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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!

user237522
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    It's probably slightly more useful to talk about $\mathbb{C}[X,Y,t]/(X - f(t), Y - g(t))$, this allows you to use algebraic geometry to conclude that the maximal ideals correspond to points in $\mathbb{C}^3$ where $X - f(Z) = Y - g(Z) = 0$ and prime ideals correspond to intersections with other solution spaces of polynomials $F(X,Y,Z) = 0$. Usually all prime ideals would be maximal, but I think the case you're interested in might be an exception. – Contravariant May 03 '18 at 01:34
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    May I suggest notation other than $R_{f,g}$, since $R_f$ normally denotes the localization of $R$ at $f$? In your specific example, note that $(R_{t^2,t^3})_{t^2}=\mathbb C[t,t^{-1}]$. Splitting into cases depending on whether the prime ideal contains $t^2$, we find they are in bijection with prime ideals of $\mathbb C[t]$. – stewbasic May 03 '18 at 01:40
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    I really like both comments, and think that they are helpful. – user237522 May 03 '18 at 01:47
  • @stewbasic: I agree that my notation is not good; I will change it. Could you please elaborate your idea, perhaps in an answer. – user237522 May 03 '18 at 01:55
  • @Contravariant: Please, could you explain 'usually prime ideals would be maximal etc." – user237522 May 03 '18 at 01:58
  • @user237522 that statement was just about $\mathbb{C}[X,Y,t](X - f(t), Y - g(t))$. The geometric intuition is that the space of points $(X,Y,t)$ where $X-f(t)=Y-g(t)=0$ has a dimension of 1. At first I had some doubts but now I actually think in this particular case that will actually be true for all $f$ and $g$. – Contravariant May 03 '18 at 10:59
  • Thank you for the explanation. Please, do you think that it is possible to somehow characterize (in terms of the coefficients and degrees) all $f$ and $g$ such that $\mathbb{C}[f,g]$ is integrally closed in its field of fractions? with further assuming that $\mathbb{C}(f,g)=\mathbb{C}(t)$ in order to exclude cases such as $f=t^4,g=t^2$. (Perhaps I will ask this in a separate question). – user237522 May 03 '18 at 15:48

1 Answers1

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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:

  • Any prime ideal of $\mathbb{C}[X,Y] / I$ is maximal
  • The maximal ideals of $\mathbb{C}[X,Y] / I$, if they exist, are of the form $(X-a, Y-b)$ where $f(a,b) = 0$ for all $f \in I$.

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 $a^3-b^2=0$.

Contravariant
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