If $F$ is a field, and $E$ a subfield of $F(X)$ containing a rational polynomial, must $E$ contain an honest-to-god polynomial?

E.g., the subfield generated by $f(X) = \frac{X}{X+1}$

Ah, so $E = F(X+\frac{1}{X})$ doesn't contain polynomials. Thanks!

Here's a cool one: Let $k$ be a perfect field and $p$ a prime. Show that there exists an algebraic extension $k^{(p)}/k$ such that each finite subextension has degree prime to $p$ and such that $k^{(p)}/k$ has no non-trivial finite extensions of degree prime to $p$.

Theorem. Let $R$ be a ring such that $R$ is a subring of a division ring $D$. If for all $d(\ne 0)\in D$ either $d\in R$ or $d^{-1}\in R$ then $R$ is a local ring.