# Talk:Algebraic integer

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Field: Number theory

The definition of algebraic integer as defined in the article does not satisfy many desirable properties of integers, i.e. that: Algebraic integers defined for arbitrarily large degrees are uncountable

Nonsense; the set of integer polynomials is countable, so the algebraic integers are a countable union of finite sets. Septentrionalis 14:32, 14 April 2006 (UTC)

Algebraic integers defined for a maximum degree of the polynomial P(x) are not closed under any operation

Depends on the operation. Septentrionalis 14:32, 14 April 2006 (UTC)

Some algebraic integers, such as I = largest real root of (x5 - x + 1) cannot be represented without using the polynomial ... Scythe33 02:28, 4 September 2005 (UTC)

I have added Schroeppel's result; since it was shown informally on a mailing list, I won't argue if someone wants to qualify "demonstrated". Septentrionalis 14:32, 14 April 2006 (UTC)

## Noetherian?

Either way, this is overlooking the obvious; but are the algebraic integers Noetherian? Septentrionalis 14:56, 14 April 2006 (UTC)

No. Consider the sequence of principal ideals generated by the elements 21/n. Joeldl 10:14, 17 February 2007 (UTC)

## Merger proposal

I propose this article be merged with Integrality, which covers a topic of which this is a special case. Joeldl 10:16, 17 February 2007 (UTC)

## √2

The text "This provides an alternative proof of the irrationality of $\sqrt{2}$" has been removed with an edit summary of "Incorrect remark removed". FWIW, I don't see anything incorrect about the remark. The full argument is as follows. Being the root of the monic polynomial $x^2-2$, $\sqrt 2$ is an algebraic integer, hence if it were rational, it would actually have to be an integer. However, $1^2=1<2<4=2^2$, and there is no other integer between 1 and 2. -- EJ 11:03, 3 December 2007 (UTC)

It looks to me like there is huge gap there. How do you go from "r = p/q for integers p and q" to "r is a root of a polynomial x+k for some integer k"?  --Lambiam 11:54, 3 December 2007 (UTC)
That's kind of the point. The full context of the quote is
• The only algebraic integers in rational numbers are the ordinary integers. In other words, the intersection of Q and A is exactly Z. The rational number a/b is not an algebraic integer unless b divides a. Note that the leading coefficient of the polynomial bx − a is the integer b. (This provides an alternative proof of the irrationality of $\sqrt{2}$)
(Emphasis mine.) Does it make sense now? -- EJ 14:11, 3 December 2007 (UTC)
Sorry if I wasn't clear, but for something to serve as an "alternative proof", it has to be a proof. Where is the proof? True or not, as phrased the statement about algebraic integers in rational numbers is a mere statement or claim. I can truthfully state: "A number of the form √p, in which p is a prime number, is irrational." It would be kind of silly to follow up with: "This provides an alternative proof of the irrationality of √2". What you can say, without being silly, is that the irrationality of √2 is a special case of this general statement.  --Lambiam 15:26, 3 December 2007 (UTC)
Aha. So no gap's involved, you're objecting to the formulation on purely stylistic grounds. Fine, I guess I can see what you mean. But then the real question remains whether it is useful to point out in the article that irrationality of $\sqrt2$ follows (call it a special case, or corollary, or what have you) from the statement that rational algebraic integers are integers, which I tend to think it is. On the one hand, it illustrates that the statement has a real mathematical contents, it is not just a formal manipulation of the definition. On the other hand, it gives a hint on how the general statement is proved. -- EJ (talk) 11:23, 5 December 2007 (UTC)
I've added something about this to the passage in question. I'm not keen on adding a statement that the proof of the general theorem is analogous to the well-known proof of the irrationality of √2, unless there is a proper reference for this. I'd hope that anyone seriously interested in algebraic integers will have no problem coming up with a proof.  --Lambiam 17:58, 5 December 2007 (UTC)
Fine with me, thanks. -- EJ (talk) 10:16, 6 December 2007 (UTC)

## "Integer" Misnomer?

It would be helpful to a casual reader to make clear right off the bat whether these are integers in the sense that the real part is in Z and the complex part is 0. If not please remark that the term is a misnomer in view of the usual notion of integer. If so please say so. Thank you. CountMacula (talk) 19:16, 15 July 2011 (UTC)

## How do you plot algebraic integers on the complex plane?

Dear Arathron. Please teach us how to plot the algebraic integers on the complex plane. It looks like Julia sets or something fractal. We want to know the logic behind it.--Enyokoyama (talk) 12:06, 28 June 2014 (UTC)

Hi Enyokoyama. I plotted the roots of the monic polinomials with integer coefficients varying from -10 to 10 and to degree 5 if I remember correctly.

I later found this very nice page about the subject, that explains a bit why it looks like a fractal: http://www.math.ucr.edu/home/baez/roots/ — Preceding unsigned comment added by Arathron (talkcontribs) 22:42, 8 July 2014 (UTC)

Dear Arathron. Thank you very much. I appreciate your answer and hope your more successes.--Enyokoyama (talk) 00:01, 9 July 2014 (UTC)

The image is nonsensical. The set of algebraic integers is dense in the plane and invariant under shifts by algebraic integers, hence whatever it is that is depicted in the image does not look anything like algebraic integers. A similarly confused image was already removed from the article once, and the reason stands.—Emil J. 09:37, 9 July 2014 (UTC)