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Does there exist an irreducible polynomial $f \in \mathbb{Z}[X]$ of degree $n \geq 2$ with a zero modulo all primes $p \equiv 3 \pmod 4$?

For example, there is such a polynomial $X^2+1$ if we choose primes $p \equiv 1 \pmod 4$. Mostly using quadratic residues, we can come up with many such polynomials for various collections of primes, but I wasn't able to find one for primes $p \equiv 3 \pmod 4$.

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Let $$f(x)= x^3-3x+4$$ It is irreducible and $$Disc(f) = 4(3)^3-27(-4)^2=-18^2$$

Let $k$ be the splitting field of $f\bmod p$. Factorize $$f(x)=\prod_{j=1}^3 (x-a_j)\in k[x]$$ Note that $$Disc(f)^{1/2}=(a_1-a_2)(a_1-a_3)(a_2-a_3)\in k, \qquad Disc(f)=\prod_{i\ne j} (a_i-a_j)$$

Because of the Frobenius automorphism, if $f\bmod p$ is irreducible then $k=\Bbb{F}_p[x]/(f(x))$ ie. $[k:\Bbb{F}_p]=3$ which implies that $k$ doesn't contain any quadratic subfield ie. $Disc(f)^{1/2}\in \Bbb{F}_p$.

And (exlucding the case $p=3$ where $f=(x+1)^3$ is reducible) since we know that $Disc(f)^{1/2}\in \Bbb{F}_p$ iff $p\not\equiv 3\bmod 4$ we get that $f$ is never irreducible when $p\equiv 3\bmod 4$, ie. $f\bmod p$ has a root.

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  • $\begingroup$ Nice idea to convert the irreducibility of a quadratic polynomial to the reducibility of a cubic one over a finite field. $\endgroup$ – Ege Erdil Jun 7 '20 at 20:11
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    $\begingroup$ It is because I played a lot with the Artin L-functions of the S_3 cubic, I remembered the character table 2 -1 -1 0 0 0 (dimension 2 multiplicity 2) , 1 1 1 -1 -1 -1, 1 1 1 1 1 1 $\endgroup$ – reuns Jun 7 '20 at 20:12
  • $\begingroup$ Thank you both! $\endgroup$ – David Popović Jun 7 '20 at 20:41
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    $\begingroup$ @DavidPopović use $x^3-3x+4$ instead which is irreducible of discriminant $-18^2$. $\endgroup$ – user760870 Jun 8 '20 at 16:20
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    $\begingroup$ By Kummer Theory, an $S_3$ extension $L/\mathbf{Q}$ which contains $\mathbf{Q}(\sqrt{-1})$ has the shape $L(\sqrt{-3}) = \mathbf{Q}(\sqrt{-3},\sqrt{-1},\alpha^{1/3})$ where $\alpha \in \mathbf{Q}(\sqrt{3})$ and the norm of $\alpha$ is a rational cube. Descending back down to the cubic field, you can take a minimal polynomial of $(a+b \sqrt{3})^{1/3} + (a-b \sqrt{3})^{1/3}$ as long as the norm of $\alpha = a + b \sqrt{3}$ has cube norm but is not a cube. This is close to equivalent (in the end) to your computation with the discriminant, except to note that $\alpha$ should not be cube itself. $\endgroup$ – user760870 Jun 9 '20 at 11:46
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Again partial answer

Not if $K/\Bbb{Q}$ is Galois where $K=\Bbb{Q}[x]/(f(x))$,

This is because $f\bmod p$ has a root (and $p$ unramified) means that $pO_K$ splits completely, and the density of primes that split completely in a Galois extension of degree $n$ can't be $\ge 1/n$, since this would contradict that $$\zeta_K(s)= F(s)\prod_{p \text{ splits completely}} \frac1{(1-p^{-s})^{\deg(f)}}$$ has a simple pole at $s=1$

(where $ F(s)$ is an Euler product which converges for $ \Re(s) > 1/2$)

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  • $\begingroup$ One can see this also as a special case of the Chebotarev density theorem, since a totally split prime in this case must have a trivial Frobenius element, and the density of such primes has to be exactly $ 1/n $ since the conjugacy class of the identity is always trivial. It is more general than my answer for degrees $ > 2 $, though it is not sharp enough by itself to prove the result in the case $ n = 2 $. $\endgroup$ – Ege Erdil Jun 7 '20 at 19:14
  • $\begingroup$ I'm not sure, but I think we can't push density methods too far, since by the same Chebotarev argument, a degree $ n $ polynomial with Galois group $ S_n $ has a root modulo $ p $ if the Frobenius element of $ p $ in its splitting field has a fixed point, and the density of permutations with fixed points in $ S_n $ tends to $ 1 - 1/e > 1/2 $ as $ n \to \infty $ (and is $ > 1/2 $ for all $ n \geq 3 $), so there's no density obstruction to the claim being true. $\endgroup$ – Ege Erdil Jun 7 '20 at 19:48
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This is not a complete answer, but under the further assumption that the Galois group of $ f $ is abelian over $ \mathbf Q $, it's easy to show with elementary class field theory that this is impossible.

By Kronecker-Weber the splitting field of $ f $ would be a subfield of some $ \mathbf Q(\zeta_n) $, and the requirement that $ f $ have a root modulo every prime $ 3 $ mod $ 4 $ (in this case, equivalent to $ f $ splitting completely modulo every prime $ 3 $ mod $ 4 $) would imply that there is a nontrivial (not equal to $ \mathbf Q $) subfield of $ \mathbf Q(\zeta_n) $ in which almost every prime $ 3 $ mod $ 4 $ (modulo finitely many exceptions coming from the discriminant of $ f $) split completely. This in turn would imply that the corresponding Frobenius elements in the Galois group $ (\mathbf Z/n \mathbf Z)^{\times} $ all fix this subfield, but it's easy to see that for all $ n $, the subgroup of $ (\mathbf Z/n \mathbf Z)^{\times} $ generated by the sufficiently large primes $ 3 $ mod $ 4 $ is in fact the whole group, implying that the only subfield of $ \mathbf Q(\zeta_n) $ in which almost every prime $ 3 $ mod $ 4 $ is split is the trivial subfield $ \mathbf Q $.

This essentially rules out all constructions based on Artin reciprocity, which in particular covers all irreducible polynomials of degree $ 2 $, since such polynomials all have Galois group $ C_2 $, which is abelian. I don't know if it's possible to cook up an example using nonabelian methods, however.

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