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Find conditions on $a$ and $b$ such that the splitting field of $x^3 +ax+b \in \mathbb Q[x]$ has degree of extension 3 over $\mathbb Q$.

I'm trying solve do this question, it seems very difficult to me, maybe because I don't know Galois Theory yet (The content of next chapters). I need help to prove this without Galois Theory.


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up vote 4 down vote accepted

Partial answer: Let $f(x)=x^3+ax+b$, let $K$ be its splitting field, and $\alpha$, $\beta$ and $\gamma$ be the roots of $f$ in $K$.

First of all, $f$ has to be irreducible, which is the same as saying it doesn't have a rational root: if it's not, and say $\alpha$ is rational, then $f(x)$ factors as $(x-\alpha)g(x)$ with $g$ a quadratic polynomial; then $K$ is also the splitting field of $g$, so it has degree $\leq2$.

So let's assume that $f$ is irreducible. Its splitting field $\mathbf Q(\alpha)$ has degree $\deg(f)=3$, so if we want $[K:\mathbf Q]=3$, we need $K=\mathbf Q(\alpha)$. In other words, we want the two other roots $\beta,\gamma$ to be in $\mathbf Q(\alpha)$. Let's look at the relation between roots and coefficients for $f$: $$ \alpha+\beta+\gamma=0\\ \alpha\beta+\beta\gamma+\gamma\alpha=a\\ \alpha\beta\gamma=-b $$ From the first and the third equation, you see that $\beta+\gamma=-\alpha$ and $\beta\gamma=-b/\alpha=\alpha^2+a$, so $\beta$ and $\gamma$ are the roots of the second degree polynomial $g(y)=y^2+\alpha y +\alpha^2+a\in\mathbf Q(\alpha)$. Those roots are in $\mathbf Q(\alpha)$ if, and only if, the discriminant $\Delta=-3\alpha^2-4a$ of $g$ is a square in $\mathbf Q(\alpha)$.

Now, the problem is that I'm not sure how to determine whether $\Delta$ is a square or not...

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How can you get these relations between roots and coeficients for f? Thank you for your answer :) – user42912 Dec 5 '12 at 21:57
@RafaelChavez By expanding $(x-\alpha)(x-\beta)(x-\gamma)$, and identifying the coefficients with $x^3+ax+b$. – jathd Dec 6 '12 at 1:02
yes, of course, thank you – user42912 Dec 6 '12 at 1:43
A non-zero polynomial $p(x) \in k[X]$ is square-free if $\textrm{gcd}(p(x),p'(x)) = 1$. Also note that since $k$ is a field (assumed to be characteristic 0), even if $p(x)$ is not square-free then it admits a square-free factorization. – Charles Boyd Dec 6 '12 at 3:01

The following conditions should give you a splitting field of degree 3 over $\mathbb{Q}$:

  1. $a,b \in \mathbb{Z}$.
  2. There exists prime $p$ such that $p \mid a$ and $p \mid b$.
  3. $p^2 \nmid b$.

Then by Eistenstein's criterion, then the polynomial is irreducible over $\mathbb{Q}[x]$ and therefore $$\big [ \mathbb{Q}(\alpha) \colon \mathbb{Q} \big ] = \textrm{Deg}(x^3 + ax + b) = 3.$$

You may be able to get a more general result by considering the case where the coefficients (again, $a,b \in \mathbb{Z}$) are coprime ($\textrm{gcd}(a,b) = 1$) and applying Gauss' lemma.

Lemma: A non-constant polynomial $p(x)$ is irreducible over $\mathbb{Z}[x]$ if and only if it is irreducible over $\mathbb{Q}[x]$ and primitive over $ \mathbb{Z}[x]$.

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You seem to be giving a criterion as to when the rupture field has degree $3$. – jathd Dec 5 '12 at 5:20
you need also that the another roots are in $\mathbb Q(\alpha)$ – user42912 Dec 5 '12 at 11:54
@CharlesBoyd Well the sign of the discriminant $\Delta$ will tell you how many roots are real. The splitting field is a real extension of $\mathbf Q$, so we need $\Delta<0$, but I don't think that's enough to know that the splitting field and the rupture field of $x^3+ax+b$ are the same. – jathd Dec 5 '12 at 21:48
If $\Delta > 0$, then $p(x)$ has 3 distinct real roots. If $\Delta = 0$, then $p(x)$ has one real root with multiplicity at least 2 (and possibly another distinct real root), if $\Delta < 0$ then $p(x)$ has one real root and a pair of complex-conjugate roots. – Charles Boyd Dec 5 '12 at 23:57
@CharlesBoyd Oh, right, I mixed up the signs. The point is that you can't have non-real roots (e.g. because the Galois group has order 3, and conjugation would give an element of order 2), so $\Delta\geq0$. And you can't have multiple roots if $x^3+ax+b$ is irreducible (hence separable), so $\Delta>0$. – jathd Dec 6 '12 at 1:08

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