Prime Ideals of Imaginary Quadratic Fields

Question

Suppose I have $\mathbb{Q}[\sqrt{d}]$, where d is negative. If $d \equiv 1$ mod 4 then the ring of integers is $\mathbb{Z}[\frac{1 + \sqrt{d}}{2}] \cong \mathbb{Z[x]}/(x^2 - x - \frac{1 - d}{4})$. It follows that $\mathbb{Z[\sqrt{d}]} \subset \mathbb{Z}[\frac{1 + \sqrt{d}}{2}]$, while the other direction is not true. Suppose p $\in \mathbb{Z}$ is prime and remains prime in $\mathbb{Z}[\frac{1 + \sqrt{d}}{2}]$. It follows by the subset relation that p remains prime in $\mathbb{Z[\sqrt{d}]} \cong \mathbb{Z[x]}/(x^2 + d)$. Does this mean that $x^2 + d$ is also irreducible in $\mathbb{Z}_{p}$? I feel like it should be because of the subset relation but I am not sure. Please let me know if my conclusion is true/valid.

Answer 1

If $p$ is an odd prime then:

$p$ will remain prime in $R=\mathbb{Z}[\frac{1 + \sqrt{d}}{2}]$

Iff

$(x^2 - x - \frac{1 - d}{4})$ is irreducible over $\Bbb Z_p$

Iff

$(x^2 - x - \frac{1 - d}{4}) \equiv 0 \pmod p$ has no solution.

Iff

$(4x^2 -4 x - ({1 - d})) \equiv 0 \pmod p)$ has no solution.

(Since $p$ is an odd prime )

Iff

$(2x-1)^2+d\equiv 0 \pmod p$ has no solution.

Hence,

$(y^2+d)$ is irreducible over $\Bbb Z_p$, where $y=2x-1.$

Answer 2

First of all, the claim obviously will run into issues if $p=2$ because $x^2 + d$ factors mod $2$ for any $d$. All we need to do is figure out when this contradicts your assumptions (namely that $2$ is inert).

Using the Dedekind factorization theorem, the claim is always true when $d=3$ mod $4$ (the conductor is 1), and even if $d=1$ mod $4$ there can only be a problem when $p=2$ because the conductor is $2$.

We therefore only risk a contradiction when $p=2$ and $d=1$ mod $4$ and $2$ is inert. Again from dedekind, 2 is inert iff the polynomial $$x^2 - x - \frac{1-d}{4}$$ is irreducible mod $2$. The only irreducible quadratic mod $2$ is $x^2 + x+ 1$ and so $2$ is inert if and only if $\frac{1-d}{4}$ is odd if and only if $d = 5$ mod $8$. This of course can occur and so there are counterexamples.

For a concrete example, just take $d=5$. We can see that $2$ is inert but $x^2 +5$ is reducible mod $2$.