# A local-global problem concerning roots of polynomials

Let $f(x)$ be a polynomial with integer coefficients, irreducible over the integers. Suppose that for all primes $p$, $f$ has a zero in the field $\mathbb{Q}_p(\sqrt{2})$. Here $\mathbb{Q}_p$ denotes the field of $p$-adic numbers. Must $f$ have a zero in the field $\mathbb{Q}(\sqrt{2})$?

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Why are you insisting $f(x)$ is irreducible over the integers (or, more or less the same, over ${\mathbf Q}$) instead of over the field ${\mathbf Q}(\sqrt{2})$? If $K$ is a number field and a polynomial $f(x) \in K[x]$ is irreducible over $K$ and has a root in every (or all but finitely many) non-archimedean completion of $K$ then $f(x)$ is linear. – KCd Nov 16 '11 at 6:12
KCd, In any case I need irreducibility over the integers otherwise I can multiply together a bunch of polynomials of the form $x^2-a$ to get a polynomial with roots on $\mathbb{Q}_p$ for all $p$ but no root in $Q(\sqrt{2})$. This is done (more or less) in Borevich and Shafarevich using quadratic reciprocity and Hensel's Lemma. The fact (about completions) that you mention is a consequence of part of Ex. 6.1 page 362 in Cassels and Frohlich. So far I have not been able to use the reasoning there to answer my question. Let me add that I am very far from an expert in Algebraic Number Theory. – gaddy Nov 16 '11 at 7:17
As KCd says, it would be more natural to ask for $f$ to be irreducible over $\mathbb{Q}$; then the answer is "yes". Also, if I replace $\mathbb{Q}(\sqrt{2})$ by other number fields, the answer is "no". Let $a$, $b$ and $c$ be the roots of $x^3-x-1$. Let $K = \mathbb{Q}(a,b,c)$, and let $f(x)=(x^2-a)(x^2-b)(x^2-c)=x^6-x^2-1$, which is irreducible over $\mathbb{Q}$. Since $abc=1$, for any odd prime $p$, one of $a$, $b$ and $c$ is square in the residue field of $K_p$. Thus, by Hensel's lemma, at least one of them is square in $K_p$. But I can't find an example with $K$ quadratic. – David Speyer Apr 14 '13 at 15:05

Corollary: Let $K = \mathbb{Q}(\sqrt{d})$ be a quadratic extension of $\mathbb{Q}$ with $d$ a square free integer and let $M$ be a proper finite extension of $K$. Then there are infinitely many prime numbers $p$ with $\left( \frac{d}{p} \right)=1$ which have no prime divisor in $M$ of first degree.
In other words, for any $f$ irreducible over $\mathbb{Q}$ which does not split completely over $K$, there are infinitely many $p$ for which $p = \pi \bar{\pi}$ in $K$, with $K_{\pi} \cong \mathbb{Q}_p$ and $f$ having no roots in $K_{\pi}$.