Talk:Gaussian integer/Archive 1

Definition of norm
The norm of a Gaussian integer is the natural number defined as


 * N(a + bi) = a2 + b2.

According to MathWorld, this definition should be deprecated. Any information on whether this is actually accepted today, or which authors use which? - Fredrik | talk 22:49, 9 July 2005 (UTC)


 * I don't see any reason to take this comment on MathWorld seriously, since it just appears to be Eric Weisstein's opinion - he doesn't cite any author to substantiate it, and it doesn't seem to make much sense in the larger context (that is, the norm used for Z[i] happens to be the square of the complex modulus, and so could be replaced by the complex modulus easily enough, but for other rings such as Z[e2&pi;i/3] the norm is unrelated to the complex modulus, and so cannot be replaced by it). Of course, if you can find a more reputable source than MathWorld for this, then that's another matter. --Zundark 09:05, 10 July 2005 (UTC)


 * It is the field norm, to be more accurate. Sometimes one wants the square root, for fairly obvious reasons. Charles Matthews 10:15, 10 July 2005 (UTC)


 * In the paragraph,


 * If the norm of a Gaussian integer z is a prime number, then z must be a Gaussian prime, since every non-trivial factorization of z would yield a non-trivial factorization of the norm. So for example 2 + 3i is a Gaussian prime since its norm is 4 + 9 = 13. This implies that since there are infinitely many ordinary primes then there must be infinitely many Gaussian primes.


 * the last sentence assumes that for every ordinary prime, there is a Gaussian integer whose norm is that prime. Surely this is nontrivial to prove. — Preceding unsigned comment added by 12.210.55.12 (talk) 11:48, 24 February 2006 (UTC)


 * No, it requires that for infinitely many (not necessarily all) primes there is a Gaussian integer whose norm is that prime. This is equivalent to "infinitely many primes are the sum of two squares", which is a trivial consequence of Fermat's theorem on sums of two squares (see link at the bottom of this article). 91.107.186.150 (talk) 01:06, 6 March 2008 (UTC)


 * doesn't that include all the ordinary primes? Gaussian integers with prime norm? :s — Preceding unsigned comment added by 129.94.6.28 (talk) 02:16, 7 August 2006 (UTC)


 * No, because 2 is an ordinary prime but is not a Gaussian prime. -- Dominus 10:40, 7 August 2006 (UTC)


 * Surely the answer is yes? 2 is a Gaussian integer with prime norm (assuming prime means rational primes, which is the usuall assumption to make, although in context the wrong one).  Maybe it should be "Gaussian integers with Gaussian prime norm" to avoid abiguity. --PhiJ 18:02, 28 October 2006 (UTC)


 * Nope. Remember that the norm of an ordinary prime is that prime squared, which is not itself prime. 91.107.186.150 (talk) 01:11, 6 March 2008 (UTC)

Image is not helpful
The second image in the page (labeled "some of the gaussian integers") is really not helpful to the understanding of the concept. Tac-Tics (talk) 05:34, 14 April 2008 (UTC)


 * While it’s purpose isn’t clear to me, note that it’s just of Gaussian primes. The Gaussian integers would merely look like a uniform grid in $$\mathbb{R}^2$$ GromXXVII (talk) 10:56, 14 April 2008 (UTC)

Clarification Needed
"This domain cannot be turned into an ordered ring, since it contains imaginary numbers."

I think this needs further explanation. I think it is worded badly as if it is true its also true for the set of pairs of integers and other structures. I don't have time to search for alternatives right now so thats why i'm leaving this note as opposed to editing it myself. Thanks JackSlash (talk) 19:50, 7 May 2008 (UTC)

Norm of a Gaussian Integer
The norm link points to field norm. It seems to me that it should point to norm (mathematics). TomyDuby (talk) 03:46, 20 July 2008 (UTC)