show/hide this revision's text 3 improved presentation

You can induct on $n$. Let $f:X\to\mathbb{P}^n$ be finite and étale.

If $H$ is a hyperplane in $\mathbb{P}^n$, there is a trivialization $\phi:f^{-1}(H)\simeq H\times F$($F$ , for a finite ) $F$, by the induction hypothesis.

If $L$ is any line in $\mathbb{P}^n$, $f^{-1}(L)$ is a finite disjoint union of $\mathbb{P}^1$'s, and you can label the components by elements of $F$ using the trivialization at the any point of $L\cap H$ (in case $L\subset H$, otherwise there is only one).

Now any fiber $f^{-1}(x)$ f^{-1}(x)$, $x\in \mathbb{P}^n$, is identified with $F$ through the labeling of the components of $f^{-1}(L)$, for any line $L$ through $x$ (this doesn't depend on the line , through $x$, their space being connected).

show/hide this revision's text 2 to index --> to label ; tentative grammar corrections

You can induct on $n$. Let $f:X\to\mathbb{P}^n$ be finite and étale. If $H$ is a hyperplane in $\mathbb{P}^n$, there is a trivialization $\phi:f^{-1}(H)\simeq H\times F$ ($F$ finite) by the induction hypothesis. If $L$ is any line in $\mathbb{P}^n$, $f^{-1}(L)$ is a finite disjoint union of $\mathbb{P}^1$'s, that and you can index label the components by elements of $F$ using the trivialization at the point $L\cap H$. Now any fiber $f^{-1}(x)$ is identified with $F$ through the indexing labeling of the components of $f^{-1}(L)$, $L$ for any line $L$ through $x$ (this doesn't depend on the line, their space being connected).

show/hide this revision's text 1

You can induct on $n$. Let $f:X\to\mathbb{P}^n$ be finite and étale. If $H$ is a hyperplane in $\mathbb{P}^n$, there is a trivialization $\phi:f^{-1}(H)\simeq H\times F$ by induction hypothesis. If $L$ is any line in $\mathbb{P}^n$, $f^{-1}(L)$ is a finite disjoint union of $\mathbb{P}^1$'s, that you can index by $F$ using the trivialization at the point $L\cap H$. Now any fiber $f^{-1}(x)$ is identified with $F$ through the indexing of the components of $f^{-1}(L)$, $L$ any line through $x$ (this doesn't depend on the line, their space being connected).