# The curves $y^2=x^{2k+1}$ are never isomorphic to each other

For every $k \in \Bbb N$, we consider the affine variety $X_k=\Bbb V(y^2-x^{2k+1}) \subset \Bbb A^2$. Gathmann's note contain the following exercise:

Show that the curves $X_k$ are pairwise nonisomorphic.

Equivalently, We need to show that the rings $$\frac{\Bbb C[x,y]}{(y^2-x^{2k+1})}$$ are pairwise nonisomorphic.

The fact that $X_1, X_0$ are not isomorphic is well -known: this is just saying that the cuspidal curve is not isomorphic to $\Bbb A^1$, since one is smooth and the other is not.

It is suggested as a hint to look at the blow-up of $X_k$ at the origin, and it surely suffices to show that these blow-ups $\tilde X_k \subset \Bbb A^2 \times \Bbb P^1$ are pairwise nonisomorphic. I'm also interested in knowing whether this is the "standard" proof, or there are other methods that work. (like the ring-isomorphism approach above.)

If we denote the coordinates of $\tilde {\Bbb A^2}$ by $((x_1,x_2),(y_1:y_2))$ then Gathmann shows that the blow-up of $X_k$ is given in $U_1=\{y_1 \neq 0\}$ by the equation $y_2^2-x_1^{2k-1}=0$. Is it enough to show that these curves are pairwise nonisomorphic in $\Bbb A^2$?

The ring isomorphism is also easy to do (blowing up is also fine). If you have an isomorphism $f:A_k\to A_l$, where $A_n=\mathbb{C}[x,y]/(y^2-x^{2n+1})$, then it induces an isomorphism of their normalizations, which is just $\mathbb{C}[t]=R$, where the map from $A_n\to R$ is, sending $x\mapsto t^2, y\mapsto t^{2n+1}$. Now, $f$ induces an automorphism of $R$, which are just linear maps, $t\mapsto at+b$, with $a\neq 0$. Of course, we may assume $k\leq l$ and both at least 1. Then following $t^2$ coming from $A_k$, which should land inside the image of $A_l$, one checks that $b=0$. Now, it is elementary to check that $k=l$, by following where $t^{2k+1}$ goes.