In this section we define Grothendieck's algebraic fundamental group. The following definition makes sense thanks to Lemma 58.5.5.
Definition 58.6.1. Let $X$ be a connected scheme. Let $\overline{x}$ be a geometric point of $X$. The fundamental group of $X$ with base point $\overline{x}$ is the group of automorphisms of the fibre functor $F_{\overline{x}} : \textit{FÉt}_ X \to \textit{Sets}$ endowed with its canonical profinite topology from Lemma 58.3.1.
\[ \pi _1(X, \overline{x}) = \text{Aut}(F_{\overline{x}}) \]
Combining the above with the material from Section 58.3 we obtain the following theorem.
Theorem 58.6.2. Let $X$ be a connected scheme. Let $\overline{x}$ be a geometric point of $X$.
The fibre functor $F_{\overline{x}}$ defines an equivalence of categories
Given a second geometric point $\overline{x}'$ of $X$ there exists an isomorphism $t : F_{\overline{x}} \to F_{\overline{x}'}$. This gives an isomorphism $\pi _1(X, \overline{x}) \to \pi _1(X, \overline{x}')$ compatible with the equivalences in (1). This isomorphism is independent of $t$ up to inner conjugation.
Given a morphism $f : X \to Y$ of connected schemes denote $\overline{y} = f \circ \overline{x}$. There is a canonical continuous homomorphism
such that the diagram
is commutative.
\[ \textit{FÉt}_ X \longrightarrow \textit{Finite-}\pi _1(X, \overline{x})\textit{-Sets} \]
\[ f_* : \pi _1(X, \overline{x}) \to \pi _1(Y, \overline{y}) \]
\[ \xymatrix{ \textit{FÉt}_ Y \ar[r]_{\text{base change}} \ar[d]_{F_{\overline{y}}} & \textit{FÉt}_ X \ar[d]^{F_{\overline{x}}} \\ \textit{Finite-}\pi _1(Y, \overline{y})\textit{-Sets} \ar[r]^{f_*} & \textit{Finite-}\pi _1(X, \overline{x})\textit{-Sets} } \]
Proof. Part (1) follows from Lemma 58.5.5 and Proposition 58.3.10. Part (2) is a special case of Lemma 58.3.11. For part (3) observe that the diagram
is commutative (actually commutative, not just $2$-commutative) because $\overline{y} = f \circ \overline{x}$. Hence we can apply Lemma 58.3.11 with the implied transformation of functors to get (3). $\square$
Lemma 58.6.3. Let $K$ be a field and set $X = \mathop{\mathrm{Spec}}(K)$. Let $\overline{K}$ be an algebraic closure and denote $\overline{x} : \mathop{\mathrm{Spec}}(\overline{K}) \to X$ the corresponding geometric point. Let $K^{sep} \subset \overline{K}$ be the separable algebraic closure.
The functor of Lemma 58.2.2 induces an equivalence
compatible with $F_{\overline{x}}$ and the functor $\textit{Finite-}\text{Gal}(K^{sep}/K)\textit{-Sets} \to \textit{Sets}$.
This induces a canonical isomorphism
of profinite topological groups.
\[ \textit{FÉt}_ X \longrightarrow \textit{Finite-}\text{Gal}(K^{sep}/K)\textit{-Sets}. \]
\[ \text{Gal}(K^{sep}/K) \longrightarrow \pi _1(X, \overline{x}) \]
Proof. The functor of Lemma 58.2.2 is the same as the functor $F_{\overline{x}}$ because for any $Y$ étale over $X$ we have
Namely, as seen in the proof of Lemma 58.2.2 we have $Y = \coprod _{i \in I} \mathop{\mathrm{Spec}}(L_ i)$ with $L_ i/K$ finite separable over $K$. Hence any $K$-algebra homomorphism $L_ i \to \overline{K}$ factors through $K^{sep}$. Also, note that $F_{\overline{x}}(Y)$ is finite if and only if $I$ is finite if and only if $Y \to X$ is finite étale. This proves (1).
Part (2) is a formal consequence of (1), Lemma 58.3.11, and Lemma 58.3.3. (Please also see the remark below.) $\square$
Remark 58.6.4. In the situation of Lemma 58.6.3 let us give a more explicit construction of the isomorphism $\text{Gal}(K^{sep}/K) \to \pi _1(X, \overline{x}) = \text{Aut}(F_{\overline{x}})$. Observe that $\text{Gal}(K^{sep}/K) = \text{Aut}(\overline{K}/K)$ as $\overline{K}$ is the perfection of $K^{sep}$. Since $F_{\overline{x}}(Y) = \mathop{\mathrm{Mor}}\nolimits _ X(\mathop{\mathrm{Spec}}(\overline{K}), Y)$ we may consider the map This is an action because The action is functorial in $Y \in \textit{FÉt}_ X$ and we obtain the desired map.
\[ \text{Aut}(\overline{K}/K) \times F_{\overline{x}}(Y) \to F_{\overline{x}}(Y), \quad (\sigma , \overline{y}) \mapsto \sigma \cdot \overline{y} = \overline{y} \circ \mathop{\mathrm{Spec}}(\sigma ) \]
\[ \sigma \tau \cdot \overline{y} = \overline{y} \circ \mathop{\mathrm{Spec}}(\sigma \tau ) = \overline{y} \circ \mathop{\mathrm{Spec}}(\tau ) \circ \mathop{\mathrm{Spec}}(\sigma ) = \sigma \cdot (\tau \cdot \overline{y}) \]
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