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Let $L/K$ be a field extension of degree $2^k$, $k\in\mathbb{N}$. $a\in L$ and $f\in K[X]$ a polynomial of degree $d$ such that $f(a)=0$, $d$ odd. Show that $a \in K$.

I know only the definition of a field extension and that of degree, this should be sufficient to find this. I have shown that $K[a]$ is a field and that $2^k=[L:K]=[L:K[a]][K[a]:K]$. I tried to show that $[K[a]:K]$ cannot be even, but all my arguments do not allow me to conclude.

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You need $f$ to be the minimal polynomial over $K$, otherwise the fact that the degree $d$ is odd is kind of pointless ; you can always make it odd by multiplying $f$ by some linear factor if required. – Patrick Da Silva Oct 16 '12 at 15:50
up vote 2 down vote accepted

This statement is incorrect.

Take $\mathbb{Q}(\sqrt[4]{5})/\mathbb{Q}$ , this is an extension of degree $4=2^{2}$ but $P(x)=(x-1)(x^{4}-5)$ is a polynomial of odd degree $(d=5)$ and $P(\sqrt[4]{5})=0$, but $\sqrt[4]{5}\not\in\mathbb{Q}$.

However, if the polynomial in question is the minimal polynomial of $a$ then since $K(a)$ is a subfield of $L/K$ and since the degree of the simple extension is $d$ (which does not divide $2^{k}$ since it is odd) you have it that $K(a)=K$ hence $a\in K$.

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Indeed! Now all is Ok, thank you. I will tell my algebra prof. his typo error. – Lucien Oct 16 '12 at 15:50
I have asked for the corrected statement of the exercice, it should be additionaly stated that $f$is irreducible. Then the exercice is of course easy. – Lucien Oct 16 '12 at 16:32

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