# Proving $\mathbb{Z}[\sqrt{-2}]$, $\mathbb{Z}[\sqrt{-1}]$, $\mathbb{Z}[\sqrt{2}]$, and $\mathbb{Z}[\sqrt{3}]$ are euclidean.

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

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

PROOF: .....

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

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

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

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

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

This proves the corollary.

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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.

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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 at 16:08
Yous should also post the theorem that the corollary refers to. –  Bill Dubuque Feb 28 at 19:22
@Pedro Tamaroff, I just added the Theorem before the Corollary. Thanks, –  A.Magnus Feb 28 at 23:50
@Bill Dubuque, I just added the Theorem before the Corollary. Thanks. –  A.Magnus Feb 28 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 at 23:55