3
$\begingroup$

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?

1 Answers 1

3

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 $\mathbb C^1\to \mathbb C^3: t\mapsto (t^5,t^3)$

  • 0
    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*](http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf) , Chapter 3, especially pages 32-33.2012-06-18