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Let $V = Z(X_0^8 + X_1^8 + X_2^8) $ and $W = Z(X_0^4 + X_1^4 + X_2^4)$ in $\mathbb P^2$. It can be shown that $V$ and $W$ are irreducible curves. Let $\phi : V \to W, (X_i) \mapsto (X_i^2)$ be a morphism.

I want to show that $V$ and $W$ are smooth, compute $\mathrm{deg}(\phi)$ and find $e_P(\phi)$ for all $P \in V$. I'm getting in a mess though, and would appreciate some feedback on the following:

First consider $V$ and work in the affine chart $\{X_0 \neq 0 \}$ i.e $V_0 = Z(1 + x^8 + y^8)$. Let $h = x^8 - (y^8 + 1)$. We see that $\frac{\partial h}{\partial x} = 8x^7$ and $\frac{\partial h}{\partial y} = 8y^7$. If $(x_0, y_0)$ is a point in $V_0$, and $x_0 \neq 0$ then $y - y_0$ is a local parameter. If $y_0 \neq 0$ then $x - x_0$ is a local parameter. $(0,0)$ is not on the curve so all points in $V_0$ are smooth.

Now work in the affine chart $\{X_1 \neq 0 \}$, i.e. $V_1 = Z(w^8 + 1 + y^8)$. It's easy to see that the exact same thing happens here: if $(w_0, y_0)$ is a point then $y - y_0$ is a local parameter if $w_0 \neq 0$ and $w - w_0$ is a local parameter if $y_0 \neq 0$. The curve is smooth at every point in $V_1$.

The same happens in $V_2$, and also in $W_0, W_1$ and $W_2$.

So suppose $p = (x_0, y_0) \in V_0$ and $\phi : V_0 \to W_0$, $(x_0,y_0) \mapsto (x_0^2, y_0^2)$. If $x_0 \neq 0$, then $x_0^2 \neq 0$ and so $t = y - y_0^2$ is a local parameter at $\phi(p)$. Then $t \phi : (x, y) \mapsto y^2 - y_0^2$. We want to find $\nu_p(t \phi)$, noting that a local parameter of $p$ is $y - y_0$. So $e_p(\phi) = \nu_p(t\phi) = \nu_p((y+y_0)(y - y_0)) = 1$. If instead $y_0 \neq 0$, then $t = x - x_0^2$ is a local parameter at $\phi(p)$ and again we get $e_p(\phi) = 1$.

Is it valid to then say that by symmetry $e_p(\phi) = 1$ for all $p \in V$? I can't see why not, but this doesn't seem to give the correct result, since (by the Finiteness Theorem) it would imply that all the sizes of the pre-images are the same, which doesn't look to be true.

Any help would be appreciated. Thanks

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up vote 0 down vote accepted

In your local computation, if $y_0=0$, then $\nu_p(t\phi)=\nu_p(y^2)=2$, so $e_p(\phi)=2$.

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