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

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

and the quadratic factor is irreducible. Thus

$\displaystyle 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.



William Stein 2012-09-24