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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, but the commutative case is probably more familiar. I expect the answers to be the same, heuristically from Koszul duality.

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    $\begingroup$ I think that for positive characteristic commutative dgas, this ought to be true, just because $d(x^p)=0$ for any $x$. Not that I've tried to write a careful proof down... $\endgroup$ Oct 3, 2014 at 20:32
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    $\begingroup$ @john Wow! I expected the opposite. I think you're right: the whole algebra is finitely generated as a module over the kernel of $d$, so the kernel of $d$ is finitely generated as an algebra. $\endgroup$ Oct 4, 2014 at 1:27

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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.

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    $\begingroup$ This has the advantage over Akhil's example of being generated in positive degrees. It is positive with respect to the cohomological convention, but if you move $z$ to degree 4, it is positive with respect to the homological convention. . . . Can't you make a free one by $dz=abc$, with $a,b,c$ odd? $\endgroup$ Oct 3, 2014 at 17:02
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    $\begingroup$ I believe that you can move $x$ to degree $2n+1$, $y$ and $z$ to degree $2n+2$, and introduce a new exterior generator $w$ in degree $2n+1$ with differential $y^2$ to make this example free and generated in whatever degrees you like. $\endgroup$ Oct 3, 2014 at 17:18
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    $\begingroup$ @TylerLawson: I think you're basically right, although the degrees are a bit off in your example, and the homology is more complicated: there is a Massey product because $xw$ and $yz$ both hit $xy^2$. But you still get an infinite list of algebra generators, for example $xz^i$. (As far as degrees go, you can put $x$ in any odd degree, $y$ in any even degree, and then deduce where $z$ and $w$ have to go.) $\endgroup$ Oct 3, 2014 at 20:20
  • $\begingroup$ @BenWieland: Yes, I think you're right about making a free example. I don't know what you mean by the "cohomological convention", but as you say, you can put $a$, $b$, and $c$ in any odd degrees. $\endgroup$ Oct 3, 2014 at 20:31
  • $\begingroup$ The homological convention is that $d$ lowers degree, while the cohomological convention is that it raises degree. You can switch between them by negating degrees, but that switches whether a DGA is positive or negative. $\endgroup$ Oct 3, 2014 at 22:18
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In the setting of cdgas, this is not true: that is, there is a simple example of a finitely presented coconnective cdga over the rational numbers whose homotopy groups are not finitely generated as an algebra.

Namely, consider the cdga $R$ of (derived) global sections of the structure sheaf ("functions") on the punctured affine plane. Then $\pi_0 R = \mathbb{Q}[x,y]$ and $\pi_{-1} R = \mathbb{Q}[x,y]/(x^\infty, y^\infty)$ where the latter refers to the cokernel of the map $\mathbb{Q}[x^{\pm}, y] \oplus \mathbb{Q}[x, y^{\pm}] \to \mathbb{Q}[x^{\pm}, y^{\pm}]$. (This is easy to check from choosing the standard cover of the punctured affine plane by the complements of the $x$ and $y$ axes, respectively.)

Clearly, the homotopy groups of $R$ are not finitely generated as an algebra, but I claim that $R$ is finitely presented as a cdga. There is in fact an explicit finite presentation of $R$ due to Bhatt and Halpern-Leinster. Namely, let $M$ be the $\mathbb{Q}[x,y]$-module given by $\mathbb{Q}[x,y]/(x,y)$ and consider the natural map $\mathbb{Q}[x,y] \to M$ and its dual $\phi: DM \to \mathbb{Q}[x,y]$, where $D$ denotes Spanier-Whitehead duality in the (derived) category of $\mathbb{Q}[x,y]$-modules. Then the presentation of $R$ is that it is the homotopy pushout of the symmetric algebra of $M$, mapping in two different ways to $\mathbb{Q}[x,y]$, once via the map extending $\phi$ and once via the map extending zero. (As a $\mathbb{Q}[x,y]$-algebra, $R$ has the following universal property: to give a map from $R$ to some other $\mathbb{Q}[x,y]$-algebra $R'$ amounts to the condition that $R'/(x,y)$ should be contractible.)

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  • $\begingroup$ Your last parenthetical sentence implies that $R$-modules are a reflective subcategory of $\mathbb Q[x,y]$-modules, thus an example of a smashing localization. This is generally true about open subschemes. This is familiar algebraic localization in codimension 1. Codimension higher than 2 isn't much different than 2. $\endgroup$ Oct 3, 2014 at 19:10
  • $\begingroup$ Do you have citations for Bhatt and Halpern-Leistner? $\endgroup$ Oct 3, 2014 at 20:07
  • $\begingroup$ @BenWieland: I think their work is forthcoming. Yes, you can construct smashing localizations in this manner for any open subscheme of a (noetherian, say) scheme. In fact, there is a one-to-one correspondence between open subschemes and compact algebras $A$ with $A \otimes A \to A$ (I learned this from Bhargav). $\endgroup$ Oct 3, 2014 at 21:09
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    $\begingroup$ @BenWieland: A proof is now in Proposition 8.9 of arxiv.org/pdf/1406.4947v3.pdf $\endgroup$ Dec 1, 2014 at 15:52

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