# Thread: Domains

1. ## Domains

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?

2. The two seem to be abelian groups for $+,$ but take a closer look at the stability under multiplication.

For example, in the second $R,\ \frac{1}{2}=\frac{1}{2}(1+0.\sqrt{2})$ belongs to it, but what about the product $\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$?

If there are no $a,b\in\mathbb{Z}$ such that $\frac{1}{4}=\frac{1}{2}(a+b\sqrt{2}),$ then $R$ is not stable under multiplication and it cannot be a ring.

3. 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.

4. $R=\{a+bc\ ;\ a,b\in\mathbb{Z}\}$ and $c$ 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 $(\mathbb{R},+,\times )$ , since $R\subseteq\mathbb{R}$ .

Proving that $(R,+)$ is a subgroup of $(\mathbb{R},+)$ is quite easy.

$1\in R$ because $1=1+0.c$

The most tricky (and here the last) part is to show that $R$ is closed under multiplication. So take two elements of $R,$ let's say $a+bc$ and $n+mc,$ and consider their product: (we want to prove it belongs to $R$)

As we're working in $(\mathbb{R},+,\times )$ which is a commutative ring, we can use the properties of addition and multiplication.

$(a+bc)(n+mc)$

$=an+amc+bcn+bcmc$ by distributivity of $\times$ over $+$

$=an+amc+bnc+bmc^2$ by commutativity of $\times$

$=an+bmc^2+amc+bnc$ by commutativity of $+$

$=an+bmc^2+(am+bn)c$ by distributivity of $\times$ over $+$

$=an+bm(c-5)+(am+bn)c$ because c root of $x^2-x+5$ means $c^2=c-5$

$=(an-5bm)+(bm+am+bn)c$ by distributivity of $\times$ over $+$ and commutativity of $+$ and $\times$

Note that $a,b,n,m\in\mathbb{Z}$ implies that $(an-5bm)$ and $(bm+am+bn)$ belong to $\mathbb{Z},$ i.e. $(an-5bm)+(bm+am+bn)c\in R,$ 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 $(\mathbb{R},+,\times )$ is a domain, we immediately have that $(R,+,\times )$ is also a domain.

Using a similar (and even easier) proof, you can show that $(\{a+b\sqrt{2}\ ;\ a,b\in\mathbb{Z}\},+,\times )$ is a domain too!