a) Show that $\displaystyle Z[\sqrt{3}]$ is a commutative ring

b) Show that if $\displaystyle a+b\sqrt{3}=c+d\sqrt{3}$ then $\displaystyle a=b$ and $\displaystyle c=d$

Answers

a) Is it sufficient to show that $\displaystyle Z[\sqrt{3}]$ satisfies additive commutative, associativity, identity and inverse and satisfies multiplicative associativity, commutativity and identity?

In that case

Noting that $\displaystyle Z$is a commutative ring

Additive commutativity

$\displaystyle (a+b\sqrt{3})+(c+d\sqrt{3})=((a+c)+(b+d)\sqrt{3})$

$\displaystyle (c+d\sqrt{3})+(a+b\sqrt{3})=((c+a)+(d+b)\sqrt{3})$

The above statements are equal to each other, then $\displaystyle Z[\sqrt{3}]$ satisfies this property.

Multiplicative commutativty

$\displaystyle (a+b\sqrt{3})*(c+d\sqrt{3})=(ac+ad\sqrt{3}+bc\sqrt {3}+3bd)$

$\displaystyle (c+d\sqrt{3})*(a+b\sqrt{3})=(ca+cb\sqrt{3}+da\sqrt {3{+3db)$

Above 2 expressions are equal to each other, then $\displaystyle Z[\sqrt{3}]$ satisfies this property.

Do I continue in this vein with the other properties to prove that $\displaystyle Z[\sqrt{3}] $is a commutative ring?

Part B

$\displaystyle a+b\sqrt{3}=c+d\sqrt{3}$

Is it sufficient to show that

$\displaystyle (a-c)+(b-d)\sqrt{3}=0+0\sqrt{3}$

This $\displaystyle a-c=0$ and $\displaystyle b-d=0$

Hence $\displaystyle a=c $and $\displaystyle b=d$

Will this suffice?

Thanks for the responses in advance