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

  • #1
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!
 
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Answers and Replies

  • #2
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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:

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]...
 
  • #3
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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  • #4
22,089
3,291
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...
 
  • #5
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?
 
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  • #6
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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