Order of Conductor f (7.1.23)

Dummit and Foote Abstract Algebra, section 7.1, exercise 23:

Define $\mathcal{O}_f=\mathbb{Z}[f\omega]=\{a+b\omega~|~a,b∈\mathbb{Z}\}$. Letting $[R]$ denote the abelian group in the ring $R$, prove:$$R = \mathcal{O}_f ⇔ R~\text{subring of}~\mathcal{O} \land 1∈R \land |[\mathcal{O}] : [R]|=f$$ Proof: Lemma 1: For $H≤G$ groups and $φ$ an isomorphism, we have $|G:H|=|φ(G):φ(H)|$. Proof: Let $n_1,...$ be a complete set of coset-distinct representatives for $H$ in $G$. We claim that their images are a complete set of coset-distinct representatives of $φ(H)$ in $φ(G)$. For any $g∈G$, we have $g=n_ih$ for some index $i$ and $h∈H$. Therefore, we have $φ(g)=φ(n_ih)=φ(n_i)φ(h)$. Assuming $φ(n_i)$ and $φ(n_j)$ are in the same coset of $φ(H)$, we have $φ(n_i)=φ(n_j)φ(h)$$=φ(n_jh)$, so that $n_i=n_jh$ and $n_i$ and $n_j$ are in the same coset of $H$, a contradiction. Thus, there are just as many coset representatives of $φ(H)$ in $φ(G)$ as $H$ in $G$.$~\square$

($⇒$) (1) All that needs to be done is to prove closure under subtraction and multiplication. We have $(a+bf\omega)-(c+df\omega)=(a-c)+(b-d)f\omega$. As well, $\omega^2=D$ or $\dfrac{D-1}{4}+\dfrac{1+\sqrt{D}}{2}$ depending on $D$ modulo 4, so$$(a+bf\omega)(c+df\omega)=(ac)+(bc+ad)f\omega+bdf^2\omega^2$$is evidently within the ring as well. (2) Clearly $1∈\mathcal{O}_f$; set $a=1$ and $b=0$. (3) Construct a mapping $φ:[\mathcal{O}]→\mathbb{Z} \times \mathbb{Z}$ by $a+b\omega \mapsto (a,b)$. This is well defined by unique $a,b$ form of the argument, clearly surjective and injective, and$$φ((a+b\omega)+(c+d\omega))=φ((a+c)+(b+d)\omega)=(a+c,b+d)$$$$=(a,b)+(c,d)=φ(a+b\omega)+φ(c+d\omega)$$so that $φ$ is homomorphic and now an isomorphism. So $[\mathcal{O}]≅\mathbb{Z} \times \mathbb{Z}$ and by definition we can tell $[\mathcal{O}_f] ≅ φ([\mathcal{O}_f])=\mathbb{Z} \times f\mathbb{Z}$. By 5.1.14, we have $(\mathbb{Z} \times \mathbb{Z})/(\mathbb{Z} \times f\mathbb{Z})≅$$(\mathbb{Z}/\mathbb{Z})\times(\mathbb{Z}/f\mathbb{Z})≅\mathbb{Z}/f\mathbb{Z}$ is a group of order $f$, so that $f=|\mathbb{Z} \times \mathbb{Z} : \mathbb{Z} \times f\mathbb{Z}|=|φ([\mathcal{O}]):φ([\mathcal{O}_f])|=|[\mathcal{O}]:[\mathcal{O}_f]|$.

Lemma 2: Any nontrivial subgroup of $\mathbb{Z}$ is of the form $n\mathbb{Z}$ for some $n∈\mathbb{Z}^+$. Proof: Let distinct $a,b$ be chosen with $n=|a-b|$ minimal; we thus have $n\mathbb{Z}$ is contained within the subgroup. Assume there is an element $x$ not within $n\mathbb{Z}$: We have $x=qn+r$ with $0 < r < n$, and since $qn∈n\mathbb{Z}$, we have $x-qn < n=|a-b|$, a contradiction. So $n\mathbb{Z}$ is precisely the subgroup in question.$~\square$

($\Leftarrow$) $R$'s identity must be $1$, as $n(a+bf\omega)≠a+bf\omega$ for any $n≠1$ and $a≠0≠b$. Therefore, we have $1\mathbb{Z} \times 0=\mathbb{Z} \times 0≤φ([R])$. For any $(a,b)∈φ([R])$, we also have $(a,0)∈φ([R])$, so that $(0,b)∈φ([R])$, implying the second factor of $φ([R])$ is a subgroup of $\mathbb{Z}$ and therefore of the form $n\mathbb{Z}$. Since $| \mathbb{Z} \times \mathbb{Z}:\mathbb{Z} \times n\mathbb{Z}|=n$, we have $n=f$ and now $φ([R])=\mathbb{Z} \times f\mathbb{Z}=φ([\mathcal{O}_f])$ and now $[R]=[\mathcal{O}_f]$. Since the specifications of multiplication are predefined in $\mathbb{C}$, we have $R=\mathcal{O}_f$.$~\square$