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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? – Ted Apr 7 '12 at 20:11
thanks ted, the original question contained an typo / error. Hopefully it is clear now. – cal7 Apr 7 '12 at 20:50
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. – David Loeffler Apr 8 '12 at 8:22
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. – cal7 Apr 8 '12 at 9:34
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? – Marc van Leeuwen Apr 8 '12 at 11:19

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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by "only finitely many prime ideals of $L$ appear in the differences" do you mean the prime ideal decomposition of the ideal $(\alpha -\beta_j)$? – cal7 Apr 9 '12 at 8:20
@cal7: yes (with varying $\beta_j$'s). – user18119 Apr 9 '12 at 10:43
thank you very much for your answer QiL. – cal7 Apr 9 '12 at 11:26
actually, I am still not quite there. What is $\alpha$ in the answer? if $\alpha \in \mathbb{Z}_p$ and $\beta_j \in L$ how do you make sense of $\alpha -\beta_j$? – cal7 Apr 9 '12 at 13:20
@cal7 I though your $\alpha$ was a lifting. In fact, you don't need to introduce $\alpha$, just suppose $\alpha_i\equiv \alpha_j \mod p$. – user18119 Apr 12 '12 at 5:47

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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thank you for this answer. – cal7 Apr 12 '12 at 8:29
Let me just spell out a consequence of this, to see if I follow. If $f_1$, $f_2$ were irreducible over $\mathbb{Q}$. Then for "p sufficiently large" either they would have no common solutions mod p (the case $f_1\neq f_2$) or they would have the same solutions (because $f_1=f_2$). right? – cal7 Apr 12 '12 at 8:50
Yes. ${}{}{}{}$ – Gerry Myerson Apr 12 '12 at 12:34

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