The Cube Root of Two

Suppose $K/\mathbf{Q}$ is not Galois. Then $e_i$, $f_i$, and $g$ are defined for each prime $p\in\mathbf{Z}$, but we need not have $e_1=\cdots=e_g$ or $f_1=\cdots =f_g$. We do still have that $\sum_{i=1}^g e_i f_i = n$, by the Chinese Remainder Theorem.

For example, let $K=\mathbf{Q}(\sqrt[3]{2})$. We know that $\O_K = \mathbf{Z}[\sqrt[3]{2}]$. Thus $2\O_K = (\sqrt[3]{2})^3$, so for $2$ we have $e=3$ and $f=g=1$.

Working modulo $5$ we have

$$x^3 - 2 = (x+2)(x^2+3x+4) \in \mathbf{F}_5[x],$$

and the quadratic factor is irreducible. Thus

$$5\O_K = (5, \sqrt[3]{2}+2)\cdot (5, \sqrt[3]{2}^2 + 3\sqrt[3]{2}+ 4).$$

Thus here $e_1=e_2=1$, $f_1=1$, $f_2=2$, and $g=2$. Thus when $K$ is not Galois we need not have that the $f_i$ are all equal.