Let $k$ be a field of characteristic zero and $A$ be a graded commutative dg-algebra over $k$ with differential of degree $+1$ satisfying $H^0(A)=k, H^i(A)=0$ for $i<0$. Denote by $\mathcal J$ a dg ideal of generated by $A^k, k<0$ ($A^k$ denotes elements of degree $k$). Is it true that $\mathcal J$ is acyclic? In other words, is the natural projection $f\_\mathcal J\colon A\to A/\mathcal J$ quasi-isomorphism?


If there is a quasi-isomorphism $A\to B$ with $B$ concentrated in non-negative degrees, then the map $f\_\mathcal J$ should be quasi-isomorphism.


If you know some general facts about the cohomology of $\mathcal J$ or when this statement is true it also would be very interesting!


It is easy to construct a counterexample when we do not pose condition $H^0(A)=k$. Let $A$ be generated by $x,y,h$ with $\deg x=-1, \deg y=1$ and $\deg h=0$ (so $x^2=y^2=0$). The differential is defined by the formula $d(x)=xyh, d(h)=d(y)=0.$ Then the element $xy\in\mathcal J$ closed but not exact. But in this case $H^0(A)$ is very big.



$A$ is a commutative connective dg-algebra satisfying $H^0(A)=k$. Is it true that a dg ideal generated by elements of negative degree acyclic?
It is not necessarily true. I can provide a counterexample but don't know any general statement about when this is true.


Consider $$A=k[x,dx,y]/\langle x^2y, ydx\rangle$$ with $x$ in degree $-2$ and $y$ in degree $3$.


The obvious map from $k[x,dx]$ to $A$ is a dg-algebra map which induces an isomorphism below degree $1$, which implies that $H^0(A)\cong k$ and $H^i(A)=0$ for $i<0$.


But the dg ideal $\mathcal{J}$ consists of $A^{<0} \oplus k[xy]$; since $[xy]$ is in degree $1$ and $\mathcal{J}$ contains nothing in degrees $0$ or $2$, we conclude that $\mathcal{J}$ cannot be acyclic.


