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I have some doubt in this post

Let $E$ be an extension field of $F$. If $a \in E$ has a minimal polynomial of odd degree over $F$, show that $F(a)=F(a^2)$.

let $n$ be the degree of the minimal polynomial $p(x)$ of $a$ over $F$ and $k$ be the degree of the minimal polynomial $q(x)$ of $a^2$ over $F$

Since $a^2 \in F(a)$, We have $F(a^2) \subset F(a)$, then $k\le n$

I don't understand why $F(a^2) \subset F(a)$?

My thinking: Take $ a \in \mathbb{R} $, $a \subset a^2 \implies F(a) \subset F(a^2)$

jasmine
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    What assumptions of the prompt haven't you used yet? The fact that you can assume the minimal polynomial is essential. – dbossaller Jan 22 '21 at 18:01
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    $F(a^2)\subset F(a)$ because $F\subset F(a)$ and $a^2\in F(a)$ and $F(a)$ is a field – J. W. Tanner Jan 22 '21 at 18:13
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    hi jasmine; do you mean to ask why $F(a^2)\subset F(a)$, or why $F(a)\subset F(a^2)$? the first of these holds without any hypotheses about the minimal polynomial of $a$, while the second requires an additional hypothesis in general. (I ask because in your "my thinking" section you seem to be trying to prove $F(a)\subset F(a^2)$, and not the other way around. I will additionally note that your argument there is not clear; what do you mean when you write $a\subset a^2$?) – Atticus Stonestrom Jan 22 '21 at 18:14
  • take $a = \mathbb{R}$ i mean for example $\mathbb{R} \subset \mathbb{R}^2$ @AtticusStonestrom – jasmine Jan 22 '21 at 18:20
  • @J.W.Tanner u mean $F(a^2)$ is not field Am i right ? – jasmine Jan 22 '21 at 18:28
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    oh, I think I see your confusion. $a$ is an element of $E$, not a subset. to get some intuition for what's going on, try to consider the following two cases: let $E=\mathbb{R}$ and $F=\mathbb{Q}$. for the first case, let $a=\sqrt{2}$. for the second case, let $a=\sqrt[3]{2}$. reflect upon the fields $F(a)$ and $F(a^2)$ for these values of $F$ and $a$.

    also, in response to your question to @J.W.Tanner, $F(a^2)$ is a field. do you know what the notation $F(b)$ means for $b\in E$?

     – Atticus Stonestrom Jan 22 '21 at 18:29
  • thanks u got it @AtticusStonestrom For $a = \sqrt 2$, $ F(a)=\mathbb{Q}({\sqrt 2})$ contain basis ${1, \sqrt 2}$ but $ F(a^2)=\mathbb{Q}(2)$ contain basis $1 \implies F(a^2) \subset F(a)$ Am i right ? – jasmine Jan 22 '21 at 18:40
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    @jasmine right! so in that case $F(a^2)\subset F(a)$ but $F(a)\not\subset F(a^2)$. now, what about when $a=\sqrt[3]{2}$? hint: for one direction, note that $[\sqrt[3]{4}]^2=2\sqrt[3]{2}$ – Atticus Stonestrom Jan 22 '21 at 18:42
  • it will be reversed $ F(a) \subset F(a^2)$ @AtticusStonestrom – jasmine Jan 22 '21 at 18:45
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    indeed, that relation holds, but we also have $F(a^2)\subset F(a)$ – can you see why? in particular, since $F(a^2)\subset F(a)$ and $F(a)\subset F(a^2)$, this means $F(a^2)=F(a)$ in this case – Atticus Stonestrom Jan 22 '21 at 18:48

1 Answers1

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Credit : Atticus Stonestrom( comment section)

For $a = \sqrt 2$, $ F(a)=\mathbb{Q}({\sqrt 2})$ contain basis $\{1, \sqrt 2\}$ but $ F(a^2)=\mathbb{Q}(2)$ contain basis $1 \implies F(a^2) \subset F(a)$

jasmine
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