# Reference or a short argument that a certain subset generates the ring of p-typical symmetric functions under plethysm

Let p be a prime, and let $\Lambda_p$ be the subring of the ring of symmetric functions $\Lambda$ (over $\mathbb{Z}$) such that $$x \in \Lambda_p$$ iff there is an $i \in \mathbb{N}$ such that $p^ix \in \Lambda$. If we let $\psi_{p^n}= \sum_{i=1} x_i^{p^n}$, then one can define $w_{p^i}$ recursively by the relation $$\psi_{p^n} = \sum_{d | p^n} dw^{p^n/d}_d.$$ It turns out that $$\Lambda_p = \mathbb{Z}[w_1,w_p,w_{p^2}, \ldots].$$

This is a somewhat standard basis, but there is another basis that I sometimes run into in the literature which I can't find a reference or argument for why it should be true. Namely, it is claimed that: $$\{w_p^{\circ i},i=0,1,\ldots\}$$ generate $\Lambda_p$ as well, where the operation is plethysm of $w_p$ with itself.

Does anyone have an argument for why this is true? A reference would be nice as well, of course!

First a quick correction. There's an error in your first definition of $\Lambda_p$. The condition on $x$ is not that $p^i x\in \Lambda$ but that $p^i x\in\mathbb{Z}[\psi_1,\psi_p,\dots]$. But this doesn't affect the rest of your question, which is that if we write $A=\mathbb{Z}[\dots,w_{p^n},\dots]$ and $A'=\mathbb{Z}[\dots,w_p^{\circ n},\dots]$, why is $A=A'$?

Briefly, the answer is the universal property of Witt vectors. Let me recall what this is. If $R$ is a $p$-torsion free ring with an endomorphism $F$ satisfying $F(x)\equiv x^p\bmod pR$, then any ring map $R\to S$ lifts uniquely to a ring map $R\to W(S)$ which commutes with the Frobenius operators. This requires a bit of clarification: the implicit map $W(S)\to S$ sends a Witt vector to its initial component, the Frobenius operator on $R$ is the given one $F$, and the Frobenius operator on $W(S)$ is the usual one in the theory of Witt vectors, also denoted $F$.

A sketch of the argument is then as follows:

1. the functor $W'(S)=\mathrm{Hom}_{\mathrm{ring}}(A',S)$ 'obviously' has this universal property
2. the Witt vector functor $W(S)$ also has this property, but non-obviously,
3. therefore $A'$ represents $W(S)$
4. but $W(S)$ is represented by $A$, and so $A$ and $A'$ agree.

Now let me give some more detail.

1: Actually to even make sense of this, we have to explain how we're viewing $W'$ as a ring-valued functor. We can think of $w_p$ as the operator $(F(x)-x^p)/p$ on $p$-torsion free rings with a Frobenius lift $F$. By iterating, we can also think of $w_p^{\circ n}$ as an operator on such rings. Further, each operator $w_{p^n}$ has 'Leibniz rules' for addition and multiplication, by which I mean that $w_p^{\circ n}(x+y)$ and $w_p^{\circ n}(xy)$ can be expressed as polynomials in the $w_p^{\circ i}(x)$ and $w_p^{\circ i}(y)$ as $i$ runs from $0$ to $n$. To show this for $w_1$ and $w_p$, you just work it out. (You'll get the usual addition and multiplication laws for Witt vectors of length 2.) Then the existence of such Leibniz rules for $w_p^{\circ n}$ follows by induction. Finally, these Leibniz rules can be interpreted as functorial addition and multiplication laws on $W'(S)$. I won't try to write this down here, but it's completely formal.

Then $A'$ has a ring endomorphism $F$ which sends $w_p^{\circ n}$ to $(w_p^{\circ n})^p + p w_p^{\circ n+1}$. The way to make sense of this is that we think $F(x)=x^p+pw_p(x)$, and so we should define $F\circ w_p^{\circ n}$ to be the previous expression. Then $F$ induces a natural transformation from $W'$ to itself, which I will confusingly also call $F$, and I claim it follows formally that $F:W'(S)\to W'(S)$ is a ring endomorphism which reduces to the Frobenius map modulo $p$. This finishes 1.

2: This is a standard lemma in the theory of Witt vectors. It often goes under the names of Dieudonné and Dwork. One reference is Michel Lazard's book Commutative formal groups'', p. 215. This is really the only non-formal part of the entire argument.

3 and 4. As stated above, these points imply that $A$ and $A'$ are isomorphic, but I really need to explain why they actually agree as subrings of $\Lambda$. The reason is the ghost components! Let $B$ denote $\mathbb{Z}[\dots,\psi_{p^n},\dots]$. Then $B$ can be viewed as a subring of both $A$ and $A'$. We want to think of $\psi_{p^n}$ as representing the ghost components, and so the inclusion $B\to A$ has to be given by the Witt polynomials $\psi_{p^n}=\sum_{i=0}^n p^i w_{p^i}^{p^{n-i}}$. On the other hand, the map $B\to A'$ is sends $\psi_{p^n}$ to the element obtain by expanding out the operators $\psi_p^{\circ n}=(w_1^p + pw_p)^{\circ n}$ (where $w_1$ is of course the identity operator). Then one has to observe that the endomorphisms $F$ on $A$ and $F$ on $A'$ induce the same endomorphism on $B$ (which is a subring of both of them). Therefore the isomorphism $A\to A'$ is the identity on the subring $B$. But since $B[1/p]$ agrees with both $A[1/p]$ and $A'[1/p]$, and $A$ and $A'$ are $p$-torsion free, $A$ and $A'$ agree as subrings of $\Lambda$.

Pheew. That was a bit longer than I thought it would be! It really is true that the only non-formal part is the Dieudonné-Dwork lemma in 2, although I fear that might not be so clear from what I wrote above...