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...


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