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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 $\operatorname{Par}\setminus\operatorname{QPar}$ can be obtained in such a way).

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).

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\Lambda_{\mathbb{Z}_{\left( q\right) }} \cap\mathbb{Q}\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 $c_{\lambda}$.

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 $w_{1}\in\Lambda_{\mathbb{Z}_{\left( q\right) } }$. 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 $\mathbb{Z}_{\left( q\right) }$.]

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 $\mathbb{Z}_{\left(q\right)}$.

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 $\mathbb{Z}_{\left( q\right) }$.]

Proof of Proposition 5. We know that $W$ is a commutative ring (since it is a subring of $\Lambda_{\mathbb{Q}}$) and contains $\Lambda_{\mathbb{Z}}$ as a subring (since $\Lambda _{\mathbb{Z}}\subseteq \Lambda _{\mathbb{Z}_{\left( q\right) }}\subseteq \Lambda_{\mathbb{Z}_{\left( q\right) }}+J=W$). 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, $\dfrac{p_{i}}{i^{k}} =\underbrace{\left( \dfrac{1}{i}\right) ^{k}}_{\in W} \underbrace{p_{i}}_{\in W}\in W$ (since $W$ is a ring). This completes our proof of Proposition 5. $\blacksquare$

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\Lambda _{\mathbb{Z}_{\left( q\right) }}\cap\mathbb{Q}\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\Lambda_{\mathbb{Z}_{\left( q\right) }} \cap\mathbb{Q}\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] $.

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\Lambda _{\mathbb{Z}_{\left( q\right) }}\cap\mathbb{Q}\left[ p_{i}\ \mid \ i\not \equiv 0\mod q\right] . \end{align*} As explained, this completes the proof of Theorem 1. $\blacksquare$

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. $\blacksquare$

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\Lambda_{\mathbb{Z}_{\left( q\right) }} \cap\mathbb{Q}\left[ p_{i}\ \mid\ i\not \equiv 0\mod q\right] $.

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. $\blacksquare$