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If $(\mathbb{Q}[x]/(f))/\mathbb{Q}$ is Galois, then $f$ splits modulo all $p$ if $Gal((\mathbb{Q}[x]/(f))/\mathbb{Q})$ is not cyclic (because the decomposition group is cyclic and thus strictly smaller than the Galois group). Inspired by this, I was interested in seeing how strongly the "if $(\mathbb{Q}[x]/(f))/\mathbb{Q}$ is Galois" condition is:

I would like to ask for an example of a polynomial $g$ irreducible over $\mathbb{Q}$ where $(\mathbb{Q}[x]/(g))/\mathbb{Q}$ is not Galois; for some reason I am having trouble constructing one, are there such examples?

Tejas Rao
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    Hint: Try a simple cubic polynomial. – lhf Oct 27 '21 at 16:32
  • Yes, a cubic polynomial with exactly one real root with have a quotient group isomorphic to a subfield of $\mathbb R,$ so it cannot have the complex roots. – Thomas Andrews Oct 27 '21 at 16:40
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    It’s actually very rare for $\mathbb Q[x]/(f)$ to be a splitting field. It happens for a few classes of polynomials, like all quadratics, the cyclotomic polynomials, and some others. – Thomas Andrews Oct 27 '21 at 16:55

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The simplest example is $f=x^3-2$. The field $\mathbb{Q}[x]/(f)$ is isomorphic to $\mathbb{Q}(\sqrt[3]{2})$ and to the fields $\mathbb{Q}(\omega\sqrt[3]{2})$, $\mathbb{Q}(\omega^2\sqrt[3]{2})$ where $\omega = \exp(2\pi i /3)$ is a 3rd root of unity.

This field cannot be Galois as $\mathbb{Q}(\sqrt[3]{2})$ only contains the real roots of $f$ and $f$ has two other complex roots.

Rdrr
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Reduction to finding a non-Galois extension

Suppose we found a finite extension $K/\Bbb Q$ that is not Galois. By the primitive element theorem, $K = \Bbb Q(\alpha)$ for some $\alpha \in K$. Let $g$ be the minimal polynomial of $\alpha$ over $\Bbb Q$. Then $\Bbb Q[x]/(g) \cong K$, so $g$ is the polynomial you seek.

So, if you know that $\Bbb Q(\sqrt[3]2)/\Bbb Q$ is not Galois (this should be an example in any introduction to Galois theory), then you know that $g = x^3-2$ works, because $g$ is the minimal polynomial of $\sqrt[3]2$.

Reduction to finding a non-normal subgroup of a finite group

Let's say you only know about Galois extensions (which can be produced by taking splitting fields of irreducible polynomials with degree $>1$). Say $L/\Bbb Q$ is Galois. Then, by Galois correspondence, subgroups of $G = \operatorname{Gal}(L/\Bbb Q)$ correspond to subextensions of $L/\Bbb Q$, and normal subgroups correspond to Galois subextensions.

Therefore, if you have a non-normal subgroup $H$ of $G$, then $L^H/\Bbb Q$ is the non-Galois extension you seek.

The simplest example would be to take the splitting field of $x^3-2$, which would have Galois group $S_3$ (the permutation group on the roots), and $C_2$ is a non-normal subgroup, and if you pick the right copy of $C_2$, then you would end up with $\Bbb Q(\sqrt[3]2)/\Bbb Q$.

How to find Galois extensions with Galois group $S_n$

This post gives an algorithm to finding Galois extensions with Galois group $S_n$, which would be helpful as they have a lot of non-normal subgroups (and this is a finite problem!)

The algorithm (in the proof of "Theorem") is to take the splitting field of $f = -15f_1 + 10f_2 + 6f_3$ where:

  • $f_1$ is a monic irreducible polynomial of degree $n$ in $(\Bbb Z/2\Bbb Z)[x]$
  • $f_2$ and $f_3$ are produced similarly by taking products of monic irreducible polynomials of the correct degrees in $(\Bbb Z/3\Bbb Z)[x]$ and $(\Bbb Z/5\Bbb Z)[x]$ respectively.

Note that these irreducible polynomials of degree $r$ in $(\Bbb Z/p\Bbb Z)[x]$ can be found by factoring $x^{p^r}-x$ mod $p$, which can be done by e.g. pari/gp, and then taking the degree $r$ factors.

Kenny Lau
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