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Let $v$ be the 3-adic valuation on $\mathbb{Q}$ and consider the subring $\mathbb{Z}_{(3)}$ of $\mathbb{Q}$ defined by $$ \mathbb{Z}_{(3)} = \{ x \in \mathbb{Q} : v(x) \geq 0 \}. $$ That is, $\mathbb{Z}_{(3)}$ is the ring of rational numbers that are integral with respect to $v$. $\mathbb{Z}_{(3)}$ is also the localization of $\mathbb{Z}$ at the prime ideal $(3)$. I know $\mathbb{Z}_{(3)}$ is integrally closed in $\mathbb{Q}$.

I want to find the integral closure of $\mathfrak{O}$ in the field $\mathbb{Q}(\sqrt{-5})$: $$ \overline{\mathbb{Z}_{(3)}} = \{x \in \mathbb{Q}(\sqrt{-5}) : x \text{ is a root of a monic irreducible polynomial with coefficients in } \mathbb{Z}_{(3)} \} $$

How can I do this? What should I be thinking about?

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Thanks to Andrea and Pete for your answers below. I ended up using a hybrid of your two solutions. –  admchrch Feb 26 '11 at 2:08

2 Answers 2

up vote 7 down vote accepted

The integral closure of $\Bbb Z$ in $K={\Bbb Q}\sqrt{-5}$ is the ring of integers ${\cal O}_K={\Bbb Z}\sqrt{-5}$, the latter equality because $-5\equiv 3\bmod 4$.

As you observed, ${\Bbb Z}_{(3)}$ is the localization of $\Bbb Z$ at the ideal $(3)$. Then, it follows from general properties (see Atiyah-Macdonald, Ch 5) that the integral closure $A$ of ${\Bbb Z}_{(3)}$ in $K$ is the localization at the ideal $(3)$ in ${\cal O}_K$.

Since $3$ splits in $K$, $A$ has two maximal ideals, namely ${\frak m}_{\pm}=(1\pm\sqrt{-5})A$.

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May I ask why 3 splits in $K$? It seems that 3 does not split into an even number of prime ideals in the 5-th cyclotomic field,(in my opinion, 3 is nonsplit in the 5-th cyclotomic field) which is the equivalent condition, and 3 is not a quadratic residue of 5, which are the reasons for me to think that 3 does not split in $K$. Thanks for any clarification. –  awllower Feb 25 '11 at 10:00
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@awllower: $3$ splits in $K$ because $-5$ is a quadratic residue modulo $3$. As to how $3$ splits in $\mathbb{Q}(\zeta_5)$ -- this is of course not a matter of opinion, but you are correct: $3$ is inert in this field, since the order of $3$ in $(\mathbb{Z}/5\mathbb{Z})^{\times}$ is equal to $4$. There is no contradiction here, because $\mathbb{Q}(\sqrt{-5})$ is not a subfield of $\mathbb{Q}(\zeta_5)$: rather the unique quadratic subfield of this cyclotomic field is $\mathbb{Q}(\sqrt{5})$. –  Pete L. Clark Feb 25 '11 at 13:48
    
Sorry, I just recall that 5* is $\sqrt5$ instead of $\sqrt{-5}$ which is why I made things go wrong, thanks very much. Although, whatever the order of 3 modulo 5 is, 3 is not congruent to 1 modulo 5, which by the theory of cyclotomic fields means that 3 is not totally split in $Q(5^*)$. –  awllower Feb 25 '11 at 15:25
    
The reference to Atiyah-Macdonald, Ch 5 was very helpful. –  admchrch Feb 26 '11 at 5:48

Here is a different and (perhaps) somewhat more elementary approach than Andrea's.

Let $R$ be the integral closure of $\mathbb{Z}_{(3)}$ in $\mathbb{Q}(\sqrt{-5})$. Clearly $R$ is a subring of $\mathbb{Q}(\sqrt{-5})$, and thus we are looking for a necessary and sufficient condition on $a,b \in \mathbb{Q}$ such that $a+b\sqrt{-5} \in R$.

Let $P(a,b)$ be the minimal polynomial of $a+b\sqrt{-5}$ over $\mathbb{Q}$, i.e., the unique monic polynomial with $\mathbb{Q}$-coefficients of least degree satisfied by $a+b\sqrt{-5}$. Certainly if $P(a,b)$ has coefficients lying in $\mathbb{Z}_{(3)}$ then $a+b\sqrt{-5}$ lies in $R$. In fact the converse is true because the ring $\mathbb{Z}_{(3)}$ is integrally closed (e.g. it is a PID and PID $\implies$ UFD $\implies$ integrally closed). For a proof of this fact, see e.g. the section on Integrally Closed Domains in these notes. (Currently this is Theorem $260$ in $\S 14.5$, but that more precise information is subject to change.)

Try out this computation for yourself: you should get that $a+b\sqrt{-5} \in R \iff a,b \in \mathbb{Z}_{(3)}$.

Now in my notes I also prove the compatibility of integral closure with localization which Andrea uses in his answer: in fact it comes about five pages earlier than the fact about minimal polynomials mentioned above. So it is certainly arguable which of these facts is "more basic". My reason for choosing the latter is because I think it is a little more concrete and amenable to computation: indeed, if you don't know this fact about minimal polynomials, it seems to me that you're going to have a hard time computing any nontrivial examples of integral closures whatsoever.

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