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link 1: https://www2.math.ethz.ch/education/bachelor/lectures/fs2014/math/algebra2/sol3

link 2:

http://www.math.washington.edu/~sullivan/4036s_wi13.pdf

I have found 2 solutions, the second one convinced me. But so far I cannot formally prove that the method given in the first link gives a reminder with its size function less then the size function of the divisor. And I think maybe there exist Gaussian integers $a+bi$, $c+di$, such that using the division algorithm in link 1 to divide $c+di$ by $a+bi$ gives a remainder whose size function value is greater than the size function value of $a+bi$.

Is it possible? If it is possible, could someone find such $a+bi$ and $c+di$?If the method in link 1 also allows us to always get a remainder with size function value less then the divisor, could someone formally prove that?

Thanks in advance!

Daniel Fischer
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Y.X.
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1 Answers1

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Yes, it is possible, because $\mathbb{Z}[i]$ is a Euclidean ring. For a proof see here on this site. It is very similar to your links, but might give you more information. The second link does the full computation for a given example. What problem do you see there? The remainder must get smaller, as is proven at the duplicate above.

Community
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Dietrich Burde
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  • Thank you! But may I ask how to prove the existence of r with its norm less then 1/2 N(b)? – Y.X. Oct 13 '16 at 13:58