# Prove that $K[x,y] / (y^2+yx^2-x(x-1))$ have infinitely many maximal ideals.

I have the following problem.

$$K$$ is an algebraically closed field. Prove that $$R=K[x,y]/(y^2+yx^2-x(x-1))$$ is an integral domain, has infinitely many maximal ideals, and decides if $$R$$ is a PID.

I have a pair of hours trying this and I have some ideas, which I will describe.

1. The first part is strainfoward. Since $$K[x,y]\cong (K[x])[y]$$ and $$K[x]$$ is a integer domain, then is sufficient show that $$J=(y^2+yx^2-x(x-1))$$ is a prime ideal, and also since J is principal and $$K[x,y]$$ is a UFD then is sufficient show that $$s(x,y)=y^2+yx^2-x(x-1)$$ is irreducible in $$(K[x])[y]$$ which is also easy via Einsentein criterion choosing the prime ideal $$P=(x)$$ in $$K[x]$$ and considering $$a_2=1$$, $$a_1=x^2$$ and $$a_0=x(x-1)$$ and writing $$s(y)=a_2y^2+a_1y-a_0\in (K[x])[y]$$.

2. In this point I think that we should use the correspondence theorem which says that the ideals of $$R$$ must be of the form $$I/(y^2+yx^2-x(x-1))$$ where $$(y^2+yx^2-x(x-1))\subset I \subset k[x,y]$$. Here I have some problems thinking about how should look maximal ideals in R. Also I have another idea that probably works, this is to use the fact that there is a one-to-one correspondence between maximal ideals in $$K[x_1,x_2,\ldots, x_n]$$ and points in $$K^n$$.

3. At this point my intuition says that if R has a lot of maximal ideals then $$R$$ should not be a PID.

¡Any suggestions and hints are welcome!

• For 3: $\mathbb{Z}$ has a lot of maximal ideals and is a PID. Commented Jun 8, 2023 at 23:48
• Thanks a lot for the counterexample. Commented Jun 8, 2023 at 23:56
• Do you know the Nullstellensatz? One consequence is that maximal ideals of your ring $R$ are in bijection with solutions $(x,y) \in K^2$ to the equation $y^2+yx^2-x(x-1) = 0$. Commented Jun 9, 2023 at 1:48
• Adding to @ViktorVaughn's comment, the equation has infinitely many solutions since for each value of $x$, the quadratic equation has at least one solution (since $K$ is algebraically closed). Commented Jun 9, 2023 at 2:25
• If you homogenise this plane affine cubic curve, you will get a non singular plane projective cubic $C$ (you can check with Sage that it is non singular) of genus one, so an elliptic curve. It has two points at infinity, so the class group of the affine curve will be a quotient of ${\bf Z}\oplus {\rm Jac}(C)(K)$ by a subgroup with two generators. This cannot be the trivial group (because ${\rm Jac}(C)(K)$ is not finitely generated) so this is not a PID. Here ${\rm Jac}(C)(K)$ is $C$ with its group law, once you have chosen a point $0$. Commented Jun 9, 2023 at 9:45

Your proof that $$R$$ is an integral domain looks great!
Let $$f(x,y)=y^2+yx^2-x(x-1)$$. To show that $$R$$ has infinitely many maximal ideals, it suffices to consider the maximal ideals $$I$$ in $$k[x,y]$$ such that $$(f)\subseteq I$$.
Applying the Nullstellensatz, write $$I=(x-a,y-b)$$, and let $$\operatorname{ev}_{(a,b)}:K[x,y]\longrightarrow K$$ be the homomorphism given by $$f(x,y)\mapsto f(a,b)$$. Suppose $$f(a,b)=0$$. Then $$\operatorname{ker}\operatorname{ev}_{(a,b)}$$ (an ideal) contains both $$(f)$$ and the maximal ideal $$I=(x-a,y-b)$$. Maximality then forces $$\operatorname{ker}\operatorname{ev}_{(a,b)}=I$$, and thus $$I$$ contains $$(f)$$.
Conversely, if $$I=(x-a,y-b)$$ is a maximal ideal containing $$(f)$$, we may express $$f\in I$$ as a combination of both $$x-a$$ and $$y-b$$; clearly then $$f$$ vanishes on $$(a,b)$$.
Since $$f$$ has infinitely many zeroes, there are infinitely many maximal ideals in $$K[x,y]$$ containing $$(f)$$.