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?