Talk:Algebraic integer
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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)
[edit] 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)
[edit] 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)
[edit] √2
The text "This provides an alternative proof of the irrationality of " 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 x2 − 2, is an algebraic integer, hence if it were rational, it would actually have to be an integer. However, 12 = 1 < 2 < 4 = 22, 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)
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- 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 )
- (Emphasis mine.) Does it make sense now? -- EJ 14:11, 3 December 2007 (UTC)
- That's kind of the point. The full context of the quote is
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- 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)
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- 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 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)
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- 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)
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