The problem states:

Let $f (x) = x^3+px+q$ be an irreducible cubic polynomial with rational coefficients and let $K$ be the splitting field of $ f(x) $ over $\mathbb{Q}$. Prove that $ [K : \mathbb{Q}] = 3 $ if and only if $ -4p^3 - 27q^2 $ is a square in $\mathbb{Q}$.

Here is how far I've come on the problem. Since $f$ is cubic and the degree of the extension is $3$, then non of the roots are in $\mathbb{Q}$ hence they are in $K$. Now, I am having trouble connecting this piece of information with the given value being a square in $\mathbb{Q}$.

Thanks in advance.


Hint: if $[K:\mathbb{Q}] = 3$, then $K$ is generated by adjoining any root $\alpha$ of $f$ to $\mathbb{Q}$. Knowing this, what can you say about the automorphisms of $K$? Using this information, what information do you get about the Galois group? Finally, what is true of the discriminant for a cubic whose Galois group is $A_{3}$? These ideas should get you started.


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