First suppose
\(\gothp = (p)\text{,}\) where
\(p \neq 2, 5\) is a rational prime that remains inert (i.e., does not split and is not ramified) in
\(K\text{.}\) This happens if and only if
\(-5\) is not a square mod
\(p\text{.}\) In this case, one of
\(-1\) and
\(5\) is a square in
\(\FF_p\text{,}\) so
\(\gotho_K/\gothp\) contains a square root of one of them, hence of both (since
\(-5\) already has a square root there). Thus the residue field does not grow when we pass to
\(L\text{,}\) that is,
\(\gothp\) is split.
Next suppose
\(p \neq 2,5\) is a rational prime that splits as
\(\gothp \overline{\gothp}\text{.}\) If
\(\gothp = (\beta)\) is principal, then the equation
\(x^2 + 5y^2 = p\) has a solution in
\(\ZZ\) (namely, for
\(x + y \sqrt{-5} = \beta\)), but this is only possible if
\(p \equiv 1 \pmod{4}\text{.}\) Then
\(p\) splits in
\(\QQ(\sqrt{-1})\) as well, so
\(p\) is totally split in
\(L\text{,}\) so
\(\gothp\) splits in
\(L\text{.}\)
Conversely, suppose
\(\gothp\) is not principal. Since there are only two ideal classes in
\(\QQ(\sqrt{-5})\text{,}\) we have
\(\gothp = \alpha (2, 1 + \sqrt{-5})\) for some
\(\alpha \in K\text{.}\) Thus
\(\Norm(\gothp) = |\Norm(\alpha)| \Norm(2, 1 + \sqrt{-5})\text{.}\) If
\(\alpha = x + y \sqrt{-5}\) for
\(x,y \in \QQ\text{,}\) we then have
\(p = 2(x^2 + 5y^2)\text{.}\) Considering things mod 4, we see that
\(2x\) and
\(2y\) must be ratios of two odd integers, and
\(p \equiv 3 \pmod{4}\text{.}\) Thus
\(p\) does not split in
\(L\text{,}\) so
\(\gothp\) cannot split in
\(L\text{.}\)
The only cases left are
\(\gothp = (2, 1+\sqrt{-5})\text{,}\) which does not split (see above), and
\(\gothp = (\sqrt{-5})\text{,}\) which does split (since
\(-1\) has a square root mod 5).