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

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


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, $Gal(f) \cong S_3$.

| cite | improve this answer | |
  • $\begingroup$ 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 :,( $\endgroup$ – abe May 11 '14 at 3:33
  • $\begingroup$ I edited my response with an example. :) $\endgroup$ – Kaj Hansen May 11 '14 at 3:34
  • $\begingroup$ Yours doesn't factor out nicely. I'm using a calculator $\endgroup$ – abe May 11 '14 at 3:37
  • $\begingroup$ OH, you want something that also factors nicely as well? Let me think for a bit. $\endgroup$ – Kaj Hansen May 11 '14 at 3:38
  • $\begingroup$ Could it be that there does not exists a nice polynomial I am looking for? $\endgroup$ – 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!!!

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.