Non-isomorphic curves with the same genus I need to show that the curves $y^7=x^2(x-1)$ and $xy^3+zx^3+yz^3=0$ are not isomorphic even if the are both of genus 3. I have tried to show that they are a quartic and a hyperelliptic curve in order to show that they are in two different class of isomorphism.
I have normalized the first one with this sequence of blow-ups and transformations:
$x=x_1y_1, y=y_1     \rightarrow x_1^3y_1-y_1^5-x_1^2=0$
$x_1=x_2y_2, y_1=y_2 \rightarrow x_2^3y_2^2-y_2^3-x_2^2=0$
$x_2=x_3y_3,y_3=y_2  \rightarrow x_3^3y_3^3-y_3-x_3^2=0$
$x_3=\frac{X_3}{Z_3},y_3=\frac{Y_3}{Z_3} \rightarrow X_3^3Y_3^3-Y_3Z_3^5-X_3^2Z_3^4=0$
$y_3'=\frac{Y_3}{X_3}, z'_3=\frac{Z_3}{X_3} \rightarrow (y'_3)^3-(y'_3)(z'_3)^5-(z'_3)^4=0$
$y_3'=y_4z_4, z'_3=z_4\rightarrow y_4^3-y_4z_4^3-z_4=0$
The last one is nonsingular but is almost exactly the other, so what is wrong?
Thanks!
 A: The curve
$$
xy^3 + yz^3 + zx^3 = 0
$$
is the famous Klein quartic, see https://en.wikipedia.org/wiki/Klein_quartic. Its automorphism group has order 168, the maximal among curves of genus 3. In particular, it has an automorphism of order 7.
On the other hand, Theorem 2(a) of [Homma, Masaaki. Automorphisms of prime order of curves. Manuscripta Math. 33 (1980/81), no. 1, 99--109] shows that a curve $C$ of genus $g \ge 2$ with an automorphism $\sigma$ of prime order $2g + 1$ is birational to one of the curves
$$
y^{2g+1} = x^a(x-1)^b,
$$
where $a,b \ge 1$ and $a + b \le g + 1$. Moreover, the proof of Proposition 3.1 in the paper shows that $a + b$ is the degree of a covering $f \colon C \to \mathbb{P}^1$ which is equivariant for $\sigma$.
All this applies easily to the Klein quartic. One can choose the morphism $f$ to be the projection
$$
(x,y,z) \mapsto (x:y),
$$
then its degree is 3, hence $a + b = 3$, hence (after a possible change of coordinates) the Klein quartic is birational to the curve
$$
y^7 = x^2(x-1).
$$
