebra.) I know it gotta be easy but I was not able to find out one. Thanks for any answer, Sereza.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Let me try to give a quick answer (but double check it). Take $R=k[x,y]_{xy}$, $S=k[x,y]$ and $I=(xy-1)$, with the obvious map $R\rightarrow S/I$. It's easy to prove that no lifting exists (any map $R\rightarrow S$ factors through $k$, so it cannot be a lifting of the given map). One more remark: $R$ discrete and cofibrant in non-positive cdga's implies $R$ formally smooth (you don't have finite presentation authomatically). Hope it works.

share|improve this answer

Your Answer

 
discard

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.