Let $F$ be a field and $n$ be any integer satisfying $(n,\text{char}(F))=1$ when $\text{char}(F)\ne 0$.

Let $\xi$ be a primitive $n$-th root of unity. I know that $\text{Gal}(F(\xi)/F)$ is isomorphic to a subgroup of $(\mathbb{Z}_n)^{\times}$ and it is isomorphic when $F=\mathbb{Q}$. (In this case, deg (irr($\xi$,$\mathbb{Q}$))=$\phi(n)$ where $\phi$ is Euler Phi-function.)

The reason (I think) why this does not hold in general is that $F$ may contain $\xi$. So if $\xi\notin F$, can we assure $|F(\xi)/F|=\phi(n)$?

In other words, is the polynomial $\prod_{1\le k \le n, \ (n,k)=1} (t-\xi^k)\in F[t]$ irreducible?

If not, please let me know some counter example.

  • $\begingroup$ A sneaky one is $\mathbb Q(\zeta_5)$ over $\mathbb Q(\sqrt 5)$. $\endgroup$ – user208649 Apr 23 '16 at 7:28

No, it is still not sufficient to assume that $\xi\notin F$. For example, if $F=\mathbb{Q}(i)$ and $$\xi=e^{2\pi i / 8}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$$ then $[F(\xi):F]=2$ but $\xi\notin F$. Observe that $\varphi(8)=4$, and $$\underbrace{[F(\xi):\mathbb{Q}]}_{4}=\underbrace{[F(\xi):F]}_{2}\underbrace{[F:\mathbb{Q}]}_{2}$$

The Galois group of the $n$th cyclotomic field is $\mathrm{Gal}(\mathbb{Q}(\xi)/\mathbb{Q}))\cong(\mathbb{Z}/n)^\times$. If (and only if) $\varphi(n)=|(\mathbb{Z}/n)^{\times}|$ is a composite number, there will be a non-trivial subgroup of $(\mathbb{Z}/n)^\times$, and hence (by the Fundamental Theorem of Galois Theory) a field $F$ with $\mathbb{Q}(\xi)\supsetneq F\supsetneq \mathbb{Q}$, which will have the property that $1<[F(\xi):F]<\varphi(n)$.

  • $\begingroup$ +1 Another small counterexample is with $F=\Bbb{Q}(\sqrt5)$ and $n=5$. The number $\xi+\overline{\xi}=2\cos(2\pi/5)$, and the formula $2\cos(2\pi/5)=(1+\sqrt5)/2$ shows that $\xi$ is of degree $2$ over $F$. $\endgroup$ – Jyrki Lahtonen Apr 23 '16 at 7:25
  • $\begingroup$ @Zev Chonoles, Thank you for letting me know this good example. It helped me very much! $\endgroup$ – user29422 Apr 23 '16 at 8:51
  • $\begingroup$ @Jyrki Lahtonen, How did you get the formula $2\cos(\frac{2\pi }{5})=\frac{1+\sqrt{5}}{2}$? $\endgroup$ – user29422 Apr 23 '16 at 9:01
  • $\begingroup$ @Jyrki Lahtonen, Oh, I see. Thank you!^^ $\endgroup$ – user29422 Apr 23 '16 at 9:08

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.