$\mathbb{Q}(\alpha)$ extension of degree 3 is galois over $\mathbb{Q}$ if and only if discriminant of minimal polynomial of $\alpha$ is square.

Question

I supposed that $\alpha$, $\beta$ and $\gamma$ are the roots of the minimal polynomial in its splitting field. So the discriminant is $(\alpha-\beta)^2(\alpha-\gamma)^2(\beta-\gamma)^2$. If every automorphism in galois group, fixes the square root of the discriminant, its a square in $\mathbb{Q}$. But now I'm really confused and I don't know what to do!

Answer 1

Let $K = \mathbb{Q}(\alpha)$, and let $f$ be the minimum polynomial of $\alpha$ over $\mathbb{Q}$.

Claim: The extension $K/\mathbb{Q}$ is Galois if and only if $K$ is the splitting field of $f$ over $\mathbb{Q}$.

Proof: The "if" direction follows because splitting fields over $\mathbb{Q}$ are Galois extensions. For the only if direction, use the fact that $K/\mathbb{Q}$ Galois $\implies$ $K/\mathbb{Q}$ normal $\implies$ $f$ splits in $K$. (qed)

Let $L$ be the splitting field of $f$ over $K$. Then $L$ is also the splitting field of $f$ over $\mathbb{Q}$. (Note that this slightly awkward definition of $L$ is necessary so that $L$ is a splitting field of $f$ over $\mathbb{Q}$ and also $K\subseteq L$). Therefore, $K/\mathbb{Q}$ is Galois if and only if $K = L$. So $K/\mathbb{Q}$ is Galois if and only if $[L:\mathbb{Q}] = 3$, which is equivalent to $\lvert\operatorname{Gal}(f)\rvert = 3$.

Since $f$ has three roots in $L$, we have $\operatorname{Gal}(f)\leq S_3$. Since $[K:\mathbb{Q}] = 3$, we have $3 \mid [L:K] = \lvert \operatorname{Gal}(f) \rvert$ by the Tower Law, which means that $A_3 \leq \operatorname{Gal}(f) \leq S_3$.

So $\lvert \operatorname{Gal}(f) \rvert = 3$ if and only if $\operatorname{Gal}(f) = A_3$, which is equivalent to the discriminant being square, by the following Theorem.

Theorem: If $f \in \mathbb{Q}[x]$ has degree $n$ and $n$ distinct roots, then $\operatorname{Gal}(f) \leq A_n$ if and only if the discriminant of $\operatorname{Gal}(f)$ is square in $\mathbb{Q}$.

Answer 2

As they told you above you need check that $\mathbb{Q}(\alpha)$ is the splitting field of the minimal polynomial of $\alpha$ over $\mathbb{Q}$. Let $f(x)=x^3+px+q \in \mathbb{Q}[x]$ irreducible, and $\alpha,\beta,\gamma$ the roots. Then the splitting field of f over $\mathbb{Q}$ is $$\mathbb{Q}(\alpha,\delta)$$ with $\delta = (x-\alpha)(x-\beta)(x-\delta)$. Because, for definition, the splitting field $\mathbb{F}$ of $f$ is $\mathbb{F}=\mathbb{Q}(\alpha,\beta,\gamma)$, but by the Vieta's formulas $\alpha+\beta+\gamma=0$, so $\mathbb{F}=\mathbb{Q}(\alpha,\beta)$, and in the same way for Vieta and some manipulations you can check that $\beta= \dfrac{2p\alpha+3q+\delta}{6\alpha^2+2p}$, so $\mathbb{F}=\mathbb{Q}(\alpha,\delta)$. Clearly $\delta^2=\triangle$, so $\triangle$ is a square in $\mathbb{Q}$ iff $\delta=\pm\sqrt{\triangle} \in \mathbb{Q} \Leftrightarrow \mathbb{F}=\mathbb{Q}(\alpha,\delta)=\mathbb{Q}(\alpha)$. $$ $$ In general, if $g(x)=x^3+ax^2+bx+c \in \mathbb{Q}[x]$, you can eliminate the quadratic term with the substitution $y=x-\frac{a}{3}$, and you get the polynomial $f(y)=y^3+py+q$, with roots $\alpha,\beta,\gamma$, and the roots of $g(x)$ are $\alpha+\frac{a}{3}$, $\beta+\frac{a}{3}$, $\gamma+\frac{a}{3}$. The roots of $f(x)$ and $g(x)$ differ by a racional constant, so the splitting field of $f$ and $g$ over $\mathbb{Q}$ are the same.