# Thread: Ring of Integers

1. ## Ring of Integers

There's a question that says, "Prove that the ring of integers O in the quadratic integer ring Q$\displaystyle \sqrt{2}$ is a Euclidean Domain.

What do they mean by ring of integers O?

I know Q$\displaystyle \sqrt{2}$={q + r$\displaystyle \sqrt{2}$;q,r in Q}
... so would the ring of integers be O = {q + 0$\displaystyle \sqrt{2}$;q in Z} or do they mean algebraic integers or something?

2. ## Re: Ring of Integers

Originally Posted by gummy_ratz
There's a question that says, "Prove that the ring of integers O in the quadratic integer ring Q$\displaystyle \sqrt{2}$ is a Euclidean Domain.

What do they mean by ring of integers O?

I know Q$\displaystyle \sqrt{2}$={q + r$\displaystyle \sqrt{2}$;q,r in Q}
... so would the ring of integers be O = {q + 0$\displaystyle \sqrt{2}$;q in Z} or do they mean algebraic integers or something?
Since $\displaystyle 2\equiv 2\text{ mod }4$ what they mean by $\displaystyle \mathcal{O}$ is the set $\displaystyle \mathbb{Z}[\sqrt{2}]$. This is a Euclidean domain with the usual norm $\displaystyle N(a+b\sqrt{2})=a^2+2b^2$. Can you prove it?

3. ## Re: Ring of Integers

Ohh okay, I think I can prove that. In another class I proved that Z[$\displaystyle \sqrt{-2}$] was a Euclidean Domain using the same norm, so I'm guessing it'll be pretty similar. Does the fact that they mentioned Q[$\displaystyle \sqrt{2}$] change anything, or do I just procede the same as if they said Z[$\displaystyle \sqrt{2}$] right off the bat?

4. ## Re: Ring of Integers

Originally Posted by gummy_ratz
Ohh okay, I think I can prove that. In another class I proved that Z[$\displaystyle \sqrt{-2}$] was a Euclidean Domain using the same norm, so I'm guessing it'll be pretty similar. Does the fact that they mentioned Q[$\displaystyle \sqrt{2}$] change anything, or do I just procede the same as if they said Z[$\displaystyle \sqrt{2}$] right off the bat?
Right off the bat!

5. ## Re: Ring of Integers

hmmm, wouldn't the standard norm for Z$\displaystyle \sqrt{2}$ be $\displaystyle a^2-2b^2$? Because if an element in Z$\displaystyle \sqrt{2}$ is $\displaystyle x=a+b\sqrt{2}$, then x multiplied by the complement of x would be $\displaystyle a^2-2b^2$.

6. ## Re: Ring of Integers

the co-domain of a euclidean function has to be a subset of the positive integers, so that we can use it to find a finite decreasing chain of "remainders".

7. ## Re: Ring of Integers

Right, I forgot the absolute value! So it's |a^2 - 2b^2|.

8. ## Re: Ring of Integers

well, that is positive, but your "norm" doesn't correspond very well to the notion of "distance from 0". for example, your norm makes 1+√2 the same "distance" away from 0 as 1.

use drexel28's norm, i assure you, he is experienced in such matters.

9. ## Re: Ring of Integers

Hmmm, okay, I'm just trying to understand where it came from. Because in all of the examples in class, we found the norm by multiplying x by its complement. And in my book they just say the norm in general is a^2 - Db^2. Here they seem to use the absolute value: The integers with the square root of 2 adjoined is a Euclidean domain « Project Crazy Project .

10. ## Re: Ring of Integers

The OP is actually correct, I misread $\displaystyle \sqrt{2}$ as $\displaystyle \sqrt{-2}$ even though I put $\displaystyle \sqrt{2}$ twice! It is a ED with norm $\displaystyle N(a+b\sqrt{2})=|a^2-2b^2|$.

Deveno, thanks for the vote of confidence though--it just seems this time I was a little off!

11. ## Re: Ring of Integers

Okay, cool. Thanks! And just one more thing, is it necessary to show N(a) < N(ab), or is it sufficient to just show N(r) < N(a)? Because on that website they don't mention N(a) < N(ab), but I see it in some definitions.

12. ## Re: Ring of Integers

Originally Posted by gummy_ratz
Okay, cool. Thanks! And just one more thing, is it necessary to show N(a) < N(ab), or is it sufficient to just show N(r) < N(a)? Because on that website they don't mention N(a) < N(ab), but I see it in some definitions.
What you are referring to is sometimes called the '$\displaystyle d$-inequality' There is no need to show it. It's a cool fact though that any Euclidean domain $\displaystyle (E,d)$ admits a Euclidean function (function which allows the Division Algorithm to work) also admits a Euclidean function $\displaystyle \widetilde{d}$ which satisfies the $\displaystyle d$-inequality. It is usually given by $\displaystyle \displaystyle \widetilde{d}(a)=\min_{b\ne 0}d(ab)$.

13. ## Re: Ring of Integers

Originally Posted by Drexel28
The OP is actually correct, I misread $\displaystyle \sqrt{2}$ as $\displaystyle \sqrt{-2}$ even though I put $\displaystyle \sqrt{2}$ twice! It is a ED with norm $\displaystyle N(a+b\sqrt{2})=|a^2-2b^2|$.

Deveno, thanks for the vote of confidence though--it just seems this time I was a little off!
you and me both! perhaps we should start a club: the "Little Bit Off" club. yearly fees could be \$9.985