Non-Galois cubic extensions Is there a necessary and sufficient condition for when a cubic extension of $\mathbb{Q}$ is not a Galois extension?
 A: 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...
A: 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. $\Box$
