Show that $R = \mathbb{C}[x,y]/(y^2-x^3-1)$ is not a PID.
My idea is to find an ideal of $\mathbb{C}[x,y]$ containing $(y^2-x^3-1)$ and show that its image is not principal.
So I have $J = (x,y+1)$, then suppose it is generated by $\alpha(x,y)$. Since we can reduce the degree of $y$ of $\alpha$, we can assume it is of the form $p(x)y+q(x)$. Then there exists polynomials $f(x,y),g(x,y),h(x,y),k(x,y)$ such that
$$x = \alpha(x,y) f(x,y) + h(x,y)(y^2-x^3-1)$$ $$y+1 = \alpha(x,y) g(x,y) + k(x,y)(y^2-x^3-1)$$
But then I am kind of stuck because I have no idea how to continue. I m not even sure if I picked the right ideal to work with
Suppose it were an UFD, then primes are irreducibles in that case. It's easier to see that $x$ and $y-1$ are irreducibles in that case and non units so that $y^2 -1=(y-1)(y+1)=x^3$, that is it wouldn't have a unique factorization. Thus $R$ is not an UFD which means that $R$ is not a PID.
Edit: As it has been said on the comments. Proving they are irreducibles is much more difficult than I thought and that this post makes it seem. Maybe there are other ways of doing this?