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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 $k\le n$

In order to prove the converse:

$q(a^2)=b_0+b_1a^2+b_2(a^2)^2\ldots+b_k(a^2)^k=0$

implies

$q(a)=b_0+b_1a^2+b_2a^4\ldots+b_ka^{2k}=0$ Then

$p(x)|q(x)$, because $p(x)$ is the minimal polynomial of $a$ over $F$.

If I prove that $n|2k$ we done, since $k$ is odd, we have $n|k$ and $n\le k$ and finally $n=k$.

So I almost finished the question I only need to know how to prove that $n|2k$ It should be only a detail, but I can't see, someone can help me please?

Thanks

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    $\begingroup$ You have $F\subset F(a^2)\subset F(a)$. What are the possibilities for the field extension degree $[F(a)\colon F(a^2)]$? $\endgroup$ – Lubin Nov 6 '12 at 0:29
  • $\begingroup$ @Lubin The only possibility I see is $[F(a):F(a^2)]$ is odd. $\endgroup$ – user42912 Nov 6 '12 at 0:38
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    $\begingroup$ Right. But $a$ satisfies a polynomial of degree 2 over $F(a^2)$, right? $\endgroup$ – Gerry Myerson Nov 6 '12 at 0:49
  • $\begingroup$ @Lubin following the answer below, we have $[F(a^2)(a):F(a^2)]=2$, what this has to do with $[F(a):F(a^2)]$? $\endgroup$ – user42912 Nov 6 '12 at 1:05
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    $\begingroup$ @user42912 What do you think the difference is between $F(a^2)(a)$ and $F(a)$? $\endgroup$ – Steven Stadnicki Nov 6 '12 at 1:17
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You know that $[F(a):F]=[F(a):F(a^2)][F(a^2):F]$. The minimal polynomial of $a$ over $F(a^2)$, assuming that $a\notin F(a^2)$, is $x^2-a^2$, so we have $[F(a):F(a^2)]=2$. Now, what is the problem with that?

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  • $\begingroup$ yes I understand that, what I don't understand is why $[F(a):F(a^2)]=2$. What we have is $[F(a^2)(a):F(a^2)]=2$ $\endgroup$ – user42912 Nov 6 '12 at 1:00
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    $\begingroup$ @user42912 : actually $F(a^2)(a)=F(a)$ $\endgroup$ – user228168 Jun 17 '16 at 4:56
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I am writing the converse part $F(a)\subset F(a^2).$ Enough to show that $a \in F(a^2).$ Let $a\notin F(a^2)$ then $ F(a^2)\subsetneq F(a). $ Since $a^2$ satisfies the polynomial $x^2-a^2\in F(a^2)$ so $[F(a):F(a^2)]=2$. Now $$ [F(a): F] = [F(a) : F(a^2)][F(a^2):F]=2[F(a^2):F].$$ which contradicts the fact that [F(a) : F] is odd. Thus $a \in F(a) $ and therefore $F(a) = F(a^2)$.

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

$[F(a):F]=n$, $n$ is odd then it is clear $F⊂ F(a^2) ⊂ F(a)$ (because $a^2∈F(a)$).

So (by law, or Theorem): $$ [F(a):F]=[F(a): F(a^2)] [F(a^2) : F] $$ If prove $[F(a): F(a^2)]=1$ then $F(a)= F(a^2)$. Conversely, suppose $[F(a): F(a^2)]≠1$. Since $[F(a): F(a^2)]$ can’t be even (because $n$ is odd) then $[F(a): F(a^2)]≥3$. $[F(a^2):F]=L$, then $$ n≥3L \tag{$*$} $$ in other hand $[F(a^2):F]=L$ implies the polynomial of degree $L$ that $a^2$ is its root.
in other way: it exist the polynomial $q(x)=b_0+b_1x+\cdots +b_L x^L$ such that $b_0+b_1a^2+\cdots +b_L a^{2L}=0$ .
it’s mean $a$ is root of the polynomial $g(x)$ of degree $2L$ and since the minimaml polynomial degree of the $a$ is $n$ we must have $$ n≤2L \tag{${*}{*}$}. $$ (because we have $p(x)|g(x)$ by theorem. so $p(x)=g(x)r(x)$ its mean : degree $p(x)$=degree $g(x)$ +degree $r(x)$ .so $n≤2L$ )

So we must have (by attention to $(*)$ and $(**)$), $3L≤n≤2L$ that is a contradiction. In general case we have : if $[F(a):F]=n$ and $m>0$ , $(n,m!)=1$ then $F(a)=F(a^m)$
Hint: if $m<n$ then $ (m!,n)\neq1$ so $m\ge n$. if suppose $[F(a^m):F]=k$ then $n=kl$.
If $k=1$ then $F(a)=F(a^m)$. since $a^m$ is root of the polynomial of degree $ml$ ,we have $n\le ml$ or $kl\le ml$ so $k\le m$ . so we have $k|m!$ and since $k|n$ , $(n,m!)=1$ we must have :$k=1$.

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    $\begingroup$ Why so complicated when it's so simple: $[F(a):F(a^2)]\le 2$ since $a$ is the root of the polynomial $X^2-a^2\in F(a^2)[X]$. It can't be $2$, so it's $1$. $\endgroup$ – user26857 Nov 28 '12 at 22:27
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Since $a^2 \in F(a)$ then $F(a^2) \subseteq F(a).$ Since $a$ is algebraic over $F$ of odd degree, then $[F(a):F]$ is odd (finite) ( and so $F(a)$ is algebraic over $F$). So,

$$[F(a):F] = [F(a):F(a^2)] x [F(a^2):F].$$

But since $f(x) = x^2 - a^2 \in F(a^2)[x]$ has $f(a) = 0,$ i.e., $a$ is a root, then $[F(a):F(a^2)]$ must divide deg$(f(x)) = 2,$ i.e., $[F(a):F(a^2)] = 1$ or $2.$ But since $[F(a):F]$ is odd, $[F(a):F(a^2)]$ cannot be $2.$ Therefore, $[F(a):F(a^2)] = 1$ implying that $F(a) = F(a^2).$ qed

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  • $\begingroup$ You forgot your money (Dollar signs in your A). $\endgroup$ – DanielWainfleet Apr 15 '16 at 4:21
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To show that $a\in F(a^2)$: The case deg $(p)=1$ is trivial as then $a$ and $a^2$ belong to $F.$ For the case deg $(p) >1$, we have $m\geq 1$, where the minimal polynomial of $a$ over $F$ is$$p(x)=\sum_{j=0}^{2 m +1}x^jf_j .$$ $$ \text {Let } \quad r(x)=\sum_{j=0}^mx^j f_{2 j+1}.$$ $$ \text { Let }\quad s(x)=\sum_{j=0}^mx^jf_{2 j}.$$ We have $0<$deg $r(x^2)=2 m<$ deg $(p)$. So $ r( a^2)\ne 0$ by the minimality of deg$(p)$. Therefore $$a=-s(a^2)/r(a^2)\in F(a^2).$$

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