Why is it that R = {a + bsqrt(2): a, b belong to Z} is a domain, but R = { (1/2)(a + bsqrt(2)): a, b belong to Z} isn't?
The two seem to be abelian groups for $\displaystyle +,$ but take a closer look at the stability under multiplication.
For example, in the second $\displaystyle R,\ \frac{1}{2}=\frac{1}{2}(1+0.\sqrt{2})$ belongs to it, but what about the product $\displaystyle \frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$?
If there are no $\displaystyle a,b\in\mathbb{Z}$ such that $\displaystyle \frac{1}{4}=\frac{1}{2}(a+b\sqrt{2}),$ then $\displaystyle R$ is not stable under multiplication and it cannot be a ring.
I understand proving cases where R is not a domain perfectly now, but I am still confused about how to prove cases where R is a domain. Another question I was assigned was to use the fact that c = (1/2)(1 + sqrt(-19)) is a root of x^2 - x + 5 to prove that R = {a + bc: a,b belong to Z} is a domain and I don't know where to start.
$\displaystyle R=\{a+bc\ ;\ a,b\in\mathbb{Z}\}$ and $\displaystyle c$ as you defined.
A good way to prove something is a ring is to show it is a subring of some ring you already know; here one can think of $\displaystyle (\mathbb{R},+,\times )$ , since $\displaystyle R\subseteq\mathbb{R}$ .
Proving that $\displaystyle (R,+)$ is a subgroup of $\displaystyle (\mathbb{R},+)$ is quite easy.
$\displaystyle 1\in R$ because $\displaystyle 1=1+0.c$
The most tricky (and here the last) part is to show that $\displaystyle R$ is closed under multiplication. So take two elements of $\displaystyle R,$ let's say $\displaystyle a+bc$ and $\displaystyle n+mc,$ and consider their product: (we want to prove it belongs to $\displaystyle R$)
As we're working in $\displaystyle (\mathbb{R},+,\times )$ which is a commutative ring, we can use the properties of addition and multiplication.
$\displaystyle (a+bc)(n+mc)$
$\displaystyle =an+amc+bcn+bcmc$ by distributivity of $\displaystyle \times$ over $\displaystyle +$
$\displaystyle =an+amc+bnc+bmc^2$ by commutativity of $\displaystyle \times$
$\displaystyle =an+bmc^2+amc+bnc$ by commutativity of $\displaystyle +$
$\displaystyle =an+bmc^2+(am+bn)c$ by distributivity of $\displaystyle \times$ over $\displaystyle +$
$\displaystyle =an+bm(c-5)+(am+bn)c$ because c root of $\displaystyle x^2-x+5$ means $\displaystyle c^2=c-5$
$\displaystyle =(an-5bm)+(bm+am+bn)c$ by distributivity of $\displaystyle \times$ over $\displaystyle +$ and commutativity of $\displaystyle +$ and $\displaystyle \times$
Note that $\displaystyle a,b,n,m\in\mathbb{Z}$ implies that $\displaystyle (an-5bm)$ and $\displaystyle (bm+am+bn)$ belong to $\displaystyle \mathbb{Z},$ i.e. $\displaystyle (an-5bm)+(bm+am+bn)c\in R,$ and we're done.
That proof of the closure of R under multiplication is quite detailled, that's why it can seem long...
I nearly forgot, you wanted to show it is a domain, i.e. the product of two non-nul elements is a non-nul element. Well a subring of a domain is a domain, so since $\displaystyle (\mathbb{R},+,\times )$ is a domain, we immediately have that $\displaystyle (R,+,\times )$ is also a domain.
Using a similar (and even easier) proof, you can show that $\displaystyle (\{a+b\sqrt{2}\ ;\ a,b\in\mathbb{Z}\},+,\times )$ is a domain too!