Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In his chapter about Hurwitz' theorem for curves, Hartshorne shows that $\mathbb{P}^1$ is simply connected, i.e. every finite étale morphism $X \to \mathbb{P}^1$ is a finite disjoint union of $\mathbb{P}^1$s. In an exercise the reader is invited to show that $\mathbb{P}^n$ is simply connected, using the result for $\mathbb{P}^1$.

I have no idea how to do this. Perhaps someone can give a hint? There are closed immersions $\mathbb{P}^1 \to \mathbb{P}^n$, along which we may pull back a finite étale morphism, but the trivializations don't have to coindice ... perhaps we can resolve this using cohomology theory? I'm a bit confused since $\mathbb{P}^n$ is $n$-dimensional, but this is in Hartshorne's chapter about curves. I don't want to use the more advanced material of SGA.

share|improve this question
This comment is a little late, since the question has been answered, but let me just say that the statement "There are n obvious closed immersions P^1→P^n, along which we may pull back a finite étale morphism, but they do not cover P^n" seems a bit strange to me. (There are a lot more than n ways to embed P^1 in P^n, and they all seem to me to be equally obvious.) Am I missing something? –  Artie Prendergast-Smith Apr 20 '11 at 7:16
add comment

7 Answers

up vote 11 down vote accepted

Here is a sketch of an argument which directly uses simple connectedness of $\mathbb P^1$, and is related to the simple connectedness of rationally connected smooth varieties mentioned by Sandor in one of his answers.

The idea is to treat the $\mathbb P^1$s in $\mathbb P^n$ as analogous to arcs in a topological space, and to make a lifting argument (just as one does in the basic topological theory of covering spaces).

Let $Y \to \mathbb P^n$ be a finite etale map. Fix a base points $x \in \mathbb P^n$ and a point $y \in Y$ lying over $X$. If $x' \in \mathbb P^n \setminus \{x\}$, there is a unique line $L$ joining $x$ and $x'$. The preimage of $L$ is a disjoint union of curves $L'$, each mapping isomorphically to $L$ (by simple connectedness of $\mathbb P^1$), and we can choose a unique $L'$ containing $y$. Now let $y'$ be the point of $L'$ lying over $x'$.

The map $x' \mapsto y'$ (and of course mapping our original point $x$ to $y$) gives a section to the given map $Y\to \mathbb P^n$, which is what we wanted.

Added: Here is one explanation of why the map $x' \mapsto y'$ is algebraic. Let $\pi:Y \to \mathbb P^n$ be our given etale map. First note that $x' \mapsto \pi^{-1}(L)$ (where $L$ is the line joining $x$ and $x'$, as above) is a morphism from $\mathbb P^n \setminus \{x\}$ to the Hilbert scheme of $Y$. Now picking out the connected component $L'$ of $\pi^{-1}(x')$ containing $y$ is a morphism from our given locally closed subset of the Hilbert scheme to the Hilbert scheme, and so altogether we see that $x' \mapsto L'$ is a morphism. Finally, mapping $L'$ to $x'$ (which can be described as forming the intersection $L' \cap \pi^{-1}(x')$) is again a morphism. So altogether we have a section $\mathbb P^n \setminus\{x\} \to Y$. One way to show that this extends as a section over all of $\mathbb P^n$ (by sending $x$ to $y$) is just to repeat the whole process for a different choice of $x$, and glue the two resulting sections.

share|improve this answer
This seems to be the most elementary approach, but still there are some details I don't understand. Namely, $x' \mapsto y'$ is first defined only as a set-theoretical map $\mathbb{P}^n \backslash x \to X \backslash x$. Why is it a morphism? And why can we extend it on $\mathbb{P}^n$? Why does it suffice to find a section? –  Martin Brandenburg Apr 21 '11 at 8:22
@Martin : as in Sandor's answer, $Y$ must be silently assumed connected, so that a section is enough. Also, I have edited my answer, which seems to be a more cumbersome version of the same idea, but seemingly not needing extension. Does it seem more airtight to you ? I surmised that such simple geometric reasoning cannot let you out of algebraic geometry over anything, although I never felt so much assured with coverings in positive characteritic. –  BS. Apr 21 '11 at 15:47
There is no need to assume that $Y$ is connected (although of course it is harmless to do so): any section of a finite etale morphism over a connected base induces an isomorphism between the base and a connected component of the cover. Thus the statement that a connected scheme is simply connected is equivalent to the statement that any finite etale morphism admits a section. –  Emerton Apr 21 '11 at 16:17
Also, the map extends to $\mathbb P^n$ just by sending $x$ to $y$. –  Emerton Apr 21 '11 at 16:20
"any section of a finite etale morphism over a connected base induces an isomorphism between the base and a connected component of the cover." Why? Also, why is your set-map (yes I knew that we map $x$ to $y$) a morphism? –  Martin Brandenburg Apr 21 '11 at 16:34
show 4 more comments

There is somewhere a theorem in Hartshorne's book saying that an ample divisor on a normal projective connected scheme of dimension at least 2 is connected. Now proceed by induction on $n$. If there is a non-trivial étale cover $X \to \mathbb P^n$, consider the inverse image of a hyperplane $\mathbb P^{n-1}$; this is connected, since the pullback of an ample divisor by a finite map is ample, and this give the required inductive step.

share|improve this answer
The theorem is III.7.9 in Hartshorne. –  Dave Anderson Apr 19 '11 at 18:53
add comment

(From SGA I, Exposé XI).

You can prove it using the following two facts:

1) A product of simply connected proper varieties is simply connected (SGA I, X, 1.7). (*)

2) The fundamental group -- so in particular, being simply connected -- is a birational invariant of proper regular varieties (SGA I, X, 3.4).

(*) I do not know whether the properness is necessary here; it is required for the more general computation of the fundamental group of a product: in positive characteristic, one of the factors needs to be proper.

share|improve this answer
@ACL: I did some minor copyediting on your answer; I hope you don't mind. (I was having trouble with the sentence in 2), which I read as saying that the fundamental group itself was simply connected...) –  Pete L. Clark Apr 20 '11 at 14:11
@Pete. Thanks a lot! –  ACL Apr 21 '11 at 6:48
Nice! so 1) gives us that n-fold product of $P^1$'s is simply connected and since product variety is birational to $P^n$, we get the required result by 2). is this correct? –  SGP Apr 24 '11 at 17:12
add comment

We may assume that $n\geq 2$. Let $f:X\to \mathbb P^n$ be a finite étale morphism where $X$ is connected and $H\subset \mathbb P^n$ a hyperplane. Then $f^*H$ is an ample divisor on $X$ and hence connected. By induction, then the restriction $f^*H\to H$ is an isomorphism, so $\deg f=1$ and $f$ is an isomorphism.

EDIT added previously silently assumed assumption that $X$ is connected.

share|improve this answer
@Sandor: You cannot conclude that $f$ is an isomorphism, just a finite disjoint union of isomorphisms.. –  Martin Brandenburg Apr 19 '11 at 20:00
@Martin: sorry, I meant to say that you can assume at the start that $X$ is connected. –  Sándor Kovács Apr 19 '11 at 21:21
How can we show that $f^* H$ is ample? And how the degree of $f$ is related with the degree of the restriction to $H$? –  Martin Brandenburg Apr 22 '11 at 20:11
@Martin: The pull back of an ample divisor via a finite map is ample. This is an exercise in Hartshorne and can be proved using pretty much any characterization of ampleness. As for the degree: $f$ is unramified, so the degree is equal to the number of preimages of any (closed) point. This remains the same for any subvariety. –  Sándor Kovács Apr 22 '11 at 22:56
add comment

I meant to add that there are other interesting ways to think about this issue. These do not conform to the request of a simple proof, but seem relevant to mention.


Every rationally connected smooth variety is simply connected (at least over $\mathbb C$) this is a result of Kollár-Miyaoka-Mori and Campana independently.


Hartshorne's conjecture, proved by Mori says that $\mathbb P^n$ is the only smooth projective variety whose tangent bundle is ample. This allows for a simple proof that $\mathbb P^n$ is simply connected: Let $f:X\to \mathbb P^n$ be a finite étale morphism and assume that $X$ is connected. Then clearly $X$ is smooth and projective and furthermore it follows that $\Omega_X\simeq f^*\Omega_{\mathbb P^n}$ and hence the tangent bundle of $X$ is also ample. By Mori's theorem it is then isomorphic to $\mathbb P^n$. However, $\mathbb P^n$ does not admit unramified self-maps of degree $d>1$ (because the induced map on the Picard group would be multiplication by $d$ and then it would imply that $\deg K_{\mathbb P^n}=0$), so $f$ has to be an isomorphism.

share|improve this answer
What does the "at least over $\mathbb C$ mean in the first point? It is true over $\mathbb C$, but what does happen over other fields? I hate "at least"! :P –  Mariano Suárez-Alvarez Apr 24 '11 at 17:23
add comment

Let me give another answer, even though it does not fit into Hartshorne's context:

Show that $\pi_1(\mathbb{P}^n)$ has to be abelian.

Use Kummer-Theory to relate coverings to torsion in $Pic (\mathbb{P}^n)=\mathbb{Z}$, see e.g. Milne's Etale Cohomology, Prop 4.11. This implies that there are no nontrivial étale coverings of degree prime to the base characteristic.

Then use Artin-Schreier theory to relate the rest of the coverings to $\Gamma(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n})/(F-1)\Gamma(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n})=0$, and $H^1(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n})^F=0$, where $F$ is the Frobenius, see e.g. Milne's, Prop 4.12.

share|improve this answer
There is one thing which has always confused me with this arugment. Since H^1(X,Z) is essentially the abelianisation of the fundamental group, how are you ruling out the fact that the fundamental group might have trivial abelianisation i.e. is a perfect group? –  Daniel Loughran Apr 20 '11 at 15:39
Oops, you are of course perfectly right. In this particular case we can be saved though, if I am not mistaken: G_m^n is an open subscheme of $\mathbb{P}^n$,so $\pi_1(\mathbb{P}^n)$ is a quotient of the abelian group $\pi_1(G_m^n)$. –  Lars Apr 20 '11 at 17:36
@Lars: so this works only in char. zero, otherwise $\pi_1(\mathbb{G}_m)$ is not abelian. –  Laurent Moret-Bailly Apr 21 '11 at 6:40
Again oops, what was I thinking. I guess I don't know of a "easy" proof that $\pi_1(\mathbb{P}^n)$ is abelian then (as $\pi_1(\mathbb{A}_k^1)^{(p)}$ is free pro-p on $\#k$ generators). –  Lars Apr 21 '11 at 10:04
add comment

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 $\phi:f^{-1}(H)\simeq H\times F$, 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 any point of $L\cap H$ (in case $L\subset H$, otherwise there is only one).

Now any fiber $f^{-1}(x)$, $x\in \mathbb{P}^n$, 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).

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.