Lattice basis for prime divisor of $(p)$

Question

Suppose that $d \equiv 2$ or $3$ modulo $4$, and that a prime $p \neq 2$ does not remain prime in $R$. Let $a$ be an integers such that $a^2 \equiv d$ modulo $p$. How would I go about showing that $(p, a + \delta)$ is a lattice basis for a prime ideal that divides $(p)$?

Answer 1

First we prove that the ideal $P = (p,a+\delta)$ has $\{p,a+\delta\}$ as a lattice basis. We just need to check that $\delta p$ and $\delta (a+\delta) = a\delta + d$ are integer linear combinations of $p,a+\delta$: indeed, $$\delta p = p(a+\delta) - a(p),$$and $$a\delta + d = a(a+\delta) - \frac{a^2-d}{p}(p).$$

Note that $\overline{\delta} = -\delta$. Then $$(N(P)) = P\overline{P} = (p,a+\delta)(p,a+\overline{\delta}) = (p^2,p(a-\delta),p(a+\delta),N(a+\delta)).$$ Then $p^2/N(P)$ is a rational algebraic integer, hence a rational integer. Furthermore, $$N(a+\delta) = (a+\delta)(a-\delta) = a^2-\delta^2 = a^2 - d$$ is an integer divisible by $p$, so $p\mid N(P)$. So $N(P)\in\{p,p^2\}$.

On the other hand, $$\alpha = \frac{p(a+\delta) - p(a-\delta)}{N(P)} = \frac{2p\delta}{N(P)}$$ is an algebraic integer, as is its conjugate $\overline{\alpha}$, so $$N(\alpha) = \alpha\overline{\alpha} = \frac{4p^2}{N(P)^2}N(\delta) = \frac{4}{(N(P)/p)^2}|d|$$ is a rational algebraic integer, hence a rational integer. Since $d$ is squarefree and $p$ is odd, we cannot have $N(P) = p^2$. Thus $N(P) = p$.

We conclude that $$(p) = (N(P)) = P\overline{P},$$ so $P$ is indeed a prime ideal dividing $(p)$, with lattice basis $($for instance$)$ $\{p,a+\delta\}$.

$($To show that $P$ is actually prime, we can note that since $(p)$ is not a prime ideal, it must be the product of a prime ideal and its conjugate; unique factorization says these conjugate ideals must be $P,\overline{P}$, which are therefore prime. Alternatively, since $N(P) = p$ is prime, $P$ must be maximal, or else, $P = UV$ for some proper ideals $U,V$ containing $P$ $($why?$)$. But then either $N(U)=1$ or $N(V)=1$; in particular, $U$ or $V$ must divide $(1)$, contradicting their properness.$)$

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One comment I would like to make is that the condition that $p$ does not remain prime is superfluous, since we are basically given that $x^2-d$ is reducible mod $p$.