# Is there a polynomial over $\mathbb{Q}$ which $K$ is its splitting field.

Let $K=\mathbb{Q}(\sqrt[11]{7},i)$. Is $K$ the splitting field of some polynomial over $\mathbb{Q}$?

My attempt: My first intuition would be no. Since if $K$ is the splitting field of some polynomial then $K/\mathbb{Q}$ must be a finite, normal extension. Considering the irreducible polynomial $x^{11}-7$ over $\mathbb{Q}$, it is clear that one of its roots, namely $\sqrt[11]{7}$ lies in $K$.

It suffices to show that some other roots of $x^{11}-7$ does not lie in $K$. But I ended up with expressions in terms of $cos\left(\frac{n\pi}{11}\right)$ and$sin\left(\frac{n\pi}{11}\right)$ whose values are nearly impossible to compute.

Is there an easier way?

• Your field is real, but there are complex roots to $x^{11}-7$ – lulu Sep 14 '17 at 11:54
• There is an $i$ adjoined to it. @lulu – thedilated Sep 14 '17 at 12:40
• Ah, so there is. Read too fast. – lulu Sep 14 '17 at 12:44

We need a primitive 11th root of unity $\zeta_{11}$ in $K$ (because all roots of $X^{11}-7$ are of the form $\zeta_{11}^k\sqrt[11]7$). Now $$22 = [K:\Bbb Q]=[K:\Bbb Q(\zeta_{11})]\cdot [\Bbb Q(\zeta_{11}):\Bbb Q]=10\cdot [K:\Bbb Q(\zeta_{11})],$$ contradiction.
• I don't understand. Doesn't $(x^{11}-7)(x^2+1)$ factorize into linear factors in $K$? – Kenny Lau Sep 14 '17 at 10:09
• @KennyLau Is $\zeta_{11}$ an element of $K$? If your polynomial factorizes in $K$, then the answer is yes. But from Hagen's proof we get that it isn't, so by contradiction the polynomial doesn't split over K. – Stefan4024 Sep 14 '17 at 10:18