Suppose $F$ is a field and the irreducible polynomial over $F$ of $x$ is of odd degree. A question I recently encountered is:
Let $F$ be a field. Let $x$ be algebraic over $F$. Suppose the minimal polynomial of $x$ is of odd degree. Show $F(x) = F(x^2)$.
We know $F(x^2)$ is a subset of $F(x)$ because $x^2$ belongs to $F(x)$.
To show the other containment, do the following:
Suppose, for a contradiction, $x$ is not an element of $F(x^2)$. Then the minimal polynomial of $x$ over $F(x^2)$ is $f(y) = y^2 - x^2$. It follows that $[F(x):F(x^2)] = 2$. But $[F(x):F(x^2)][F(x^2):F] = 2[F(x^2):F]$ is odd, a contradiction.
Is there any hole in my logic? For some reason I'm not quite sure if this is correct.
 A: Your proof is fine, but there's a more direct one.
Write $f(X)=g(X^2)+h(X^2)X$, where $f(X)$ is the minimal polynomial of $x$; in other words, group the even degree terms in $g(X^2)$ and the odd degree terms in $h(X^2)$ (after collecting $X$ in the odd degree terms). Note that $h(X^2)\ne0$, because $f$ has odd degree and that $h(x^2)\ne0$, because $h(X^2)$ has degree smaller than $f(X)$. Then $f(x)=0$ gives
$$
x=-\frac{g(x^2)}{h(x^2)}
$$
A: That looks fine. 
A more direct proof, without Galois theory
Let $p(X)$ be the minimal polynomial for $x$ and $q(X)$ the minimal polynomial for $x^2$. Then $q(X^2)$ has roots $\pm x$, an so must be divisible by $p(X)$ and $p(-X)$. Is it possible for $p(X)$ and $p(-X)$ to have common factors? No. (The argument for this is where we use that $p$ is of odd degree and irreducible.)
So $p(X)p(-X)$ must be a divisor of $q(X^2)$, so the degree of $q$ must be at least the degree of $p$.
Caveat: Oh, I just realized this argument requires $-x\neq x$, so $F$ would have to not be of characteristic $2$.
