# Proving 20-th cyclotomic field has class number one.

(This is related to one of my previous questions; reading is not required though)

I'm still banging my head on the following exercise.

Consider the primitive $n$-th root of unity $\zeta_{n} := exp(\frac{2\pi i}{n})$. Show that the number field $K := \mathbb{Q}(\zeta_{20})$ has class number one.

In the exercise, the following hint is given:

Show that it suffices to show that any prime ideal above the primes $2,3,5,7,11$ is principal. We know that the quadratic subfields of $\mathbb{Q}(\zeta_{20})$ are $\mathbb{Q}(\sqrt{-5})$, $\mathbb{Q}(\sqrt{5})$, $\mathbb{Q}(i)$. The prime 2 may be treated via $\mathbb{Q}(i)$. For 3 and 7, observe that $\omega_1^2 + \omega_2^2 = 3$ and $\omega_1^4 + \omega_2^4 = 7$, where $\omega_1 := (1+\sqrt{5})/2$ and $\omega_2 := (1-\sqrt{5})/2$. For 5 show that the norm from $\mathbb{Q}(\zeta_{20})$ to $\mathbb{Q}(\zeta_{5})$ of $(\zeta_5 + \zeta_5^{-1})+\zeta_5^2\cdot i$ is $1-\zeta_5$. For 11, first determine its prime factors in $\mathbb{Q}(\zeta_5)$.

What I have accomplished so far:

I have shown via Minkowski-bound that it suffices to show that any prime ideal above the primes $2,3,5,7,11$ is principal.
I also managed to show that any prime ideal above $2$ or $11$ must be principal.

So it remains to show that for the primes $3,5,7$, obviously using the hints above. However I don't really seem to get to the point where the hints make sense to me.

If it helps, I also computed the inertia/decomposition fields: Given a prime number $p$, let $r$ denote number of prime ideals of $\mathbb{Z}[\zeta_{20}]$ above $p$. Then let $e$ be ramification index and $f$ the inertia degree. For primes $3,5,7$ get:

• Prime 3: $r=2, e=1, f=4$. Decomposition field = $\mathbb{Q}(\sqrt{-5})$, inertia field = $K$
• Prime 5: $r=2, e=4, f=1$. Decomposition field = inertia field = $\mathbb{Q}(i)$
• Prime 7: $r=2, e=1, f=1$. Decomposition field = $\mathbb{Q}(\sqrt{-5})$, inertia field = $K$

Thanks for any help in advance, I'd really like to close this chapter.

## 1 Answer

For the prime 3 you have that $3=w_1^2+w_2^2=(w_1+iw_2)(w_1-iw_2)$ where the decomposition is to algebraic integers (since $i,w_1,w_2\in\mathcal{O}_K$), so that as ideals you have the decomposition $<3>=<w_1+iw_2><w_1-iw_2>$ (because they are conjugate, neither of them can be $\mathcal{O}_K$, since then you will get that 3 is an integral unit). The factorization of $<3>$ to prime ideals will be the product of the factorizations of $<w_1\pm iw_2>$. Since you already computed that there are only 2 prime ideals over 3, then this must be the decomposition and these ideals are principal. For $p=7$ you can do a similar process by noting that $w_1^4+w_2^4=(w_1^2+iw_2^2)(w_1^2-iw_2^2)$.

For $p=5$, note that $\prod_1^4(1-\zeta_5^i)=5$ is the norm of $(1-\zeta_5)$ from $\mathbb{Q}[\zeta_5]$ to $\mathbb{Q}$. Hence, by the hint in the question you get that $5=Norm((\zeta_5 + \zeta_5^{-1})+\zeta_5\cdot i)$. It follows that each of the conjugates of $(\zeta_5 + \zeta_5^{-1})+\zeta_5\cdot i$ is not invertible (norm not $\pm 1$), so you get a decomposition of $<5>$ to nontrivial 8 principal ideals, so each one of them must be prime.

• Seems so clear once you see it. Super sweet. – johnnycrab Jan 18 '17 at 22:46