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 possibilties:

• 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, we will later show that 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. This happens 50% of the time, which is suggested by quadratic reciprocity (but not proved this way), 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 50% of the time.

Suppose, in particular, that $K=\mathbf{Q}(\sqrt{5})$, so $\O _K=\mathbf{Z}[\gamma]$, where $\gamma=(1+\sqrt{5})/2$. Then $p=5$ is ramified, since $p\O _K = (\sqrt{5})^2$. 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 $5$.