Find the degree of the splitting field of the polynomial $(x^2+x+1)(x^3+x+1)$ over $\mathbb{Q}$.

Attempt at Solution:

I know that $(x^2+x+1)$ is irreducible over $\mathbb{Q}$ and I think that $(x^3+x+1)$ is irreducible over $\mathbb{Q}$ since two of its roots are complex. Then I think I can say that the splitting field has subfields. These subfields have the same degrees as their respective polynomials, 2 & 3. I don't know if this is even relevant.

I don't know where to go from here or if I'm even on the right track.

Any help/hints would be greatly appreciated.

  • $\begingroup$ do you some how think degree should be $6$? $\endgroup$ – user87543 Dec 6 '13 at 5:19
  • $\begingroup$ I know its the answer. I just don't know why. Is it just that $2\cdot3=6$ since we have the polynomial $(x^2+x+1)\cdot(x^3+x+1)$, so we multiply the subfield degrees? $\endgroup$ – Desperate Fluffy Dec 6 '13 at 5:20
  • $\begingroup$ I did not realize that degree of splitting field of $x^3+x+1$ is $6$ and it is not $3$.. please see below answer for more details... its more informative. $\endgroup$ – user87543 Dec 6 '13 at 5:28
  • $\begingroup$ That only shows the degree is at least 6. $\endgroup$ – LASV Dec 6 '13 at 6:55

Your idea is solid, but you got the wrong degree for the splitting field of $x^3 + x + 1$. In fact, it is 6, not 3, because after adjoining the single real root, the polynomial does not factor completely. An easy way of seeing this is that since the other roots are (as you said) complex but non-real, after adding a real root to $\mathbb{Q}$ we still have a field of real numbers that consequently can't contain the complex roots. So we have an irreducible quadratic factor left over and have to adjoin another root to split it, giving another extension of relative degree 2.

This suggests, but does not completely prove, that the splitting field you want has degree 12. To actually prove this, you'll need to show that $x^2 + x + 1$ doesn't factor over the splitting field of $x^3 + x + 1$; I'll let you figure out how to verify this.


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.