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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.

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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.

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