How to find PID?

Question

$\mathbb{R}[x,y]/(x^3-y^2)$ is a PlD? Where $\mathbb{R}$ is a real numbers. Now is there are any general formula to fund that $R[x,y]/(x^p-y^q)$ is PID or not PID? Where $\mathbb{R}$ is real numbers and $p$ and $q$ are prime numbers.

Answer 1

Consider the ring homomorphism $\phi$

$$\mathbb{R}[X,Y]\to \mathbb{R}[X]\\ \phi_{|\mathbb{R}}=\text{Id}\\ X\mapsto X^2\\ Y\mapsto X^3 $$

It is then easy to see that $\ker(\phi)=(X^3-Y^2);\ \phi(\mathbb{R}[X,Y])=\mathbb{R}_{\text{deg}\neq 1}[X] $.

By the first homomorphism theorem

$$\mathbb{R}[X,Y]/(X^3-Y^2)\simeq \mathbb{R}_{\text{deg}\neq 1}[X] $$

The latter is not a PID, since $(X^2,X^3+X^2)$ is not principal

This can easily be generalized to any prime numbers such that $p<q<2p$

Answer 2

Let's replace $\mathbb{R}$ by an arbitrary field $k$. Let $p$ and $q$ be two distinct primes with $q < p$. Observe that, \begin{equation} k[X,Y] / (X^p - Y^q) \simeq k[t^q, t^p]. \end{equation} Consider the ideal $I = (t^q, t^p) \subset R = k[t^q, t^p]$. Observe that $I$ is a subset of all polynomials $f(t) \in k[t]$ with every monomial of degree at least $q$ and $f(0) = 0$. Suppose if possible, $I$ is a principal ideal of $R$, say $I = (f(t^q, t^p))$. Then $f(0) = 0$ and $f$ divides $t^q$ in $R$. Thus we have $f = c t^q$ for some $c \in k \setminus \{0\}$.

Now $f = c t^q$ divides $t^p$ in $R$. Suppose $t^p = f g(t^q, t^p)$. Since $R \subset k(t)$, this gives us $c^{-1} t^{p-q} = g(t^p, t^q) \in R$. Since deg $g \geq q$ we note that we arrive at a contradiction whenever $p - q < q$.

Thus for any arbitrary field $k$ and distinct primes $p$ and $q$ satisfying $q < p < 2q$, we see that $R$ is not a PID.