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
