The Cube Roots 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 $f=g=1$. To factor $3\O _K$, we note that working modulo $3$ we have

$$x^3 - 2 = (x-2)(x^2+2x+1) = (x-2)(x+1)^2 \in \mathbf{F}_3[x],$$

so

$$3\O _K = (3, \sqrt[3]{2}- 2)\cdot (3, \sqrt[3]{2}+ 1)^2.$$

Thus $e_1=1$, $e_2=2$, $f_1=f_2=1$, and $g=2$. Next, 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 $g=2$.

Next we consider what happens in the Galois closure of $K$. Since the three embeddings of $\sqrt[3]{2}$ in $\mathbf{C}$ are $\sqrt[3]{2}$, $\zeta_3\sqrt[3]{2}$, and $\zeta_3^2\sqrt[3]{2}$, we have

$$M=K^{\gc } = \mathbf{Q}(\sqrt[3]{2}, \zeta_3) = K.L,$$

where $L=\mathbf{Q}(\zeta_3)=\mathbf{Q}(\sqrt{-3})$, since $\zeta_3=(-1+\sqrt{-3})/2$ is a primitive cube root of unity. The notation $K.L$ means the ``compositum of $K$ and $L$'', which is the smallest field generated by $K$ and $L$.

Let's figure out $e$, $f$, and $g$ for the prime $p=3$ relative to the degree six Galois field $M/\mathbf{Q}$ by using Theorem 13.2.2 and what we can easily determine about $K$ and $L$. First, we know that $efg=6$. We have $3\O _K=\mathfrak{p}_1\mathfrak{p}_2^2$, so $3\O _M=\mathfrak{p}_1\O _M\cdot (\mathfrak{p}_2 \O _M)^2$, and the prime factors of $\mathfrak{p}_1\O _M$ are disjoint from the prime factors of $\mathfrak{p}_2\O _M$. Thus $e>1$ is even and also $g>1$. The only possibility for $e,f,g$ satisfying these two conditions is $e=2$, $f=1$, $g=3$, so we conclude that $3\O _M = \mathfrak{q}_1^2 \mathfrak{q}_2^2 \mathfrak{q}_3^2$ without doing any further work, and without actually knowing the $\mathfrak{q}_i$ explicitly.

Here's another interesting deduction that we can make ``by hand''. Suppose for the moment that $\O _M = \mathbf{Z}[\sqrt[3]{2},\zeta_3]$ (this will turn out to be false). Then the factorization of $(\sqrt{-3})\subset \O _L$ in $\O _M$ would be exactly reflected by the factorization of $x^3-2$ in $\mathbf{F}_3=\O _L/(\sqrt{-3})$. Modulo $3$ we have $x^3-2=x^3+1 = (x+1)^3$, which would imply that $(\sqrt{-3}) = \mathfrak{q}^3$ for some prime $\mathfrak{q}$ of $\O _M$, i.e., that $e=6$ and $f=g=1$, which is incorrect. Thus $\O _M \neq\mathbf{Z}[\sqrt[3]{2},\zeta_3]$. Indeed, this conclusion agrees with the following computation, which asserts that $[\O _M : \mathbf{Z}[\sqrt[3]{2},\zeta_3]] = 24$:

   > R<x> := PolynomialRing(RationalField());
   > K := NumberField(x^3-2);
   > L := NumberField(x^2+3);
   > M := CompositeFields(K,L)[1];
   > O_M := MaximalOrder(M);
   > a := M!K.1;
   > b := M!L.1;
   > O := Order([a,b]);
   > Index(O_M,O);
   24