Structure of Hopf algebras - trouble understanding an old paper 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 $\tilde{f}(RB\cap IB^{p+1})\neq RA\cap IA^{p+1}$.
 A: 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.  
A: 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.  
