Let \(\mathcal{O}\) be the quadratic ring of integers in \(\mathbb{Q}(\sqrt{D})\). For any positive integer \(f\) prove that the set \(\mathcal{O}_f = \{a + bf\omega \mid a,b \in \mathbb{Z}\}\) is a subring of \(\mathcal{O}\) containing the identity, and that \(\left[\mathcal{O} : \mathcal{O}_f\right] = f\). Prove conversely that a subring of \(\mathcal{O}\) containing the identity and having finite index \(f\) (as additive group) is equal to \(\mathcal{O}_f\).
\(\implies\): Evident.
\(\impliedby\):Lemma. Let \(G = \mathbb{Z} \times \mathbb{Z}\) and \(H \le G\). If \(\left|G : H\right|\) is finite, then \(H \cong a\mathbb{Z} \times b\mathbb{Z}\) for positive integers \(a,b\).
Proof. Suppose there are nonzero integers \(m,n\) such that \((m,0),(0,n) \in H\). We can take \(m,n\) to be the smallest such positive integer, which implies \(H \cong m\mathbb{Z} \times n\mathbb{Z}\). Now suppose \((m,0) \not\in H\) for all \(m \in \mathbb{Z}-\{0\}\). Then any two cosets \((i,0) + H\) and \((j,0) + H\) with \(i,j \in \mathbb{Z}-\{0\}\), \(i \neq j\) are distinct, since \((i,0)-(j,0) \not\in H\). This contradicts the fact that \(\left|G : H\right|\) is finite. The same argument can be applied if \((0,n) \not\in H\) for all \(n \in \mathbb{Z}-\{0\}\). \(\blacksquare\)
As an additive group, \(\mathcal{O}\) is isomorphic to \(\mathbb{Z} \times \mathbb{Z}\). Any subgroup \(H\) of index \(f\) is of the form \(m\mathbb{Z} \times n\mathbb{Z}\), where \(m,n\) are the smallest positive integers appearing in their respective components and \(mn = f\). However by assumption \(H\) contains the multiplicative identity and so \(m = 1\), forcing \(H = \mathbb{Z} \times f\mathbb{Z}\). As a subring, \(H\) is precisely \(\mathcal{O}_f\).
No comments:
Post a Comment