Find a minimal polynomial of a primitive 5-th root of unity over $\mathbb{Q}_3(\sqrt{-7})$

Question

Let $K=\mathbb{Q}_3(\sqrt{-7})$ be the base field and $L=K(\zeta_5)$ where $\zeta_5$ is a primitive $5$-th root of unity. I know that the extension $L/K$ is unramified of degree $2$.

Now I would like to find a minimal polynomial of $\zeta_5$ over $K$. I know that such a minimal polynomial must divide $x^4 + x^3 + x^2 + x + 1 = \frac{x^5-1}{x-1}$, i.e. it will split into two quadratic polynomials (which both are potential minimal polynomials of $\zeta_5$). But aside from that, I don't know how to compute the actual factors.

Could you please help me with this problem?

Answer 1

Given a prime p not dividing n and a p-adic local field $M$ with residual field $\mathbf F_q$, with $q=p^f$, it is classically known that $M(\zeta_n)/M$ is unramified cyclic, with Galois group generated by the Frobenius automorphism $s$ determined by $s(\zeta_n)= {\zeta_n}^q$ . In particular $[M(\zeta_n):M]=r$, the smallest integer $r\ge 1$ s.t. $q^r \equiv 1$ mod $n$. See e.g. Serre's "Local Fields", chap. IV,§4, prop. 16. Applying this theorem with your notations $K=\mathbf Q_3 (\sqrt {-7})$ and $L=K(\zeta_5)$, and accepting that $K/\mathbf Q_3$ is quadratic unramified, we find here that $\mathbf Q_3(\zeta_5)/\mathbf Q_3$ and $K(\zeta_5)/K$ are unramified (cyclic) of respective degrees degree $4$ and $2$. But a fixed algebraic closure of $\mathbf Q_3$ contains a unique unramified extension of $\mathbf Q_3$ of given degree, so that $\mathbf Q_3(\zeta_5)=K(\zeta_5)$. In particular, $K$ coincides with the unique quadratic subfield of $\mathbf Q_3(\zeta_5)/\mathbf Q_3$, namely $K=\mathbf Q_3(\zeta_5 +{\zeta_5}^{-1})$, and the minimal polynomial of $K(\zeta_5)/K$ is $X^2 -(\zeta_5 +{\zeta_5}^{-1})X + 1$.

Answer 2

$$(1+2\zeta_5+2\zeta_5^9)^2=5$$ (expand the LHS then use $\sum_{n=0}^4 \zeta_5^n=0$)

Thus the minimal polynomial is $$(x-\zeta_5)(x-\zeta_5^9)=x^2-\frac{1+\sqrt5}{2} x+1$$ Where $$\sqrt5 = \sqrt{-7} \sqrt{-5/7}=\sqrt{-7} \sum_{k=0}^\infty {1/2\choose k}(-1-5/7)^k$$ $$=\sqrt{-7} \sum_{k=0}^\infty {1/2\choose k}(-4/7)^k 3^k \in \Bbb{Z}_3[\sqrt{-7}]$$

Note: in fact it is $x^2-\frac{1 \pm \sqrt5}{2} x+1$ and $\sqrt5= \pm \sqrt{-7} \sqrt{-5/7}$ and which pair of 5-th root of unity I chose is implied by the sign I chose