Talk:Gaussian integer

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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[e2πi/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.

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

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)

Definition[edit]

May I query this: "A Gaussian integer is a complex number whose real and imaginary part are both integers." That can't be right. The imaginary part is by definition some multiple of i. Even where b = 0, the imaginary part is a natural number, but still not an integer. Isn't it more correct to say "a complex number where the real part and the argument of the imaginary part are both integers"? JackofOz 01:53, 13 March 2006 (UTC)

Be careful. The imaginary part of 2+3i is 3, not 3i. Dmharvey 01:55, 13 March 2006 (UTC)
Also, but 0 is an integer, you may be confusing the integers with the positive integers. And you may be confusing modulus with argument, as the argument of any non-zero integer multiple of i is π/2, while its modulus is that integer which you multiplied by i. --PhiJ 16:37, 5 November 2006 (UTC)

Hi, I am confused by the Latex definition given on the page "Formally, Gaussian integers are the set "

\mathbb{Z}[i]=\{a+bi \mid a,b\in \mathbb{Z} \}.

Shouldn't a,b \in \mathbb{I} ? I thought every complex number can be represented as a tuple of two real numbers. And these Gaussian integers ought to belong to \mathbb{R} or to be precise the set \mathbb{I}. Also, the above query/talk/point/discussion mentions that both a, b are integers, while the Latex definition, as per my understanding, puts them in \mathbb{Z}. My apologies if I am wrong here as I dont understand this math concept. However, its precisely the reason for my doubt. I hope somebody with proper understanding clarifies this. -- wadkar <AT> gmail <DOT> com — Preceding unsigned comment added by 125.63.107.5 (talk) 22:03, 17 April 2012 (UTC)

I don't know what you mean by \mathbb{I}, but \mathbb{Z} is standard notation for the set of (real) integers. It is short for the German word "Zahlen". —David Eppstein (talk) 22:46, 17 April 2012 (UTC)
Thanks, I got confused by the notation and their meaning. — Preceding unsigned comment added by 125.63.107.34 (talk) 17:07, 27 April 2012 (UTC)

Image is not helpful[edit]

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[edit]

"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[edit]

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)

Other kinds of "complex integers"[edit]

I am only an amateur mathematician, and I don't know the term for numbers of the form a + b \sqrt{-2} \ \, (or more generally a + b \sqrt{-k} \ \, ) with integers a and b (and k), but I do know that they are significant. I suggest that these two steps be taken:

  • Someone more knowledgeable than myself should write an article about this ring (or such rings) in case such articles don't already exist.
  • A link to this article should be provided within the article about Gaussian integers, at the very least in the "See also" section.

Thanks! 188.169.229.30 (talk) 13:14, 3 January 2012 (UTC)

That is a good suggestion, and the article already has links to the numbers you mentioned, Eisenstein integers and quadratic integers, in the "See also" section and elsewhere. Perhaps the "See also" links need brief explanatory notes? —Mark Dominus (talk) 14:37, 3 January 2012 (UTC)

Relation to quadratic integers[edit]

I tried to change "The Gaussian integers are a special case of the quadratic integers." I got it wrong. Thank you David Eppstein for reverting.

However I still think the original version is badly put. It is a messy mixture of singular and plural. Surely we are not trying to say merely "Every Gaussian integer is a quadratic integer": we are trying to say something about the Gaussian integers as a structure. So I've had a second go at editing it, by saying, "The Gaussian integers form a commutative ring, being a particular case of a commutative ring of quadratic integers." I hope I've got it right this time, but if not, perhaps someone could reword it to say something clear about the Gaussian integers as a structure, rather than just reverting. Prim Ethics (talk) 15:12, 15 December 2013 (UTC)