Construct an irreducible polynomial of degree $3$ over $\mathbb{Q}$

I'm trying to construct an irreducible polynomial of degree 3 with rational coefficients such that this irreducible polynomial has one root in the real numbers and has two complex roots. Everything I try is a failure :(. For ex, $(x-\sqrt 2)(x-i)(x+i)$ doesn't give me an irreducible polynomial over rationals. Can anyone construct one simple one for me? But ** it has to factor out nicely.....**I'm trying to see that this Galois group is nonabelian. Thanks :D

• Try $x^3-2$, for example. – André Nicolas May 11 '14 at 3:42
• @AndréNicolas: Yesss! $x^3 - 2$! – Robert Lewis May 11 '14 at 4:03

Hint:

Any cubic polynomial with negative discriminant will have two complex roots and one real root.

Note that the discriminant is given by:

$$\Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$$

It is a theorem that, if the discriminant of an irreducible cubic is not a perfect square, then $Gal(f) \cong S_3$, which is not abelian.

Here's an example:

$$f(x) = x^3 - x - 1$$

This is an irreducible polynomial with discriminant $\Delta = -23$. Therefore, $Gal(f) \cong S_3$.

• Thanks, but the discriminant doesn't help make a simple cubic polynomial for example, $(2x^2+x^2+x+1)=(2x+1.47797)(x^2-0.238984x+.0676605)$ it's not nice :,( – abe May 11 '14 at 3:33
• I edited my response with an example. :) – Kaj Hansen May 11 '14 at 3:34
• Yours doesn't factor out nicely. I'm using a calculator – abe May 11 '14 at 3:37
• OH, you want something that also factors nicely as well? Let me think for a bit. – Kaj Hansen May 11 '14 at 3:38
• Could it be that there does not exists a nice polynomial I am looking for? – abe May 11 '14 at 3:38

Boy will I be embarrassed if this is wrong, BUT what about $x^3 - 2$? The roots are $\sqrt[3]2$, $\sqrt[3]2 e^{2\pi i / 3}$, $\sqrt[3]2 e^{4\pi i /3}$. One real, and a complex conjugate pair.

$x^3 -2 = (x - \sqrt[3]2)(x^2 + \sqrt[3]2 x + \sqrt[3]4)$ $= (x - \sqrt[3]2)(x - \sqrt[3]2 e^{2\pi i / 3})(x - \sqrt[3]2 e^{4\pi i / 3}), \tag{1}$

the good friend of Galois theorists everywhere! Come on down!

Obviously can be generalized to $x^3 - \alpha$, where $\alpha$ is not a cube in $\Bbb Q$:

$x^3 - \alpha = (x - \sqrt[3] \alpha)(x^2 + \sqrt[3] \alpha x + \sqrt[3] {\alpha^2})$ $= (x - \sqrt[3]\alpha)(x - \sqrt[3]\alpha e^{2\pi i / 3})(x - \sqrt[3]\alpha e^{4\pi i / 3}), \tag{1\2}$

Then shift things in $x$ by setting $x = y - \beta$ for $\beta \in \Bbb Q$:

$x^3 - \alpha \rightarrow y^3 -3\beta y^2 + 3\beta^2 y - \beta^3 - \alpha; \tag{3}$

you get even more cubics meeting your needs! And Spirit of Cardano looks down from Heaven, smiles, and winks at the Ghost of Galois! You three make quite a "group"!

Hope this helps! Cheers,

and as always,

Fiat Lux!!!

• The answer which put me over $10,400$! an important personal goal! Whoopee! – Robert Lewis May 11 '14 at 4:48