# Commutative ring testing

• September 18th 2010, 02:53 PM
I-Think
Commutative ring testing
Consider the set
$Z[\sqrt{3}]=(a+b\sqrt{3} | a, b \in{Z})$

a) Show that $Z[\sqrt{3}]$ is a commutative ring
b) Show that if $a+b\sqrt{3}=c+d\sqrt{3}$ then $a=b$ and $c=d$

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

Noting that $Z$is a commutative ring

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

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

Multiplicative commutativty
$(a+b\sqrt{3})*(c+d\sqrt{3})=(ac+ad\sqrt{3}+bc\sqrt {3}+3bd)$
$(c+d\sqrt{3})*(a+b\sqrt{3})=(ca+cb\sqrt{3}+da\sqrt {3{+3db)$

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

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

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

Is it sufficient to show that
$(a-c)+(b-d)\sqrt{3}=0+0\sqrt{3}$
This $a-c=0$ and $b-d=0$
Hence $a=c$and $b=d$
Will this suffice?

Thanks for the responses in advance
• September 18th 2010, 05:51 PM
rualin
That is what I would do. It's a boring exercise to get you learning the ring axioms.