15
$\begingroup$

I'm trying to get a better understanding of étale fundamental groups, and I think that the overall idea -- the big picture -- is beginning to become clear, but my computational ability seems to be essentially nonexistent.

I think that I understand some of the simpler examples (like $\pi(\text{Spec }K)$, when $K$ is a field), but could someone please explain a slightly more complicated (at least in my view) example? Perhaps (to move away from examples involving fields) something like $\pi(\text{Spec }\mathbb{Z}[X])$ or $\pi(\text{Spec }\mathbb{Z}[1/p])$?

$\endgroup$
9
$\begingroup$

I believe that $\pi(\text{Spec } \mathbf Z[1/p])$ is the Galois group of the maximal extension of $\mathbf Q$ unramified outside $p$.

On the other hand, I believe that $\pi(\text{Spec }\mathbf Z[X])$ is trivial. If $Y$ were finite étale over $\mathbf Z[X]$, then for each $n$, the base-change $Y_n$ of $Y$ under the map $$\mathbf Z[X] \to \mathbf Z : X \mapsto n$$ would be finite étale over $\mathbf Z$, hence trivial (as you surely know that $\pi(\mathbf Z) = 0$). It seems that the coefficients of a polynomial defining $Y$ would have to vanish at every integer...

It's no reason that the étale fundamental group is hard or even impossible to calculate. Absolute Galois groups, the simplest case, are very difficult to calculate! (What does it even mean to "calculate" the absolute Galois group of $\mathbf Q$? I don't know!)

Another nice example comes from elliptic curves. If $E/\mathbf Q$ is an elliptic curve, then $\pi(E) = TE,$ the 'global' Tate module of $E$ : $$TE = \varprojlim_n E[n],$$ where the integers are ordered by divisibility ($E[n]$ denotes the $n$-torsion of $E$ over $\overline{\mathbf Q}$). It is a free $\widehat{\mathbf Z}$-module of rank $2$ (note that it's abelian, a rare feat for a fundamental group!). This is the "$\text{GL}_2$" analogue of the fact (which you probably also know) that $\pi(\mathbf G_m) = \widehat{\mathbf Z}$, where $\mathbf G_m = \text{Spec }\mathbf Q[X, X^{-1}]$ is the multiplicative group over $\mathbf Q$. The proof goes as follows: if $f: E' \to E$ is a finite étale covering of degree $n$, then $E'$ must have genus $1$ by Riemann-Hurwiz. By possibly extending the base field, it acquires a point, in such a way that $f$ becomes an isogeny. By pre-composing $f$ with the dual isogeny, we get the map $E \to E' \to E$ which is just multiplication by $n$ on $E$ (which is defined over $\mathbf Q$!). Thus the multiplication-by-$n$ maps form a cofinal system in the category of finite étale coverings, and the result follows.

Also, it is no coincidence that the fundamental group (in the usual sense) of a torus is $\mathbf Z^2$!

$\endgroup$
  • $\begingroup$ This is very nice. I have a couple elementary questions: is the p-adic Tate module of $E$ also an étale fundamental group, perhaps of $E\times \operatorname{Spec}\mathbb{Q}_p$? And is there a short answer to the question of where $GL_2$ comes in? $\endgroup$ – Slade Oct 18 '13 at 5:19
  • $\begingroup$ @user33433 Thanks! The étale fundamental group of the base-change of $E$ to $\mathbf Q_p$ will also be $\widehat{\mathbf Z}^2$; the exact same argument carries over (it is true over any field of char. 0). I don't think that $T_pE$ can be immediately realized as an étale fundamental group, though I may be wrong. It is rather like the $p$-Sylow subgroup. As for "$GL_2$", it comes from the action of the absolute Galois group $G_{\mathbf Q}$ on the Tate module: after choosing a basis for the Tate module, we get a representation $G_{\mathbf Q} \to GL_2({\widehat{\mathbf Z}})$. $\endgroup$ – Bruno Joyal Oct 18 '13 at 5:29
  • $\begingroup$ I second user33433: this is very nice. Your argument regarding $\pi(\text{Spec }\mathbb{Z})[X]$ looks right to me. I'm also willing to buy what you said about $\pi(\text{Spec }\mathbb{Z}[1/p])$, but I'll have to spend some time working out the details on that one. :D $\endgroup$ – user101616 Oct 23 '13 at 16:33
  • $\begingroup$ @BrunoJoyal: I've been working my way through Lenstra's Galois Theory for Schemes, and so some of these examples have come up again. As such, I was going back through what you said here, but I have a question: what did you mean by "a polynomial defining $Y$"? I'm not terribly well-verses in scheme theory, so I hoping this is just a simple definition… but it's not clear to me. Thanks! $\endgroup$ – user101616 Dec 28 '13 at 10:03
  • $\begingroup$ @BrunoJoyal I believe $T_pE$ is the maximal pro-p quotient of $\pi_1(E)$, as opposed the to maximal p-power subgroup. $\endgroup$ – Jesse Mar 25 '14 at 2:59
5
$\begingroup$

Lenstra's introduction to Galois theory for schemes is very gentle and discusses some basic examples:

Proposition (Corollary 6.17): If $X$ is a normal integral scheme with function field $K$, and $M$ is the composite of all finite separable extensions $L$ of $K$ in some fixed algebraic closure $\overline{K}$ such that $X$ is unramified in $L$, then $\pi_1(X)$ is isomorphic to the Galois group $\mathrm{Gal}(M/K)$.

In particular (Exercise 6.27): If $X$ is a smooth curve with function field $K$, then $\pi_1(X)$ is the quotient $\mathrm{Gal}(K^{sep}/K) / N$, where $N$ is the closed subgroup generated by all inertia subgroups $I_x$ with $x \in X$ closed.

Examples (6.18, 6.22, 6.23, 6.24)

  • If $K$ is a number field and $A=\mathcal{O}_K$, $a \in A$, $X=\mathrm{Spec}(A[1/a])$, then $M/K$ is the largest Galois extension which is umramified at all non-zero primes not dividing $a$.
  • $\pi_1(\mathrm{Spec}(\mathbb{Z}))=\{1\}$ (since Minkowski's Theorem implies that every proper algebraic extension of $\mathbb{Q}$ ramifies at some prime number).
  • $\pi_1(\mathrm{Spec}(\mathbb{Z}_p)) = \widehat{\mathbb{Z}}$
  • If $K$ is a field, then $\pi_1(\mathbb{P}^1_K)=\pi_1(\mathrm{Spec}(K))$
  • If $K$ is a field of characteristic $0$, then $\pi_1(\mathbb{A}^1_K)=\pi_1(\mathrm{Spec}(K))$
  • If $A$ is a finite ring, then $\pi_1(\mathrm{Spec}(A)) \cong \prod_{\mathfrak{m} \in \mathrm{Spec}(A)} \hat{\mathbb{Z}}$.

Exercises (6.31 $-$ 6.40)

  • $\pi_1(\mathbb{A}^1_{\mathbb{Z}})$ and $\pi_1(\mathbb{P}^1_{\mathbb{Z}})$ are trivial.
  • The ring homomorphism $\mathbb{Z}_p \to \mathbb{Z}/p^n$ induces an isomorphism $\pi_1(\mathrm{Spec}(\mathbb{Z}/p^n)) \to \pi_1(\mathrm{Spec}(\mathbb{Z}_p))$.
  • If $A$ is a complete local ring with residue class field $k$, then $\pi_1(\mathrm{Spec}(A)) \cong \pi_1(\mathrm{Spec}(k))$.
  • $\pi_1(\mathrm{Spec}(\mathbb{Z}[i]))$ and $\pi_1(\mathrm{Spec}(\mathbb{Z}[(1 + \sqrt{-3})/2]))$ are trivial.
  • $\pi_1(\mathrm{Spec}(\mathbb{Z}[\zeta_{20}]))$ and $\pi_1(\mathrm{Spec}(\mathbb{Z}[(1 + \sqrt{-163})/2]))$ are trivial.
  • $\pi_1(\mathrm{Spec}(\mathbb{Z}[\sqrt{-5}]))$ has order two.
  • $\pi_1(\mathrm{Spec}(\mathbb{Z} \times_{\mathbb{Z}/(6)} \mathbb{Z})) \cong \widehat{\mathbb{Z}}$
  • $\pi_1(\mathrm{Spec}(\mathbb{Z}[x]/(x^6-1))) \cong \widehat{\mathbb{Z}}$
  • $\pi_1(\mathrm{Spec}(\mathbb{Z}[\sqrt{-3}]))$ has order two.

Let me finish with the following remark: Every profinite group arises as the fundamental group of some connected scheme. See here.

$\endgroup$
  • $\begingroup$ Thanks! I actually did have Lenstra's introduction, I just hadn't gotten around to reading it yet. :D I suppose now is a good time to work through it. Those examples/exercises look like they are exactly what I need right now. $\endgroup$ – user101616 Oct 23 '13 at 16:28
  • $\begingroup$ One of your examples states $\pi_1(Spec(\Bbb Z))=1$ because of the Minkowski theorem. Do you have $\pi_1(Spec(\mathcal{O}_K))\simeq Cl_K$ for any number field $K$ ($Cl_K$ is the class group of $K$)? $\endgroup$ – Ángel Valencia Jun 14 '16 at 21:08
  • $\begingroup$ @ÁngelValencia: This is discussed in Lenstra's notes, if I remember correctly. $\endgroup$ – Martin Brandenburg Oct 21 '16 at 14:10
  • $\begingroup$ I'll check them, thank you. $\endgroup$ – Ángel Valencia Oct 22 '16 at 19:58
  • 2
    $\begingroup$ @ÁngelValencia No, by the way. You might guess that $\text{Cl}(K)=\pi_1(\text{Spec}(\mathcal{O}_K))^\text{ab}$ since $\text{Cl}(K)$ is the Galois group of the maximal abelian unramified extension of $K$, but this is not even true since the Hilbert class field is unramified also at the infinite place of $K$. What's true is that $\pi_1(\text{Spec}(\mathcal{O}_K))^\text{ab}$ is isomorphic to the narrow class group almost by definition (it's the Galois group of the maximal abelian unramified at all finite places extensions of $K$). $\endgroup$ – Alex Youcis Nov 24 '16 at 12:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy