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I am trying to solve the following exercise:

Let $\alpha$ be algebraic over $F$. Suppose the degree of $\alpha$ over $F$ is odd. Show that $F(\alpha)=F(\alpha^2)$

It's not clear to me how to solve this. I've been thinking and I guess that $F(\alpha^2)\subset F(\alpha)$ but I don't know about the other direction. Can you help me?

Red Banana
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    For your guess: what do you need to show to verify that $F(\alpha^2) \subset F(\alpha)$? – Greg Martin Mar 26 '21 at 01:08
  • The possible degrees the extension $F(a^2)\subset F(a)$ is either 2 or 1. But if it were 2 you would have that $[F(a):F]=[F(a):F(a^2)]\cdot [F(a^2):F]=2\cdot [F(a^2):F]$ – MIO Mar 26 '21 at 01:09
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    @GregMartin I thought the following: $F(\alpha^2)$ is the smallest field that contains $\alpha^2$. $F(\alpha)$ is the smallest field that contains $\alpha$, as this is a field, then it contains $\alpha^2$ and hence, contains $F(\alpha^2)$. Does that makes sense? – Red Banana Mar 26 '21 at 01:18

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$[F(\alpha):F]=[F(\alpha):F(\alpha^2)][F(\alpha^2):F]$ since $[F(\alpha):F]$ is odd, we deduce that $[F(\alpha):F(\alpha^2)]$ is odd and is $1$ since the fact $\alpha$ is a solution of $X^2-\alpha^2$ implies that $[F(\alpha):F(\alpha^2)]\leq 2$.

Tsemo Aristide
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