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In a 1957 paper (, Tate shows that if $I \subset R$ is an ideal of the noetherian ring R then there is a graded commutative DGA $X$ over $R$ with $H_i X=0$ except $H_0 X= R/I$ (I guess R should be noetherian). Further, $X$ is a free $R$ module in each degree. Is it known if there is a similar result for any other classes of commutative $R$ algebra?

If such an extension of the result is false, I would very much appreciate a counterexample.

Thanks for your time, Sean

EDIT: (by other classes I mean commutative R algebras (Not DGAs) that are not of the form $R/I$). Also, I am happy with counter examples where the DGA is not level-wise free but instead projective or flat (although it shouldn't end up mattering for my purposes).

I am not concerned with relaxing the noetherian hypothesis but in resolving a different commutative $R$-algebra (not in the sense of Tate).

An extension of the above result would be equivalent to saying that there are no level wise free/projective/flat $E_\infty$ algebras with homology a given commutative $R$-algebra $A$ that are not strictly commutative (in the graded sense).

I worry that I am making things more confusing, sorry if that is the case, and thank you for the answers so far.

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You don't impose conditions on $R$, so what do you meain by "other classes"? – Fernando Muro Aug 30 '11 at 23:05
I suspect the "commutative $R$ algebra" in the question refers not to $R$, but to the graded commutative DGA, which Tate calls an $R$-algebra. – Graham Leuschke Aug 31 '11 at 1:42
@Fernando Muro: does that clarify things? Thanks for the suggestions so far. – Sean Tilson Aug 31 '11 at 2:44

Alternative examples can be constructed using bar resolution and shuffle product:

Let $k$ be a commutative ring, $R$ a commutative $k$-algebra with augementation $\epsilon: R \to k$ and let $B_\ast(R)$ be the bar resolution of $k$ over $R$. That is $$B_n(R) = R \otimes R^{\otimes n}$$ (tensor product is taken over $k$) with $R$-operation $$r \cdot (r_0 \otimes r_1 \otimes \cdots \otimes r_n) = (rr_0) \otimes r_1 \otimes \cdots \otimes r_n$$ and differential $d = \sum_{i=0}^n (-1)^id_i$ with $$d_i(r_0 \otimes \cdots \otimes r_n) = r_0 \otimes \cdots \otimes r_ir_{i+1} \otimes \cdots \otimes r_n \hspace{6pt} (i < n)$$ $$d_n(r_0 \otimes \cdots \otimes r_n) = \epsilon(r_n) \cdot r_0 \otimes \cdots \otimes r_{n-1}$$

(see the book MacLane: Homology, X.2). Then $H_0(B_\ast(R))=k$, $H_i(B_\ast(R))=0$ $(i>0)$ holds and if $R$ is projective (resp. free) as $k$-module than $B_n(R)$ is a projective (resp. free) $R$-module. Eventually the shuffle product $$B_n(R) \times B_m(R) \to B_{n+m}(R)$$ turns $B_\ast(R)$ into a DGA over $R$ that is strictly (graded) commutative (see MacLane X.12.1, 12.3).

If $R = \mathbb{Z}G$ with an abelian group $G$, the example above gives just the usual construction of the Pontryagin product in group homology (see the book Brown: Cohomology of Groups, V.5).

Concerning the statement:

An extension of the above result would be equivalent to saying that there are no level wise free/projective/flat $E_\infty$ algebras with homology a given commutative $R$-algebra $A$ that are not strictly commutative (in the graded sense).

I think there is a misconception: Projective/free resolutions are unique only up to homotopy. So, even if there is a resolution with strictly commutative DGA-structure there may also be other resolutions (of the same algebra) those DGA-structure is not strictly commutative. I think the best result possible is to show that (under suitable assumptions) each DGA is homotopy commutative.

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Suppose I want to look at operations in the homology of a DGA coming from the $E_\infty$ structure. If the given DGA is htopy equivalent (quasi isomorphic) then all of the operations much vanish right? This follows from the comparison theorem that you cite above. I am interested in whether or not there are in fact non-zero operations. Thanks for pointing out the Bar construction, I don't know why I didn't think of that. – Sean Tilson Sep 1 '11 at 14:54

one result inspired by Tate's paper is

S. Lichtenbaum, On the vanishing of ${\rm Tor}$ in regular local rings. Illinois J. Math. 10 1966 220--226.

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On the assumption that the OP is interested in relaxing the Noetherian condition on R, the article "On the homology theory of general commutative rings" by Northcott (1961), may be helpful. The review states, in part:

[...] [T]he author extends the machinery of Tate to study Tor$_R$(K,K) as an algebra, where K=R/M, M an (arbitrary proper) ideal of R. The main idea is to obtain algebraic resolutions of R/M and the results of Tate along these lines are extended to the non-noetherian case by transfinite methods.
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The edit makes it clear that relaxing the Noetherian condition was not what the OP was looking for. I'll leave it up in case it is of interest to others. – B R Aug 31 '11 at 4:18
up vote 0 down vote accepted

I am not sure what the protocol is on answering ones own question so I am making it community wiki.

What I was most curious about is the following: When does the projective resolution of a commutative algebra as a module over a commutative ring have the structure of a graded commutative (skew in Tate's paper) DGA?

As Mariano pointed out, when I asked a different version on MSE, this is always the case. Tate showed that this is true for algebras of the form $R/I$ where $I$ is an ideal of $R$ (the commutative ring we are working over). As Mariano points out, every finitely generated (I think this can probably be dropped) commutative $R$ algebra is a quotient of $R[x_1, x_2, ... x_n]$. So there is a graded commutative DGA that is a free resolution of $X$ as an $R[x_1, x_2, ... x_n]$-module by Tate's result. You simply forget from $R[x_1, x_2, ... x_n]$-modules to $R$-modules and you still have a graded commutative DGA that is a free resolution of $X$ as an $R$-module.

Anyway this is essentially what I was looking for.

(Mariano, if you want to post something like this, I will un-accept this answer and accept yours.)

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