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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 $$v_2(X^\lambda_\rho(-1))\ge v_2(z_\rho)+\ell(\lambda)-\ell(\rho).$$

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    $\begingroup$ 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 $v_2(b_\rho^\lambda)\geq 0$. $\endgroup$ Sep 4, 2021 at 18:21
  • $\begingroup$ 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 $\Lambda \cap \mathbb{Q}[p_1,p_3,p_5,\cdots]$. $\endgroup$ Sep 4, 2021 at 18:43
  • $\begingroup$ 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? $\endgroup$ Sep 5, 2021 at 2:10
  • $\begingroup$ 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 ... $\endgroup$ Sep 5, 2021 at 2:13
  • $\begingroup$ ... 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. $\endgroup$ Sep 5, 2021 at 2:16

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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 $i<j$ 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 $i<j$. 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 $\left\vert \lambda\right\vert +\ell\left( \lambda\right) <N$. 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 \begin{equation} 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} \end{equation}

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 $<v$. In other words, if $\mu\in\operatorname{Par}_{v}$ satisfies $\mu\neq\lambda$, then $\ell\left( \mu\right) <v$. 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) }_{<v=\ell\left( \lambda\right) }<\left\vert \lambda\right\vert +\ell\left( \lambda\right) =N \end{align*} 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 \begin{equation} 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} \end{equation}

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, \begin{equation} \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} \end{equation} 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 \begin{equation} z_{\alpha\sqcup\beta}=z_{\alpha}z_{\beta}. \label{darij1.pf.t2.c3.zaub} \tag{5} \end{equation} Moreover, if $\alpha$ and $\beta$ are any two partitions, then \begin{equation} p_{\alpha\sqcup\beta}=p_{\alpha}p_{\beta}. \label{darij1.pf.t2.c3.paub} \tag{6} \end{equation}

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 \begin{equation} f=\sum_{\lambda\in\operatorname{QPar}}c_{\lambda}p_{\lambda} \label{darij1.pf.t1.f=} \tag{7} \end{equation} 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, \begin{equation} \left\langle p_{\lambda},z_{\mu}^{-1}p_{\mu}\right\rangle =\delta_{\lambda ,\mu} \label{darij1.pf.t1.dualbases} \tag{8} \end{equation} for any $\lambda\in\operatorname{Par}$ and $\mu\in\operatorname{Par}$ (where the $\delta_{\lambda,\mu}$ is a Kronecker delta). In other words, \begin{equation} \left\langle p_{\lambda},p_{\mu}\right\rangle =z_{\mu}\delta_{\lambda,\mu } \label{darij1.pf.t1.dualbases2} \tag{9} \end{equation} 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 \begin{equation} w_{2}=\sum_{\beta\in\operatorname{Par}\setminus\operatorname{QPar}}d_{\beta }p_{\beta} \label{darij1.pf.t1.w2=} \tag{10} \end{equation} 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. $\blacksquare$

$\endgroup$
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  • $\begingroup$ 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). $\endgroup$ Sep 11, 2021 at 15:11

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