# Talk:Irreducible polynomial

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## Monic or not?

Some authors I've seen seem to assume that all irreducible polynomials are necessarily monic, but this doesn't follow from the definition here (e.g. 2x^2 - 4 does not factor over the rationals as a product of two non-constant polynomials). Could we clarify this point? Thanks. Dcoetzee 12:16, 13 July 2010 (UTC)

## Irreducibility over the integers

The definition that the article gives for reducibility over the integers is not universal. For example Stewart's Galois Theory has Definition 3.10 which says that the polynomial is reducible in a subring of C if it can be factored into two polynomials of lesser degree. Shouldn't both definitions be included? — Preceding unsigned comment added by 184.78.155.40 (talk) 19:59, 17 November 2013 (UTC)

Stewart's book is about Galois theory, and in Galois theory all polynomials are or may be supposed to be monic. In the case of monic polynomials, both definitions are equivalent. On the other hand, considering that irreducibility over the integers and irreducibility over the rationals are the same, as it results from Stewart's definition, would contradicts the general definition given in the section "Generalization". Moreover, the primitive part–content factorization is an important concept in polynomial factorization, because of the theorem "a primitive polynomial is irreducible over the rationals if and only if it is irreducible over the integers". How could it be stated with Stewart's restrictive definition? Note also that the modern approach of polynomial factorization, generated by the needs of computer algebra was certainly ignored by Stewart, when he wrote his book published in 1989.
Nevertheless, I would agree that a section "Irreducible polynomial over the integers or a unique factorization domain" is needed to avoid confusion. It should (and even must) contain a discussion to explain why old definitions like Steward's one are not convenient.
D.Lazard (talk) 14:05, 18 November 2013 (UTC)

## What is a "non-trivial polynomial"?

The first sentence says that the factors must be "non-trivial polynomials", but "non-trivial polynomial" is not linked or defined. The article on Triviality (mathematics) does not mention polynomials, so it does not help. --50.53.53.71 (talk) 11:49, 21 September 2014 (UTC) Fixed D.Lazard (talk) 12:40, 21 September 2014 (UTC)
Thanks. --50.53.53.71 (talk) 17:41, 21 September 2014 (UTC)

## Definition

The main definition now is "a polynomial is said to be irreducible if it cannot be factored into the product of two (or more) polynomials of positive degree, whose coefficients are of a specified type".

• First of all it is wrong: the constants are NOT irreducible polynomials.
• Then, it is ambiguous: what does "a specified type" means?
• The part "(or more)" is not needed — it complicates the definition.
• Finally, in all the books authors define irreducible polynomials by using the notion of field. Almost anybody who need the notion of irreducible polynomial knows what is field (at least main examples).
• Anyway we cannot leave the definition invented by a user. Or somebody can give a reference?

In the case of $\mathbb {Z}$ or unique factorization domain the definition of irreducible polynomial as an irreducible element is strange: by this definition the constant polynomial 2 is irreducible over $\mathbb {Z}$ o, while 2x is not. By definition of some authors it is not true. So, one needs to add a reference. NoKo (talk) 14:40, 12 October 2014 (UTC)

"Almost anybody who need the notion of irreducible polynomial knows what is field": Polynomial long division is commonly taught before fields and ring, and this suffices to introduce divisibility and irreducible polynomials. Thus your quoted assertion is wrong.
I agree with your other remarks, but, as this article does not satisfies the guidelines of WP: Manual of style, it is difficult to implement them in a way that improve the article. In particular, it is correct that the first sentence of the lead contains an informal definition, but, normally it should be clear that this definition is informal, and an accurate definition should appear in the body. Also, the lead is far too long. Having an informal lead and several accurate sections (one for each nature of coefficients) would allow to distinguish the cases, keeping the article accurate and easier to read for everybody. Also for polynomials over a ring, two definitions are used in the literature. Firstly, irreducible over the field of fractions, in the case of an integral domain, and secondly irreducible as an element, in the case of an unique factorization domain. Both definitions are equivalent for primitive polynomials over a unique factorization domain (this should appear in the article).
Are you willing to restructure the article in this way? D.Lazard (talk) 15:22, 12 October 2014 (UTC)
So, one need to rearrange the text in the beginning by adding the section "Definition" with formal definition first "over a field", then maybe over some class of rings. But even an informal definition should be correct. At least reader should see from the first line that the (invertible) constants and zero are NOT irreducible polynomials. This is a point, which many of them want to look in the article. If you are afraid of a contradiction with the definition "as irreducible element" maybe you can use some phrases like "non-trivial polynomial" and "specified type" to give the informal definition. This informalities are acceptable, I think, if the formal definition will appear in the section "Definition". To explain the situation in cases more general than the case "over a filed" (might be with references) is also important. NoKo (talk) 23:11, 12 October 2014 (UTC)

## Is p3(x) really reducible over the integers?

In the text as it is now, it says that p3(x) (=9x^2-3\,=3(3x^2-1)\,=3(x\sqrt{3}-1)(x\sqrt{3}+1) ) is one of the "first three polynomials" reducible over the integers and that "(the third one is reducible because the factor 3 is not invertible in the integers)".

It seems to me that p3(x) is not reducible over the integers because the factor sqrt{3} is not an integer.

Am I correct? Wandering-teacher (talk) 15:05, 22 March 2016 (UTC)

There is an irritating distinction being made in the article between irreducible polynomials over a field (like the rational numbers, where every coefficient has a multiplicative inverse) and irreducible polynomials over a ring (like the integers, where coefficients do not generally have multiplicative inverses). The claim made in the article is that in the latter sense, the polynomial is irreducible because it can be factored as 3 * (3x^2 - 1), and the factor 3 "counts" as a factor (because it does not have a reciprocal in the integers). The usual notion of irreducible polynomial (i.e., the one that I learned de facto in school) is irreducibility over either the rational numbers or the real numbers; both of these are fields, so you don't end up with silly distinctions like this.
The accessibility of the article would probably be improved by focusing more on the field case, at least in the first few sections. --JBL (talk) 15:24, 22 March 2016 (UTC)

## Phrasing of coefficient constraint

Joel B. Lewis, 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)

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)
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)
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)
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)
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)

## Contradiction

Sentence 1 in the lead says

In mathematics, an irreducible polynomial is, roughly speaking, a non-constant polynomial that cannot be factored into the product of two non-constant polynomials.

The section "Simple examples" says

Over the integers, the first three polynomials are reducible (the third one is reducible because the factor 3 is not invertible in the integers)...

where the third one is

$p_{3}(x)=9x^{2}-3\,=3(3x^{2}-1)\,=3(x{\sqrt {3}}-1)(x{\sqrt {3}}+1)$ This seems contradictory, as the first quote implies that p3 is irreducible over the integers. Loraof (talk) 17:29, 11 October 2017 (UTC)

This issue is related to the sections Irreducibility over the integers, What is a "non-trivial polynomial"?, Definition, and Is p3(x) really reducible over the integers? above. Ultimately the problem stems from trying to write an article that covers the cases of coefficients in a field and in other rings simultaneously. It does not seem like an easy problem to solve, but any attempt should probably begin by reading those talk-page sections. --JBL (talk) 21:40, 11 October 2017 (UTC)
If I understand correctly, $p_{3}(x)$ is irreducible according to one definition, but reducible according to another definition.
First definition in Irreducible polynomial#Definition:

A polynomial with integer coefficients, or, more generally, with coefficients in a unique factorization domain R, is sometimes said to be irreducible (or irreducible over R) if it is an irreducible element of the polynomial ring, that is, it is not invertible, not zero, and cannot be factored into the product of two non-invertible polynomials with coefficients in R.

$p_{3}(x)$ can be factored as $3(3x^{2}-1)$ , and since $3$ is non-invertible, $p_{3}(x)$ is reducible over the integers according to this definition.
Second definition in Irreducible polynomial#Definition:

Another definition is frequently used, saying that a polynomial is irreducible over R if it is irreducible over the field of fractions of R (the field of rational numbers, if R is the integers).

$p_{3}(x)$ is irreducible over the rationals, so according to this definition it is also irreducible over the integers.
Is this analysis correct? If yes, we should probably refine the claims about irreducibility of $p_{3}(x)$ : We should state that it is irreducible according to one definition, but reducible according to another, and (briefly) explain why. Chrisahn (talk) 13:21, 6 December 2018 (UTC)
Good point. I have edited the article for clarifying that the second definition is not used in this article. D.Lazard (talk) 18:04, 6 December 2018 (UTC)

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## "Non-constant"

From Irreducible polynomial#Definition (regarding polynomials with integer coefficients):

Both definitions generalize the definition given for the case of coefficients in a field, because, in this case, the non-constant polynomials are exactly the polynomials that are non-invertible and non-zero.

If I understand correctly, this means that $p(x)=3$ would be considered a non-constant polynomial over the integers. I find it hard to reconcile this with any reasonable definition of constant / non-constant. Can we fix this? Chrisahn (talk) 13:34, 6 December 2018 (UTC)

This was ambiguous, and is (I guess) now clarified.
@D.Lazard: Thanks! I still find it strange that $p(x)=3$ would be considered a "non-constant" polynomial. What do you think? Chrisahn (talk) 18:25, 6 December 2018 (UTC)
$p(x)=3$ is never considered as non-constant, and I do not see any sentence of the article that could implies that it could be considered as non-constant. D.Lazard (talk) 18:39, 6 December 2018 (UTC)
@D.Lazard: But it's a non-invertible and non-zero polynomial over the integers, right? If that is correct, then it's also non-constant according to this sentence: "...in this case, the non-constant polynomials are exactly the polynomials that are non-invertible and non-zero." I agree that $p(x)=3$ should not be considered non-constant. I'd say that sentence is misleading. Chrisahn (talk) 22:51, 6 December 2018 (UTC)
The referent of "this case" in that sentence is "the case when the coefficients come from a field". That could be clearer. --JBL (talk) 00:35, 7 December 2018 (UTC)
Clarified. D.Lazard (talk) 02:48, 7 December 2018 (UTC)
Yeah, I thought "this case" referred to the case described in the previous sentence. Thanks for clearing this up! Chrisahn (talk) 16:49, 10 December 2018 (UTC)

## Some awkwardness in the introduction

The introduction seems to make strong assumptions about the ring the polynomial is defined over. In particular, an irreducible polynomial is not necessarily the same thing as a prime element over a general integral domain, and the statement "a non-constant polynomial that cannot be factored into the product of two non-constant polynomials" is obviously only true for fields. While the definition section clarifies this, it seems unnecessarily misleading to the casual reader.

(Note: this is my first time using Wikipedia's talk pages.) Mathematician-at-heart (talk) 03:49, 4 October 2020 (UTC)

Your first post follows perfectly the recommendations of the manual of style.
I agree that "prime" and "irreducible" are not always equivalent. Moreover, "prime polynomial" is rarely used, except when proving the equivalence over a field. Thus, I have edited the parenthesis accordingly. On the other hand the statement "a non-constant polynomial that cannot be factored into the product of two non-constant polynomials" being a definition cannot be true or false. The precise definition, when the coefficients do not belong to a field, is too technical and not enough commonly used for belonging to the lead. This a reason for having "roughly" in the first sentence. The other reason for this "roughly" is the ambiguity of "cannot be factored", explained in the next sentences. D.Lazard (talk) 08:33, 4 October 2020 (UTC)