Let $X:=(y^2-x^3)\subset \mathbb{C}^2$ be the vanishing of the polynomial $f(x,y)=y^2-x^3.$ I have proved an exercise in Hartshorne: If $\varphi:\mathbb{C} \to X, \ t \mapsto (t^2,t^3)$ is the parametrization for the cusp defined above, then $\varphi$ is a bijective morphism of affine algebraic varieties but is not an isomorphism. The fact that $\varphi$ is a bijective morphism can be easily checked. To show that it is not an isomorphism, I first consider a homomorphism $\varphi^*:A(X)\to A(\mathbb{C})$ induced at the level of coordinate rings of $X$ and $\mathbb{C}$. Since isomorphism of affine algebraic varieties is equivalent to isomorphism of coordinate rings, it's enough to show $\varphi^*$ is not surjective. We know $A(X)=\mathbb{C}[x,y]/I(X)$ where $I(X)$ is the radical ideal of $X$, and $A(\mathbb{C})=\mathbb{C}[t].$ Given the definition of $\varphi^*$, which is $x\mapsto t^2$ and $y\mapsto t^3,$ one observes that $t\in \mathbb{C}[t]$ does not lie in the image of $\varphi^*.$ This shows $\varphi$ is not an isomorphism.

I read a text somewhere that says that in fact there is no isomorphism between $X$ and $\mathbb{C}.$ I need some guidance to prove this.

At this stage, I know very little of algebraic geometry. So I tried to use the idea as above. If $X$ and $C$ were isomorphic, then their coordinate rings would also be. The goal now would be to arrive at a contradiction. I know that the maximal ideals of $A(X)$ and $A(\mathbb{C})$ correspond to points of $X$ and $\mathbb{C},$ respectively. But I don't know if this might lead me anywhere. If it does, how can I proceed? Thanks!

  • $\begingroup$ Show that if m is a maximal ideal in coordinate ring of the line, then m^2 has codimension 2 in the ring. Show then that this is not so in the cusp. It follows from this that the two rings are not isomorphic as C-algebras. $\endgroup$ May 26, 2016 at 22:35
  • $\begingroup$ @monomorphic Both answers are good. Is that hard for you to upvote them and accept one? $\endgroup$
    – user26857
    May 31, 2016 at 6:32

2 Answers 2


The coordinate ring of $\mathbb A^1$ is $k[x]$, which is integrally closed because it is a UFD.

The coordinate ring of the cusp is isomorphic to $k[t^2, t^3]$ which is not integrally closed (look at $t$).


The cusp prevents the given curve from being smooth. In dimension 1, smoothness or regularity is equivalent to normality. One can prove that the co-ordinate ring of the cusp is not normal, whereas the co-ordinate ring of a line clearly is.


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