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 $\tilde{f}(RB\cap IB^{p+1})\neq RA\cap IA^{p+1}$.
