# Understanding a proof in Washington's “Cyclotomic Fields”

I'm working through Washington's "Cyclotomic Fields" and having a problem with the proof of Proposition 3.8, which states: Given an abelian group G, there is an everywhere-unramified extension of number fields $L/K$ such that $\text{Gal}(L/K)\cong G$.

The proof proceeds (roughly) as follows: Write $G=C_{n_1}\times\cdots\times C_{n_r}$ (a product of cyclic groups). Choose distinct primes $p_i\equiv 1\pmod{2n_i}$.

We may choose Dirichlet characters $\psi_i$ with conductor $p_i$ and order $p_i-1$, and define $\chi_i=\psi_i^{(p_i-1)/n_i}$. Since $(p_i-1)/n_i$ is even, we also have $\chi_i(-1)=1$. Additionally, we pick another odd prime $p_{r+1}$ and an odd Dirichlet character $\chi_{r+1}$ of conductor $p_{r+1}$. We now define a character: $$\chi=\chi_1\cdots\chi_{r+1},$$ and a group of characters $$X=\langle \chi_1,...,\chi_{r+1}\rangle$$ (as I understand, a subgroup of characters mod $p_1\cdots p_{r+1}$). Now $\chi$ has an associated field which we call $K$, and $X$ has a corresponding field $L$ containing $K$. The main observations are:

1. $\chi(-1)=-1$, and it follows that $K$ has no real embeddings. Thus the archimedean primes of $K$ do not ramify in $L$.
2. By a prior theorem, $L$ and $K$ have the same ramification indices at all primes $p \in \mathbf{Z}$. Thus $L/K$ is everywhere unramified.

I agree with the proof up to this point, but I don't understand why $\text{Gal}(L/K) \cong G$. According to Washington: $X$ is generated by $\{\chi_1,...,\chi_r,\chi\}$, so $$\text{Gal}(L/K) \cong X/\langle \chi\rangle \cong \langle \chi_1,...,\chi_r\rangle$$ which is then isomorphic to $G$. However, the last isomorphism doesn't make sense to me: if the $\chi_i$ have coprime orders, then $\chi$ generates $X$, which makes the quotient trivial.

Here's a specific example I'm working with: take $G$ cyclic of order 3, $p_1=7$, and pick $p_2=3$. We can easily obtain an even character $\chi_1$ mod 7 of order 3, and we only have one choice for $\chi_2$. Then $\chi$ clearly has order 6 and generates $X=\langle \chi_1,\chi_2 \rangle$. Thus we have $L=K$.

I'm sure there's a flaw in my understanding somewhere...could someone point me in the right direction?

• Washington pg. 27 line 2: "We assume that the order $n_{r+1}$ of $\chi_{r+1}$ is divisible by $n_1, \dots, n_r$." – John M Jun 10 '16 at 20:20
• @JohnM got it, thank you! This appears to be missing from my edition – user346992 Jun 10 '16 at 20:31

The 2nd edition of Washington states on pg. 27, line 2, "We assume that the order $n_{r+1}$ of $\chi_{r+1}$ is divisible by $n_1, \dots, n_r$." This is missing from the 1st edition.