A cubic polynomial over $\mathbb{Q}$ with a degree three splitting field?

Question

I'm wondering if it's possible to have a cubic polynomial $f(x)\in\mathbb{Q}$ with three distinct roots (in $\mathbb{C}$) that has a degree three splitting field?

If $f(x)=(x-a)(x-b)(x-c)$, with $a,b,c$ distinct, then I believe they must all be real, because otherwise we would get an automorphism of order two (complex conjugation) in $\operatorname{Gal}(f)\cong \mathbb{Z}/3\mathbb{Z}$, contradicting Lagrange's theorem. Now, I can't cook up an example that works, but I'm not sure how to prove nothing does work either. What am I missing?

$\Bbb{Q}(\cos(2\pi / 7))$ (embed in the abelian extension $\Bbb{Q}(\zeta_7)/\Bbb{Q}$) — reuns, Mar 06 '21 at 17:52
See also https://math.stackexchange.com/questions/1767252/expressing-the-roots-of-a-cubic-as-polynomials-in-one-root — lhf, Mar 06 '21 at 18:12

score 3 · Accepted Answer · answered Mar 06 '21 at 17:53

3

All you need is a cubic polynomial whose discriminant is a square.

E.g. $f(x) = x^3 - 3x + 1$.

Reference: the wiki page on cubic equation

answered Mar 06 '21 at 17:53

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score 2 · Answer 2 · answered Mar 06 '21 at 17:54

Such fields are known as cyclic cubic fields, because their Galois group is cyclic of order 3. Here's the LMFDB's list of cyclic cubic fields, which includes over four thousand examples.

One easy example (and the first entry in the list) is the cubic subfield of the cyclotomic field $\mathbb{Q}(e^{2\pi i/7})$, which has Galois group $\mathbb{Z}/6\mathbb{Z}$ and therefore has a unique cubic subfield.

A cubic polynomial over $\mathbb{Q}$ with a degree three splitting field?

2 Answers2