$\textbf{Exercise}[7.3.12]$ (Parts (a) and (b)) Let $D \in{\mathbb{Z}}$ that is not a perfect square and let
$$S=\{\begin{pmatrix}
a&b\\
Db&a
\end{pmatrix} \mid a,b\in{\mathbb{Z}}\}$$
(a) Prove that $S$ is a subring of $M_{2 \times 2}(\mathbb{Z})$
(b) If $D$ is not a perfect square in $\mathbb{Z}$ prove that the map $\varphi: \mathbb{Z}[\sqrt{D}] \longrightarrow S$ defined by
$$\varphi(a+b\sqrt{D})=\begin{pmatrix}
a&b\\
Db&a
\end{pmatrix}$$
is a ring isomorphism.
$\textit{Proof:}$ To show that $S$ is a subring of $M_{2 \times 2}(\mathbb{Z})$, we need to show that $S \neq \emptyset $, $S$ is closed under subtraction, and $S$ is closed under multiplication. $S$ is clearly non-empty. For example, $\begin{pmatrix}
1&0\\
0&1
\end{pmatrix} \in{S}$
Now let $A,B\in{S}$. $$A=\begin{pmatrix}
a&b\\
Db&a
\end{pmatrix}, B=\begin{pmatrix}
c&d\\
Dd&c
\end{pmatrix}$$
Then, $$A-B=\begin{pmatrix}
a&b\\
Db&a
\end{pmatrix}-\begin{pmatrix}
c&d\\
Dd&c
\end{pmatrix}=\begin{pmatrix}
a-c&b-d\\
D(b-d)&a-c
\end{pmatrix} \in{S}$$
The product,
$$AB=\begin{pmatrix}
a&b\\
Db&a
\end{pmatrix}\begin{pmatrix}
c&d\\
Dd&c
\end{pmatrix}=\begin{pmatrix}
ac+bdD&ad+bc\\
cbD+adD&Dbd+ac
\end{pmatrix}=\begin{pmatrix}
ac+bdD&ad+bc\\
D(ad+bc)&ac+bdD
\end{pmatrix}\in{S}$$
Therefore, $S$ is a subring.
We now need to show that $\varphi$ is well defined, a bijection, and a ring homomorphism. To check that $\varphi$ is well defined, we pick two equal representatives from $\mathbb{Z}[\sqrt{D}]$ and show that their images in $S$ are equal. So, let $x=a+b\sqrt{D}$ and $y=c+d\sqrt{D}$. Note that the assumption that $D$ is square-free is necessary. For example, let $D=4=2^2$. Consider $0+1 \cdot \sqrt{4}=2$ and $2+0\cdot \sqrt{4}=2$. These are 2 elements in $\mathbb{Z}[\sqrt{4}]$ that are equal, yet have a different representation. Then, $\varphi(0+1 \cdot \sqrt{4})=\begin{pmatrix}
0&1\\
2&0
\end{pmatrix}$ however, $\varphi(1+0 \cdot \sqrt{4})=\begin{pmatrix}
1&0\\
0&1
\end{pmatrix}$. This gives us 2 different matrices being mapped from the same element. So, if $D$ has a square in its factorization, then $\varphi$ is ill-defined.
Let $D$ be square-free. Now, $\varphi(x)=\begin{pmatrix}
a&b\\
Db&a
\end{pmatrix}$ and $\varphi(y)=\begin{pmatrix}
c&d\\
Dd&c
\end{pmatrix}$. Since $a,b,c,d\in{\mathbb{Z}}$ and $D$ is square-free, $x=y \Longleftrightarrow a=c$ and $b=d$ thus
$\varphi(x)=\begin{pmatrix}
a&b\\
Db&a
\end{pmatrix}=\begin{pmatrix}
c&d\\
Dd&c
\end{pmatrix}=\varphi(y)$. Reversing this process, if $\varphi(x)=\varphi(y)$ then each entry in the above matrices must be equal, and we get that $a=c$ and $b=d$ so $\varphi$ is an injective map.
$\varphi$ is surjective: Let $M\in{S}$ where $M=\begin{pmatrix}
a&b\\
Db&a
\end{pmatrix}$. Picking $x\in{\mathbb{Z}[\sqrt{D}]}$ where $x=a+b\sqrt{D}$, then $\varphi(x)=M$.
$\varphi$ is a ring homomorphism since it preserves both operations. Again letting $x=a+b\sqrt{D}$ and $y=c+d\sqrt{D}$, we see $\varphi$ preserves addition since:
$$\varphi(x+y)=\varphi((a+b\sqrt{D})+(c+d\sqrt{D}))=\varphi((a+c)+(b+d)\sqrt{D})$$
$$=\begin{pmatrix}
a+c&b+d\\
D(b+d)&a+c
\end{pmatrix}=\begin{pmatrix}
a&b\\
Db&a
\end{pmatrix}+\begin{pmatrix}
c&d\\
Dd&c
\end{pmatrix}=\varphi(x)+\varphi(y)$$
$$\varphi(xy)=\varphi((a+b\sqrt{D})(c+d\sqrt{D}))=\varphi(ac+ad\sqrt{D}+bc\sqrt{D}+bdD)=\varphi(ac+(ad+bc)\sqrt{D}+bdD)$$
$$=\varphi(ac+bdD+(ad+bc)\sqrt{D})=\begin{pmatrix}
ac+bdD&ad+bc\\
D(ad+bc)&ac+bdD
\end{pmatrix}$$
$$\varphi(x)\varphi(y)=\begin{pmatrix}
a&b\\
Db&a
\end{pmatrix}
\begin{pmatrix}
c&d\\
Dd&c
\end{pmatrix}=\begin{pmatrix}
ac+bdD&ad+bc\\
D(ad+bc)&ac+bdD
\end{pmatrix}=\varphi(xy)$$
Thus, $\varphi$ is a well-defined, bijective ring homomorphism i.e. a ring isomorphism.
$\therefore \quad \mathbb{Z}[\sqrt{D}] \cong S$ $\blacksquare$
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