Use (2.5.3) to show that \(\PP^n\) is simply connected.
Solution
Here simply connected means that every finite étale cover is trivial.
1. It is enough to prove that every connected finite étale cover\[ f:Y\longrightarrow\PP^n \]is an isomorphism.
A finite étale cover has only finitely many connected components. The image of each nonempty connected component is open because an étale morphism is open, and closed because a finite morphism is closed. Since \(\PP^n\) is connected, each component maps surjectively to \(\PP^n\) and is itself a connected finite étale cover.
Thus, if every connected cover is an isomorphism, an arbitrary finite étale cover is a finite disjoint union of copies of \(\PP^n\), which is exactly a trivial cover.
2. The assertion holds for \(n=1\).
This is Hartshorne IV.2.5.3: \(\PP^1\) has no nontrivial connected finite étale cover. Equivalently, the Riemann--Hurwitz computation for $\PP^1$ shows that a connected finite étale cover of \(\PP^1\) has degree one and is an isomorphism.
3. Assume \(n\ge2\) and that \(\PP^{n-1}\) is simply connected. Let\[ f:Y\longrightarrow\PP^n \]be a connected finite étale cover. Then \(Y\) is a normal projectiveintegral variety of dimension \(n\).
Since \(f\) is finite and \(\PP^n\) is projective, \(Y\) is projective. Since \(f\) is étale and \(\PP^n\) is smooth over \(k\), the scheme \(Y\) is smooth, hence regular and normal.
A regular scheme has pairwise disjoint irreducible components: at an intersection point the local ring would have more than one minimal prime, whereas a regular local ring is a domain. Thus its irreducible components are open and closed. Because \(Y\) is connected, it has only one irreducible component. Hence \(Y\) is integral. Étale morphisms preserve dimension, so \(\dim Y=n\).
4. If\[ H\cong\PP^{n-1}\subseteq\PP^n \]is a hyperplane and\[ D=f^{-1}(H), \]then \(D\) is connected.
The hyperplane \(H\) is an effective Cartier divisor with \[ \mco_{\PP^n}(H)\cong\mco_{\PP^n}(1). \] Since \(f\) is flat, its pullback \[ D=f^*H \] is an effective Cartier divisor on \(Y\), and \[ \mco_Y(D) \cong f^*\mco_{\PP^n}(1). \] A finite pullback of an ample invertible sheaf is ample, so \(\mco_Y(D)\) is ample.
By step 3, \(Y\) is a normal projective variety of dimension at least two. Hartshorne III.7.9, the Enriques–Severi–Zariski connectedness theorem, therefore says that the support of the effective ample divisor \(D\) is connected. Thus \(D\) is connected.
5. The restriction\[ f|_D:D\longrightarrow H\cong\PP^{n-1} \]is an isomorphism.
The square \[ \begin{CD} D @>>> Y\\ @V{f|_D}VV @VV{f}V\\ H @>>> \PP^n \end{CD} \] is cartesian. Hence \(f|_D\) is the base change of the finite étale morphism \(f\), so it is finite étale. Step 4 says that \(D\) is connected. By the induction hypothesis, every connected finite étale cover of \(\PP^{n-1}\) is an isomorphism. Therefore \(f|_D\) is an isomorphism.
6. The morphism \(f\) has degree one and is an isomorphism.
A finite étale morphism is finite locally free of constant rank on the connected target \(\PP^n\). This rank is its degree and is preserved by base change. Hence \[ \deg f = \deg(f|_D) = 1 \] by step 5.
Since \(Y\) and \(\PP^n\) are integral, \(f\) induces an extension of function fields of degree one, so it is birational. It is also finite, while \(\PP^n\) is normal. A finite birational morphism onto a normal integral scheme is an isomorphism. Thus \(f\) is an isomorphism.
7. For every \(n\ge1\), \(\PP^n\) is simply connected.
Step 2 is the base case. Steps <1>3–<1>6 give the induction step from \(n-1\) to \(n\). Step 1 then upgrades the connected-cover statement to all finite étale covers.
8. Q.E.D.
Step 7 is exactly the required conclusion.