1. ## Commutative ring testing

Consider the set
$\displaystyle Z[\sqrt{3}]=(a+b\sqrt{3} | a, b \in{Z})$

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 $\displaystyle c=d$

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

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

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

2. That is what I would do. It's a boring exercise to get you learning the ring axioms.