# Thread:

1. ## Splitting Field

Let $F$be a field of characteristic $p$. Prove that $x^p-a$either is irreducible or splits in $F$...

2. if you prove one thing you are through ...

all roots of this polynomial are equal (in some appropriate extension)

3. Originally Posted by ynj
Let $F$be a field of characteristic $p$. Prove that $x^p-a$ either is irreducible or splits in $F$...
i'll prove something more than what the problem is asking! i'll show that either $x^p -a$ is irreducible or $x^p - a = (x + c)^p,$ for some $c \in F$:

suppose $x^p - a$ is not irreducible. then $x^p - a = g(x)(f(x))^m,$ for some polynomials $f(x), g(x) \in F[x],$ where $f(x)$ is irreducible with $1 \leq \deg f(x) < p, \ 1 \leq m \leq p,$ and $\gcd(f(x),g(x))=1.$

differentiating (formally) will give us: $0=px^{p-1}=g'(x)(f(x))^m + mg(x)f'(x)(f(x))^{m-1}.$ hence $mg(x)f'(x)=-g'(x)f(x),$ which is impossible unless $m=p, \ \deg g(x) = 0$ and $\deg f(x)= 1$

because $\gcd(f'(x),f(x))=\gcd(f(x),g(x))=1.$ so suppose $m=p, \ g(x) = \alpha$ and $f(x)=\beta x + \gamma.$ then $x^p-a = \alpha (\beta x + \gamma)^p=\alpha(\beta^p x^p + \gamma^p)$ and hence $\alpha \beta^p = 1, \ \alpha \gamma^p = -a.$ letting $c = \gamma \beta^{-1}$

we'll get $x^p - a = (x + c)^p. \ \Box$
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