Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have this short class note from my graduate number theory:

THEOREM: Assume that $\vert N(x + y \sqrt d)\vert < 1$ for any two rational numbers $x$ and $y$ with $\vert x \vert \leq 1/2$ and $\vert y \vert \leq 1/2$. Define $\delta : \mathbb Z[\sqrt d] \setminus \{0\} \to \mathbb N$ by $z \mapsto \vert N(z)\vert$, then $\mathbb Z[\sqrt d]$ is euclidean with regards to $\delta$.

PROOF: Omitted.

COROLLARY: The integral domains $\mathbb{Z}\left[\sqrt{-2}\right]$, $\mathbb{Z}\left[\sqrt{-1}\right]$, $\mathbb{Z}\left[\sqrt{2}\right]$, and $\mathbb{Z}\left[\sqrt{3}\right]$ are euclidean.

PROOF: Let x and y be rational numbers with $|x| \leq 1/2$ and $|y| \leq 1/2$. Then

$$|N\left(x + y\sqrt{-2}\right)| = |x^2 + 2y^2| \leq 3/4 < 1,$$

$$|N\left(x + y\sqrt{-1}\right)| = |x^2 + y^2| \leq 1/2 < 1,$$

$$|N\left(x + y\sqrt{2}\right)| = |x^2 - 2y^2| \leq 1/2 < 1,$$

$$|N\left(x + y\sqrt{3}\right)| = |x^2 - 3y^2| \leq 3/4 < 1.$$

This proves the corollary.

My questions are about the Corollary:

  1. Why did my professor make the assumption that $|x| \leq 1/2$ and $|y| \leq 1/2$, when it is about $\mathbb{Z}$ the integer? (Post Script: Never mind about this question, I got it now from the Theorem before the Corollary.)
  2. There should be missing explanations before he suddenly jumped to "This proves the corollary." What are they?

Any help would be very much appreciated. Thank you for your time.

share|cite|improve this question
Your professor probably approximates integers by rationals to be $1/2$ a unit away, and carries usual fractional division, then mends this and uses those norms are $<1$ to show the algorithm is actually Euclidean although the division is integral. If you provided the details of what your professor does, one could help more. – Pedro Tamaroff Feb 28 '14 at 16:08
Yous should also post the theorem that the corollary refers to. – Bill Dubuque Feb 28 '14 at 19:22
@Pedro Tamaroff, I just added the Theorem before the Corollary. Thanks, – A.Magnus Feb 28 '14 at 23:50
@Bill Dubuque, I just added the Theorem before the Corollary. Thanks. – A.Magnus Feb 28 '14 at 23:51
@LoveMath The corollary is an immediate application of the the Theorem. What is not clear about that? Do you lack a proof of the theorem? – Bill Dubuque Feb 28 '14 at 23:55

1 Answer 1

up vote 1 down vote accepted

For (1), this is a question about the theorem you mentioned, it has to do with choosing the integer closest to the rational number you chose. For any rational number you can choose the closest integer to it, and it will be less than or equal to $\displaystyle \frac12$ away.

You end up choosing $\alpha, \beta \in \mathbb Z[\sqrt d]$ where $\beta \neq 0$ and then consider $\displaystyle \frac\alpha\beta$ which can be written as $r + s \sqrt d$ where $r, s \in \mathbb Q$, then choose $x, y$ integers such that $x$ is closest integer to $r$ and $y$ is closest integer to $s$.

For (2), you have the theorem at your disposal that says if (some hypothesis are satisfied on $\mathbb Z[\sqrt d]$) then $\mathbb Z[\sqrt d]$ is a Euclidean domain (The proof of that theorem should somehow show it is Euclidean, along the lines of $a = bq + r$ with $r = 0$ or $N(r) < N(b)$.)

In your corollary, you show that each of $\mathbb Z[\sqrt -2], \mathbb Z[\sqrt -1], \mathbb Z[\sqrt 2], \mathbb Z[\sqrt 3]$ satisfy the given hypothesis, hence you can conclude that each of those are Euclidean domains.

share|cite|improve this answer
Thanks! I was alerted belatedly by email. I have up-voted and green check-marked. Thanks again. – A.Magnus Feb 24 at 18:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.