$2$-adic valuation of Schur $P$-functions in the power-sum basis

For a partition $$\lambda$$, let $$P_\lambda$$ be the Schur $$P$$-functions (case $$t=-1$$ of Hall-Littlewood symmetric functions) and let $$p_\lambda=p_{\lambda_1}p_{\lambda_1}\cdots p_{\lambda_k}$$ be the power-sum symmetric functions.

It is known that the space $$\Gamma$$ spanned by the $$P_\lambda$$ for $$\lambda$$ with distinct parts is the same as the one spanned by the $$p_\rho$$ for $$\rho$$ with odd parts (see e.g. Macdonald's Symmetric Functions and Hall Polynomials). I noticed that the coefficients of the transition matrix between the two bases for $$\Gamma$$ are all in $$\mathbb{Z}_{(2)}$$; that is, if $$\lambda$$ is a partition with distinct parts and $$P_\lambda = \sum_\rho a^\lambda_\rho p_\rho$$ then $$v_2(a^\lambda_\rho) \ge 0$$. Is this fact known?

Note that we have $$a^\lambda_\rho=2^{\ell(\rho)-\ell(\lambda)}z_\rho^{-1}X^\lambda_\rho(-1)$$, where $$\ell(\cdot)$$ is the number of parts of a partition and $$X^\lambda_\rho \in \mathbb{Z}[t]$$ are the Green polynomials (see Macdonald chapter III.7). In particular, the question comes down to showing that $$v_2(X^\lambda_\rho(-1))\ge v_2(z_\rho)+\ell(\lambda)-\ell(\rho).$$

• A symmetric function $f$ is integral, denoted $f\in\Lambda$, if it is an integral linear combination of monomial symmetric functions. It is well-known that $P_\lambda\in\Lambda$ and (as stated by the proposer) $P_\lambda\in \mathbb{Q}[p_1,p_3,p_5,\dots]$. The following stronger result seems to be true: if $f\in\Lambda\cap \mathbb{Q}[p_1,p_3,p_5,\dots]$ and $f=\sum_\rho b^\lambda_\rho p_\rho$, then $v_2(b_\rho^\lambda)\geq 0$. Sep 4, 2021 at 18:21
• Thanks for your comment! That seems equivalent to my question actually, since unless I'm mistaken the $P_\lambda$ (for arbitrary $\lambda$) form a $\mathbb{Z}$-basis for $\Lambda$, and in particular those with $\lambda$ having distinct parts form a $\mathbb{Z}$-basis for $\Lambda \cap \mathbb{Q}[p_1,p_3,p_5,\cdots]$. Sep 4, 2021 at 18:43
• I'm not sure if the fact is known, but I think I can prove it (modulo some results about Hall-Littlewood functions from Steven Sam's notes that I haven't properly read but have no specific reasons to distrust). Are you looking for a proof or a reference? Sep 5, 2021 at 2:10
• There are two main ingredients: (1) Proving that $\mathbb{Z}_{\left(2\right)} \left[q_k/2 \mid k \text{ is odd}\right] = \mathbb{Z}_{\left(2\right)} \left[p_k \mid k \text{ is odd}\right]$, where the $q_k$ are defined by $\sum\limits_{k \in \mathbb{N}} q_k u^k = \prod\limits_j \dfrac{1+x_ju}{1-x_ju}$. This equality follows from Lemma 9.1.3 in Steven Sam's Math 740 notes (Spring 2017), because the powers of $2$ in the denominator of $z_\lambda^{-1}$ are cancelled out by the $2^{\ell\left(\lambda\right)}$ with a slack of $1$ (so dividing ... Sep 5, 2021 at 2:13
• ... by $2$ still gives an integer). (2) Proving that $P_\lambda \in \mathbb{Z}\left[q_k / 2 \mid k \text{ is odd} \right]$. This follows from Theorem 8.1.6 in op. cit. (applied to $t = -1$), since each $q_\mu / 2^{\ell\left(\mu\right)}$ is clearly a product of $q_k / 2$s for various $k$s. Here one needs to be a bit careful, since some of the $k$s will be even, but the formula (9.1.1) in op. cit. shows (by induction) that the $q_k / 2$s for even $k$ still belong to $\mathbb{Z}\left[q_k / 2 \mid k \text{ is odd} \right]$. Here is hoping I didn't get confused. Sep 5, 2021 at 2:16

Here is a proof of the generalization suggested by Richard Stanley in the comments and even of a more general result (with "odd" replaced by "not divisible by a given prime $$q$$"). It is completely different from the argument I sketched in the comments, and is completely elementary (using no Macdonald polynomials). Unfortunately, it is also somewhat awkward and way too long (much of it devoted to a fight with notations).

For any commutative ring $$R$$, we let $$\Lambda_{R}$$ be the ring of symmetric functions over $$R$$; this is a commutative $$R$$-algebra. Let $$\operatorname{Par}$$ be the set of all partitions. We shall use the standard notation $$p_{\lambda}$$ for the power-sum symmetric function indexed by a partition $$\lambda$$.

Fix a prime $$q$$. Let $$\mathbb{Z}_{\left( q\right) }$$ denote the ring of all rational numbers that can be written in the form $$\dfrac{a}{b}$$ for two integers $$a$$ and $$b$$ such that $$b$$ is coprime to $$q$$. These numbers are known as $$q$$-integers. Obviously, $$\mathbb{Z}_{\left( q\right) }$$ is a subring of $$\mathbb{Q}$$, so that $$\Lambda_{\mathbb{Z}_{\left( q\right) }}$$ is a subring of $$\Lambda_{\mathbb{Q}}$$.

Now, Richard Stanley's generalization (generalized a bit further) claims:

Theorem 1. We have \begin{align*} \Lambda_{\mathbb{Z}_{\left( q\right) }}\cap\mathbb{Q}\left[ p_{i} \ \mid\ i\not \equiv 0\mod q\right] =\mathbb{Z}_{\left( q\right) }\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] . \end{align*}

The proof requires some preparations, which include showing some results of independent interest.

First, we introduce a few more classical notations from symmetric function theory: For any partition $$\lambda$$ and any $$i\geq1$$, we let $$m_{i}\left( \lambda\right)$$ denote the multiplicity of $$i$$ in $$\lambda$$ (that is, the number of parts of $$\lambda$$ that equal $$i$$). For any partition $$\lambda$$, we define the positive integer \begin{align*} z_{\lambda}:=\prod_{i=1}^{\infty}\left( \left( m_{i}\left( \lambda\right) \right) !\cdot i^{m_{i}\left( \lambda\right) }\right) . \end{align*}

Let $$V$$ be the $$\Lambda_{\mathbb{Z}}$$-subalgebra of $$\Lambda_{\mathbb{Q}}$$ generated by the fractions $$\dfrac{p_{i}}{i^{k}}$$ for all positive integers $$i$$ and all nonnegative integers $$k$$. In other words, let \begin{align*} V=\Lambda_{\mathbb{Z}}\left[ \dfrac{p_{i}}{i^{k}}\ \mid\ i>0\text{ and } k\geq0\right] \subseteq\Lambda_{\mathbb{Q}}. \end{align*}

Now, we claim the following:

Theorem 2. We have \begin{align*} z_{\lambda}^{-1}p_{\lambda}\in V \end{align*} for each partition $$\lambda$$.

To prove this, we need a simple arithmetic lemma:

Lemma 3. Let $$c$$ and $$d$$ be two integers with $$d\neq0$$. Then, there exist some integers $$a$$ and $$b$$ and some nonnegative integer $$i$$ such that $$c^i =ad+bc^{i+1}$$.

Proof of Lemma 3. The ring $$\mathbb{Z}/d\mathbb{Z}$$ is finite (since $$d\neq0$$). For any integer $$m$$, we let $$\overline{m}\in\mathbb{Z}/d\mathbb{Z}$$ denote the residue class of $$m$$ in this ring. The infinitely many residue classes $$\overline{c^{0}},\overline{c^{1}},\overline{c^{2}},\ldots$$ all belong to the finite ring $$\mathbb{Z}/d\mathbb{Z}$$, and thus cannot all be distinct (by the pigeonhole principle). In other words, there exist two nonnegative integers $$i$$ and $$j$$ satisfying $$i and $$\overline{c^i }=\overline{c^j }$$. Consider these $$i$$ and $$j$$. We have $$\overline{c^i }=\overline{c^j }$$; in other words, $$c^i \equiv c^j \mod d$$. Hence, $$d\mid c^i -c^j$$. In other words, $$c^i -c^j =ad$$ for some integer $$a$$. Consider this $$a$$. However, recall that $$i. Thus, $$j=i+w$$ for some positive integer $$w$$. Consider this $$w$$. The integer $$w-1$$ is nonnegative (since $$w$$ is positive); thus, $$c^{w-1}$$ is an integer. Hence, we can define an integer $$b$$ by $$b=c^{w-1}$$. From $$j = i+w = \left(w-1\right) + \left(i+1\right)$$, we obtain $$c^j = c^{\left(w-1\right) + \left(i+1\right)} = \underbrace{c^{w-1}}_{= b} c^{i+1} = bc^{i+1}$$. Now, from $$c^i -c^j =ad$$, we obtain \begin{align*} c^i & =ad+\underbrace{c^j }_{= bc^{i+1}} =ad+bc^{i+1}. \end{align*} This proves Lemma 3. $$\blacksquare$$

Proof of Theorem 2. We shall use the notation $$\ell\left( \lambda\right)$$ for the length of a partition $$\lambda$$ (that is, the number of all parts of $$\lambda$$). For instance, $$\ell\left( \left( 5,2,2\right) \right) =3$$. We shall also use the notation $$\left\vert \lambda\right\vert$$ for the size of a partition $$\lambda$$ (that is, the sum of all parts of $$\lambda$$). We shall prove Theorem 2 by strong induction on $$\left\vert \lambda\right\vert +\ell\left( \lambda\right)$$. Thus, we fix some $$N\in\mathbb{N}$$, and we assume (as the induction hypothesis) that Theorem 2 holds for all $$\lambda$$ with $$\left\vert \lambda\right\vert +\ell\left( \lambda\right) . We now must prove Theorem 2 for all $$\lambda$$ with $$\left\vert \lambda\right\vert +\ell\left( \lambda\right) =N$$.

So let $$\lambda$$ be a partition satisfying $$\left\vert \lambda\right\vert +\ell\left( \lambda\right) =N$$. We must prove that $$z_{\lambda} ^{-1}p_{\lambda}\in V$$.

If $$\lambda=\varnothing$$, then this is obvious (since $$z_{\lambda} ^{-1}p_{\lambda}=1$$ in this case). Thus, for the rest of this induction step, we WLOG assume that $$\lambda\neq\varnothing$$.

We are in one of the following three cases:

Case 1: All parts of $$\lambda$$ are equal to $$1$$.

Case 2: All parts of $$\lambda$$ are equal, but not equal to $$1$$.

Case 3: Not all parts of $$\lambda$$ are equal.

Let us consider Case 1 first. In this case, all parts of $$\lambda$$ are equal to $$1$$. In other words, $$\lambda=\underbrace{\left( 1,1,\ldots,1\right) }_{v\text{ entries}}$$ for some positive integer $$v$$ (since $$\lambda \neq\varnothing$$). Consider this $$v$$. Thus, $$p_{\lambda}=p_{1}^v$$ and $$\left\vert \lambda\right\vert =v$$ and $$\ell\left( \lambda\right) =v$$. Let $$\operatorname{Par}_{v}$$ denote the set of all partitions of $$v$$. Thus, $$\lambda\in\operatorname{Par}_{v}$$. Now, let $$h_v \in \Lambda_{\mathbb{Z}}$$ denote the $$v$$-th complete homogeneous symmetric function. A well-known formula (e.g., (2.5.17) in Grinberg/Reiner, arXiv:1409.8356v7) yields \begin{align*} h_{v}=\sum_{\mu\in\operatorname{Par}_{v}}z_{\mu}^{-1}p_{\mu }=z_{\lambda}^{-1}p_{\lambda}+\sum_{\substack{\mu\in\operatorname{Par}_{v}; \\\mu\neq\lambda}}z_{\mu}^{-1}p_{\mu}, \end{align*} so that $$$$z_{\lambda}^{-1}p_{\lambda}=h_v -\sum_{\substack{\mu\in \operatorname{Par}_{v};\\\mu\neq\lambda}}z_{\mu}^{-1}p_{\mu}. \label{darij1.pf.t2.c1.2} \tag{1}$$$$

However, the only partition of $$v$$ that has length $$\geq v$$ is the partition $$\underbrace{\left( 1,1,\ldots,1\right) }_{v\text{ entries}}=\lambda$$. Thus, if $$\mu$$ is a partition of $$v$$ distinct from $$\lambda$$, then $$\mu$$ has length $$. In other words, if $$\mu\in\operatorname{Par}_{v}$$ satisfies $$\mu\neq\lambda$$, then $$\ell\left( \mu\right) . Hence, if $$\mu \in\operatorname{Par}_{v}$$ satisfies $$\mu\neq\lambda$$, then \begin{align*} \underbrace{\left\vert \mu\right\vert }_{=v=\left\vert \lambda\right\vert }+\underbrace{\ell\left( \mu\right) }_{ and therefore $$z_{\mu}^{-1}p_{\mu}\in V$$ (by our induction hypothesis, applied to $$\mu$$ instead of $$\lambda$$). Thus, \eqref{darij1.pf.t2.c1.2} becomes \begin{align*} z_{\lambda}^{-1}p_{\lambda}=\underbrace{h_v}_{\in\Lambda_{\mathbb{Z} }\subseteq V}-\sum_{\substack{\mu\in\operatorname{Par}_{v};\\\mu \neq\lambda}}\underbrace{z_{\mu}^{-1}p_{\mu}}_{\in V}\in V-\sum_{\substack{\mu \in\operatorname{Par}_{v};\\\mu\neq\lambda}}V\subseteq V. \end{align*} Hence, $$z_{\lambda}^{-1}p_{\lambda}\in V$$ has been proved in Case 1.

Let us next consider Case 2. In this case, all parts of $$\lambda$$ are equal, but not equal to $$1$$. In other words, $$\lambda=\underbrace{\left( n,n,\ldots,n\right) }_{v\text{ entries}}$$ for some positive integers $$n\neq1$$ and $$v$$ (since $$\lambda\neq\varnothing$$). Consider these $$n$$ and $$v$$. Thus, $$p_{\lambda}=p_n^v$$ and $$z_{\lambda}=v!\cdot n^v$$ and $$\left\vert \lambda\right\vert =nv$$ and $$\ell\left( \lambda\right) =v$$. From $$n\neq1$$, we obtain $$n>1$$ (since $$n$$ is a positive integer). Thus, $$nv > v$$ (since $$v > 0$$). In other words, $$v < nv$$.

From $$p_{\lambda}=p_n^v$$ and $$z_{\lambda}=v!\cdot n^v$$, we obtain $$$$z_{\lambda}^{-1}p_{\lambda}=\left( v!\cdot n^v \right) ^{-1}p_n ^v =\dfrac{p_n^v }{v!\cdot n^v }. \label{darij1.pf.t2.c2.1} \tag{2}$$$$

We must prove that $$z_{\lambda}^{-1}p_{\lambda}\in V$$. In other words, we must prove that $$\dfrac{p_n^v }{v!\cdot n^v }\in V$$ (since $$z_{\lambda} ^{-1}p_{\lambda}=\dfrac{p_n^v }{v!\cdot n^v }$$).

We shall now use Lemma 3 to decompose $$\dfrac{p_n^v }{v!\cdot n^v }$$ into a sum of partial fractions -- one with a denominator of $$v!$$ and another with a power of $$n$$ in the denominator. We will then prove that both of these fractions belong to $$V$$.

Indeed, Lemma 3 (applied to $$c=n^v$$ and $$d=v!$$) yields that there exist some integers $$a$$ and $$b$$ and some nonnegative integer $$i$$ such that $$\left( n^v \right) ^i =a\cdot v!+b\left( n^v \right) ^{i+1}$$. Consider these $$a$$, $$b$$ and $$i$$. Multiplying both sides of the equality $$\left( n^v \right) ^i =a\cdot v!+b\left( n^v \right) ^{i+1}$$ by $$\dfrac {p_n^v }{\left( n^v \right) ^{i+1}\cdot v!}$$, we obtain \begin{align} \dfrac{p_n^v }{v!\cdot n^v } &=\left(a\cdot v!+b\left( n^v \right) ^{i+1}\right) \cdot \dfrac{p_n^v }{\left( n^v \right) ^{i+1}\cdot v!} \\ &=a\cdot\dfrac{p_n^v }{\left( n^v \right) ^{i+1}}+b\cdot\dfrac{p_n^v }{v!} . \label{darij1.pf.t2.c2.parfrac} \tag{3} \end{align} Thus, in order to prove that $$\dfrac{p_n^v }{v!\cdot n^v }\in V$$, it suffices to show that $$\dfrac{p_n^v }{\left( n^v \right) ^{i+1}}\in V$$ and $$\dfrac{p_n^v }{v!}\in V$$ (because $$a$$ and $$b$$ are integers, and $$V$$ is a ring).

The first of these two claims is easy: We have $$\dfrac{p_n}{n^{i+1}}\in V$$ (since $$\dfrac{p_n}{n^{i+1}}$$ is one of the designated generators of the $$\Lambda_{\mathbb{Z}}$$-algebra $$V$$). Hence, $$\left( \dfrac{p_n}{n^{i+1} }\right) ^v \in V$$ (since $$V$$ is a ring). In other words, $$\dfrac{p_n^v }{\left( n^v \right) ^{i+1}}\in V$$ (since $$\left( \dfrac{p_n}{n^{i+1} }\right) ^v =\dfrac{p_n^v }{\left( n^v \right) ^{i+1}}$$).

It remains to prove that $$\dfrac{p_n^v }{v!}\in V$$. To do this, we will need the Frobenius endomorphism $$\mathbf{f}_n$$. It is defined as follows: For any commutative ring $$R$$, we let \begin{align*} \mathbf{f}_n:\Lambda_{R}\rightarrow\Lambda_{R} \end{align*} be the $$R$$-algebra homomorphism that sends each symmetric function $$f$$ to $$f\left( x_{1}^{n},x_{2}^{n},x_{3}^{n},\ldots\right)$$ (where we regard $$f$$ as a symmetric formal power series in countably many indeterminates $$x_{1},x_{2} ,x_{3},\ldots$$). This homomorphism $$\mathbf{f}_n$$ is called the $$n$$-th Frobenius endomorphism and is functorial in $$R$$ (that is, it commutes with the morphisms $$\Lambda_{R}\rightarrow\Lambda_{S}$$ induced by ring homomorphisms $$R\rightarrow S$$). If you like to think in terms of plethysm, $$\mathbf{f}_n$$ can be described as sending each $$f\in\Lambda_{R}$$ to the plethysm $$f\left[ p_n\right]$$.

The functoriality of $$\mathbf{f}_n$$ in $$R$$ entails that the $$\mathbf{f}_n$$ defined for $$R=\mathbb{Z}$$ is a restriction of the $$\mathbf{f}_n$$ defined for $$R=\mathbb{Q}$$. Thus, we can safely denote both of these maps by $$\mathbf{f}_n$$ without risking confusion. They both are ring homomorphisms (since they are $$R$$-algebra homomorphisms for appropriate $$R$$). Of course, $$\mathbf{f}_n\left( \Lambda_{\mathbb{Z}}\right) \subseteq\Lambda _{\mathbb{Z}}$$ (since the $$\mathbf{f}_n$$ defined for $$R=\mathbb{Z}$$ is a restriction of the $$\mathbf{f}_n$$ defined for $$R=\mathbb{Q}$$).

It is easy to see that $$\mathbf{f}_n\left( p_{i}\right) =p_{in}$$ for each $$i>0$$. Hence, for each positive integer $$i$$ and each nonnegative integer $$k$$, we have \begin{align*} \mathbf{f}_n\left( \dfrac{p_{i}}{i^{k}}\right) =\dfrac{p_{in}}{i^{k} }=n^{k}\cdot\dfrac{p_{in}}{\left( in\right) ^{k}}\in V \end{align*} (since $$\dfrac{p_{in}}{\left( in\right) ^{k}}$$ is one of the designated generators of the $$\Lambda_{\mathbb{Z}}$$-algebra $$V$$). Therefore, \begin{align*} \Lambda_{\mathbb{Z}}\left[ \mathbf{f}_n\left( \dfrac{p_{i}}{i^{k}}\right) \ \mid\ i>0\text{ and }k\geq0\right] \subseteq V \end{align*} (since $$V$$ is a $$\Lambda_{\mathbb{Z}}$$-algebra).

Now, from $$V=\Lambda_{\mathbb{Z}}\left[ \dfrac{p_{i}}{i^{k}}\ \mid\ i>0\text{ and }k\geq0\right]$$, we obtain \begin{align*} \mathbf{f}_n\left( V\right) & =\mathbf{f}_n\left( \Lambda _{\mathbb{Z}}\left[ \dfrac{p_{i}}{i^{k}}\ \mid\ i>0\text{ and }k\geq0\right] \right) \\ & =\underbrace{\left( \mathbf{f}_n\left( \Lambda_{\mathbb{Z}}\right) \right) }_{\subseteq\Lambda_{\mathbb{Z}}}\left[ \mathbf{f}_n\left( \dfrac{p_{i}}{i^{k}}\right) \ \mid\ i>0\text{ and }k\geq0\right] \\ & \qquad\left( \text{since }\mathbf{f}_n\text{ is a ring homomorphism} \right) \\ & \subseteq\Lambda_{\mathbb{Z}}\left[ \mathbf{f}_n\left( \dfrac{p_{i} }{i^{k}}\right) \ \mid\ i>0\text{ and }k\geq0\right] \subseteq V. \end{align*}

Let $$\mu$$ be the partition $$\underbrace{\left( 1,1,\ldots,1\right) }_{v\text{ entries}}$$. Then, $$p_{\mu}=p_{1}^v$$ and $$z_{\mu}=v!\cdot \underbrace{1^v }_{=1}=v!$$ and $$\left\vert \mu\right\vert =v$$ and $$\ell\left( \mu\right) =v$$. Hence, \begin{align*} \underbrace{\left\vert \mu\right\vert }_{=v}+\underbrace{\ell\left( \mu\right) }_{=v} & =\underbrace{v}_{< nv} + v\\ & <\underbrace{nv}_{=\left\vert \lambda\right\vert }+\underbrace{v} _{=\ell\left( \lambda\right) }=\left\vert \lambda\right\vert +\ell\left( \lambda\right) =N. \end{align*} Hence, $$z_{\mu}^{-1}p_{\mu}\in V$$ (by our induction hypothesis, applied to $$\mu$$ instead of $$\lambda$$). In view of $$z_{\mu}=v!$$ and $$p_{\mu}=p_{1}^v$$, this rewrites as $$v!^{-1}\cdot p_{1}^v \in V$$. Hence, $$$$\mathbf{f}_n\left( v!^{-1}\cdot p_{1}^v \right) \in \mathbf{f} _n\left( V\right) \subseteq V. \label{darij1.pf.t2.c2.5} \tag{4}$$$$ However, since $$\mathbf{f}_n$$ is a $$\mathbb{Q}$$-algebra homomorphism, we have \begin{align*} \mathbf{f}_n\left( v!^{-1}\cdot p_{1}^v \right) =v!^{-1}\cdot\left( \mathbf{f}_n\left( p_{1}\right) \right) ^v =\dfrac{\left( \mathbf{f}_n\left( p_{1}\right) \right) ^v }{v!}=\dfrac{p_n^v }{v!} \end{align*} (since $$\mathbf{f}_n\left( p_{1}\right) =p_n$$). Thus, \eqref{darij1.pf.t2.c2.5} rewrites as $$\dfrac{p_n^v }{v!}\in V$$. Hence, \eqref{darij1.pf.t2.c2.parfrac} becomes \begin{align*} \dfrac{p_n^v }{v!\cdot n^v }=a\cdot\underbrace{\dfrac{p_n^v }{\left( n^v \right) ^{i+1}}}_{\in V}+b\cdot\underbrace{\dfrac{p_n^v }{v!}}_{\in V}\in V \end{align*} (since $$V$$ is a ring and since $$a$$ and $$b$$ are integers). In view of \eqref{darij1.pf.t2.c2.1}, this rewrites as $$z_{\lambda}^{-1}p_{\lambda}\in V$$. Hence, $$z_{\lambda}^{-1}p_{\lambda}\in V$$ has been proved in Case 2.

Let us finally consider Case 3. In this case, not all parts of $$\lambda$$ are equal.

We need another notation: If $$\alpha=\left( \alpha_{1},\alpha_{2} ,\ldots,\alpha_{m}\right)$$ and $$\beta=\left( \beta_{1},\beta_{2} ,\ldots,\beta_n\right)$$ are two partitions, then $$\alpha\sqcup\beta$$ shall denote the partition obtained by sorting the tuple $$\left( \alpha_{1} ,\alpha_{2},\ldots,\alpha_{m},\beta_{1},\beta_{2},\ldots,\beta_n\right)$$ in weakly decreasing order. For instance, $$\left( 3,2,2\right) \sqcup\left( 5,3,2\right) =\left( 5,3,3,2,2,2\right)$$.

It is easy to see that if $$\alpha$$ and $$\beta$$ are two partitions that have no part in common, then $$$$z_{\alpha\sqcup\beta}=z_{\alpha}z_{\beta}. \label{darij1.pf.t2.c3.zaub} \tag{5}$$$$ Moreover, if $$\alpha$$ and $$\beta$$ are any two partitions, then $$$$p_{\alpha\sqcup\beta}=p_{\alpha}p_{\beta}. \label{darij1.pf.t2.c3.paub} \tag{6}$$$$

Now, recall that not all parts of $$\lambda$$ are equal. Hence, we can write $$\lambda$$ in the form $$\lambda=\alpha\sqcup\beta$$ where $$\alpha$$ and $$\beta$$ are two nonempty partitions that have no part in common. (Indeed, we can define $$\alpha$$ and $$\beta$$ by choosing an arbitrary part $$i$$ of $$\lambda$$, then letting $$\alpha$$ be the partition consisting of all parts of $$\lambda$$ equal to $$i$$, while $$\beta$$ is the partition consisting of all remaining parts of $$\lambda$$.) Consider these $$\alpha$$ and $$\beta$$. It is easy to see that $$\left\vert \alpha\right\vert <\left\vert \alpha\sqcup\beta\right\vert$$ (since $$\beta$$ is nonempty) and $$\ell\left( \alpha\right) <\ell\left( \alpha\sqcup\beta\right)$$ (for the same reason). Since $$\alpha\sqcup \beta=\lambda$$, these two inequalities rewrite as $$\left\vert \alpha \right\vert <\left\vert \lambda\right\vert$$ and $$\ell\left( \alpha\right) <\ell\left( \lambda\right)$$. Adding these two inequalities together, we obtain \begin{align*} \left\vert \alpha\right\vert + \ell\left( \alpha\right) <\left\vert \lambda\right\vert +\ell\left( \lambda\right) =N. \end{align*} Hence, $$z_{\alpha}^{-1}p_{\alpha}\in V$$ (by our induction hypothesis, applied to $$\alpha$$ instead of $$\lambda$$). Similarly, $$z_{\beta}^{-1}p_{\beta}\in V$$. However, from \eqref{darij1.pf.t2.c3.zaub} and \eqref{darij1.pf.t2.c3.paub}, we obtain \begin{align*} z_{\alpha\sqcup\beta}^{-1}p_{\alpha\sqcup\beta}=\left( z_{\alpha}z_{\beta }\right) ^{-1}p_{\alpha}p_{\beta}=\underbrace{z_{\alpha}^{-1}p_{\alpha}}_{\in V}\cdot\underbrace{z_{\beta}^{-1}p_{\beta}}_{\in V}\in V \end{align*} (since $$V$$ is a ring). In view of $$\alpha\sqcup\beta=\lambda$$, this rewrites as $$z_{\lambda}^{-1}p_{\lambda}\in V$$. Hence, $$z_{\lambda}^{-1}p_{\lambda}\in V$$ has been proved in Case 3.

We have now proved $$z_{\lambda}^{-1}p_{\lambda}\in V$$ in all three cases 1, 2 and 3. Thus, the induction step is complete.

Thus, Theorem 2 is proved by induction. $$\blacksquare$$

We are not quite ready to prove Theorem 1 yet. We first need some more notations.

We let $$\operatorname{QPar}$$ be the set of all partitions that have no part divisible by $$q$$. (If $$q=2$$, these are precisely the partitions into odd parts.)

We let $$J$$ be the ideal of the ring $$\Lambda_{\mathbb{Q}}$$ generated by the $$p_{i}$$ with $$i\equiv0\mod q$$. In other words, $$J=\sum\limits_{i=1}^{\infty}p_{iq}\Lambda_{\mathbb{Q}}$$. Recall that the family $$\left( p_{\lambda}\right) _{\lambda\in\operatorname{Par}}$$ is a basis of the $$\mathbb{Q}$$-vector space $$\Lambda_{\mathbb{Q}}$$. Thus, the $$\mathbb{Q}$$-vector subspace $$J$$ of $$\Lambda_{\mathbb{Q}}$$ has basis $$\left( p_{\lambda }\right) _{\lambda\in\operatorname{Par}\setminus\operatorname{QPar}}$$ (because multiplying any $$p_{\mu}$$ by a $$p_{iq}$$ yields a $$p_{\lambda}$$ with $$\lambda\in\operatorname{Par}\setminus\operatorname{QPar}$$, and conversely, any $$p_{\lambda}$$ with $$\lambda \in \operatorname{Par}\setminus\operatorname{QPar}$$ can be obtained in such a way).

A well-known fact (or easy exercise) in abstract algebra says the following:

Lemma 4. Let $$B$$ be a subring of a ring $$A$$. Let $$I$$ be a (two-sided) ideal of $$A$$. Then, $$B+I$$ is a subring of $$A$$.

Applying Lemma 4 to $$A=\Lambda_{\mathbb{Q}}$$, $$B=\Lambda_{\mathbb{Z}_{\left( q\right) }}$$ and $$I=J$$, we conclude that $$\Lambda_{\mathbb{Z}_{\left( q\right) }}+J$$ is a subring of $$\Lambda_{\mathbb{Q}}$$. We denote this subring $$\Lambda_{\mathbb{Z}_{\left( q\right) }}+J$$ by $$W$$. We note that $$W$$ is furthermore a $$\mathbb{Z}_{\left( q\right) }$$-subalgebra of $$\Lambda _{\mathbb{Q}}$$ (since $$W$$ is a subring of $$\Lambda_{\mathbb{Q}}$$ and is preserved under scaling by $$\mathbb{Z}_{\left( q\right) }$$).

Next, we observe:

Proposition 5. We have $$V\subseteq W$$.

Proof of Proposition 5. We have \begin{align} \Lambda _{\mathbb{Z}}\subseteq \Lambda _{\mathbb{Z}_{\left( q\right) }}\subseteq \Lambda_{\mathbb{Z}_{\left( q\right) }}+J=W . \end{align} Now, $$W$$ is a commutative ring (since it is a subring of $$\Lambda_{\mathbb{Q}}$$) and contains $$\Lambda_{\mathbb{Z}}$$ as a subring (since $$\Lambda _{\mathbb{Z}}\subseteq W$$). Hence, $$W$$ is a $$\Lambda_{\mathbb{Z}}$$-algebra. Thus, in order to prove that $$V\subseteq W$$, it suffices to show that $$\dfrac{p_{i}}{i^{k}}\in W$$ for each positive integer $$i$$ and each nonnegative integer $$k$$ (by the definition of $$V$$).

So let us show this. Fix a positive integer $$i$$ and a nonnegative integer $$k$$. We must prove that $$\dfrac{p_{i}}{i^{k}}\in W$$. If $$i\equiv0\mod q$$, then this follows from the obvious fact that $$\dfrac{p_{i}}{i^{k}}\in J\subseteq\Lambda_{\mathbb{Z}_{\left( q\right) }}+J=W$$. Thus, we WLOG assume that $$i\not \equiv 0\mod q$$. Hence, $$i$$ is coprime to $$q$$. Hence, $$\dfrac{1}{i}\in \mathbb{Z}_{\left( q\right) }$$, so that $$\dfrac{1}{i^{k}}\in\mathbb{Z}_{\left( q\right) }\subseteq W$$. Now, $$\dfrac{p_{i}}{i^{k}} =\underbrace{\left( \dfrac{1}{i}\right) ^{k}}_{\in W} \underbrace{p_{i}}_{\in \Lambda_{\mathbb{Z}} \subseteq W}\in W$$ (since $$W$$ is a ring). This completes our proof of Proposition 5. $$\blacksquare$$

At last, we can prove Theorem 1.

Proof of Theorem 1. It is clear that $$\mathbb{Z}_{\left( q\right) }\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] \subseteq \Lambda_{\mathbb{Z}_{\left( q\right) }}\cap\mathbb{Q}\left[ p_{i} \ \mid\ i\not \equiv 0\mod q\right]$$. Hence, it suffices to prove the reverse inclusion, i.e., to prove that \begin{align*} \Lambda_{\mathbb{Z}_{\left( q\right) }}\cap\mathbb{Q}\left[ p_{i} \ \mid\ i\not \equiv 0\mod q\right] \subseteq \mathbb{Z}_{\left( q\right) }\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] . \end{align*} Thus, we fix an arbitrary $$f\in\Lambda_{\mathbb{Z}_{\left( q\right) }} \cap\mathbb{Q}\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right]$$. We must prove that $$f\in \mathbb{Z}_{\left( q\right) }\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right]$$.

We have $$f\in\Lambda_{\mathbb{Z}_{\left( q\right) }}\cap\mathbb{Q}\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] \subseteq \mathbb{Q}\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right]$$. Hence, $$f$$ is a $$\mathbb{Q}$$-linear combination of the family $$\left( p_{\lambda}\right) _{\lambda\in\operatorname{QPar}}$$ (since this family $$\left( p_{\lambda}\right) _{\lambda\in\operatorname{QPar}}$$ is a basis of the $$\mathbb{Q}$$-vector space $$\mathbb{Q}\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right]$$). In other words, we can write $$f$$ in the form $$$$f=\sum_{\lambda\in\operatorname{QPar}}c_{\lambda}p_{\lambda} \label{darij1.pf.t1.f=} \tag{7}$$$$ for some $$c_{\lambda}\in\mathbb{Q}$$. Consider these $$c_{\lambda}$$. We shall prove that they all belong to $$\mathbb{Z}_{\left(q\right)}$$.

We let $$\left\langle \cdot,\cdot\right\rangle$$ denote the Hall inner product on $$\Lambda_{\mathbb{Q}}$$. This is a $$\mathbb{Q}$$-bilinear form sending $$\Lambda_{\mathbb{Q}}\times\Lambda_{\mathbb{Q}}$$ to $$\mathbb{Q}$$ and sending $$\Lambda_{\mathbb{Z}_{\left( q\right) }}\times\Lambda_{\mathbb{Z}_{\left( q\right) }}$$ to $$\mathbb{Z}_{\left( q\right) }$$ (since its restriction to $$\Lambda_{\mathbb{Z}_{\left( q\right) }}\times\Lambda_{\mathbb{Z}_{\left( q\right) }}$$ is the Hall inner product on $$\Lambda_{\mathbb{Z}_{\left( q\right) }}$$).

It is well-known (see, e.g., Corollary 2.5.17(b) in Grinberg/Reiner, arXiv:1409.8356v7) that the families $$\left( p_{\lambda}\right) _{\lambda\in\operatorname{Par}}$$ and $$\left( z_{\lambda}^{-1}p_{\lambda }\right) _{\lambda\in\operatorname{Par}}$$ are dual bases of $$\Lambda _{\mathbb{Q}}$$ with respect to the Hall inner product. Hence, $$$$\left\langle p_{\lambda},z_{\mu}^{-1}p_{\mu}\right\rangle =\delta_{\lambda ,\mu} \label{darij1.pf.t1.dualbases} \tag{8}$$$$ for any $$\lambda\in\operatorname{Par}$$ and $$\mu\in\operatorname{Par}$$ (where the $$\delta_{\lambda,\mu}$$ is a Kronecker delta). In other words, $$$$\left\langle p_{\lambda},p_{\mu}\right\rangle =z_{\mu}\delta_{\lambda,\mu } \label{darij1.pf.t1.dualbases2} \tag{9}$$$$ for any $$\lambda\in\operatorname{Par}$$ and $$\mu\in\operatorname{Par}$$.

Now, fix a partition $$\mu\in\operatorname{QPar}$$. Then, Theorem 2 (applied to $$\lambda=\mu$$) yields $$z_{\mu}^{-1}p_{\mu}\in V\subseteq W$$ (by Proposition 5). Hence, $$z_{\mu}^{-1}p_{\mu}\in W=\Lambda_{\mathbb{Z}_{\left( q\right) } }+J$$. In other words, we can write $$z_{\mu}^{-1}p_{\mu}$$ in the form $$z_{\mu }^{-1}p_{\mu}=w_{1}+w_{2}$$ for some $$w_{1}\in\Lambda_{\mathbb{Z}_{\left( q\right) }}$$ and some $$w_{2}\in J$$. Consider these $$w_{1}$$ and $$w_{2}$$.

Now, it is easy to see that $$\left\langle f,w_{1}\right\rangle \in \mathbb{Z}_{\left( q\right) }$$. [Proof: We have $$f\in\Lambda _{\mathbb{Z}_{\left( q\right) }}\cap\mathbb{Q}\left[ p_{i}\ \mid \ i\not \equiv 0\mod q\right] \subseteq\Lambda_{\mathbb{Z} _{\left( q\right) }}$$ and $$w_{1}\in\Lambda_{\mathbb{Z}_{\left( q\right) }}$$. Thus, $$\left(f, w_1\right) \in \Lambda_{\mathbb{Z}_{\left( q\right) }}\times\Lambda_{\mathbb{Z}_{\left( q\right) }}$$. Hence, $$\left\langle f,w_{1}\right\rangle \in\mathbb{Z}_{\left( q\right) }$$, because the Hall inner product $$\left\langle \cdot,\cdot\right\rangle$$ sends $$\Lambda_{\mathbb{Z}_{\left( q\right) }}\times\Lambda_{\mathbb{Z} _{\left( q\right) }}$$ to $$\mathbb{Z}_{\left( q\right) }$$.]

Furthermore, it is easy to see that $$\left\langle f,w_{2}\right\rangle =0$$. [Proof: We have $$w_{2}\in J$$; thus, we can write $$w_{2}$$ as a $$\mathbb{Q}$$-linear combination of the family $$\left( p_{\lambda}\right) _{\lambda \in\operatorname{Par}\setminus\operatorname{QPar}}$$ (since this family is a basis of the $$\mathbb{Q}$$-vector space $$J$$). In other words, we can write $$w_{2}$$ in the form $$$$w_{2}=\sum_{\beta\in\operatorname{Par}\setminus\operatorname{QPar}}d_{\beta }p_{\beta} \label{darij1.pf.t1.w2=} \tag{10}$$$$ for some coefficients $$d_{\beta}\in\mathbb{Q}$$. Consider these $$d_{\beta}$$. From \eqref{darij1.pf.t1.f=} and \eqref{darij1.pf.t1.w2=}, we obtain \begin{align*} \left\langle f,w_{2}\right\rangle & =\left\langle \sum_{\lambda \in\operatorname{QPar}}c_{\lambda}p_{\lambda},\sum_{\beta\in \operatorname{Par}\setminus\operatorname{QPar}}d_{\beta}p_{\beta }\right\rangle \\ & =\sum_{\lambda\in\operatorname{QPar}}\ \ \sum_{\beta\in\operatorname{Par} \setminus\operatorname{QPar}}c_{\lambda}d_{\beta}\underbrace{\left\langle p_{\lambda},p_{\beta}\right\rangle }_{\substack{=z_{\beta}\delta _{\lambda,\beta}\\\text{(by \eqref{darij1.pf.t1.dualbases2})}}}\\ & =\sum_{\lambda\in\operatorname{QPar}}\ \ \sum_{\beta\in\operatorname{Par} \setminus\operatorname{QPar}}c_{\lambda}d_{\beta}z_{\beta}\underbrace{\delta _{\lambda,\beta}}_{\substack{=0\\\text{(since }\lambda\neq\beta \\\text{(because }\lambda\in\operatorname{QPar}\\\text{whereas }\beta \in\operatorname{Par}\setminus\operatorname{QPar}\text{))}}}\\ & =\sum_{\lambda\in\operatorname{QPar}}\ \ \sum_{\beta\in\operatorname{Par} \setminus\operatorname{QPar}}c_{\lambda}d_{\beta}z_{\beta}0=0, \end{align*} qed.]

From $$z_{\mu}^{-1}p_{\mu}=w_{1}+w_{2}$$, we obtain \begin{align*} \left\langle f,z_{\mu}^{-1}p_{\mu}\right\rangle =\left\langle f,w_{1} +w_{2}\right\rangle =\underbrace{\left\langle f,w_{1}\right\rangle } _{\in\mathbb{Z}_{\left( q\right) }}+\underbrace{\left\langle f,w_{2} \right\rangle }_{=0}\in\mathbb{Z}_{\left( q\right) }. \end{align*}

On the other hand, from \eqref{darij1.pf.t1.f=}, we obtain \begin{align*} \left\langle f,z_{\mu}^{-1}p_{\mu}\right\rangle & =\left\langle \sum _{\lambda\in\operatorname{QPar}}c_{\lambda}p_{\lambda},z_{\mu}^{-1}p_{\mu }\right\rangle =\sum_{\lambda\in\operatorname{QPar}}c_{\lambda} \underbrace{\left\langle p_{\lambda},z_{\mu}^{-1}p_{\mu}\right\rangle }_{\substack{=\delta_{\lambda,\mu}\\\text{(by \eqref{darij1.pf.t1.dualbases})} }}=\sum_{\lambda\in\operatorname{QPar}}c_{\lambda}\delta_{\lambda,\mu}\\ & =c_{\mu}. \end{align*} Hence, \begin{align*} c_{\mu}=\left\langle f,z_{\mu}^{-1}p_{\mu}\right\rangle \in\mathbb{Z}_{\left( q\right) }. \end{align*}

Forget that we fixed $$\mu$$. We thus have shown that $$c_{\mu}\in\mathbb{Z} _{\left( q\right) }$$ for each $$\mu\in\operatorname{QPar}$$. In other words, $$c_{\lambda}\in\mathbb{Z}_{\left( q\right) }$$ for each $$\lambda \in\operatorname{QPar}$$. Hence, \eqref{darij1.pf.t1.f=} becomes \begin{align*} f=\sum_{\lambda\in\operatorname{QPar}}\underbrace{c_{\lambda}}_{\in \mathbb{Z}_{\left( q\right) }}p_{\lambda}\in\sum_{\lambda\in \operatorname{QPar}}\mathbb{Z}_{\left( q\right) }p_{\lambda}=\mathbb{Z} _{\left( q\right) }\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] \end{align*} (by the definition of $$\operatorname{QPar}$$).

Forget that we fixed $$f$$. We thus have shown that $$f\in\mathbb{Z}_{\left( q\right) }\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right]$$ for each $$f\in\Lambda_{\mathbb{Z}_{\left( q\right) }}\cap\mathbb{Q}\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right]$$. In other words, \begin{align*} \Lambda_{\mathbb{Z}_{\left( q\right) }}\cap\mathbb{Q}\left[ p_{i} \ \mid\ i\not \equiv 0\mod q\right] \subseteq \mathbb{Z}_{\left( q\right) }\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] . \end{align*} As explained, this completes the proof of Theorem 1. $$\blacksquare$$

• Wow, that's a nice generalization, and your elementary proof is pretty neat (it's long because you wrote absolutely all the small details, but the big idea is quite simple). Sep 11, 2021 at 15:11