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Let $L/K$ a finite extension and $f(x)\in K[x]$ a non-linear irreducible polynomial. Prove that if $\mathrm{gcd}\left( \mathrm{deg}(f) , \left[ L:K \right] \right)=1$ then $f(x)$ has no roots in $L$.


Added: (Solution based on the answer below)

Suppose $f(x)$ has a root in $L$, namely $\alpha$ and consider the extension $K(\alpha)/K$. Since $f$ is irreducible we have that $[K(\alpha) : K] = \mathrm{deg}(f) > 1$. On the other hand we have that $[L:K]=[L:K(\alpha)][K(\alpha):K]$. Then $[L:K]=[L:K(\alpha)](\mathrm{deg}(f))$ but this is imposible since $\mathrm{gcd}\left( \mathrm{deg}(f) , \left[ L:K \right] \right)=1$ and $\mathrm{deg}(f) > 1$.

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    I would just add "and $\mathrm{deg}(f)\gt 1$" at the end; otherwise, perfect. (It my even be okay without it; I just like hammering the nails in solidly).2011-01-12

1 Answers 1

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  1. What do you know about the degree $[K(\alpha):K]$ of an extension when $\alpha$ is a root of an irreducible polynomial $g(x)\in K[x]$?

  2. What do you know about the degrees $[L:K]$, $[K:F]$, and $[L:F]$ of extensions when you have a tower $F\subset K\subset L$ ?

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    Thank you so much for your time. I'll do that. :)2011-01-12