Corollary 1   Let $K$ be a number field, $h_K$ is its class number and let $p$ be a prime. Then $K$ has an everywhere unramified Galois extension of degree $p$ if and only if $h_K$ is divisible by $p$ .

Theorem 1   Let $F/K$ be a Galois extension of number fields, let $h_F$ and $h_K$ be their respective class numbers and let $p$ be a prime number such that $p$ does not divide $[F:K]$ , the degree of the extension. Then, if $p$ divides $h_K$ , the class number of $F$ , $h_F$ , is also divisible by $p$ .

Proof. Let $F$ , $K$ and $p$ be as in the statement of the theorem. Assume that $p| h_K$ . Thus, by the corollary above, there exists an unramified Galois extension field $E$ of $K$ of degree $p$ . Notice that the fact that $p$ does not divide the degree of the extension $F/K$ implies that $F\cap E=K$ . In particular, the compositum $FE$ is a Galois extension of $F$ and $$\Gal(FE/F)\cong \Gal(E/E\cap F)\cong \Gal(E/K).$$ Thus, the extension $FE/F$ is of degree $p$ , Galois, and therefore abelian. By the corollary above, in order to prove the theorem it suffices to show that the extension $FE/F$ is unramified. Suppose for a contradiction that $\mathcal{Q}_F$ is a prime ideal which ramifies in the extension $FE/F$ . Let $\mathcal{Q}_{EF}$ be a prime lying above $\mathcal{Q}_F$ and let $\mathcal{Q}_K$ be a prime of $K$ such that $\mathcal{Q}_F$ lies above it. Similarly, let $\mathcal{Q}_E$ be a prime of $E$ lying above $\mathcal{Q}_K$ and such that the prime $\mathcal{Q}_{EF}$ lies above $\mathcal{Q}_E$ . For an arbitrary extension $A/B$ , the ramification index of a prime $\mathcal{Q}_A$ is denoted by $e(\mathcal{Q}_B|\mathcal{Q}_A)$ . Then, by the multiplicativity of the ramification index in towers, we have: $$e(\mathcal{Q}_{EF}|\mathcal{Q}_K)=e(\mathcal{Q}_{EF}|\mathcal{Q}_F)\cdot e(\mathcal{Q}_{F}|\mathcal{Q}_K)=e(\mathcal{Q}_{EF}|\mathcal{Q}_E)\cdot e(\mathcal{Q}_{E}|\mathcal{Q}_K)$$ Since we assumed that $\mathcal{Q}_F$ is ramified in $EF/F$ , and the degree of the extension is $p$ , we must have $e(\mathcal{Q}_{EF}|\mathcal{Q}_F)=p$ . Therefore, by the equality above, $p$ divides $e(\mathcal{Q}_{EF}|\mathcal{Q}_E)\cdot e(\mathcal{Q}_{E}|\mathcal{Q}_K)$ . Notice that the extension $E/K$ is everywhere unramified, therefore $e(\mathcal{Q}_E|\mathcal{Q}_K)=1$ . Also, $[EF:E]=[F:K]$ which, by hypothesis, is relatively prime to $p$ . Thus $e(\mathcal{Q}_{EF}|\mathcal{Q}_E)$ is also relatively prime to $p$ , and so, $p$ is not a divisor of $e(\mathcal{Q}_{EF}|\mathcal{Q}_E)\cdot e(\mathcal{Q}_{E}|\mathcal{Q}_K)$ , which leads to the desired contradiction, finishing the proof of the theorem.