# 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.

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.