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

In summary, the polynomial X^2+1 in \mathbb{R} has two roots, but neither of them is in \mathbb{R}. The polynomial X^2+1 in \mathbb{F}_2 has a root at infinity, but neither of its other roots are in \mathbb{F}_2.
  • #1
rukawakaede
59
0
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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  • #2
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]...
 
  • #3
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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  • #4
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
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?
 
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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! :)
 

1. What is a coefficient of a polynomial in K[x]?

A coefficient of a polynomial in K[x] is the numerical value that is multiplied by the variable in each term of the polynomial. In other words, it is the number that appears in front of the variable.

2. How is the coefficient affected by the field characteristic?

The field characteristic, denoted by p, is the number of times the identity element must be added to itself to equal 0. In a field of characteristic p, the coefficient of a polynomial can only take on p possible values, as any number added to itself p times will equal 0. This means that the coefficients in a polynomial in K[x] are limited to p distinct values.

3. Can the coefficient of a polynomial be 0 in a field of characteristic p?

Yes, the coefficient can be 0 in a field of characteristic p. This is because the identity element, which is added to itself p times to equal 0, is also equal to 0. Therefore, a coefficient of 0 is a valid value in a field of characteristic p.

4. How does the coefficient affect the degree of a polynomial?

The degree of a polynomial is determined by the highest power of the variable present. The coefficient does not affect the degree, as even if the coefficient is 0, the variable with the highest power is still present. However, the coefficient does affect the term with the highest power, as it is multiplied by the variable.

5. Can the coefficient of a polynomial be negative in a field of characteristic p?

Yes, the coefficient can be negative in a field of characteristic p. The sign of the coefficient is not affected by the field characteristic, as the identity element added to itself p times is still equal to 0 regardless of the sign. Therefore, negative coefficients are possible in a polynomial in K[x].

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