Quadratic Extensions

Suppose $K/\mathbf{Q}$ is a quadratic field. Then $K$ is Galois, so for each prime $p\in\mathbf{Z}$ we have $2=efg$. There are exactly three possibilities:

• Ramified: $e=2$, $f=g=1$: The prime $p$ ramifies in $\O_K$, so $p\O_K = \mathfrak{p}^2$. There are only finitely many such primes, since if $f(x)$ is the minimal polynomial of a generator for $\O_K$, then $p$ ramifies if and only if $f(x)$ has a multiple root modulo $p$. However, $f(x)$ has a multiple root modulo $p$ if and only if $p$ divides the discriminant of $f(x)$, which is nonzero because $f(x)$ is irreducible over $\mathbf {Z}$. (This argument shows there are only finitely many ramified primes in any number field. In fact, the ramified primes are exactly the ones that divide the discriminant.)
• Inert: $e=1$, $f=2$, $g=1$: The prime $p$ is inert in $\O_K$, so $p\O_K = \mathfrak{p}$ is prime. It is a nontrivial theorem that this happens half of the time, as we will see illustrated below for a particular example.
• Split: $e=f=1$, $g=2$: The prime $p$ splits in $\O_K$, in the sense that $p\O_K = \mathfrak{p}_1\mathfrak{p}_2$ with $\mathfrak{p}_1\neq \mathfrak{p}_2$. This happens the other half of the time.

For example, let $K=\mathbf{Q}(\sqrt{5})$, so $\O_K=\mathbf{Z}[\gamma]$, where $\gamma=(1+\sqrt{5})/2$. Then $p=5$ is ramified, since $5\O_K = (\sqrt{5})^2$. More generally, the order $\mathbf{Z}[\sqrt{5}]$ has index $2$ in $\O_K$, so for any prime $p\neq 2$ we can determine the factorization of $p$ in $\O_K$ by finding the factorization of the polynomial $x^2-5\in \mathbf{F}_p[x]$. The polynomial $x^2-5$ splits as a product of two distinct factors in $\mathbf{F}_p[x]$ if and only if $e=f=1$ and $g=2$. For $p\neq 2,5$ this is the case if and only if $5$ is a square in $\mathbf{F}_p$, i.e., if $\left(\frac{5}{p}\right) = 1$, where $\left(\frac{5}{p}\right)$ is $+1$ if $5$ is a square mod $p$ and $-1$ if $5$ is not. By quadratic reciprocity,

$$\left(\frac{5}{p}\right) = (-1)^{\frac{5-1}{2}\cdot \frac{p-1}{2}... ... if } p\equiv \pm 1\pmod{5} -1&\text{ if } p \equiv \pm 2\pmod{5}.\end{cases}$$

Thus whether $p$ splits or is inert in $\O_K$ is determined by the residue class of $p$ modulo $5$. It is a theorem of Dirichlet, which was massively generalized by Chebotarev, that $p\equiv \pm 1$ half the time and $p \equiv \pm 2$ the other half the time.