UPDATE: I am grateful to Peter May for the accepted answer, which makes most of the details below irrelevant. However, I will leave them in place for the record.

I am trying to understand the proof of Theorem 2 in the 1969 paper Remarks on the Structure of Hopf Algebras by Peter May (because I would like to prove a similar result with different hypotheses). The statement is given in the discussion below, and the proof contains the following claim:

Since $\tilde{f}$ is the identity on $RA$, $\tilde{f}(F_sRA)=RA\cap F_sA$

I do not see why this should be true; I hope that someone will enlighten me (or suggest a workaround, or an alternative proof or reference for the same theorem).

The context is as follows:

  • $A$ is a connected, graded, commutative algebra over a perfect field $K$ of characteristic $p>0$. You can assume if you like that it is concentrated in even degrees. Later, we will assume that $A$ is a Hopf algebra, but we do not need that for the moment.
  • $IA$ is the augmentation ideal, and $QA=IA/IA^2$.
  • $f\colon QA\to IA$ is a section of the natural projection.
  • $\xi\colon A\to A$ is the $p$'th power map. I am interested in the case where this is injective, so you can assume that if you like.
  • $RA=\sum_{i\geq 0}\xi^i(f(QA))$.
  • $B=V(RA)$ is the commutative graded $K$-algebra freely generated by $RA$ subject to the relation $\xi(r)=r^p$ for all $r\in RA$. (In the case where $\xi$ is injective, this is just the free commutative algebra on $f(QA)$.)
  • $\tilde{f}\colon B\to A$ is the unique homomorphism of $K$-algebras that acts as the identity on $RA$. This is easily seen to be surjective.
  • The statement of the theorem is that if $A$ is a Hopf algebra, then $\tilde{f}$ is an isomorphism.
  • We will write $RB$ for the obvious copy of $RA$ inside $B$, so $\tilde{f}\colon RB\to RA$ is an isomorphism.
  • The statement that I am worried about says that $\tilde{f}(RB\cap IB^t)=RA\cap IA^t$ for all $t\geq 0$.
  • If $r\in RB$ and $r$ can be written as a sum of terms like $u_1\dotsb u_t$ with $u_i\in IB$, then we can apply $\tilde{f}$ to express $\tilde{f}(r)$ as a sum of terms in $IA^t$. Thus $\tilde{f}(RB\cap IB^t)\subseteq RA\cap IA^t$.
  • Suppose instead that $s\in RA$ and $s$ can be written as a sum of terms like $v_1\dotsb v_t$ with $v_i\in IA$. As $\tilde{f}\colon RB\to RA$ is an isomorphism, there is a unique $r\in RB$ with $\tilde{f}(r)=s$; we want to show that this lies in $IB^t$. As $\tilde{f}\colon IB\to IA$ is surjective, we can choose $u_i\in IB$ with $\tilde{f}(u_i)=v_i$. This gives an element $r'\in IB^t$ with $\tilde{f}(r')=\tilde{f}(r)$, so $\tilde{f}(r-r')=0$. However, we only know that $\tilde{f}$ is injective on $RB$, and we do not know that $r'\in RB$, so we cannot conclude that $r=r'$. Thus, we do not seem to have a proof that $f(RB\cap IB^t)\supseteq RA\cap IA^t$.
  • Suppose that the element $r$ above does not lie in $IB^t$, but only in $IB^{t-1}$. Then $r$ represents a nonzero element in the associated graded group $E^0RB$ in filtration degree $t-1$, whose image in $E^0RA$ is zero. Thus, the map $\tilde{f}\colon E^0RB\to E^0RA$ is not injective, so we cannot proceed with the next step.

The proof in the paper does not seem to use the coproduct in any essential way when reaching the statement that I have questioned. Thus, we should ask whether it is true in the absence of a coproduct. Consider the following case (where $p$ is an odd prime):

  • $A=\mathbb{F}_p[x,y]/(x^{p+1}-y^p)$, with $|x|=2p$ and $|y|=2p+2$
  • $QA=\mathbb{F}_p\{x,y\}$
  • $RA=\mathbb{F}_p\{x^{p^k},y^{p^k}:k\geq 0\}$, with $y^{p^k}=x^{p^k(p+1)}$ for $k>0$
  • $B=\mathbb{F}_p[x,y]$
  • $RB=\mathbb{F}_p\{x^{p^k},y^{p^k}:k\geq 0\}$

If we consider $y^p$ as an element of $RB$, then it is equal to $x^{p+1}$ and so lies in $RA\cap IA^{p+1}$. However, the corresponding element of $RB$ does not lie in $IB^{p+1}$, so $\tilde{f}(RB\cap IB^{p+1})\neq RA\cap IA^{p+1}$.

  • 1
    $\begingroup$ Since you are happy to suppose the $p$-th power is injective, $A$ is polynomial. So if $\overline{f}$ is surjective, it is split epi as a map of algebras. So if $f$ is injective on $Q(A)$ it is injective. $\endgroup$
    – user43326
    Sep 14, 2015 at 15:47
  • $\begingroup$ @user43326 How do you know that $A$ is polynomial? That is a theorem of Borel, but a large part of the point of May's paper is to give a sharpened version and alternative proof of Borel's result, so I am not happy to quote it. $\endgroup$ Sep 14, 2015 at 15:56
  • $\begingroup$ Isn't this in Milnor-Moore? $\endgroup$
    – user43326
    Sep 14, 2015 at 15:58
  • $\begingroup$ If you aren't happy with Milnor-Moore Theorem 7.11, you can always use Dieudonne modules. The injectivity of the $p$-th power implies that Dieudonne module is torsion-free, and the connectedness hypothesis implies that there is no infinitely divisible elements. $\endgroup$
    – user43326
    Sep 14, 2015 at 16:12
  • 2
    $\begingroup$ I need to adapt these results to a different context, where $A$ is not graded and connected, but I have other assumptions that may do the same job. May's approach looks promising for that purpose, if the issue that I described can be resolved. I should probably try to adapt the Milnor-Moore argument as well, but for the moment I am looking at May's paper. I do not have an antipode, which makes an approach via Dieudonne modules less promising. $\endgroup$ Sep 14, 2015 at 16:25

2 Answers 2


Ah, Neil, my apologies, sins of a forgetful old man. You are quite right to object to Theorem 3 in that 1969 paper, because it is false as stated. In fact, Paul Goerss gave me a counterexample ages ago. That appears as Example 23.4.2 in More Concise Algebraic Topology, by Kate Ponto and myself, which is available here: http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf (and at booksellers everywhere). Advertising that book, it reworks Milnor and Moore, among other things, in Chapters 20-23. In response to Paul's counterexample, it also reworks the relevant part of the 1969 paper, proving a correct variant of the theorem you object to, namely Theorem 23.4.1. Work over a perfect field $K$.

Theorem. Let $A$ be a connected, commutative, and associative quasi Hopf algebra. For a morphism of $K$-modules $\sigma: QA \rightarrow IA$ such that $\pi\sigma = id$, where $\pi: IA \rightarrow QA$ is the quotient map, let $R(A;\sigma)$ be the sub abelian restricted Lie algebra of $A$ generated by the image of $\sigma$. For a suitable choice of $\sigma$, the morphism of algebras $f: V(R(A;\sigma)) \rightarrow A$ induced by the inclusion of $R(A;\sigma)$ in $A$ is an isomorphism.

Quasi here just means not necessarily coassociative. The proof is an adaptation of the proof of Theorem 4 in the 1969 paper, which is correct. The point is that you must choose the section carefully to make it go.


Well, Peter May can try. Sins of my youth, Neil, and I agree the writing is obscure (that is a seven page paper, excess concision came early). The relevant claim there is in a sentence with two parts. In your notation, the first part says that $\tilde{f}(RA\cap I(V(RA))^t) = RA\cap IA^t$. That is a statement about the image of the surjection and is not a monomorphism claim. We are thinking of $V(RA)$ and $A$ as two entities with a map $\tilde f$ between them, so, a priori, it does not make logical sense to ask for the equality you ask for.

The second part of the sentence is that $E^0\tilde f$ restricted to $E^0RA$ is a monomorphism. (Associated graded with respect to the filtrations given by powers of the augmentation ideals). That is, we only need injectivity when we restrict $E^0\tilde f$ to $E^0RA$, and we do have injectivity (indeed identity) before passage to the associated graded. There is an omitted verification of the second claim, but I think it works. It amounts to saying that if $x\in RA\cap I(V(RA))^t$, then it is not the case that $\tilde{f}(x)$ is a linear combination of elements of $IA^{t'}$ with $t'>t$. Since $A$ is connected, I think we can induct by degree. There are no spare generators around to muck things up in the way you envisioned.

In sum, I claim that I never in the structure of the proof needed to prove the equality that you are worried about a priori, rather it falls out of the conclusion: I am relying on the fact that injectivity need only be shown on primitives in the associated graded.

  • $\begingroup$ Thanks, Peter, for your response, but I think I am still not convinced. I have added further explanation to the question. $\endgroup$ Sep 15, 2015 at 21:42
  • $\begingroup$ I'll get back to this asap. I'm swamped right now. But look at the sketched monomorphism argument using induction, remembering that that is all about primitives. I agree that you have to use the coproduct structure for the statement you question, but logically we don't need it a priori. $\endgroup$
    – Peter May
    Sep 16, 2015 at 1:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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