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This has been asked before, but I have a different solution and I would like to check it.

Let $F$ be a field and $u$ be algebraic over $F$ of odd degree. Prove that $F(u) = F(u^2)$.

Of course $F(u^2) \subset F(u)$. If we can show that $u \in F(u^2)$, we are through.

Let $$f(x) = x^{2n+1} + a_{2n}x^{2n} + \cdots + a_1 x + a_0$$ be the minimum polynomial of $u$ over $F$. Then:

$$u g(u) + h(u) = 0$$

where

$$g(x) = x^{2n} + a_{2n-1}bx^{2n-2} + a_{2n-3} x^{n-2} + \cdots + a_3 x^2 + a_1$$

and

$$h(x) = a_{2n}x^{2n} + a_{2n-2} x^{2n-2} + \cdots + a_2 x^2 + a_0$$

We can't have $g(u) = 0$ because the degree of $u$ is $2n+1$. So $g(u)$ is a unit in $F(u^2)$. Hence $u = - h(u)g(u)^{-1} \in F(u^2)$.

Does this look good?

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    $\begingroup$ @reuns doesn't that say $k\leq 2$? $\endgroup$ – sharding4 Jul 25 '17 at 1:31
  • $\begingroup$ Yes I meant $k$ $\endgroup$ – reuns Jul 25 '17 at 1:39
  • $\begingroup$ I think both are nice little arguments. $\endgroup$ – sharding4 Jul 25 '17 at 1:43
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Not sure why you want a confirmation.

  • Your proof

Let $f(x) = g(x^2)+xh(x^2)$ the minimal polynomial of $u$ over $F$, with $deg(f) = 2n+1,deg(h)=n, deg(g) \le n$.

$h(u^2) \ne 0$ since otherwise $u$ would be a root of $h(x^2)$ whose degree is $< 2n+1$.

Thus $f(u) = 0 \implies u = \frac{-g(u^2)}{h(u^2)} \in F(u^2)$.

  • The traditional one

$[F(u):F] =[F(u):F(u^2)][F(u^2):F]$. Since $[F(u):F(u^2)] \le 2$ and $2 \nmid [F(u):F]$ we obtain $[F(u):F(u^2)]=1$ and hence $F(u) = F(u^2)$.

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