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I am wondering about primes $p$ in $\mathbb Z$ that are split in $\mathcal O_{K}$, $K=\mathbb Q(\sqrt d)$. Let $\omega=\sqrt d$ if $d \equiv 2,3 \mod 4$ and $\omega=\frac{1+\sqrt d}{2}$ if $d \equiv 1 \mod 4$ and let $f(x)$ be the monic polynomial with coefficients in $\mathbb Z$ of least degree with $\omega$ as a root. Hence $$\mathcal O_{K}\cong \mathbb Z[\omega]\cong \mathbb Z[x]/(f(x))$$ Let's say that my definition of $p$ being ramified, split and inert, is that the corresponding $\bar f(x)\in \mathbb Z/p\mathbb Z[x]$ has a repeated root, two distinct roots and no roots (i.e. is irreducible), respectively. Then $$\mathcal O_K /(p)\cong \frac{\mathbb Z[x]/(f(x))}{(p)} \cong \frac{\mathbb Z[x]/(f(x))}{(p,f(x))/(f(x))} \cong \frac{\mathbb Z[x]}{(p, f(x))} \cong \frac{\mathbb Z[x]/p \mathbb Z[x]}{(p, f(x))/(p)}\cong \frac{\mathbb F_p[x]}{(\bar f(x))}$$ So for the ramified case $$\frac{\mathcal O_K}{(p)}\cong \frac{ \mathbb F_p[x]}{((x-\alpha)^2)}\cong \frac{ \mathbb F_p[x]}{(x^2)}$$ and for the inert case $f(x)$ is irreducible, so $\mathcal O_K \cong \mathbb F_{p^2}$. But I don't know how to deduce that $\mathcal O_K \cong \mathbb F_{p}^2$ in the split case. I get $$\frac{\mathcal O_K}{(p) } \cong \frac{\mathbb F_p[x]}{((x-\alpha)(x-\beta))} $$ What manipulations can I do to this in order to simplify it?

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If $\alpha\neq\beta$, then $x-\alpha$ and $x-\beta$ are comaximal in $\mathbb F_p[x]$ (that is $(x-\alpha)+(x-\beta)=(1)$. So Chinese Remainder Theorem implies $\frac{\mathbb F_p[x]}{(x-\alpha)(x-\beta)}\cong\frac{\mathbb F_p[x]}{(x-\alpha)}\times\frac{\mathbb F_p[x]}{(x-\beta)}\cong\mathbb F_p\times\mathbb F_p$.

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  • $\begingroup$ Thanks!! I didn't even realise that the Chinese remainder theorem applied outside of the integers. But a quick google search gave this set of notes that proved it. I also had never heard of comaximal ideas before. It is a nice generalisation of coprime elements. $\endgroup$ – James Apr 24 '15 at 10:02

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