Show that if $F(a)=F(a^2)$, then $a$ is algebraic over $F$. Let $K/F$ be a field extension and let $a \in K$. Show that if $F(a)=F(a^2)$, then $a$ is algebraic over $F$.
I have tried but unable to show it.
 A: $a \in F(a^2)$ means there is a rational function $g$ over the field $F$ such that $a = g(a^2)$.  Multiplying by the denominator gives us a polynomial equation for $a$. 
A: It's not as elegant as the solution by Robert Israel, but if $a$ is an indeterminate, then it not possible for $a$ to be of the form $f(a^{2})/g(a^{2})$, for polynomials $f, g$, as then in $f(a^{2}) = a g(a^{2})$ the degree on the left is even, the one on the right is odd.
This argument is used for instance when one takes $F$ to be the field with $2$ elements. The argument shows that $F(a^{2}) \subsetneq F(a)$, and then  $\lvert F(a) : F(a^{2}) \rvert = 2$, as $a$ is a root of $h(x) = (x - a)^{2} = x^{2} - a^{2}\in F(a^{2})[x]$. The polynomial $h(x)\in F(a^{2})[x]$ is thus irreducible, but it has the double root $a$.
A: $F(a) =F(a^2)$ thus both $a,a^2 \in F(a)$ thus also $a,a^2,a^3,\cdots,a^n,\cdots \in F(a)$ and there exists a $n$ such that the set $\{1,a,a^2,a^3,\cdots,a^n\}$ is linearly dependent. So there exists $c_0,c_1,c_2,\cdots , c_n \in F$ such that $c_0+c_1a+c_2a^2+\cdots+c_na^n=0$ but not all $c_i$ is $0$. Thus consider the corresponding polynomial $f(x)=c_0+c_1x+\cdots+c_nx^n$. Here $f(a)=0$ and thus $a$ is algebraic over $F$.
