1

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!

Alexey
  • 91
  • 1
    All non-zero prime ideals of this ring are maximal. – Angina Seng Jul 05 '20 at 19:10
  • It is not true that all prime ideals of $\mathbb C[x, y] / (y^2 - x^3 + x)$ are maximal. What is true is that all nonzero prime ideals of $\mathbb C[x, y] / (y^2 - x^3 + x)$ are maximal. – Dylan C. Beck Jul 05 '20 at 22:14

2 Answers2

1

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 $y^2=x^3-x$.

Angina Seng
  • 158,341
0

Consider the ring $R = \mathbb C[x, y] / (y^2 - x^3 + x).$ Like you have already established, $R$ is an integral domain. We claim that $\dim R = 1$ so that every nonzero prime ideal of $R$ is maximal. Considering that $S = \mathbb C[x, y]$ is a finitely generated $\mathbb C$-algebra, it follows that $\operatorname{height} I + \dim(S/I) = \dim S$ for every ideal $I$ of $S.$ Consequently, it suffices to show that $\operatorname{height}(y^2 - x^3 + x) = 1.$

By Krull's Height Theorem, we have that $\operatorname{height}(y^2 - x^3 + x) \leq 1.$ On the other hand, as $S$ is a domain, $0$ is the unique minimal prime ideal of $S,$ i.e., we have that $\operatorname{height}(y^2 - x^3 + x) \geq 1.$

Dylan C. Beck
  • 4,811
  • Great! Thank you for the hint, Carlo! – Alexey Jul 05 '20 at 20:58
  • Oh, no! I was again too fast on jubilation! Isn't the ideal that I proposed, $(f, y^2-x^3+x)$, where $f$ is an irreducible polynomial in $\mathbb C[x]$ and another one is irreducible mod $f$, isn't it maximal? Yes, indeed, because the max ideals of $\mathbb C[x,y]$ have dimension 2, so any ideal containing $(f, y^2-x^3+x)$ would have the dimension 3 - a contradiction. – Alexey Jul 05 '20 at 21:34
  • I fear that you misunderstand. Every one-dimensional domain has the property that every nonzero prime ideal is maximal. We claim that $R = \mathbb C[x, y]/(y^2 - x^3 + x)$ is a one-dimensional domain. Considering that you have already established that it is a domain, it suffices to show that it is one-dimensional. And it is because the height of the ideal $I = (y^2 - x^3 + x)$ is one. – Dylan C. Beck Jul 05 '20 at 21:57
  • Yes, exactly. You are correct. If there were a prime ideal containing $(f, y^2 - x^2 + x),$ then it would have height $3;$ but this is impossible because the height of an ideal in $\mathbb C[x, y]$ is at most $2.$ – Dylan C. Beck Jul 05 '20 at 22:13
  • Thank you, Carlo! – Alexey Jul 05 '20 at 22:14