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Define:

$X_{1}:=\{(x,y,z) \in \mathbb{C}^{3}: x^{3}-y^{5}=0\}$

$X_{2}:=\{(x,y,z) \in \mathbb{C}^{3}: z=0,x=y^{2}\} \cup \{(x,y,z) \in \mathbb{C}^{3}: x=0,y=z^{2}\}$.

$X_{3}:=\{(x,y,z) \in \mathbb{C}^{3}: z=0,x=y^{2}\} \cup \{(x,y,z) \in \mathbb{C}^{3}: z=0,y=x^{2}\}$.

Question 1:

Prove or disprove: $X_{3} \cong X_{2}$ as varieties.

I think yes, can we simply let $Y=\{(x,y,z) \in \mathbb{C}^{3}: z=0,x=y^{2}\}$ and $W=\{(x,y,z) \in \mathbb{C}^{3}: x=0,y=z^{2}\}$ then map $X_{2} \rightarrow X_{3}$ as follows: if the point $(x,y,z) \in Y$ then let the map be the identity. Otherwise send $(x,y,z)$ to $(z,y,x)$. Is this OK? Any easier way?

Question 2:

Prove or disprove: $X_{3} \cong X_{1}$.

I think no. To argue we count the number of irreducible components. First note that that only irreducible component of $X_{1}$ is $X_{1}$ itself. Now I claim $X_{3}$ has two irreducible components, so it suffices to show that:

$\{(x,y,z) \in \mathbb{C}^{3}: z=0,x=y^{2}\}$ and $\{(x,y,z) \in \mathbb{C}^{3}: z=0,y=x^{2}\}$ are both irreducible (clearly they are closed).

Well the first one is equal to $\{(y^{2},y,0): y \in \mathbb{C}\} \cong \mathbb{C}$ so irreducible.

The second one is equal to $\{(x,x^{2},0): x \in \mathbb{C}\} \cong \mathbb{C}$ so irreducible.

Therefore $X_{3}$ has two irreducible components so no such isomorphism exists. Is this OK?

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1 Answer

up vote 3 down vote accepted

a) No, this is unfortunately not OK because your map $X_2\to X_3$ is not injective: it sends both $(1,1,0)$ and $(0,1,1)$ to $(1,1,0)$

The sad truth is that you can't repair this: the varieties $X_2$ and $X_3$ are not isomorphic because $X_2$ has one singular point, namely $(0,0,0)$, whereas $X_3$ has four and isomorphic varieties have the same number of singular points.
The four singular points of $X_3$ are $(0,0,0),(1,1,0), (j,j^2,0), (j^2,j,0)$ where $j=e^{\frac {2i\pi}{3}}$

b) What you write about the non-isomorphism of $X_1$ and $ X_3$ is correct but you have not justified that $X_1$ is irreducible.
For that you can either notice that $x^3-y^5$ is an irreducible polynomial or notice that $X_1$ is the image of $\mathbb C^1\to \mathbb C^3: t\mapsto (t^5,t^3)$

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thanks, a question: how did you find the singular points of $X_{3}$? since each piece of the union of $X_{3}$ is isomorphic to $\mathbb{C}$ it suffices to consider the intersection no? however I get that $x=1$ but also that $x$ can be a third root of unity. Can you please show how did you get $(1,1,0)$? I mean indeed showing that $X_{3}$ has more than one singular point is enough, however I want to find all singular points of $X_{3}$. –  user10 Jun 18 '12 at 18:22
    
also how to check indeed $(1,1,0)$ is singular we need to compute $I(C_{3})$ no and then use Jacobian? is there a shorter way? –  user10 Jun 18 '12 at 18:44
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Dear user, since the variety $X_3$ is the union of two smooth curves, its singular points are exactly given by the intersection points of the two curves. So we have to solve the system $ y=x^2, x=y^2, z=0$. There are 4 solutions and you are right that I forgot the two obtained via the primitive cubic roots of unity. I have corrected my answer : thanks and congratulations for your vigilance! –  Georges Elencwajg Jun 18 '12 at 20:07
    
thanks to you! is this a "standard" fact or is it obvious: "its singular points are exactly given by the intersection points of the two curves"? –  user10 Jun 18 '12 at 20:21
    
In general: the singular points of a curve are the singular points of its irreducible components and the points lying on at least two irreducible components. The singular points of the irreducible components can be found by the jacobian criterion. In the plane case, this boils down to the fact that if you multiply two polynomials without constant term, then the product has zero linear term. Have a look at Fulton's online Algebraic curves , Chapter 3, especially pages 32-33. –  Georges Elencwajg Jun 18 '12 at 20:54
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