# Non-Galois cubic extensions

Is there a necessary and sufficient condition for when a cubic extension of $\mathbb{Q}$ is not a Galois extension?

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It is sufficient that the extension is with a real root of $x^3 - m$, with $m \neq 0,1$ without cube factors. (Since the extension is not normal in this case). OK, I now see you wanted "iff" conditions, I didn't notice that until now. Sorry, I'll leave this comment anyway... – Andy Jul 7 '12 at 12:38
@QRH you should consider accepting an answer to this question – Stella Biderman Jan 2 at 19:00

Theorem. Let $\alpha$ be a primitive element of a cubic extension $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$. Then $\mathbb{Q}(\alpha)$ is Galois over $\mathbb{Q}$ if and only if the discriminant of the irreducible polynomial of $\alpha$ is a rational square.

Proof. Let $f(x)$ be the monic irreducible of $\alpha$, which must be a cubic. If $\alpha=\alpha_1$, $\alpha_2$, and $\alpha_3$ are the roots of $f(x)$, then the discriminant of $f(x)$ is equal to $$\Delta = (r_1-r_2)^2(r_1-r_3)^2(r_2-r_3)^2;$$ the splitting field contains $\sqrt{\Delta}=(r_1-r_2)(r_1-r_3)(r_2-r_3)$. If the extension if Galois, then all three roots lie in $\mathbb{Q}(\alpha)$, hence so does $\sqrt{\Delta}$. Since $[\mathbb{Q}[\sqrt{\Delta}]\colon\mathbb{Q}]$ is either $1$ or $2$, and would have to divide $[\mathbb{Q}(\alpha)\colon\mathbb{Q}]=3$, it follows that $\sqrt{\Delta}\in\mathbb{Q}$; that is $\Delta$ is a rational square as claimed.

Conversely, suppose that $\mathbb{Q}(\alpha)$ is not Galois. Then the splitting field of $\alpha$ over $\mathbb{Q}$ has degree $6$, and the cubic extension is not normal. Hence, the Galois group of $f(x)$ over $\mathbb{Q}$ is isomorphic to $S_3$ (it cannot be cyclic of order $6$, because then the unique subextension $\mathbb{Q}(\alpha)$ of degree $3$ would be Galois). But that means that there is an automorphism of the splitting field $K$ that transposes $r_1$ and $r_2$ and fixes $r_3$; this automorphism does not fix $\sqrt{\Delta}=(r_1-r_2)(r_1-r_3)(r_2-r_3)$; therefore, $\sqrt{\Delta}\notin\mathbb{Q}$, so $\Delta$ is not a rational square, as claimed. $\Box$

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This proof generalizes with $\mathbb{Q}$ replaced with any field of characteristic not equal to $2$. In characteristic $2$ the discriminant is always a square... but see math.uconn.edu/~kconrad/blurbs/galoistheory/… . – Qiaochu Yuan Jul 7 '12 at 18:13
@QiaochuYuan: Thanks for the comment! – Arturo Magidin Jul 7 '12 at 18:18
This is explained at the beginning of chapter 4 [here][1]. [1]:jmilne.org/math/CourseNotes/FT.pdf – DonAntonio Jul 8 '12 at 2:36

In a rather trivial way, a cubic extension $\,\Bbb Q(\alpha)/\Bbb Q\,$ is not Galois iff it is not normal iff the minimal polynomial of $\,\alpha\,$ in $\,\Bbb Q[x]\,\,,\,\,p(x)\,$ say, has only one root in $\,\Bbb Q(\alpha)\,$, namely $\,\alpha\,$ itself, iff the quadratic $$\frac{p(x)}{x-\alpha}\in\Bbb Q(\alpha)[x]$$ is irreducible there...

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