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

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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Hi rukawakaede!

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

$$f(X)=(X-\alpha_1)(X-\alpha_2)$$

but we won't have that $(X-\alpha_i)$ in K[X] (otherwise $\alpha_i\in \mathbb{F}_2$ 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 $X^2+1$ in $\mathbb{R}$. The sum of the roots is i-i=0, but neither i nor -i are in $\mathbb{R}$...

Hi rukawakaede!

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

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

$$f(X)=(X-\alpha_1)(X-\alpha_2)$$

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

No, see my previous example. Another example which is not in char 2, is the polynomial $X^2+1$ in $\mathbb{R}$. The sum of the roots is i-i=0, but neither i nor -i are in $\mathbb{R}$...
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 $X^p-X+\alpha$ and if $\beta$ is one root of the polynomia (in some field extension of K)l, then we know all the roots:

$$\beta,\beta+1,...,\beta+p-1$$

this is essentially Fermat's little theorem. So let

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

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...

Ok, rukawakaede, the polynomials you mention are called "Artin-Schreier polynomials" and are very well studied.

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

$$\beta,\beta+1,...,\beta+p-1$$

this is essentially Fermat's little theorem. So let

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

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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micromass thanks!
I am now fully understood.
Thank you very much, especially for telling me the name of the polynomial. Learned extra stuff! :)