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I'm having trouble determining if the algebraic sets $Z(x^2-y^3)\subset \mathbb{A}^2$ and $Z(y^2-x^3-x^2)\subset\mathbb{A}^2$ are isomorphic over $\mathbb{C}$. My guess is that this boils down to determining if $\mathbb{C}[x,y]/(x^2-y^3)$ is isomorphic to $\mathbb{C}[x,y]/(y^2-x^3-x^2)$ but then again, I'm stuck.

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Isomorphic as sets? as abelian groups? as rings? –  Gerry Myerson Oct 24 '12 at 3:05
    
Isomorphic as affine varieties. –  Victor Oct 24 '12 at 3:10
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This is a dublicate of my question math.stackexchange.com/questions/128918/…. There I also give an alternative proof. –  Martin Brandenburg Oct 24 '12 at 8:02
    
Thanks for pointing it out! –  Victor Oct 24 '12 at 8:08
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up vote 3 down vote accepted

Thinking geometrically, we expect these varieties are not isomorphic, due to the fact that the first is a cusp, while the second is a node. One way to verify this is to consider the tangent cone of each. In the first case, we get $TC_{(0,0)}=V(x^2)$ which is interpreted as the line $x=0$ with multiplicity $2.$ In the second case, we get $TC_{(0,0)}=V(y^2-x^2)=V((y-x)(y+x))$ which is two distinct lines.

Since the tangent cone is an invariant under isomorphism, we see that there is no point on the first variety to correspond with the origin in the second, and vice-versa.

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Is there a more elementary way to see this? (i.e. without using the tangent cone) –  Victor Oct 24 '12 at 3:33
    
Dear @Victor, actually, I think you could do essentially the same thing with the tangent space rather than tangent cone, but I actually found this to be easier to calculate! –  Andrew Oct 24 '12 at 3:47
    
@Victor, actually, you should ignore my previous remark; we cannot use the tangent space instead, since both defining polynomials have no linear terms. Thus both curves have tangent space $\cong\Bbb A^2,$ so the tangent cone is quite convenient here. –  Andrew Oct 24 '12 at 3:56
    
I've been trying to do this question too. I'm interested if there is a solution that is basic in the sense that one shows it by showing the rings are not isomorphic or something. –  Aleks Vlasev Oct 24 '12 at 5:08
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There are elementary proofs, but they are longer and quite silly compared to geometric arguments. This is really what algebraic geometry is about ... –  Martin Brandenburg Oct 24 '12 at 7:57
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