## 58.16 Specialization maps

In this section we construct specialization maps. Let $f : X \to S$ be a proper morphism of schemes with geometrically connected fibres. Let $s' \leadsto s$ be a specialization of points in $S$. Let $\overline{s}$ and $\overline{s}'$ be geometric points lying over $s$ and $s'$. Then there is a specialization map

\[ sp : \pi _1(X_{\overline{s}'}) \longrightarrow \pi _1(X_{\overline{s}}) \]

The construction of this map is as follows. Let $A$ be the strict henselization of $\mathcal{O}_{S, s}$ with respect to $\kappa (s) \subset \kappa (s)^{sep} \subset \kappa (\overline{s})$, see Algebra, Definition 10.155.3. Since $s' \leadsto s$ the point $s'$ corresponds to a point of $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ and hence there is at least one point (and potentially many points) of $\mathop{\mathrm{Spec}}(A)$ over $s'$ whose residue field is a separable algebraic extension of $\kappa (s')$. Since $\kappa (\overline{s}')$ is algebraically closed we can choose a morphism $\varphi : \overline{s}' \to \mathop{\mathrm{Spec}}(A)$ giving rise to a commutative diagram

\[ \xymatrix{ \overline{s}' \ar[r]_-\varphi \ar[rd] & \mathop{\mathrm{Spec}}(A) \ar[d] & \overline{s} \ar[l] \ar[ld] \\ & S } \]

The specialization map is the composition

\[ \pi _1(X_{\overline{s}'}) \longrightarrow \pi _1(X_ A) = \pi _1(X_{\kappa (s)^{sep}}) = \pi _1(X_{\overline{s}}) \]

where the first equality is Lemma 58.9.1 and the second follows from Lemmas 58.14.2 and 58.9.3. By construction the specialization map fits into a commutative diagram

\[ \xymatrix{ \pi _1(X_{\overline{s}'}) \ar[rr]_{sp} \ar[rd] & & \pi _1(X_{\overline{s}}) \ar[ld] \\ & \pi _1(X) } \]

provided that $X$ is connected. The specialization map depends on the choice of $\varphi : \overline{s}' \to \mathop{\mathrm{Spec}}(A)$ above and we will write $sp_\varphi $ if we want to indicate this.

Lemma 58.16.1. Consider a commutative diagram

\[ \xymatrix{ Y \ar[d]_ g \ar[r] & X \ar[d]^ f \\ T \ar[r] & S } \]

of schemes where $f$ and $g$ are proper with geometrically connected fibres. Let $t' \leadsto t$ be a specialization of points in $T$ and consider a specialization map $sp : \pi _1(Y_{\overline{t}'}) \to \pi _1(Y_{\overline{t}})$ as above. Then there is a commutative diagram

\[ \xymatrix{ \pi _1(Y_{\overline{t}'}) \ar[r]_{sp} \ar[d] & \pi _1(Y_{\overline{t}}) \ar[d] \\ \pi _1(X_{\overline{s}'}) \ar[r]^{sp} & \pi _1(X_{\overline{s}}) } \]

of specialization maps where $\overline{s}$ and $\overline{s}'$ are the images of $\overline{t}$ and $\overline{t}'$.

**Proof.**
Let $B$ be the strict henselization of $\mathcal{O}_{T, t}$ with respect to $\kappa (t) \subset \kappa (t)^{sep} \subset \kappa (\overline{t})$. Pick $\psi : \overline{t}' \to \mathop{\mathrm{Spec}}(B)$ lifting $\overline{t}' \to T$ as in the construction of the specialization map. Let $s$ and $s'$ denote the images of $t$ and $t'$ in $S$. Let $A$ be the strict henselization of $\mathcal{O}_{S, s}$ with respect to $\kappa (s) \subset \kappa (s)^{sep} \subset \kappa (\overline{s})$. Since $\kappa (\overline{s}) = \kappa (\overline{t})$, by the functoriality of strict henselization (Algebra, Lemma 10.155.10) we obtain a ring map $A \to B$ fitting into the commutative diagram

\[ \xymatrix{ \overline{t}' \ar[r]_-\psi \ar[d] & \mathop{\mathrm{Spec}}(B) \ar[d] \ar[r] & T \ar[d] \\ \overline{s}' \ar[r]^-\varphi & \mathop{\mathrm{Spec}}(A) \ar[r] & S } \]

Here the morphism $\varphi : \overline{s}' \to \mathop{\mathrm{Spec}}(A)$ is simply taken to be the composition $\overline{t}' \to \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$. Applying base change we obtain a commutative diagram

\[ \xymatrix{ Y_{\overline{t}'} \ar[r] \ar[d] & Y_ B \ar[d] \\ X_{\overline{s}'} \ar[r] & X_ A } \]

and from the construction of the specialization map the commutativity of this diagram implies the commutativity of the diagram of the lemma.
$\square$

Lemma 58.16.2. Let $f : X \to S$ be a proper morphism with geometrically connected fibres. Let $s'' \leadsto s' \leadsto s$ be specializations of points of $S$. A composition of specialization maps $\pi _1(X_{\overline{s}''}) \to \pi _1(X_{\overline{s}'}) \to \pi _1(X_{\overline{s}})$ is a specialization map $\pi _1(X_{\overline{s}''}) \to \pi _1(X_{\overline{s}})$.

**Proof.**
Let $\mathcal{O}_{S, s} \to A$ be the strict henselization constructed using $\kappa (s) \to \kappa (\overline{s})$. Let $A \to \kappa (\overline{s}')$ be the map used to construct the first specialization map. Let $\mathcal{O}_{S, s'} \to A'$ be the strict henselization constructed using $\kappa (s') \subset \kappa (\overline{s}')$. By functoriality of strict henselization, there is a map $A \to A'$ such that the composition with $A' \to \kappa (\overline{s}')$ is the given map (Algebra, Lemma 10.154.6). Next, let $A' \to \kappa (\overline{s}'')$ be the map used to construct the second specialization map. Then it is clear that the composition of the first and second specialization maps is the specialization map $\pi _1(X_{\overline{s}''}) \to \pi _1(X_{\overline{s}})$ constructed using $A \to A' \to \kappa (\overline{s}'')$.
$\square$

Let $X \to S$ be a proper morphism with geometrically connected fibres. Let $R$ be a strictly henselian valuation ring with algebraically closed fraction field and let $\mathop{\mathrm{Spec}}(R) \to S$ be a morphism. Let $\eta , s \in \mathop{\mathrm{Spec}}(R)$ be the generic and closed point. Then we can consider the specialization map

\[ sp_ R : \pi _1(X_\eta ) \to \pi _1(X_ s) \]

for the base change $X_ R/\mathop{\mathrm{Spec}}(R)$. Note that this makes sense as both $\eta $ and $s$ have algebraically closed residue fields.

Lemma 58.16.3. Let $f : X \to S$ be a proper morphism with geometrically connected fibres. Let $s' \leadsto s$ be a specialization of points of $S$ and let $sp : \pi _1(X_{\overline{s}'}) \to \pi _1(X_{\overline{s}})$ be a specialization map. Then there exists a strictly henselian valuation ring $R$ over $S$ with algebraically closed fraction field such that $sp$ is isomorphic to $sp_ R$ defined above.

**Proof.**
Let $\mathcal{O}_{S, s} \to A$ be the strict henselization constructed using $\kappa (s) \to \kappa (\overline{s})$. Let $A \to \kappa (\overline{s}')$ be the map used to construct $sp$. Let $R \subset \kappa (\overline{s}')$ be a valuation ring with fraction field $\kappa (\overline{s}')$ dominating the image of $A$. See Algebra, Lemma 10.50.2. Observe that $R$ is strictly henselian for example by Lemma 58.12.2 and Algebra, Lemma 10.50.10. Then the lemma is clear.
$\square$

Let $X \to S$ be a proper morphism with geometrically connected fibres. Let $R$ be a strictly henselian discrete valuation ring and let $\mathop{\mathrm{Spec}}(R) \to S$ be a morphism. Let $\eta , s \in \mathop{\mathrm{Spec}}(R)$ be the generic and closed point. Then we can consider the specialization map

\[ sp_ R : \pi _1(X_{\overline{\eta }}) \to \pi _1(X_ s) \]

for the base change $X_ R/\mathop{\mathrm{Spec}}(R)$. Note that this makes sense as $s$ has algebraically closed residue field.

Lemma 58.16.4. Let $f : X \to S$ be a proper morphism with geometrically connected fibres. Let $s' \leadsto s$ be a specialization of points of $S$ and let $sp : \pi _1(X_{\overline{s}'}) \to \pi _1(X_{\overline{s}})$ be a specialization map. If $S$ is Noetherian, then there exists a strictly henselian discrete valuation ring $R$ over $S$ such that $sp$ is isomorphic to $sp_ R$ defined above.

**Proof.**
Let $\mathcal{O}_{S, s} \to A$ be the strict henselization constructed using $\kappa (s) \to \kappa (\overline{s})$. Let $A \to \kappa (\overline{s}')$ be the map used to construct $sp$. Let $R \subset \kappa (\overline{s}')$ be a discrete valuation ring dominating the image of $A$, see Algebra, Lemma 10.119.13. Choose a diagram of fields

\[ \xymatrix{ \kappa (\overline{s}) \ar[r] & k \\ A/\mathfrak m_ A \ar[r] \ar[u] & R/\mathfrak m_ R \ar[u] } \]

with $k$ algebraically closed. Let $R^{sh}$ be the strict henselization of $R$ constructed using $R \to k$. Then $R^{sh}$ is a discrete valuation ring by More on Algebra, Lemma 15.45.11. Denote $\eta , o$ the generic and closed point of $\mathop{\mathrm{Spec}}(R^{sh})$. Since the diagram of schemes

\[ \xymatrix{ \overline{\eta } \ar[d] \ar[r] & \mathop{\mathrm{Spec}}(R^{sh}) \ar[d] & \mathop{\mathrm{Spec}}(k) \ar[d] \ar[l] \\ \overline{s}' \ar[r] & \mathop{\mathrm{Spec}}(A) & \overline{s} \ar[l] } \]

commutes, we obtain a commutative diagram

\[ \xymatrix{ \pi _1(X_{\overline{\eta }}) \ar[d] \ar[r]_{sp_{R^{sh}}} & \pi _1(X_ o) \ar[d] \\ \pi _1(X_{\overline{s}'}) \ar[r]^{sp} & X_{\overline{s}} } \]

of specialization maps by the construction of these maps. Since the vertical arrows are isomorphisms (Lemma 58.9.3), this proves the lemma.
$\square$

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