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Wikipedia states :

If an univariate polynomial $p$ has a root (in some field extension) which is also a root of an irreducible polynomial $q$, then $p$ is a multiple of $q$, and thus all roots of $q$ are roots of $p$; this is Abel's irreducibility theorem. This implies that the roots of an irreducible polynomial may not be distinguished through algebraic relations.

The reason the roots of an irreducible polynomial may not be distinguished through algebraic relations ("written out in factored form") is not because of the first sentence above, it is because the polynomial is irreducible in field $F$ therefore has no roots expressible in field $F$.

If $F$ is the algebraic numbers where all polynomial roots are contained and the polynomial is irreducible over $F$ then it has no roots anywhere and is linear.

In an appropriate field, distinguishing the roots is simply a matter of writing out the factors using algebraic relations, with coefficients in that field. So the reason an irreducible polynomial's roots can not be written out in factored form ("distinguished by algebraic relations") is not Abel's irreducibility theorem, it is whether or not the polynomial is irreducible or not in the field in question.

Is this understanding correct or is Wikipedia correct and if so why?

Hayden
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Cris
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  • The statement "this implies that the roots of an irreducible polynomial may not be distinguished through algebraic relations" is, I believe, not asking whether you can write the factors out. It can be taken to say something more of the quality "If $p$ is irreducible in a field $F$, then $F$ cannot make any distinction between the roots of $p$ in any field extension $K$ (assuming they exist)", in which case Abel's Irreducibility theorem indeed makes this clear – Hayden Jul 06 '14 at 13:13
  • Obviously a field F can't distinguish between roots (since the roots, if extant, are in K are not in F). So you think "algebraic relations" means something about the relation of F to K? And not algebraic relations as in writing the roots out algebraically? I disagree and I don't see how Abel's theorem says anything about different fields, it says something about the relationship of polynomials within a field. – Cris Jul 06 '14 at 13:20
  • Yes, that is exactly what I think; this is along the same reasoning as Nishant's answer: there is always an $F$-automorphism of a splitting field $K$ of an irreducible polynomial $p$ that permutes the roots. The fact that "$F$ does not witness the distinction" is that there is an $F$-automorphism. Abel's Irreducibility Theorem obviously talks about different fields, since talking about having roots in another field talk about, well, another field. – Hayden Jul 06 '14 at 13:26
  • The question works. Anonymous down vote is incorrect. – Cris Jul 06 '14 at 13:30
  • @Hayden and it doesn't talk about having roots in another field. http://en.wikipedia.org/wiki/Abel%27s_irreducibility_theorem ? – Cris Jul 06 '14 at 13:33
  • Perhaps you care to share how one would talk about the root of an irreducible polynomial without that root lying in a field extension? – Hayden Jul 06 '14 at 13:34
  • @Hayden ok I get that. Two questions tho : 1) how do F-automorphisms permute things in K which F doesn't have -- what is the map from an element in F to the extension elements in K? I don't get that. – Cris Jul 06 '14 at 13:38
  • They're $K$ automorphisms fixing $F$, so they can permute things not in $F$, such as the roots of a polynomial. – Nishant Jul 06 '14 at 13:40
  • @Hayden 2) How is Abel needed in order to say that roots are algebraically indistinguishable. The very idea of irreducibility in one field implies that field can not split that polynomial? Abel unnecessary. – Cris Jul 06 '14 at 13:42
  • @Cris what I'm trying to say is that making the phrase "algebraically indistinguishable" is to claim the existence of an $F$-automorphism of a splitting field $K$ that permutes the roots. – Hayden Jul 06 '14 at 13:45
  • Anonymous down voter: Your message is unclear. Would you have something about the question change? If so, you are free to alter it - and yet it is clear. Is the question not a fit for this site IYO? You may vote to close it - and yet it is a fit. Is there something else which is troubling you? I assure you that is not me causing your trouble! So, anonymous down voter, message is unclear and it seems incorrect. – Cris Jul 06 '14 at 13:46
  • @Hayden I see. But isn't that obvious and why need to invoke that here? It doesn't explain why the polynomials are irreducible. The minimum length description of that is simply because they don't have roots in F. So it's inessential to invoke Abel here - and "algebraic relations" makes more sense to mean algebra operations (which can sometimes distinguish roots), than some type of relations between fields which by definition never distinguish roots anyway. – Cris Jul 06 '14 at 13:56
  • @Cris It might be important to note that 'irreducible' does not actually mean 'does not have roots'. It means that there is no proper polynomial factor. For example, (x^2+1)^2$ has no roots in $\mathbb{R}$ but is not irreducible over it. I think the best thing to take from the discussion is that wikipedia is using 'algebraically indistinguishable' in a different manner from your intended meaning. – Hayden Jul 06 '14 at 14:01
  • @Hayden you should note that, yes. Abel is not required to form the implication that the roots are indistinguishable in the field where their polynomial is irreducible, that follows from irreducibility. Nor is it required under the Galoisian interpretation of indistinguishability, that follows from the definition of field extension. There may be some logical trace you can make through Abel, to get back to either of these, but it is unnecessary and unclear -- not a direct "implication" as Wikipedia states. Unless, as my still unsatisfied question asks, there is more to it. – Cris Jul 06 '14 at 15:04

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Well, first of all, if you have an irreducible polynomial $f(x)\in F[x]$, then that polynomial is the minimal polynomial over $F$ of any of its roots. If another polynomial in $F[x]$ has a root in common, then by doing the division algorithm, you see that the remainder must be $0$ to not contradict the minimality of $f$, and so $f$ divides the new polynomial.

The reason the roots are "algebraically indistinguishable" is that, given any two roots, there is always an automorphism of the splitting field that sends one root to the other, i.e., the "Galois" group acts transitively on the roots (quotes not needed if the $f$ is separable).

Nishant
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  • If there is a splitting field E then the roots are distinguishable by definition in it, so how can the roots be both distinguishable and indistinguishable in a splitting field E -- why is there "always" this automorphism that makes them indistinguishable? So E/F is Galois extension -- why even need to invoke Galois here? Also -- how does this answer my question? – Cris Jul 06 '14 at 13:29
  • Sorry, I meant that they're indistinguishable over $F$. Of course, they are indistinguishable over $E$. My point was that an automorphism of $E$ fixing $F$ exists which takes any root to any other root, which is why you can't use polynomials in $F[x]$ to distinguish them. – Nishant Jul 06 '14 at 13:39
  • To distinguish the roots? Ok, obviously. That's one way of saying that's why polynomials in F[x] that are irreducible don't have roots in F -- but you don't need to invoke that to say that. They simply don't have roots in F, or equivalently, they're irreducible. So why does Abel need to be included? It doesn't explain it -- and "algebraic relations" makes more sense to mean algebra operations (which can sometimes distinguish roots), than some type of relations between fields which by definition never distinguish roots anyway. – Cris Jul 06 '14 at 13:53
  • Any "algebraic relation" can be turned into a polynomial, so since any polynomial with one of $f$'s roots as a root has ALL of $f$'s roots as roots, there is no way to distinguish them. This is exactly what Abel's theorem says. – Nishant Jul 06 '14 at 14:20
  • Abel is correct, yes, and among his conclusions is not "there is no way to distinguish roots", but "f can be used to factor polynomials it shares roots with". To use Able to make the implication, "the roots of an irreducible polynomial are indistinguishable in F" is incorrect, and unnecessary -- this is implied by its very irreducibility. My question, "Is there something I was missing that made Abel essential to implying that an irreducible polynomials roots are indistinguishable" is still unsatisfied. – Cris Jul 06 '14 at 15:04
  • Well, Abel's irreduciblity theorem is implied by the polynomial's irreducibility...in particular, the fact that an irreducible polynomial is the minimal polynomial for its root (modulo units). – Nishant Jul 06 '14 at 17:14
  • It is. And to say irreducibility is implied by Abel is to be guilty of circular reasoning. Or worse, unnecessary, exological verbosity. – Cris Jul 06 '14 at 21:59