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QUESTION: Suppose you are given a polynomial with integer coefficients. And assume it only has complex roots(no real roots). Does it necessarily follow that in a finite field(say mod $p$) that it cannot have roots? Why or why not?

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Nope. Consider $x^{2}+1$ over $\mathbb{F}_{2}$.

More generally, $x^{2^{n}}+1$ has no real roots, but splits completely over $\mathbb{F}_{2}$ as $(x+1)^{2^{n}}$ by Frobenius. You can cook up similar examples without this fact, too: $x^{2}+x+1$ has a root over $\mathbb{F}_{3}$ (namely, $1$), but we can easily see that it has no real roots via the quadratic formula.

If we consider the more general question of how irreducibility over $\mathbb{Q}$ and $\mathbb{R}$ affects irreducibility over finite fields, the results are also interesting. In fact, being irreducible over $\mathbb{Q}$ exerts little control over irreducibility over finite fields. There's the famous example of $f(x) = x^{4}+1 \in \mathbb{Q}[x]$. One can show that $f(x)$ is in fact irreducible over $\mathbb{Q}$, and $f$ clearly has no real roots; however, $f(x)$ is reducible modulo every prime $p$. You can find many references to this fact on this site. Of course, $f(x)$ factors as $(x^{2}-\sqrt{2}x+1)(x^{2}+\sqrt{2}x+1)$ over $\mathbb{R}[x]$, so $f$ is not irreducible over $\mathbb{R}$.

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