= Solution
If a monic <polynomial> $f$ has roots $\alpha_1,\ldots,\alpha_n$ in a <splitting field>, its <polynomial discriminant> is
$$
\operatorname{disc}(f)
=\prod_{1\leq i<j\leq n}(\alpha_i-\alpha_j)^2.
$$
It is a symmetric polynomial in the roots, so it belongs to the base <field>; it is nonzero exactly when $f$ is a <separable polynomial>.
Put
$$
D=\prod_{i<j}(\alpha_i-\alpha_j),
$$
the <Vandermonde determinant>. A root permutation $\sigma$ sends $D$ to $\operatorname{sgn}(\sigma)D$. Hence the <Galois group of a polynomial> is contained in the <alternating group> $A_n$ exactly when every Galois automorphism fixes $D$, which by the fixed-field property is equivalent to $D\in K$. This implies that $\operatorname{disc}(f)=D^2$ is a square in $K$. Conversely, if $\operatorname{disc}(f)=d^2$ for $d\in K$, then $D^2=d^2$, so $D=\pm d\in K$. The Galois group therefore fixes $D$ and consists of even permutations. This proves the <discriminant criterion for an alternating Galois group>.
Now let
$$
f(T)=T^3-2T+2.
$$
The <rational root theorem> shows that none of $\pm1,\pm2$ is a root, so this cubic is irreducible over $\mathbb Q$. The <discriminant of a depressed cubic> is
$$
\operatorname{disc}(f)
=-4(-2)^3-27(2)^2
=32-108=-76=-4\cdot19.
$$
This is not a square in $\mathbb Q$, so the <Galois group of an irreducible cubic> is $S_3$ over $\mathbb Q$.
Over $K=\mathbb Q(\sqrt{-19})$, the discriminant is the square
$$
-76=(2\sqrt{-19})^2.
$$
The cubic remains irreducible: a root in the degree-two extension $K/\mathbb Q$ would generate over $\mathbb Q$ both a degree-three field, by irreducibility, and a subfield of a degree-two field, contradicting the <tower law for field extensions>. Its Galois group over $K$ is consequently $A_3\cong C_3$.
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