Talk:Irreducible polynomial

message: , maybe a discussion about this will be interesting.  The wording "... is irreducible if the coefficients 1 and −2 are considered as integers" has, in my interpretation, two problems.  The first is the obvious one: that it appears to try to draw a distinction between the integers as elements of the ring of integers, and the integers as elements of a subring of the reals.  This is like claiming that the set of integers cannot be regarded as a subset of the reals (or, to belabour the point, that integers cannot be considered as being real numbers).  The second is more subtle, and is my primary objection: the factorization constraint specifically applies to the coefficients of the factor polynomials (by definition, also to the polynomial to be factored, but that does not make a difference here).  The wording as it stands states that constraint applies specifically and only to the specific coefficients of the polynomial to be factored, which would only be useful if the first problem did not exist.  Do you at least see where I'm coming from?  —Quondum 02:05, 7 October 2017 (UTC)
replies: text: I'll change "if the coefficients 1 and −2 are considered as integers" into "if the polynomial is considered as a polynomial with integer coefficients". The formulation "if the polynomial is considered as a polynomial over the integers" is also fine for me, but may be considered as too technical or jargon. D.Lazard (talk) 08:21, 7 October 2017 (UTC)
replies: text: I'm not sure the concern is properly addressed.  One should not assume that the reader makes the "standard" assumptions that someone familiar with the topic does.  The person familiar topic might read "considered as a polynomial with integer coefficients" to mean "all operations being in the ring of polynomials with integer coefficients", which is what is really meant.  The person who makes no such implicit inferences might simply interpret it at face value: that the coefficients of the initial polynomial are integers.  Face it, "considered as" is not a formally defined phrase in mathematics (and has even less meaning to the newbie); it is a cue to look for unstated inferences.  What about "... is irreducible if it is to be factored as polynomials with integer coefficients"?  —Quondum 12:25, 7 October 2017 (UTC)
replies: text: In fact, the concern is not really about the sentence that has been discussed, but about the preceding one. I'll try to address this. D.Lazard (talk) 12:57, 7 October 2017 (UTC)

text: I have edited the lead in an attempt to clarify this. Feel free to improve my wording. This discussion is, in fact, about the distinction between irreducibility and absolute irreducibility, and it appears that absolutely irreducible was not linked here, I have fixed this. D.Lazard (talk) 13:42, 7 October 2017 (UTC)
replies: text: My primary concern is essentially addressed by the clarification "the field or ring to which the coefficients of the polynomial and its possible factors are supposed to belong".  I'm not convinced that absolute irreducibility is relevant to the discussion at all.  As I understand it, all that is being said is that an irreducible polynomial has exactly one "nontrivial" (more technically, "prime") factor in the polynomial ring under consideration.  The definition section is pretty clear on this, and has reasonably concise and understandable language.  However, my primary concern of possibly incorrect initial interpretation has at least been addressed.  —Quondum 14:10, 7 October 2017 (UTC)
thread_title:  Phrasing of coefficient constraint 

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message: I think it's worth talking about rewriting the lede - someone I was helping a little while ago was also thrown off by this and it looks like it's caused confusion at various times. I would prefer talking about the existence of a "non-trivial factorisation" and then clarifying what this means for integral domains and fields. Changes have been talked about above but I still think it's all unclear, at the very least it should be pointed out that \mathbb Z is an ID not a field and so we are looking at the other criterion. Wanted to discuss here before making any major changes. --Caliburn · (Talk · Contribs · CentralAuth · Log) 19:45, 24 December 2021 (UTC)
replies: text: I have added a paragraph in the lead for clarifying this. D.Lazard (talk) 17:27, 25 December 2021 (UTC)
thread_title:  Again about the lede 

message: The section  claims the above.
This is true for the second definition of irreducibility, where reducibles factor into two non-constant polynomials, but unless I'm mistaken about something, it seems false for the first definition.
Counter example:
3x+3 is reducible over the integers because it equals 3(x+1), and 3 is not a unit.
However, it is not reducible over F_5, because 3 is a unit there. FourierIsDeathier (talk) 09:33, 10 April 2022 (UTC)
replies: text: D.Lazard (talk) 10:12, 10 April 2022 (UTC)
thread_title:  "Irreducible over F_p implies irreducible over Z" 

message: It rambles around the idea that roots might have "no explicit algebraic expression". This is not what Abel-Ruffini Theorem says. Note that $\sqrt{2}$ is as algebraic as the root of every other polynomial.
The section has no meaning unless it states correctly the type of expression that is claiming might or might not exist and which existence or not does not affect irreducibility. It seems that the intent of the section was to refer to expressions in radicals.
It is questionable whether the clarification offered by the section is needed. The definition(s) state where the coefficients are taken from and if the coefficients are the complex numbers, then the definition allows all complex numbers in the factorization, potentially even using non-algebraic numbers.
Anyway, if the section is to exist, it should at least have the correct statement of what is claiming. 208.127.178.195 (talk) 19:08, 5 January 2023 (UTC)
thread_title:  The section Nature of a factor has ambiguous/incorrect language 

message: , could you check this statement in the lead for correctness?
::"Most often, a polynomial over an integral domain  is said to be irreducible if it is not the product of two polynomials that have their coefficients in , and are not unit in ."
The last constraint seems to exclude the factor polynomials that have even one unit coefficient, but this can't be right.  I think the intention might have been "... and which are not unit in the polynomial ring", since this would have fitted the Definition section more closely.  Note also that it is not really sensible to treat a constant polynomial an element of  as may have been intended, despite the natural embedding of  into the polynomial ring.  —Quondum 21:25, 25 March 2023 (UTC)
replies: text: As, over an integral domain, the total degree of a product of polynomials is the sum of the total degrees of the factors, the units of a polynomial ring are exactly the same as the units of the integral domain of coefficients. So, both formulations are equivalent. The one of the article is simpler, but if you find it confusing, be free of changing it. D.Lazard (talk) 21:46, 25 March 2023 (UTC)
replies: text: I've tweaked the main thing that was bothering me.  The secondary one (that you have responded to) may not be worth addressing as being natural enough.  —Quondum 22:59, 25 March 2023 (UTC)
thread_title:  Error in lead? 

message: The polynomial X^4 + 1 is given as a “simple” example of irreductible poly over \mathbb{Z} and reducible over every finite field \mathbb{F}_b. Is that so?
The previous section gives the factorization \left(x^2 + \sqrt{2}x + 1\right)\left(x^2 - \sqrt{2}x + 1\right), which is only possible if 2 is a square in \mathbb{F}_p (a quadratic residue modulo p), that is p = 1 or 7 mod 8. Another possible factorization is \left(x^2 + \sqrt{-2}x - 1\right)\left(x^2 - \sqrt{-2}x - 1\right), which is possible if -2 is a quadratic residue modulo p, that is p = 1 or 3 mod 8. The case p = 5 mod 8 is not resolved. Is X^4 + 1 polynomial really reducible in \mathbb{F}_5, \mathbb{F}_{13}, \mathbb{F}_{29}…?
: self-reply Answer is YES. If p = 5 mod 8, (implying p = 1 mod 4) then -1 is a quadratic residue. Let's call r = √-1. \left(X^2-r\right)\left(X^2+r\right) = X^4 - r^2. Et voilà. 2A01:E0A:D8B:7AF0:3E97:EFF:FEBC:B87B (talk) 18:51, 1 September 2023 (UTC)
replies: text: There is no need to consider the congruences modulo : if -1 or 2 is a quadratic residue, then the first or the third factorization holds. If none is a quadratic residue, then their product –2 is a quadratic residue, and the second factorization holds. D.Lazard (talk) 18:59, 1 September 2023 (UTC)
thread_title:  Reducible over every finite field?