# Is $R$ an integral domain?

Let $R = \{a + b\alpha |\ a,b \in \mathbb{Z}\}\subseteq \mathbb{C}$ where $\alpha = \frac{1}{2}(1+\sqrt{-19})$

Is $R$ an integral domain?

To show whether or not $R$ is an integral domain, letting $x = \{a+b\alpha\}$ and $y = \{c+d\alpha\}$, where $a,b,c,d \in \mathbb{Z}$ and showing $xy = yx \in R$ would suffice without going into further, since an integral domain is a commutative ring at first?

Hint $\$ One easily verifies that it is a subring of $\,\Bbb C\,$ since it satisfies the subring test, i.e. it contains $\,1\,$ and is closed under subtraction and multiplication (use $\,\alpha^2 = m\alpha + n\,$ for some $\,m,n\in\Bbb Z).\,$ Therefore, being a subring of a field, it is an integral domain.

• That's not quite a hint, more of a direct answer! Good call though, that's a bit easier than the elementary number theory involved in my answer. – walkar Feb 15 '15 at 17:31
• @Bill Dubquqe Thanks for your hint/direct answer. – Jellyfish Feb 15 '15 at 17:36
• @walker, this problem is just some random problem that I am working on to prepare for my test, not some hw problem :) – Jellyfish Feb 15 '15 at 17:37
• @abcd1234 That's exactly why it contains $1$! Not arbitrary at all, it's an honest-to-goodness element of $R$. – walkar Feb 15 '15 at 17:45
• @abcd1234  As you note, $\,1\in R\,$ follows immediately from the definition of $\,R\ \$ – Bill Dubuque Feb 15 '15 at 17:48

Hint: An integral domain is a commutative ring with no zero divisors. Write out an arbitrary product of elements in $R$ and set it equal to zero, and see what you can do from there.

It does not suffice to show simply that $R$ is a commutative ring -- that is necessary but not sufficient to being an integral domain.

• That's where I was confused. – Jellyfish Feb 15 '15 at 17:00
• A subring of a commutative ring is commutative, so you get that part for free anyway. – Max Feb 15 '15 at 17:01
• A subring of an integral domain (such as the field $\mathbf C$) is an integraldomain. – Bernard Feb 15 '15 at 17:07
• @abcd1234 Write $(a+b\alpha)(c+d\alpha) = ac+bd\alpha^2 +(ad+bc)\alpha = 0$ and consider all the cases. You will end up with one of $a+b\alpha$ or $c+d\alpha$ must be $0$, or some contradictions with $\alpha \in \mathbb{Z}$. – walkar Feb 15 '15 at 17:07
• Ok, so if $\alpha^2 = \alpha - 5$ then, $ac + bd(\alpha - 5) +(ad +bc)\alpha = (ac - 5bd) + (bd + ad + bc)\alpha =$ some $p + q\alpha$, since $a,b,c,d \in \mathbb{Z}$. Now, what are we concluding here? – Jellyfish Feb 15 '15 at 17:23