Characterizing all non-integrally closed subrings of $\mathbb{C}[t]$ P. M. Cohn proved the following: A subalgebra of $\mathbb{C}[t]$ is free if and only if it is integrally closed.
(Actually, Cohn's result deals with $k[t]$, $k$ any field, but I do not mind to take $k=\mathbb{C}$).
Let $f=f(t),g=g(t) \in \mathbb{C}[t]$, with $\deg(f)>\deg(g)> 1$,
and consider $\mathbb{C}[f,g]$.
For example: $f=t^3$, $g=t^2$, and then $\mathbb{C}[t^3,t^2] \subsetneq \mathbb{C}[t]$ is not integrally closed, since $t \in \mathbb{C}(t)=\mathbb{C}(t^3,t^2)$ is of course integral over $\mathbb{C}[t^3,t^2]$ ($t$ is a root of $T^2-t^2$, and $t$ is also a root of $T^3-t^3$) but $t \notin \mathbb{C}[t^3,t^2]$.

Is it possible to characterize all (non-)integrally closed subrings of $\mathbb{C}[t]$ of the form $\mathbb{C}[f,g]$ as above? 

It would be nice if it is possible to find a general form of $f$ and $g$ in terms of their coefficients and degrees, so denote:
$f=a_nt^n+\cdots+a_1t+a_0$ and $g=b_mt^m+\cdots+b_1t+b_0$,
where $n > m > 1$ and $a_i,b_i \in \mathbb{C}$.

What about characterizing all free subrings $F$ of $\mathbb{C}[t]$ generated by two elements? (see remark (2) below; I wish to exclude $\mathbb{C}[t^4,t^2]$). If the field of fractions of a free $2$-generated subring $F$ equals $\mathbb{C}(t)$ is necessarily $F=\mathbb{C}[t]$? 

Remarks: 
(1) The assumption $m>1$ excludes the case $g=b_1t+b_0$ and then $\mathbb{C}[f,g]=\mathbb{C}[t]$, so it is integrally closed.
(2) I think that I better further assume that the field of fractions of $\mathbb{C}[f,g]$ equals $\mathbb{C}(t)$, in order to exclude cases such as $f=t^4, g=t^2$, which I am not interested in.
(3) Abhyankar-Moh-Suzuki theorem says that if $\mathbb{C}[f,g]=\mathbb{C}[t]$ then $m$ divides $n$. Therefore, if $m$ does not divide $n$, then $\mathbb{C}[f,g] \subsetneq \mathbb{C}[t]$; in this case, is it still possible for $\mathbb{C}[f,g]$ to be integrally closed, under the further assumption that $\mathbb{C}(f,g)=\mathbb{C}(t)$? 
(4) If $\mathbb{C}[f,g] \subseteq \mathbb{C}[t]$ is flat and $\mathbb{C}(f,g)=\mathbb{C}(t)$, then $\mathbb{C}[f,g]=\mathbb{C}[t]$ and we are done (= $\mathbb{C}[f,g]$ is integrally closed). This follows from the second answer to this question
(trivially, $\mathbb{C}[f,g] \subseteq \mathbb{C}[t]$ is integral). Is there a criterion for flatness in terms of the coefficients of $f$ and $g$ and their degrees?
(5) Additional relevant questions and papers are: Necessary and sufficient condition that a localization of an integral domain is integrally closed 1; 
Generalized quotient rings (Richman) 2;
(Some remarks on Richman simple extensions of an integral domain 3).
Thank you very much for any help! 
 A: If $A\subset \mathbb{C}[t]$ is an integrally closed $\mathbb{C}$-subalgebra with $A\neq \mathbb{C}$, then $A=\mathbb{C}[u]$ for some $u=p(t)$. Thus, if $A=\mathbb{C}[f,g]\subset\mathbb{C}[t]$ and $\deg f,\deg g$ neither dividing the other, then $A$ is not integrally closed by Abhyankar-Moh-Suzuki. Of course, this is not a characterization, since even if $\deg f$ diivided $\deg g$, $A$ still might be not integrally closed, like $f=t^3, g=t^2+t^6$.
A: Question $(3)$ is easily answered . . .

Let $m,n$ be positive integers with $m \le n$.

Suppose $f,g\in \mathbb{C}[t]$ are such that


*

*$\deg(f)=m$.$\\[4pt]$

*$\deg(g)=n$.$\\[4pt]$

*$\mathbb{C}(f,g)=\mathbb{C}(t)$.


Claim:$\;$If $\mathbb{C}[f,g]$ is integrally closed, then $m{\,\mid\,}n$.

Proof:

Assume $f,g,m,n$ satisfy the specified conditions, and suppose $\mathbb{C}[f,g]$ is integrally closed.

Since $f\in \mathbb{C}[t]$ is non-constant, it follows from the identity $f(t) - f=0$, that $t$ is integral over $\mathbb{C}[f]$, hence $t$ is integral over $\mathbb{C}[f,g]$.

From $t \in \mathbb{C}(t)=\mathbb{C}(f,g)$, we get that $t$ is in the field of fractions of $\mathbb{C}[f,g]$, hence, since $\mathbb{C}[f,g]$ is integrally closed, it follows that $t\in \mathbb{C}[f,g]$, so $\mathbb{C}[f,g]=C[t]$

By the results of Abhyankar-Moh-Suzuki, it follows that $m{\,\mid\,}n$, which proves the claim.
