5

Consider the ideal in $\mathbb C[x,y]$ given by $ \mathfrak{a} = \langle y^2 - x^3 \rangle$. We can easily prove that $\mathfrak{a}$ is prime by considering the morphism of rings $$ \mathbb C[x,y] \to \mathbb C[t] $$ given by $x \mapsto t^2$ and $y \mapsto t^3$. Thus, $A := \mathbb C[x,y] / \mathfrak{a}$ is a domain. However, this domain is not integrally closed as $$ (y/x)^2 - x =0 $$ is an integral dependence realtion for $y/x$ and $y/x \notin A$. My first question is

Is there any relation (geometric) between the curve $y^2 = x^3$ and the fact that $y/x$ suffices an integral dependence realtion in $A$ but it is not in $A$?

Moreover, since $A$ is not integrally closed, we know one of its localization at maximal ideals (a point, since $\mathbb C$ is algebraically closed) is not integrally closed (since $A = \bigcap_{\mathfrak{m}} A_\mathfrak{m}$). By the graph of the curve we could guess it is the origin (when I saw de the graph I immediatly thought it was the origin, as it appears to be weird, what's the correct term for this weirdness?). Thus, I believe $$ \mathbb C[x,y]/\langle y^2 - x^3 \rangle_{(x,y)} \simeq \mathbb C[t^2,t^3 ]_{(t^2,t^3)} $$ is not integrally closed, but I was not able to simplify the expression. Then, my second question question comes

What's the relation between this singularity at the point, integrality and in spirit of the first question $y/x$?

I believe I lack the geometric (and algebraic) intuition to completely understand what is happening.

  • It has a cusp at the origin. – Eric Towers Mar 21 '18 at 02:44
  • 2
    Related – Viktor Vaughn Mar 21 '18 at 05:33
  • 1
    I think your same polynomial equation (y/x)^2 - x shows that the localization you asked about is not integrally closed. The only thing you have to mention is that (y/x) still isn't in that localization (unlike localizations at any other point you might have tried). I hope someone replies to you with a good geometric explanation for what's going on! – CJD Mar 21 '18 at 05:37
  • 1
    You should have a look at section 4.3 in Eisenbud's Commutative Algebra book for a geometric interpretation. About the cusp, it is indeed the point where the local ring is not integrally closed. The relation you're asking for is the one that says that regular rings are integrally closed. Therefore, the local ring can only fail to be closed if the point is singular (it has a weirdness, as you so well put it). In the one dimensional case, like yours, the relation is actually strict: any point where the local ring is integrally closed is also regular. – Artur Araujo Mar 23 '18 at 01:21
  • 1
    +1 to all the above. And the interesting thing for curves is that $y/x$ is always the missing element to look for (though sometimes it hides). What you are doing by introducing $y/x$ is constructing a patch of the blow-up, and enough iterations of this process will resolve the singular point (give you a smooth curve). – John Brevik Mar 23 '18 at 11:54

0 Answers0