Given two (monic) polynomials $f=f(t),g=g(t) \in \mathbb{C}[t]$, consider the ring $\mathbb{C}[f,g]$.
Is it possible to describe all prime ideals and maximal ideals in $\mathbb{C}[f,g]$?
Of course, if $P$ is a prime ideal in $\mathbb{C}[f,g]$, then $\mathbb{C}[f,g]/P$ is an integral domain, and if $M$ is a maximal ideal in $\mathbb{C}[f,g]$, then $\mathbb{C}[f,g]/M$ is a field. (Also, though I am not sure if this helps, $\mathbb{C}[f,g]$ may not be a principal ideal domain).
In particular, what is the answer for $f(t)=t^2$ and $g(t)=t^3$? This question is perhaps relevant.
Edit: See also this question and this paper (maybe they are relevant).
Thank you very much for any hints and comments!
Continuing the construction I mentioned in the comments, consider the map $\phi : \mathbb{C}[X,Y] \to \mathbb{C}[t]$, letting $I = \ker \phi$ then it can easily be shown that $ \mathbb{C}[f(t), g(t)] \cong \mathbb{C}[X,Y] / I$. It can also be proven that $I = (X -f(t), Y -g(t)) \cap \mathbb{C}[X,Y]$. Which, if necessary, can be calculated using Groebner bases.
Now we can use that there is a bijection between ideals of $\mathbb{C}[X,Y] / I$ and ideals of $\mathbb{C}[X,Y]$ that contain $I$, and that an ideal in $\mathbb{C}[X,Y] / I$ is prime/maximal precisely when the corresponding ideal in $\mathbb{C}[X,Y]$ is.
Furthermore assuming that $I$ itself is prime we can use the fact that $C[X,Y]$ has krull dimension 2 to prove the following:
In particular in the case $f(t) = t^2,\, g(t) = t^3$ we get $\ker\phi = (X^3 - Y^2)$, and so the prime and maximal ideals of $ \mathbb{C}[t^2, t^3] \cong \mathbb{C}[X,Y] / (X^3 - Y^2)$ are precisely those of the form $(X-a, Y-b)$ with $a,b \in \mathbb{C}$ such that $a^3-b^2=0$.