MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a sensible notion of blowing up in any of the available frameworks for derived algebraic geometry? I am happy to remain in the affine setting, where I think the right question to ask is "what does it mean to take powers of an ideal?" in say, a commutative differential graded algebra.

share|cite|improve this question
I've heard from experts that the answer is no: there is not a good notion of powers of an ideal. (Say, invariant under quasi-isomorphism, or definable in Lurie's framework). So there's no $n$th infinitesimal neighborhoods in DAG, only formal completions. – Moosbrugger Nov 29 '11 at 3:48
Moosbrugger: this is exactly what I feared. Maybe we can entice somebody to post an example of what is going wrong? I'm planning to complete my rings at the end of the day anyway, so a little hope remains. – Thel Seraphim Nov 29 '11 at 5:09
I'm not sure what to say -- it's hard to argue why a notion does not exist. If you look in Sections 4 and 5 of DAG XII, you can see Lurie avoiding the notion of $n$th power of an ideal, but I don't see anywhere that he explicitly says that it doesn't exist or why. The essential notion which is missing here is image of an ideal under a map, which doesn't exist since image isn't a notion in a stable $\infty$-category. By the way, even the notion of ideal is fraught in the derived setting: e.g., the only invariant notion of quotient I know of is a map $A\to B$ which is surjective on $\pi_0$. – Moosbrugger Nov 29 '11 at 20:19
With that said, I don't see any problem formulating the universal property for blow-ups in the derived setting -- the notion of Cartier divisor is no problem. However, the usual proof of representability doesn't go through, and it may take some fiddling with deformation theory to show that it is representable. – Moosbrugger Nov 29 '11 at 20:22

I assume that you're working in the DG category, so I take "ideal" to mean "DG ideal". Let $R$ be a commutative DG algebra and $I$ an ideal of $R$. Then for each $n\geq 1$, I set $I^n$ equal to the intersection of all the DG ideals of $R$ that contain the set $S(n)$ consisting of all elements of the form $a_1\cdots a_n$ such that $a_1,\ldots,a_n\in I$. In other words, $I^n$ is the DG ideal of $R$ generated by $S(n)$.

share|cite|improve this answer
This seems like the right thing to do, and is the approach I'm taking in my own work. I'll accept the answer in a few days if nobody else comes along. Your $I^n$ is the image of $mult: I^{\otimes n} \to R$, and I am worried about the homotopy invariance of the concept of "image" here. Maybe this concern is neurotic. – Thel Seraphim Nov 29 '11 at 3:58

I think there was is a problem with characteristics. What is I^n? Is exactly the nth symmetric power of I. In characteristic 0, the symmetric algebra is a direct summand of the tensor algebra, but in positive characteristic I do not think so.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.