Coefficient of a polynomial in K[x] where K is a field of characteristic p

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Discussion Overview

The discussion revolves around properties of polynomials in the context of a field K of characteristic p, specifically focusing on the implications of coefficients and roots of polynomials. Participants explore questions related to the existence of roots in K and the structure of polynomials, particularly Artin-Schreier polynomials.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants question whether the polynomial g(x) can be concluded to be in K[x] given that f(x) is in K[x].
  • It is noted that if ck-1 is in K, it does not necessarily imply that the roots α1, α2, ..., αk are also in K, with examples provided from characteristic 2 and real numbers.
  • One participant mentions the specific case of the polynomial f(x) = xp - x - a and expresses a desire to show that ck-1 ∈ K implies f(x) has a root in K, but feels stuck.
  • Another participant introduces Artin-Schreier polynomials and discusses their properties, including the relationship between roots and field extensions.
  • There is a suggestion to consider the implications of ck-1 being in K in relation to the splitting of f(x) in K[x].

Areas of Agreement / Disagreement

Participants express differing views on the implications of coefficients being in K and the existence of roots in K. There is no consensus on whether ck-1 ∈ K guarantees roots in K, as examples challenge this assumption.

Contextual Notes

Participants reference specific examples and properties of polynomials that may not be universally applicable, indicating limitations in the assumptions made about roots and coefficients in various fields.

Who May Find This Useful

This discussion may be of interest to those studying polynomial theory, field extensions, and the properties of polynomials in finite fields, particularly in the context of characteristic p.

rukawakaede
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Let K be a field of characteristic p.

Suppose f(x)=(xk+ck-1xk-1+...+c0)(xp-k+...) in K[x] with 1≤k≤p-1.

My question is:

1. since f(x) in K[x], can I conclude g(x)=xk+ck-1xk+...+c0 in K[x] as well?

2. We see that in general if g(x)=xk+ck-1xk-1+...+c0 then ck-1=-(α12+...+αk) where α11,...,αk are the root of g(x).

Now if ck-1∈K, can I conclude that α11,...,αk∈K as well?

very appreciated if someone could solve my confusion.
Thank you!
 
Last edited:
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Hi rukawakaede! :smile:

I'm not sure where these questions come from, so I don't know if I interpreted them correctly:

rukawakaede said:
Let K be a field of characteristic p.

Suppose f(x)=(xk+ck-1xk+...+c0)(xp-k+...) in K[x] with 1≤k≤p-1.

My question is:

1. since f(x) in K[x], can I conclude g(x)=xk+ck-1xk+...+c0 in K[x] as well?

Take [itex]K=\mathbb{F}_2[/itex] and take f(X)=X2+X+1. Then f splits in a field extension of K, that is, we have

[tex]f(X)=(X-\alpha_1)(X-\alpha_2)[/tex]

but we won't have that [itex](X-\alpha_i)[/itex] in K[X] (otherwise [itex]\alpha_i\in \mathbb{F}_2[/itex] which cannot be).

2. We see that in general if g(x)=xk+ck-1xk+...+c0 then ck-1=-(α12+...+αk) where α11,...,αk are the root of g(x).

Now if ck-1∈K, can I conclude that α11,...,αk∈K as well?

No, see my previous example. Another example which is not in char 2, is the polynomial [itex]X^2+1[/itex] in [itex]\mathbb{R}[/itex]. The sum of the roots is i-i=0, but neither i nor -i are in [itex]\mathbb{R}[/itex]...
 
micromass said:
Hi rukawakaede! :smile:

I'm not sure where these questions come from, so I don't know if I interpreted them correctly:
Take [itex]K=\mathbb{F}_2[/itex] and take f(X)=X2+X+1. Then f splits in a field extension of K, that is, we have

[tex]f(X)=(X-\alpha_1)(X-\alpha_2)[/tex]

but we won't have that [itex](X-\alpha_i)[/itex] in K[X] (otherwise [itex]\alpha_i\in \mathbb{F}_2[/itex] which cannot be).
No, see my previous example. Another example which is not in char 2, is the polynomial [itex]X^2+1[/itex] in [itex]\mathbb{R}[/itex]. The sum of the roots is i-i=0, but neither i nor -i are in [itex]\mathbb{R}[/itex]...

Thanks micromass!

I understand what you said.

In fact the polynomial is given as f(x)=xp-x-a where a∈K.

I want to show that ck-1∈K implies f(x) has a root in K but stuck here.

Since p=2 case is trivial. does that means we can show this by induction?Could you please give some suggestions?
 
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Ok, rukawakaede, the polynomials you mention are called "Artin-Schreier polynomials" and are very well studied.

In general, if we have the polynomial [itex]X^p-X+\alpha[/itex] and if [itex]\beta[/itex] is one root of the polynomia (in some field extension of K)l, then we know all the roots:

[tex]\beta,\beta+1,...,\beta+p-1[/tex]

this is essentially Fermat's little theorem. So let

[tex]L=K[x]/(X^p-X+\alpha)[/tex]

then L is the field extension which contains all the roots of the polynomial. Can you solve your question now? Tell me if you need more hints...
 
micromass said:
Ok, rukawakaede, the polynomials you mention are called "Artin-Schreier polynomials" and are very well studied.

In general, if we have the polynomial [itex]X^p-X+\alpha[/itex] and if [itex]\beta[/itex] is one root of the polynomia (in some field extension of K)l, then we know all the roots:

[tex]\beta,\beta+1,...,\beta+p-1[/tex]

this is essentially Fermat's little theorem. So let

[tex]L=K[x]/(X^p-X+\alpha)[/tex]

then L is the field extension which contains all the roots of the polynomial. Can you solve your question now? Tell me if you need more hints...

I know if K adjoint a root of f(x), called it α, then in fact K(α) consists all of its roots, i.e. K(α,α+1,...,α+(p-1))=K(α) and so K(α) is the splitting field of f(x).

Now I have question:
if ck-1, as defined above, is in K, then we said the polynomial f(x) splits in K[x]?I am not fully understand, in particular, how can I show the implication ck-1∈K implies f(x) has a root in K?
 
Last edited:
micromass thanks!
I am now fully understood.
Thank you very much, especially for telling me the name of the polynomial. Learned extra stuff! :)
 

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