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

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- Aug 25th 2009, 10:42 PMynjSplitting Field
Let $\displaystyle F$be a field of characteristic $\displaystyle p$. Prove that $\displaystyle x^p-a$either is irreducible or splits in $\displaystyle F$...

- Aug 26th 2009, 09:00 AMnirax
if you prove one thing you are through ...

all roots of this polynomial are equal (in some appropriate extension) - Aug 26th 2009, 09:14 AMNonCommAlg
i'll prove something more than what the problem is asking! i'll show that either $\displaystyle x^p -a$ is irreducible or $\displaystyle x^p - a = (x + c)^p,$ for some $\displaystyle c \in F$:

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

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

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

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