how to show a polynomial is reducible over a field F

[tim656]

Please help. let F be a field of characteristic p (prime), let f(x) = (x - b)^p. show that if f(x) is reducible, then b must be in F.

 

[daon2]

Please help. let F be a field of characteristic p (prime), let f(x) = (x - b)^p. show that if f(x) is reducible, then b must be in F.

This proof is easy if you have the tools. Show some work, please... where are you stuck?

If you are unsure where to start, use the binomial expansion of f(x).

 

[tim656]

so i have: x^p - (p^C₁)x^p-1b +(p^C₂)x^p-1b² - ............+ (p^C_p-1)xb^p-1 - b^p
= (-1)(b - x)^p for p is prime

so f(x)=(b-x)^pg(x)

 

[daon2]

so i have: x^p - (p^C₁)x^p-1b +(p^C₂)x^p-1b² - ............+ (p^C_p-1)xb^p-1 - b^p
= (-1)(b - x)^p for p is prime

so f(x)=(b-x)^pg(x)

Okay, that -1 seems unnecessary. But notice p divides (p choose i) for all i strictly between 0 and p

 

[tim656]

so f(x) is not primitive then.

 

[daon2]

What I was getting at is \(\displaystyle f(x)=x^p-b^p\).

If \(\displaystyle f(x)\) is reducible over F, then write \(\displaystyle f(x)=x^p-b^p=h(x)g(x)\) where \(\displaystyle h(x)=(x-b)^s\) \(\displaystyle g(x)=(x-b)^t\) and \(\displaystyle 1\le s,t<p\).

The constant term of \(\displaystyle h(x)\) divides the constant term of \(\displaystyle f(x)\), right? See if you can get anywhere with this.

 

[tim656]

the constant term of h(x) is a multiple of p, so p^2 divides the constant term of f(x).

 

[daon2]

the constant term of h(x) is a multiple of p, so p^2 divides the constant term of f(x).

No, it isn't. What is the constant term of (x-b)^2? (x-b)^3?.... (x-b)^s?

 

[tim656]

-b^s

 

[daon2]

Here is what we have so far: \(\displaystyle h(x)\in \mathbb{F}_{p^n}[x]\) has constant term \(\displaystyle b^s\in \mathbb{F}_{p^n}\), where \(\displaystyle 1\le s < p\). Now you must show \(\displaystyle b\in \mathbb{F}_{p^n}\). I am reluctant to do any more.