Compute the inverse of an ideal in a given algebraic number field

Recall that if K is an algebraic number field with ring of integers \mathcal{O} and if A \subseteq \mathcal{O} is an ideal, then A^{-1} = \{r \in K \ |\ rA \subseteq \mathcal{O}\}. Compute A^{-1} for each of the following ideals A in their respective field K: (1) A = (2) in K = \mathbb{Q} with \mathcal{O} = \mathbb{Z}, (2) A = (1+i) in K = \mathbb{Q}(i) with \mathcal{O} = \mathbb{Z}[i], and (3) A = (3, 1+2\sqrt{-5}) in K = \mathbb{Q}(\sqrt{-5}) with \mathcal{O} = \mathbb{Z}[\sqrt{-5}].

  1. We claim that (2)^{-1} = \{ \frac{a}{2} \ |\ a \in \mathbb{Z} \} in \mathbb{Q}. (\subseteq) Suppose r \in (2)^{-1} in \mathbb{Q}. In particular, 2r \in \mathbb{Z}, so 2r = a for some a \in \mathbb{Z}. So r = \frac{a}{2} with a \in \mathbb{Z}. (\supseteq) Suppose a \in \mathbb{Z} and consider \frac{a}{2}. If 2k \in (2) (in \mathbb{Z}) is arbitrary, then \frac{a}{2} \cdot 2k = ak \in \mathbb{Z}. So \frac{a}{2}(2) \subseteq \mathbb{Z}, and thus \frac{a}{2} \in (2)^{-1}.
  2. We claim that (1+i)^{-1} = \{ \frac{a}{2} + \frac{b}{2}i \ |\ a,b \in \mathbb{Z}, a \equiv b \mod 2 \} in \mathbb{Q}(i). (\subseteq) Suppose s+ti \in (1+i)^{-1} in \mathbb{Q}(i). In particular, (s+ti)(1+i) = (s-t)+(s+t)i \in \mathbb{Z}[i], so that s-t,s+t \in \mathbb{Z}. Adding, we have 2s \in \mathbb{Z}, so that 2s = a for some a \in \mathbb{Z}, and so s = \frac{a}{2}. Now \frac{a}{2} + t \in \mathbb{Z}, so that \frac{a}{2} + t = b for some b \in \mathbb{Z}. Now t = \frac{2b-a}{2}. Note that a \equiv 2b-a mod 2. So we have s+ti = \frac{h}{2} + \frac{k}{2}i where h,k \in \mathbb{Z} and h \equiv k mod 2, as desired. (\supseteq) Suppose a,b \in \mathbb{Z} such that a \equiv b mod 2 and consider \eta = \frac{a}{2} + \frac{b}{2}i. Note that \eta(1+i) = \frac{a-b}{2} + \frac{a+b}{2}i \in \mathbb{Z}[i], since a+b \equiv a-b \equiv 0 mod 2. Thus \eta\zeta \in \mathbb{Z}[i] for all \zeta \in (1+i), so we have \eta(1+i) \subseteq \mathbb{Z}[i], and thus \eta \in (1+i)^{-1}.
  3. We claim that (3,1+2\sqrt{-5})^{-1} = \{ \frac{a}{3} + \frac{b}{3}\sqrt{-5}\ |\ a,b \in \mathbb{Z}, a-b \equiv 0 \mod 3 \} in \mathbb{Q}(\sqrt{-5}). (\subseteq) Suppose s+t\sqrt{-5} \in (3,1+2\sqrt{-5})^{-1}. In particular, 3(s+t\sqrt{-5}) \in \mathbb{Z}[\sqrt{-5}] so that (comparing coefficients) s = \frac{a}{3} and t = \frac{b}{3} for some a, b\ in \mathbb{Z}. Moreover, we have (s+t\sqrt{-5})(1+2\sqrt{-5}) \in \mathbb{Z}[\sqrt{-5}], so that (comparing coefficients) s-10t \in \mathbb{Z}. Say \frac{a}{3} - \frac{10b}{3} = h, so that a-10b = 3h. Mod 3, we have a-b \equiv 0, as desired. (\supseteq) Consider \mu = \frac{a}{3} + \frac{b}{3}\sqrt{-5} where a-b \equiv 0 mod 3. Now 3\mu \in \mathbb{Z}[\sqrt{-5}] and (1+2\sqrt{-5})\mu = \frac{a-10b}{3} + \frac{2a+b}{3}\sqrt{-5} \in \mathbb{Z}[\sqrt{-5}] since a-10b \equiv 0 mod 3 and 2a+b \equiv -a+b \equiv -(a-b) \equiv 0 mod 3. So \mu\zeta \in \mathbb{Z}[\sqrt{-5}] for all \zeta \in (3,1+2\sqrt{-5}), and thus \mu \in (3,1+2\sqrt{-5})^{-1}.
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