For a,b∈Z, let B(a,b) ∈ M(2,Z) be defined by B(a,b)=[a3bba]. Let S={B(a,b);a,b∈Z}⊆ M(2,Z). Show that S≅Z[3]={a+b3;a,b∈Z}

S={B(a,b):a,b∈Z}⊆ M(2,Z) B=(a,b)=[a3bba] Z[3]={a+b3:a,b∈z} Show that S≅Z(3) Defined A function f: S→z (3) by f(B(a,b))=a+b(3) ie f(B(a,b))=f([a3bba])=a+b3 Clearly f is well defined. f is one-one Kerf={B(a,b)∣f(B(a,b))=0} ={B(a,b)∣a+b3=0}={B(a,b)∣a=0Bb=0} {B(0,0)}={[03x000]}={[0000]}={0} Kerf={0} since Kerf={0}. Then f is one-one. f is onto:- det p+q3∈z[3], p,q∈z ∋B(p,q)=[d3qqp]∈◼S such that f(B(p,q))=p+q3 This show that f is onto. q3 f:S→z(3) by f(B(a,b))=a+b3 is esomorphism. ∴ S≅z(3)

"For a,b∈Z, let B(a,b) ∈ M(2,Z) be defined..." solved and explained

High School
Algebra
Algebra I
Algebra foundations

he298c

Answered question

2021-09-14

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

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

Answer & Explanation

nitruraviX

Skilled2021-09-15Added 101 answers

$S = {B (a, b) : a, b \in Z} \subseteq$ M(2,Z) $B = (a, b) = [\begin{matrix} a & 3 b \\ b & a \end{matrix}]$

$Z [\sqrt{3}] = {a + b \sqrt{3} : a, b \in z}$

Show that $S ≅ Z (\sqrt{3})$

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

ie $f (B (a, b)) = f ([\begin{matrix} a & 3 b \\ b & a \end{matrix}]) = a + b \sqrt{3}$

Clearly f is well defined.

f is one-one

$K e r f = {B (a, b) ∣ f (B (a, b)) = 0}$

$= {B (a, b) ∣ a + b \sqrt{3} = 0} = {B (a, b) ∣ a = 0 B b = 0}$

${B (0, 0)} = {[\begin{matrix} 0 & 3 x 0 \\ 0 & 0 \end{matrix}]} = {[\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]} = {0}$

$K e r f = {0}$

since $K e r f = {0}$ . Then f is one-one.

f is onto:- det p $+ q \sqrt{3} \in z [\sqrt{3}]$ , $p, q \in z$

$∋ B (p, q) = [\begin{matrix} d & 3 q \\ q & p \end{matrix}] \in ◼ S$ such that $f (B (p, q)) = p + q \sqrt{3}$

This show that f is onto.

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

$∴ S ≅ z (\sqrt{3})$

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