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After reading comments on an earlier version I have decided to completely restate my question:

Let $f_1(x),\dots, f_n(x) \in \mathbb{Z}[X]$ be polynomials. These are fixed. Let $p$ be a prime. If $\alpha$ is a simple root of $f_1(x),\dots, f_n(x)$. Then for each $i$ with $1 \leq i \leq n$, $\alpha$ 'lifts' to a unique $\alpha_i \in \mathbb{Z}_p$ such $f_i(\alpha_i)=0$ and each $\alpha_i$ is congruent to $\alpha \mod p$.

Now, in general there is no reason why the $\alpha_i$ should be equal. My question is, are they equal if $p$ is sufficiently large, given that we have fixed $f_1,\dots,f_n$?

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    @MarcvanLeeuwen: I think maybe the OP means that $\alpha$ is a simultaneous root of all $f_i(x)$ as polynomials in $\mathbb{Z}[x]$: that is, $f_i(\alpha) = 0$ in $\mathbb{Z}$, not just modulo $p$. In that case the answer to the OP's question is yes if \alpha >0: for any p > \alpha, each $\alpha_i$ is exactly the same as $\alpha$ (of course, $\alpha_i$ is in $\mathbb{Z} / {p_i}\mathbb{Z}$, i.e., is not unique in $\mathbb{Z}$, while $\alpha \in \mathbb{Z}$, so it's a bit wrong to say they are equal, but if we take $\alpha_i$ to be the smallest one...).2012-04-08

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Edit add the statement I want to prove:

Fix non-zero polynomials $f_1, \dots, f_n\in \mathbb Z[X]$. There exists $N>0$ such that for all $p>N$, if $\alpha_1, \dots, \alpha_n\in \mathbb Z_p$ satisfy $f_i(\alpha_i)=0$ for all $i\le n$, then for any pair $i, j\le n$, the condition $\alpha_i\equiv \alpha_j \mod p$ implies $\alpha_i=\alpha_j$.

Each $f_i(x)$ has finite many zeros in $\overline{\mathbb Q}$. Let $L$ be a finite extension of $\mathbb Q$ containing all these zeros $\beta_j$ (when $i$ varies). Only finitely many prime ideals of $L$ appear in the differences $\beta_\ell-\beta_j$. So for $p$ big enough (coprime with the above prime ideals), if $\beta_\ell\ne \beta_j$, then $v_p(\beta_\ell -\beta_j)=0$ and $\beta_\ell\not\equiv \beta_j \mod p$. As $\alpha_1, \dots, \alpha_n$ are among the $\beta_j$'s (one can embed $\mathbb Q[\alpha_1, \dots, \alpha_n]$ in $L$), the statement is proved.

Remark All these have little to do with $f_1,\dots, f_n$ and one don't really have to suppose $\alpha_i\in \mathbb Z_p$ (the condition $\alpha_i\equiv \alpha_j \mod p$ then means that $(\alpha_i-\alpha_j)/p$ is integral over $\mathbb Z_p$).

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    @cal7 I though your $\alpha$ was a lifting. In fact, you don't need to introduce $\alpha$, just suppose $\alpha_i\equiv \alpha_j \mod p$.2012-04-12
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If $f_1$ and $f_2$ have no common factor over the rationals, and $\alpha$ is a root of both of them mod $p$, then their resultant is a non-zero multiple of $p$. So if $f_1$ and $f_2$ are fixed, then there are only finitely many primes $p$ for which they have a common zero. Thus, "$p$ sufficiently large" just doesn't happen, unless the polynomials have a common factor over the rationals, and in that case the question is trivial, isn't it?

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    Yes. ${}{}{}{}$2012-04-12