4

Let $\mathbb{Q}(\gamma):\mathbb{Q}$ be the extension where $\gamma$ is a zero of $p(t) = t^3 - 3t^2 +3$

I'm having trouble figuring out if this extension is normal or not. I know that if I can show $$p(t) =t^3-3t^2+3 = (t-\gamma)(t^2+(\gamma-3)t+(\gamma^2-3\gamma))$$ does not split over $\mathbb{Q}(\gamma):\mathbb{Q}$, then I'm done (since if $\mathbb{Q}(\gamma):\mathbb{Q}$ was normal, then since $p(t)$ has a zero in $\mathbb{Q}(\gamma):\mathbb{Q}$, $p(t)$ must split) or alternatively, if it does split, $\mathbb{Q}(\gamma):\mathbb{Q}$ is the splitting field of $p(t)$ and since the extension is finite, $\mathbb{Q}(\gamma):\mathbb{Q}$ is then normal. I'm just stuck on proving if it splits or not. I tried to plug in the quadratic into the quadritic formula but it didn't seem to lead me anywhere.

Thank you!

klamont15
  • 477
  • 3
  • 11
  • 4
    Have you computed the discriminant of $p(t)$? – Angina Seng Mar 08 '18 at 07:34
  • 3
    If the importance of Lord Shark's suggestion is not clear to you, you can, instead, try and show that if $\gamma$ is one zero of $p(t)$ then $\gamma^2-2\gamma$ is another. – Jyrki Lahtonen Mar 08 '18 at 19:54
  • 2
    @JyrkiLahtonen, how did you find that polynomial? (see also https://math.stackexchange.com/questions/1767252/expressing-the-roots-of-a-cubic-as-polynomials-in-one-root) – lhf Mar 09 '18 at 20:29
  • 1
    @lhf The discriminant was a "giveaway" to the following: $$p(x+1)=x^3-3x+1.$$ That I happened to know to be the minimal polynomial of $2\cos(2\pi/9)$, and therefore subject to the "famous" rule: if $r$ is a zero, then so is $r^2-2$. So I simply "translated by one" the automorphism $r\mapsto r^2-2$. The zeros of $p(t)$ are $1+2\cos(2k\pi/9)$ with $k=1,2,4$. Admittedly I was also lucky. Possibly there are many elements from this same field sharing the discriminant of their minimal polynomial with the field. – Jyrki Lahtonen Mar 09 '18 at 20:47

0 Answers0