For any curve \(X\), the algebraic fundamental group \(\pi_1(X)\) is defined as \(\cocolim \operatorname{Gal}(K'/K)\), where \(K\) is the function field of \(X\), and \(K'\) runs over all Galois extensions of \(K\) such that the corresponding curve \(X'\) is étale over \(X\) (III, Ex. 10.3).
Thus, for example, \(\pi_1(\PP^1)=1\). (See 2.5.3)
Show that for an elliptic curve \(X\), \[ \pi_1(X) = \begin{cases} \Prod_{\ell \text{ prime}} \ZZ_\ell \times \ZZ_\ell & \characteristic k = 0, \\ \\ \Prod_{\ell\neq p} \ZZ_\ell \times \ZZ_\ell & \characteristic k = p \text{ and } \Hasse X = 0, \\ \\ \ZZ_p \times \Prod_{\ell\neq p} \ZZ_\ell \times \ZZ_\ell & \characteristic k = p \text{ and } \Hasse X \neq 0 \end{cases} ,\] where \(\ZZ_\ell=\varprojlim \ZZ/\ell^n\) is the \(\ell\)-adic integers.
Hints: Any Galois étale cover \(X'\) of an elliptic curve is again an elliptic curve. If the degree of \(X'\) over \(X\) is relatively prime to \(p\), then \(X'\) can be dominated by the cover \(n_X: X \to X\) for some integer \(n\) with \((n, p)=1\). The Galois group of the covering \(n_X\) is \(\ZZ/n \times \ZZ/n\). Étale covers of degree divisible by \(p\) can occur only if the Hasse invariant of \(X\) is not zero.
Note: More generally, Grothendieck has shown (SGA 1, X, 2.6) that the algebraic fundamental group of any curve of genus \(g\) is isomorphic to a quotient of the completion, with respect to subgroups of finite index, of the ordinary topological fundamental group of a compact Riemann surface of genus \(g\), i.e., a group with \(2g\) generators \(a_1, \ldots, a_g, b_1, \ldots, b_g\) and the relation \(\qty(a_1 b_1 a_1^{-1} b_1^{-1}) \cdots \qty(a_g b_g a_g^{-1} b_g^{-1})=1\).
Solution
Fix the origin \(O\in X(k)\). We compute the inverse limit over connected finite Galois etale covers of \(X\).
1. Every connected finite etale cover\[ \phi:Y\longrightarrow X \]is a genus-one curve; after choosing \(O_Y\in\phi^{-1}(O)\), it is an ellipticcurve and \(\phi:(Y,O_Y)\to(X,O)\) is an etale isogeny.
Let \(d=\deg\phi\). Since \(\phi\) is etale, Riemann–Hurwitz gives \[ 2g(Y)-2=d\bigl(2g(X)-2\bigr)=0. \] Thus \(g(Y)=1\). The point \(O_Y\) makes \(Y\) an elliptic curve, and a morphism of elliptic curves that sends the origin to the origin is a group homomorphism. Hence \(\phi\) is an isogeny. Because \(\phi\) is etale, it is separable and \[ \#\ker\phi(k)=d. \] Translations by the points of \(\ker\phi(k)\) are exactly the deck transformations, so \[ \Gal(Y/X)\cong\ker\phi(k). \]
2. If \(\phi:Y\to X\) is an etale isogeny of degree \(d\), then the dualisogeny satisfies\[ \phi\circ\hat\phi=[d]_X. \]In particular, whenever \([d]_X\) is etale, the cover \([d]_X:X\to X\)dominates \(\phi\).
This is Exercise IV.4.7(c). The displayed factorization is a morphism of covers \[ \begin{aligned} X&\xrightarrow{\hat\phi}Y\xrightarrow{\phi}X,\\ X&\xrightarrow{[d]_X}X. \end{aligned} \] Thus \([d]_X\) dominates \(\phi\) whenever it belongs to the etale covering system.
3. If \(m\) is prime to \(p=\characteristic k\) (with no restriction when\(\characteristic k=0\)), then\[ [m]_X:X\longrightarrow X \]is an etale Galois cover with\[ \boxed{\Gal([m]_X)\cong(\ZZ/m\ZZ)^2.} \]
The multiplication map has degree \(m^2\). When \(m\) is invertible in \(k\), it is separable, hence etale on the elliptic curve. Its kernel is the full \(m\)-torsion group \[ X[m](k)\cong(\ZZ/m\ZZ)^2. \] By step 1, this kernel is the deck-transformation group.
4. Suppose \(\characteristic k=p>0\) and the Hasse invariant of \(X\) iszero. Then every finite etale cover of \(X\) has degree prime to \(p\).
Suppose that an etale isogeny \(\phi:Y\to X\) has degree \[ d=p^r m, \qquad (m,p)=1. \] By step 2, \[ \phi\circ\hat\phi=[d]_X. \] Since \(\phi\) is separable, the inseparable degree of the left side is the inseparable degree of \(\hat\phi\).
When the Hasse invariant is zero, \(X\) is supersingular and \([p]_X\) is purely inseparable of degree \(p^2\). Hence the inseparable degree of \([d]_X\) is \(p^{2r}\). It follows that the inseparable degree of \(\hat\phi\) is \(p^{2r}\). But \[ \deg\hat\phi=\deg\phi=p^r m \] by Exercise IV.4.7(f), whose \(p\)-adic valuation is only \(r\). Therefore \(r=0\).
5. In the Hasse-zero case, the covers \([m]_X\) with \((m,p)=1\) arecofinal, and therefore\[ \boxed{ \pi_1(X) \cong \prod_{\ell\ne p}(\ZZ_\ell\times\ZZ_\ell). } \]
By step 4 every connected finite etale Galois cover has degree \(d\) prime to \(p\). Step 2 says that it is dominated by \([d]_X\), and step 3 computes the Galois group of this standard cover. Hence the standard covers form a cofinal system. Taking the inverse limit over integers prime to \(p\) gives \[ \varprojlim_{(m,p)=1}(\ZZ/m\ZZ)^2 \cong \prod_{\ell\ne p}\ZZ_\ell^2. \]
6. Suppose \(\characteristic k=p>0\) and the Hasse invariant of \(X\) isnonzero. For \(r\ge0\) let\[ F_X^{(r)}:X\longrightarrow X^{(p^r)} \]be the \(r\)-fold relative Frobenius and let\[ V_X^{(r)}:X^{(p^r)}\longrightarrow X \]be the corresponding Verschiebung, so that\[ [p^r]_X=V_X^{(r)}\circ F_X^{(r)}. \]For \((m,p)=1\), put\[ \psi_{r,m}=[m]_X\circ V_X^{(r)}:X^{(p^r)}\longrightarrow X. \]Then \(\psi_{r,m}\) is etale Galois and\[ \boxed{ \Gal(\psi_{r,m}) \cong \ZZ/p^r\ZZ\times(\ZZ/m\ZZ)^2. } \]
Nonzero Hasse invariant is the ordinary case. Thus \(F_X^{(r)}\) has degree \(p^r\) and is purely inseparable, while \(V_X^{(r)}\) is separable of degree \(p^r\). Hence \(V_X^{(r)}\) is etale. Step 3 shows that \([m]_X\) is also etale, so \(\psi_{r,m}\) is etale.
For \(r=0\) the assertion about the \(p\)-primary kernel is trivial, so assume \(r\geq1\). The second duality identity is \[ F_X^{(r)}\circ V_X^{(r)} =[p^r]_{X^{(p^r)}}. \] Hence \[ \ker V_X^{(r)}(k) \subseteq X^{(p^r)}[p^r](k). \] The Frobenius twist is again ordinary. Therefore \([p^r]\) has separable degree \(p^r\), so \[ \#X^{(p^r)}[p^r](k)=p^r. \] Moreover \([p]\) has separable degree \(p\), so the subgroup of these points killed by \(p\) has exactly \(p\) elements. By the structure theorem for finite abelian groups, a finite abelian \(p\)-group with exactly \(p\) elements killed by \(p\) is cyclic. Thus \[ X^{(p^r)}[p^r](k)\cong\ZZ/p^r\ZZ. \] Since the separable isogeny \(V_X^{(r)}\) has degree \(p^r\), its kernel also has \(p^r\) geometric points. The preceding inclusion is therefore an equality, and \[ \ker V_X^{(r)}(k)\cong\ZZ/p^r\ZZ. \] Also \[ X^{(p^r)}[m](k)\cong(\ZZ/m\ZZ)^2 \] lies in \(\ker\psi_{r,m}\). These subgroups have coprime orders and their product has order \(p^r m^2=\deg\psi_{r,m}\), so they give the entire kernel. Step 1 identifies this kernel with the Galois group.
7. The covers \(\psi_{r,m}\) of step 6 are cofinal among connectedfinite etale Galois covers of \(X\).
Let \(\phi:Y\to X\) be an etale isogeny and write \[ \deg\phi=p^r m, \qquad (m,p)=1. \] By step 2, \[ \phi\circ\hat\phi=[p^r m]_X. \] In the ordinary case the inseparable degree of \([p^r m]_X\) is \(p^r\). Since \(\phi\) is separable, the inseparable degree of \(\hat\phi\) is also \(p^r\). Also \[ \deg\hat\phi=\deg\phi=p^r m \] by Exercise IV.4.7(f), so the separable degree of \(\hat\phi\) is \(m\).
The separable–inseparable factorization of \(\hat\phi\) has the form \[ \hat\phi=u\circ F_X^{(r)} \] for a separable isogeny \(u:X^{(p^r)}\to Y\) of degree \(m\).
To justify the Frobenius factor explicitly, put \[ K=k(X),\qquad L=k(Y), \] using the inclusion \(L\subseteq K\) induced by \(\hat\phi\), and let \(M\) be the maximal subfield of \(K\) separable over \(L\). Then \[ [K:M]=p^r. \] The extension \(K/M\) is purely inseparable, so every \(a\in K\) satisfies \(a^{p^r}\in M\); hence \(K^{p^r}\subseteq M\).
Because \(k\) is algebraically closed, it is perfect. Choose a separating transcendence element \(t\) for the one-variable function field \(K/k\) and write \(d=[K:k(t)]\). Frobenius gives \[ [K^p:k(t^p)]=d, \] while \([k(t):k(t^p)]=p\). Consequently \[ [K:K^p] = \frac{[K:k(t^p)]}{[K^p:k(t^p)]} =p, \] and iteration gives \[ [K:K^{p^r}]=p^r. \] Thus \(K^{p^r}=M\). It follows that \(M/L\) is separable of degree \(m\), and the corresponding factorization of curves is precisely the \(r\)-fold relative Frobenius followed by the separable map \(u\) displayed above.
Using \([p^r]_X=V_X^{(r)}F_X^{(r)}\), we get \[ \phi\circ u\circ F_X^{(r)} = [m]_X\circ V_X^{(r)}\circ F_X^{(r)}. \] The relative Frobenius of the smooth curve is finite faithfully flat, hence is an epimorphism for morphisms of schemes. Cancellation therefore gives \[ \phi\circ u = [m]_X\circ V_X^{(r)} = \psi_{r,m}. \] Thus \(\psi_{r,m}\) dominates \(\phi\).
8. In the ordinary positive-characteristic case,\[ \boxed{ \pi_1(X) \cong \ZZ_p\times\prod_{\ell\ne p}(\ZZ_\ell\times\ZZ_\ell). } \]
6. Hence\[ \begin{aligned} \pi_1(X) &\cong \varprojlim_{r,(m,p)=1} \left(\ZZ/p^r\ZZ\times(\ZZ/m\ZZ)^2\right)\\ &\cong \ZZ_p\times\prod_{\ell\ne p}(\ZZ_\ell\times\ZZ_\ell). \end{aligned} \]
9. If \(\characteristic k=0\), then\[ \boxed{ \pi_1(X) \cong \prod_{\ell\text{ prime}}(\ZZ_\ell\times\ZZ_\ell). } \]
Every multiplication map \([m]_X\) is etale in characteristic zero. For a connected finite etale Galois cover \(\phi\) of degree \(d\), step 2 shows that \([d]_X\) dominates \(\phi\). Thus the multiplication maps are cofinal. By step 3, \[ \begin{aligned} \pi_1(X) &\cong \varprojlim_m(\ZZ/m\ZZ)^2\\ &\cong \prod_{\ell\text{ prime}}(\ZZ_\ell\times\ZZ_\ell). \end{aligned} \]
10. Q.E.D.
Step 9 gives the characteristic-zero formula, step 5 gives the Hasse-zero formula in characteristic \(p\), and step 8 gives the nonzero Hasse-invariant formula in characteristic \(p\).