# Field Extensions

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). –  Arturo Magidin Jan 12 '11 at 4:14

## 1 Answer

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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WooW! I guess I was complicating a lot this problem. Thank a lot for your time. And sorry for the dumb question. –  Chu Jan 12 '11 at 3:57
@Chu: Don't forget to formally accept an answer (the one you find most helpful) when you are satisfied. I might also suggest that you edit your question to add your solution if you with; that way we can help you with that too. –  Arturo Magidin Jan 12 '11 at 4:00
Ok. Do i post the solution within the space where the question go? –  Chu Jan 12 '11 at 4:02
@Chu: You should have an edit button for the question. I would suggest editing it, and adding the solution at the bottom; perhaps divide it with a horizontal line <hr/> from the original text, mark it as an addition (say, by adding *Edit:* or *Added:*) and then write your solution. Or you can add your solution as an Answer if you prefer. –  Arturo Magidin Jan 12 '11 at 4:05
Thank you so much for your time. I'll do that. :) –  Chu Jan 12 '11 at 4:06