Is $F(u) =F(u^2)$?

Question

Let $K=F(u) $, where $u \in K$ algebric over $F$ and $[K:F] $ is odd. How can I show that $K=F(u^2)$?

I think that I have to use some linear algebra, but I don't know how.

Answer 1

Notice that $F \subseteq F(u^2) \subseteq F(u)=K$ (a rude looking chain if you ask me).

The degree formula tells you that $[K:F] = [K:F(u^2)][F(u^2):F]$ so if $[K:F]$ is odd, then both $[K:F(u^2)]$ and $[F(u^2):F]$ are odd.

Consider $x^2-u^2$. This polynomial has coefficients in $F(u^2)$. So if $u \not\in F(u^2)$, it has no roots in $F(u^2)$ and thus is irreducible over this field. But then $[K:F(u^2)]$ (note: $K=F(u)=F(u,u^2)$) is a degree 2 extension (contradiction).

Therefore, $u \in F(u^2)$ so $F(u^2)=K$.

Pedagogy note: When designing a problem, using the letters $F$, $u$, and $K$ together (as is tempting during a $u$-substitution problem) will inevitably result in you writing something which looks mildly obscene on the board. Choose your letters wisely my friends.