Is this a valid equivalent definition of an algebraically closed field?
Let $ F $ be a field. Define a function $ f(x): F\longrightarrow F $ as a finite combination of the operations in $ F $ (addition, multiplication and their inverses). If the range of the function $ f $ is $ F $, then $ F $ is an algebraically closed field.
My reasoning is that such function $ f $ should be a rational function $ f(x)=\dfrac{p(x)}{q(x)} $ where $ p(x) $ and $ q(x) $ are polynomials with coefficients in $ F $, so for any fixed element $ y\in F $ I can write
$ f(x)=y $
$ \dfrac{p(x)}{q(x)}=y $
$ p(x)=yq(x) $
$ p(x)-yq(x)=0 $
As $ p(x)-yq(x) $ is a polynomial with coefficients in $ F $, if $ F $ is truly algebraically closed then there must be some root $ r\in F $ an so
$ f(r)=y $
As $ y $ is an arbitrary element of $ F $, then the range $ R $ of the function $ f $ satisfies $ R=F $
This should work for the field of complex numbers $ \mathbb{C} $ but does this work for any algebraically closed field?