0

I want to find the splitting field of $f(x)=x^4+2$ over the rationals, $\mathbb{Q}$; and the degree of that splitting field over $\mathbb{Q}$.

So I first solved the equation $x^4=-2$, and get the roots are $x_k=2^{1/4}\exp(i\pi/4)\exp(i\pi\cdot k/2)$. So my splitting field over the rationals is $\mathbb{Q}(2^{1/4}\exp(i\pi/4)\exp(i\pi/2))$; I am not sure how to calculate $[\mathbb{Q}(2^{1/4}\exp(i\pi/4)\exp(i\pi/2)):\mathbb{Q}]$?

Any hints? thanks!

MathematicalPhysicist
  • 4,191
  • Hint: your polynomial's roots are $;\pm\frac1{\sqrt[4]2}(1+i);,;;\pm\frac1{\sqrt[4]2}(-1+i);$ , so you then need to have in your splitting field both $;\sqrt[4]2;,;;i;$ ... – DonAntonio Mar 04 '22 at 13:12
  • @DonAntonio ah OK. so the Splitting field is $\mathbb{Q}(2^{1/4},i)$ and its degree is $[\mathbb{Q}(2^{1/4},i):\mathbb{Q}]=[\mathbb{Q}(2^{1/4},i):\mathbb{Q}(i)]\cdot [\mathbb{Q}(i):\mathbb{Q}]=4\cdot 2=8$, am I correct? – MathematicalPhysicist Mar 04 '22 at 13:19
  • Looks good. I would personally choose to do the $i$ expansion last, because to me it's more immediately obvious that $[\Bbb Q(i,\sqrt[4]2):\Bbb Q(\sqrt[4]2)]=2$ than $[\Bbb Q(i,\sqrt[4]2):\Bbb Q(i)]=4$. You also have to actually prove that $i$ and $\sqrt[4]2$ are in the splitting field, though. Not just take DonAntonio's word for it. – Arthur Mar 04 '22 at 13:22
  • Exactly what Arthur said...though you're already pretty close. – DonAntonio Mar 04 '22 at 13:24
  • For the splitting field and the degree in general for $x^n-a$ see this duplicate. In your case it is $4\phi(4)=8$. – Dietrich Burde Mar 04 '22 at 13:27

1 Answers1

0

Hint: A solution to this equation is provided by a primitive 8th unit of unity $\xi$, since $\xi^4=-1$. Thus $(2^{1/4}\xi)^4 = -2$. The further solutions refer to $\xi^3,\xi^5,\xi^7$ which are the other primitive 8th roots of unity.

Wuestenfux
  • 20,964
  • 1
    A solution to this equation is not a primitive 8th root precisely because $\xi^4=-1\neq-2$. The phrasing here needs some work. – Arthur Mar 04 '22 at 13:35