Why isn't R a domain for \frac{1}{2}(a+b\sqrt{2})?

[Artusartos]

Homework Statement

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

Homework Equations

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?

Thanks in advance

 

[HallsofIvy]

You haven't explained this very well! I take it that you do NOT mean "R= (1/2)(a+ b\sqrt{2})", a single number, but rather "R= \{(1/2)(a+ b\sqrt{2})\}", the <b>set</b> of all such numbers with a and b integers. Yes, if a= 2 and b= 0, we have (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. <br /> <br /> 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 <b>integral</b> domain?

 

[Artusartos]

You haven't explained this very well! I take it that you do NOT mean "R= (1/2)(a+ b\sqrt{2})", a single number, but rather "R= \{(1/2)(a+ b\sqrt{2})\}", the <b>set</b> of all such numbers with a and b integers. Yes, if a= 2 and b= 0, we have (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. <br /> <br /> 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.<br /> <br /> <br /> Now, what makes you believe this is not an <b>integral</b> domain?

<br /> <br /> 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...

 

[Borek]

I actually did write R={...}, but the brackets automatically got deleted.

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