I recently came across the following fact from this list of counterexamples:

There are no polynomials of degree $< 5$ that have a root modulo every prime but no root in $\mathbb{Q}$.

Furthermore, one such example is given: $(x^2+31)(x^3+x+1)$ but I have not been able to prove that this does has that property above. How can such polynomials be generated and can we identify a family of them?


marked as duplicate by Grigory M, Davide Giraudo, Avitus, Ali Caglayan, Mark Fantini Dec 31 '14 at 11:32

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    $\begingroup$ Do you know Galois theory of finite fields? If $p$ is an odd prime and $f(x)$ is a monic irreducible of degree $3$ in $\mathbf F_p[x]$ then the splitting field of $f(x)$ over $\mathbf F_p$ has degree $3$, so the Galois group must be $A_3$ and thus the discriminant of $f(x)$ is a square in $\mathbf F_p$. This implies if $F(x)$ is a monic irreducible cubic in $\mathbf Z[x]$ with discriminant $D$ that is not a square in $\mathbf Z$ then $F(x)(x^2-D)$ has a root mod $p$ for every prime number $p$. For example, $x^3+x+1$ has discriminant $-31$, which is your example. $\endgroup$ – KCd Dec 30 '14 at 23:18
  • $\begingroup$ related: is there an irreducible polynomial that has a root modulo every prime? $\endgroup$ – Grigory M Dec 31 '14 at 10:32
  • $\begingroup$ I have added an answer at math.stackexchange.com/questions/608919/… that explains why your example has a root modulo every prime. $\endgroup$ – Hurkyl Jun 14 '16 at 10:31

If you just want an easy example of polynomial that has root modulo every prime but not in $\mathbb Q$ — just take e.g. $$ (x^2-2)(x^2-3)(x^2-6) $$ (it has this property since the product of two non-squares mod p is a square mod p).

One more interesting example is $x^8-16$ (standard proof uses quadratic reciprocity).

As for possibility of complete description of all such polynomials — I'm skeptical.

  • $\begingroup$ Thanks. Also why can't we find another polynomial of degree $< 5$? KCd gave some sort of an answer in the comments but it did not quite answer the question. $\endgroup$ – Sandeep Silwal Dec 31 '14 at 1:24
  • $\begingroup$ Actually, I don't know (beside quadratic case). You might want to ask a separate question. $\endgroup$ – Grigory M Dec 31 '14 at 12:35

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