In fact I'm trying to prove that $\mathbb C[x,y]/(y^2-x^3+x)$ is a Dedekind domain. Till now I believe I was able to show that it is a Noetherian integral domain (easy) which is an integrally closed domain. If I prove that all prime ideals of $\mathbb C[x,y]/(y^2-x^3+x)$ are maximal, the job is done.
We know that prime ideals of $\mathbb C[x,y]/(y^2-x^3+x)$ are in bijective, inclusion-preserving correspondence with prime ideals of $\mathbb C[x,y]$ which contain $y^2-x^3+x$. If I'm not wrong, these may be in the form $(f, y^2-x^3+x)$, where $f$ is an irreducible polynomial in $\mathbb C[x]$ (so it should be in form $x-\alpha$, where $\alpha \in \mathbb C$ and $\alpha \neq 1, -1$, right?) and such that $y^2-x^3+x$ is irreducible mod $f$. But maximal ideals of $\mathbb C[x,y]$ are in the form $(x-a,y-b)$, where $a$ and $b$ are from $\mathbb C$. Can we somehow represent $(f, y^2-x^3+x)$ in such a way?
Correction that I realized later: we don't need to represent this ideal differently! It is indeed maximal since $(f)$, for $f$ irreducible in $\mathbb C[x]$, is a prime ideal of $\mathbb C[x,y]$, and then $(0) \subset (f) \subset (f, y^2-x^3+x))$, so $(f, y^2-x^3+x)$ has the height 2, then any other prime (maximal) ideal containing that ideal would have the height 3 – a contradiction! Krull dimension of $\mathbb C[x,y]$ is 2 and all its maximal ideals have height 2. Thus it is a maximal ideal and all prime ideals of $\mathbb C[x,y]/(y^2-x^3+x)$ are maximal.
Do you find any mistake? Thank you!
Best Answer
Write $R$ for $\Bbb C[x,y]/(y^2-x^3+x)$. As you say, maximal ideals of $R$ correspond to maximal ideals of $\Bbb C[x,y]$ which contain $(y^2-x^3+x)$. The maximal ideas of $\Bbb C[x,y]$ have the form $(x-a,y-b)$ and this contains $(y^2-x^3+x)$ iff $b^2-a^3+a$. The reason is that $(x-a,y-b)$ is the kernel of the homomorphism $\Bbb C[x,y]\to\Bbb C$ taking $x$ to $a$ and $y$ to $b$. So the maximal ideals of $R$ correspond to the "finite" points on the elliptic curve $y^2=x^3-x$.