# Divided powers

Consider a differential graded algebra (A, d) sitting in homological degrees 0, 1, 2, … and with d : A_n —> A_{n – 1}. Then the cohomology H(A) is also a differential graded algebra (with zero differential of course).

We say that (A, d) is strictly commutative if xy = (-1)^e yx with e = deg(x)deg(y) and x^2 = 0 when x has odd degree. In this case H(A) is a strictly commutative differential graded algebra.

We say that (A, d) is a strictly commutative differential graded algebra endowed with divided powers if for every homogeneous element x of A in even degree d we have divided powers γ_n(x) of degree nd satisfying the usual rules for divided powers, and satisfying the compatibility

d(γ_n(x)) = d(x) γ_{n – 1}(x), for all n > 1

with the differential. Then H(A) is a strictly commutative differential graded algebra endowed with divided powers, right?

Wrong! Can you spot the mistake?

## 3 thoughts on “Divided powers”

1. Pingback: Computing Tor « Stacks Project Blog

• See this paper Definition 2.1. No for your last question.