Is every subalgebra of $F[x]$ one-dimensional? Let $F$ be a field. Let $A$ be a subalgebra of the polynomial ring $F[x]$. Does $A$ necessarily have Krull dimension $\leq 1$?
Alternative proof: WLOG let $A\neq F$, since otherwise the claim is trivial. Then, $A$ contains a nonconstant polynomial $p$. As a consequence, $p\left(X\right) = p$ is a nonzero polynomial equation with $A$-coefficients satisfied by $X=x$. Thus, $F\left[x\right]$ is an integral ring extension of $A$. Since Krull dimension is invariant under integral extensions (e. g. Theorem 3 in http://www.math.rwth-aachen.de/~Eva.Zerz/ast10/dim1.pdf ), this yields $\dim A = \dim F = 1$.
Such an algebra is a domain. If it is finitely generated, then its Krull dimension is the same as its trascendence degree, and then it is clear that it is at most that of $k[x]$.