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I have a question about simple field extensions.

For a field $F$, if $[F(a):F]$ is odd, then why is $F(a)=F(a^2)$?

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Firstly, $F(a^2)\subseteq F(a)$. If the inclusion is strict, then $[F(a):F(a^2)]\neq 1$. Now $a$ satisfies the polynomial $X^2-a^2\in F(a^2)[X]$, so the minimal polynomial $m_{a,F(a^2)}(x)$ has degree at most $2$, and this is necessarily $2$, so $[F(a):F(a^2)]=2$.

Then since the degree of field extensions is multiplicative, you get $2\mid [F(a):F]$, a contradiction.

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Note that $F(a)\supseteq F(a^2)\supseteq F$, hence $$\text{odd }=[F(a):F]=[F(a):F(a^2)][F(a^2):F].$$ What can $[F(a):F(a^2)]$ be? What, therefore, must it be?

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Since $[F(a):F]$ is odd the minimal polynomial of $a$ is an odd degree polynomial, say $p(x)=b_{0}x^{2k+1}+b_1x^{2k}+...b_{2k+1}$, now since $a$ satisfies $p(x)$ we have: $b_{0}a^{2k+1}+b_1a^{2k}+...b_{2k+1}=0$ $\implies$ $a(b_0a^{2k}+b_2a^{2k-2}+...b_{2k})+b_1a^{2k}+b_3a^{2k-2}+...b_{2k+1}=0$ $\implies$ $a= -(b_1a^{2k}+b_3a^{2k-2}+...+b_{2k+1})/(b_0a^{2k}+b_2a^{2k-2}+...b_{2k})$ $\implies$ $a$ is in $F(a^2)$,so obviously $F(a^2)=F(a)$

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I don't see where the argument breaks down if the degree of $a$ is even. – Gerry Myerson Nov 1 '11 at 11:48
The argument breaks down when there is a possibility that the coefficients of all the odd degree monomials are zero(when you take $p(x)$ is of even degree), where it is an impossibility in the case where $p(x)$ of odd degree since $b_0$ should never be zero. – Dinesh Nov 1 '11 at 13:08
Thanks. So the only time it doesn't work for even degree is when the minimal polynomial has no terms of odd degree. – Gerry Myerson Nov 1 '11 at 21:41
@GerryMyerson exactly – Dinesh Nov 2 '11 at 2:31

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