# Why isn't this a domain?

1. Dec 27, 2012

### Artusartos

1. The problem statement, all variables and given/known data

If we have $$R = \{ \frac{1}{2}(a+b\sqrt{2}) \}$$.

2. Relevant equations

3. The attempt at a solution

For the identity, we have a=2 and b=0. So the identity is not equal to zero. Also, there can't be any zero divisors, because...
$$[\frac{1}{2}(a+b\sqrt{2})][\frac{1}{2}(a'+b'\sqrt{2})]$$ = $$\frac{1}{4}(aa' + ab'\sqrt{2} + a'b\sqrt{2} + 2bb')$$. If a,a',b, b' are not zero, then this can't be zero.

So why isn't it a domain?

Last edited by a moderator: Dec 27, 2012
2. Dec 27, 2012

### HallsofIvy

Staff Emeritus
You haven't explained this very well! I take it that you do NOT mean "$R= (1/2)(a+ b\sqrt{2})[itex]", a single number, but rather "[itex]R= \{(1/2)(a+ b\sqrt{2})\}[itex]", the set of all such numbers with a and b integers. Yes, if a= 2 and b= 0, we have [itex](1/2)(2+ 0\sqrt{2})= 1$ as multiplicative identity which is not 0, the additive indentity, so this is a ring with more than one member.

Yes, in order that $[(1/2)(a+ b\sqrt{2}][(1/2)(a'+ b'\sqrt{2})]= (1/4)(aa'+ ab'\sqrt{2}+ a'b\sqrt{2}+ 2bb')$ be 0, we must have aa'+ 2bb'= 0 and ab'+ a'b= 0. The second gives ab'= -a'b or a/b= -a'/b'. The first gives a/b= -2b'/a'. that is, -a'/b'= -2b'/a' or $a'^2= 2b'^2$ which is impossible.

Now, what makes you believe this is not an integral domain?

3. Dec 27, 2012

### Artusartos

Thanks. I actually did write R={...}, but the brackets automatically got deleted. This is a question in my textbook. It tells us to prove that this set is not a domain, and I'm confused about that...

4. Dec 27, 2012

### Staff: Mentor

You may want to learn about the difference between {} and \{\} in LaTeX