# For a,b \inZ, let B(a,b) \inM(2,Z) be defined by B(a,b)

For $$a,b \in Z$$, let B(a,b) $$\in$$ M(2,Z) be defined by $$B(a,b) =\begin{bmatrix}a & 3b \\b & a \end{bmatrix}$$.

Let $$S =\left\{B(a,b);a,b\in Z \right\}\subseteq$$ M(2,Z). Show that $$S\cong Z\left[ \sqrt{3}\right]=\left\{a+b\sqrt{3};a,b\in Z\right\}$$

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nitruraviX

$$S=\left\{B(a,b):a,b\in Z\right\} \subseteq$$ 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.

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