\(S=\left\{B(a,b):a,b\in Z\right\} \subseteq\) M(2,Z) \(B=(a,b)=\begin{bmatrix}a & 3b \\b & a \end{bmatrix}\)

\(Z\left[ \sqrt{3}\right]=\left\{a+b\sqrt{3}:a,b\in z\right\}\)

Show that \(S\cong Z(\sqrt{3})\)

Defined A function f: \(S\rightarrow z\) \((\sqrt{3})\) by \(f(B(a,b))=a+b(\sqrt{3})\)

ie \(f(B(a,b))=f\left(\begin{bmatrix}a & 3b \\b & a \end{bmatrix}\right)=a+b\sqrt{3}\)

Clearly f is well defined.

f is one-one

\(Kerf=\left\{B(a,b)\mid f(B(a,b))=0\right\}\)

\(=\left\{B(a,b)\mid a+b\sqrt{3}=0\right\}=\left\{B(a,b)\mid a=0 B b=0\right\}\)

\(\left\{B(0,0)\right\}=\left\{\begin{bmatrix}0 & 3x0 \\0 & 0 \end{bmatrix}\right\}=\left\{\begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}\right\}=\left\{0\right\}\)

\(Kerf=\left\{0\right\} \)

since \(Kerf=\left\{0\right\}\). Then f is one-one.

f is onto:- det p\(+q\sqrt{3}\in z\left[\sqrt{3}\right]\), \(p,q \in z\)

\(\ni B(p,q)=\begin{bmatrix}d & 3q \\q & p \end{bmatrix}\in\blacksquare S\) such that \(f(B(p,q))=p+q\sqrt{3}\)

This show that f is onto.

\(q\sqrt{3}\ \ \ f:S\rightarrow z(\sqrt{3})\) by \(f(B(a,b))=a+b\sqrt{3}\) is esomorphism.

\(\therefore \ \ \ S\cong z(\sqrt{3})\)