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Let $\mathbb{Z}$ be the set of all integers, and $R=\mathbb{Z}[\sqrt{-5}]=\{a+\sqrt{-5}b\mid a, b\in\mathbb{Z}\}$. Is there any characterization for maximal ideals of $R$ with respect to maximal ideals of $\mathbb{Z}$ or other description for them? Thanks for any help.

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This is very similar to the phenomenon that happens in $\mathbb{Z}[i]$ where each integer prime is either a square in the Gaussian integer ($2=i(1+i)^2$), it decomposes to to primes (for example $5=(1+2i)(1-2i)$) or it remains a prime (for example 3), and moreover, these cases are determined by the residue of the prime mod 4. These sort of methods appear in any book on algebraic number theory.

First of all, since $R$ is the ring of integers of $\mathbb{Q}[\sqrt{-5}]$, it is a Dedekind domain and in particular its maximal ideals are exactly the nonzero prime ideals.

Suppose that $P$ is a nonzero prime ideal in $R$. Then $P\cap \mathbb{Z}$ is a nonzero prime ideal in $\mathbb{Z}$, namely $P=p\mathbb{Z}$. Thus, it is enough to look for prime ideals in $R$ that contain $pR$ where $p$ runs over the primes in $\mathbb{Z}$. This is the same as finding the primes in the ring $R/pR\cong \mathbb{F}_p[x]/(x^2+5)$ where $\mathbb{F}_p$ is the field with $p$ elements.

Now you have 3 cases. If $x^2+5$ has two distinct solutions mod $p$, then by the Chinese reminder theorem you have that $R/pR\cong \mathbb{F}_p \times \mathbb{F}_p$, and this ring has exactly two primes (the two copies of $\mathbb{F}_p$). Going back to the ring $R$, the corresponding ideals are $\left<p,a-\sqrt{-5}\right>$ and $\left<p,b-\sqrt{-5}\right>$ where $a,b$ are roots of $x^2+5$ mod $p$. For example, for $p=3$ you have $1,2$ as distinct roots of $x^2+5$, so that the primes over $3$ are $\left<3,1-\sqrt{-5}\right>$ and $\left<3,2-\sqrt{-5}\right>$.

The second case is that $x^2+5$ is irreducible mod $p$, so that $R/pR$ is a field, hence $pR$ is prime.

The third case is when $x^2+5$ has only one root mod $p$ (with multiplicity), for example if $p=5$, you only have the root $x=0$, or when $p=2$, you only have the root $x=1$. In this case you have $x^2+5=(x-a)^2 \pmod p$ so that $a-\sqrt{-5}$ must be in your prime ideal, and you should check that the prime ideal must be then $\left<p,a-\sqrt{5}\right>$.

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  • $\begingroup$ One should probably say "quadratic reciprocity" at some point in this answer. $\endgroup$ – Ege Erdil Dec 13 '16 at 12:18
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Suppose that $p \in \mathbb{Z}$ is prime. Either the ideal generated by $p$ in $\mathbb{Z}[\sqrt{-5}]$ is prime in that ring too, or it is the square of a prime ideal, or it is a product of two distinct prime ideals. The ring $\mathbb{Z}[\sqrt{-5}]$ is a Dedekind domain so an ideal is maximal if and only if it is prime. Every prime ideal of $\mathbb{Z}[\sqrt{-5}]$ arises from some prime ideal $(p)\subseteq\mathbb{Z}$ in one of the ways I described. You might like to see if you can work out criteria for when each of the three cases happens.

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