If a differential graded algebra is finitely generated as an algebra, is its homology finitely generated as an algebra?


Is it easier if we impose any of the three conditions: characteristic zero; free as an algebra; generated in positive degree? generated in negative degree?


What about commutative algebras or Lie algebras? (is anything else sensible to ask?) What if we reduce the grading from $\mathbb Z$ to $\mathbb Z/2$ (ie, differential super-algebras)?


What I'm really interested in is the case of $\mathbb Q$-DGLAs freely generated in homologically positive degree, because that corresponds to [homotopy groups of finite complexes](https://mathoverflow.net/questions/168486/finite-complexes-whose-homotopy-groups-are-not-finitely-generated), but the commutative case is probably more familiar. I expect the answers to be the same, heuristically from Koszul duality.



Is homology finitely generated as an algebra?
Another counterexample: let $A$ be the algebra $\mathbb{Q}[y,z]/(y^2) \otimes \bigwedge(x)$ with $x$ in degree 1, $y$ and $z$ in degree 2. Put a differential on this by $z \mapsto xy$. This is a commutative dga in characteristic 0 generated in positive degrees, but of course it's not free. Its homology is spanned by the classes $xz^i$ and $yz^i$ for all $i \geq 0$, and the product on the homology algebra is trivial. So it is infinitely generated as an algebra.


