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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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    Not sure I understand the question. You're already assuming $\alpha$ is a root of each $f_i$, and the lift of $\overline{\alpha}$ is unique, so the lift must be $\alpha$, no?2012-04-07
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    thanks ted, the original question contained an typo / error. Hopefully it is clear now.2012-04-07
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    Are you asking if you can lift $\alpha$ to an $\tilde \alpha$ which is a root of all the $f_i$ simultaneously? The answer is no: for instance, consider $f_1(x) = x$ and $f_2(x) = x - p$. Then $\alpha = 0$ is a root of $\bar f_1$ and $\bar f_2$ but you have no hope of lifting it to a simultaneous root of $f_1$ and $f_2$, because there aren't any.2012-04-08
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    Hi David. No, I am asking whether for fixed polynomials, in the case where there is a simultaneous root mod p does that root lift to the same root in $\mathbb{Z}_p$ if p >> 0. In the example you give, If q is a large prime (larger than p) then there are no simultaneous solutions mod q.2012-04-08
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    The question does not seem well formulated to me. If $\alpha$ is a simultaneous root mod $p$, then $\alpha$ implicitly fixes $p$; one cannot interpret it as a root modulo any other prime. So what does it mean that $p\gg0$? Very large with respect to what exactly?2012-04-08
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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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