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I'm new to algebraic number theory and have the following questions: consider the quadratic field $K = \mathbb{Q}(\sqrt{2})$. Let $p \in \mathbb{Z}$ be a prime number, and $(p)$ be the ideal generated by $p$ in the ring $O_K$ of algebraic integers in $K$.

A) Prove that if $p = 2$ then $(p)$ splits as the product of two equal prime ideals.

B) Prove that if $p$ is an odd prime and $\frac{p^2−1}{8}$ is odd then $(p)$ is a prime ideal.

C) Prove that if $p$ is an odd prime and $\frac{p^2−1}{8}$ is even then $(p)$ splits as the product of two distinct prime ideals.

My thoughts so far: I know we have an isomorphism $O_K \simeq \mathbb{Z}[x]/(x^2 −2)$, which I think should help but I'm not sure how.

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Let $R = \mathcal{O}_{K}$. As you note, $R = \mathbb{Z}[\sqrt{2}] \cong \mathbb{Z}[X]/\langle X^{2}-2 \rangle$. Note that for any prime $p$, $R/pR \cong \mathbb{Z}[X]/\langle p, X^{2}-2 \rangle \cong (\mathbb{Z}/p\mathbb{Z})[X]/\langle X^{2}-2 \rangle$. (In the last isomorphism, I am being a bit sloppy and identifying $X^{2}-2$ with its residue class in $(\mathbb{Z}/p\mathbb{Z})[X]$.)

The factorization of $p$ into a product of prime ideals of $R$ can consequently be understood in terms of how the polynomial $X^{2}-2$ factors over $\mathbb{Z}/p\mathbb{Z}$. There is a general theorem at work here, namely Dedekind's theorem, but you can attack things quite directly. (I would personally view this problem as a nice way to see how Dedekind's theorem works in a simple case.) Here are some hints for each part:

(A) See if you can prove that $\sqrt{2}R$ is a prime ideal; it shouldn't be hard from there to show that $2R = (\sqrt{2}R)^{2}$. (Hint: $R/\sqrt{2}R \cong \mathbb{Z}[X]/\langle X, X^{2}-2\rangle \cong \ldots$). This reflects the fact that $R/2R \cong (\mathbb{Z}/2\mathbb{Z})[X]/\langle X^{2} \rangle$.

(B), (C): The conditions on $p$ here control whether $2$ is a quadratic residue mod $p$, i.e. whether $X^{2}-2$ splits over $\mathbb{Z}/p\mathbb{Z}$. You should try to show that in (B), $2$ is NOT a quadratic residue mod $p$, and so $X^{2}-2$ is irreducible over $\mathbb{Z}/p\mathbb{Z}$, in which case $pR$ is prime. Likewise, in (C), $2$ IS a quadratic residue mod $p$. Since $p$ is odd, $X^{2}-2$ splits into distinct factors over $\mathbb{Z}/p\mathbb{Z}$. Try to use this factorization (and maybe the Chinese Remainder Theorem) to construct the prime factorization of $pR$.

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  • $\begingroup$ Thanks for your response. I can follow most of it, but am not sure on a couple of points: for A, I don't know how to prove that $\sqrt{2}R$ is a prime ideal. For C, I'm not sure how to find the distinct factors of $X^2-2$ to construct the prime factorisation of $pR$. $\endgroup$ – user766821 Apr 3 '20 at 20:57

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