= Solution
Use the customary squarefree convention for the quadratic radicand $d$. For $d\equiv2,3\pmod4$, the <ring of integers of a number field> is $\mathbb Z[\sqrt d]$, so
$$
\mathcal O_{\mathbb Q(\sqrt d)}/(p)\cong\mathbb F_p[X]/(X^2-d).
$$
The ideal $(p)$ is prime exactly when this quotient is an <integral domain>. A <polynomial> quotient of a field is an <integral domain> exactly when the defining <polynomial> is irreducible. This proves the assertion, including the possibility of ramified primes where the reduction has a repeated root.
Squarefreeness is necessary for the printed assertion as written: $d=18\equiv2\pmod4$ defines the same field as $d=2$. The prime $3$ is inert in $\mathbb Q(\sqrt2)$ because $X^2-2$ is irreducible over $\mathbb F_3$, although $X^2-18=X^2$ is reducible. Thus \b[the result requires a squarefree quadratic radicand].
Back to article page
