In the introduction to HAGII Toen and Vezzosi write that in brave new algebraic geometry (that is, algebraic geometry over the category of symmetric spectra) Z[T] is not smooth over Z.

I am told that this is due to the fact we allow negative homotopy groups (which do not occur with connective spectra, a setting for `ordinary' derived (but cowardly) algebraic geometry).

Can someone with only a vague understanding of derived geometry (like me) understand why this phenomenon occurs? Is it simply a case of extending the notion of smoothness incorrectly?

p.s. there seems to de a tag "derived algebraic geometr" can it be corrected?

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    $\begingroup$ I would not think that this has anything to do with negative homotopy groups. As I see it, "brave" means "replacing chain complexes by spectra". It doesn't mean "replacing connective spectra by spectra". (Chain complexes can be non-connective, too.) $\endgroup$ Nov 17, 2011 at 17:19
  • $\begingroup$ I think there is just a limit to the length of tags, and they get cut off after that point. Unfortunately, changing that is above my pay grade (and I say that as a moderator). $\endgroup$
    – Ben Webster
    Nov 17, 2011 at 17:26
  • $\begingroup$ @Tom Goodwillie: I think derived `usually' means bounded above (co)chain complexes. Taking unbounded complexes is called by TV complicial geometry. While working with symmetric spectra is the brave new geometry. Have I got my terminology confused? $\endgroup$ Nov 17, 2011 at 17:32
  • $\begingroup$ In making a new kind of (derived) algebraic geometry, you could (a) drop the boundedness, or you could (b) replace chain complexes by spectra [or you could (a+b) do both]. Surely "brave new" means (b), not (a). And when people say that Z[T] is not smooth over Z, I am sure that they are doing (b), not (a). $\endgroup$ Nov 18, 2011 at 1:05
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    $\begingroup$ smooth means that the relative cotangent complex is locally free, whatever these words mean. $\endgroup$ Jun 25, 2012 at 3:08

1 Answer 1


I think the answer to your question is "yes". Toen-Vezzosi go over this in Proposition 2.4.15, but here is some version of why.

Away from characteristic zero, there's a sharp difference between being a "free" algebra (meaning, having some kind of universal mapping property) and looking like a polynomial algebra. This is because basically every construction has to be replaced with the derived version in order for them to give sensible answers.

For example, in characteristic zero you might take a vector space $V$ over $\mathbb{Q}$ and form the "free" algebra $$ Sym(V) = \bigoplus_{n \geq 0} Sym^k(V) $$ and this has the property that maps of commutative ring objects $Sym(V) \to A$ are the same as maps $V \to A$ of underlying objects. This still works even if $V$ is a chain complex.

However, the symmetric power functors aren't well-behaved integrally: there are maps of chain complexes $C \to D$ which are weak equivalences such that $Sym^k(C) \to Sym^k(D)$ isn't a weak equivalence. (These examples aren't particularly hard to find, either.) This means that we have to take the symmetric power $$ Sym^k(V) = (V^{\otimes k})_{\Sigma_k} $$ and replace it with the derived version: derived tensor product over whatever your base is, and derived functors of $\Sigma_k$-coinvariants.

Toen-Vezzosi mention the example which they call $\mathbb{F}_p[T]$, which is free on a 1-dimensional vector space over $\mathbb{F}_p$. Here, the derived functors of tensor product don't intrude, but the derived functors of coinvariants do, and they contribute a large amount (namely, the homology of the symmetric groups). In degree zero you're just getting a polynomial algebra on a single generator, but there is extra stuff in positive (homological) degree coming from the Dyer-Lashof operations.

EDIT: Sorry, this object is the "smooth" object.

The thing that looks like a "polynomial" object is the monoid algebra $\mathbb F_p[\mathbb N]$. On homology groups this looks like a polynomial algebra, but it doesn't have simple mapping properties that connect with the notion of smoothness. In particular, they give a description of the derived cotangent complex (which is some measure of how hard it is to build this object out of "free" objects) and it's rather large (certainly not concentrated in degree 0).

  • $\begingroup$ thanks for the answer. before trying to understand properly what you wrote, can I ask you if in characteristic zero the line is smooth? from what you wrote it naively seems to me that this is a problem of positive or mixed characteristic. $\endgroup$ Nov 17, 2011 at 17:33
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    $\begingroup$ In characteristic zero, yes, it is smooth. $\endgroup$ Nov 17, 2011 at 17:37

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