independence of $p$-adic valuations

We prove the following particular case:

Proposition 1.

Let $p_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{p}_n\mathrm{\in }\mathrm{Z}$ be distinct prime numbers and let $\mathrm{\mid }\mathrm{\cdot }{\mathrm{\mid }}_{p_i}$ be the corresponding $p$-adic valuations of $\mathrm{Q}$. Let $a_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{a}_n\mathrm{\in }\mathrm{Z}$ and let ${ϵ}_i$ be arbitrary positive real numbers, then there exists $y\mathrm{\in }\mathrm{Z}$ such that for all $i\mathrm{=}\mathrm{1}\mathrm{,}\mathrm{\dots }\mathrm{,}n$:

Proof.

Let $p$ be an arbitrary prime, and let $ϵ$ be an arbitrary positive real number. Notice that $\mathbb{Z}$ injects into ${\mathbb{Z}}_p=\underleftarrow{\lim }\mathbb{Z}/p^n\mathbb{Z}$, the $p$-adic integers. For any $b\in \mathbb{Z}$, we also write $b$ for its image in ${\mathbb{Z}}_p$, and it can be written as a sequence $b=(b_j)$ with $b\equiv b_{j}mod{p}^j$. Let $n=n_{p,ϵ}\in \mathbb{N}$ be such that (and thus for any other $c\in \mathbb{Z}$ such that $c\equiv b_{n}mod{p}^n$ we have ).

Now, for the proof of the proposition, let $n_i=n_{p_i,{ϵ}_i}$ and recall that by the Chinese Remainder Theorem we have an isomorphism:

$$\prod _{i=1}^n\mathbb{Z}/p_i^{n_i}\mathbb{Z}\equiv \mathbb{Z}/(\prod p_i^{n_i})\mathbb{Z}$$

Therefore we can find an element $\tilde{y}$ of $\mathbb{Z}/(\prod p_i^{n_i})\mathbb{Z}$ (and thus a lift $y$ of $\tilde{y}$ to $\mathbb{Z}$) such that $y\equiv a_{i}mod{p}_i^{n_i}$ for all $i=1,\mathrm{\dots },n$. Hence:

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