starting i.5

trees/sga1/sga1-i/sga1-i.5.tree

@@ -0,0 +1,67 @@
+\title{Fundamental property of étale morphisms}
+\taxon{section}
+\number{I.5}
+\parent{sga1-i}
+
+\import{sga1-macros}
+\put\transclude/numbered{false}
+
+\subtree[sga1-i.5.1]{
+  \taxon{theorem}
+  \number{I.5.1}
+
+  \p{
+    Let #{f\colon X\to Y} be a morphism of finite type.
+    For #{f} to be an open immersion, it is necessary and sufficient for it to be \em{étale} and \em{radicial}.
+  }
+
+  \proof{
+    \p{
+      Recall what "radicial" means: injective, with radicial residual extensions (recall also that it means that the morphism remains injective under any base extension).
+      The necessity is trivial, and the sufficiency remains to be shown.
+      We are going to give two different proofs: the first is shorter, the second is more elementary.
+    }
+
+    \ol{
+      \li{
+        A flat morphism is open, and so we can suppose (by replacing #{Y} with #{f(X)}) that #{f} is an onto \em{homeomorphism}.
+        For any base extension, it remains true that #{f} is flat, radicial, and surjective, thus a homeomorphism, and a fortiori closed.
+        Thus #{f} is \em{proper}.
+        Thus #{f} is \em{finite} (reference: Chevalley's theorem), defined by a coherent sheaf #{\sh{B}} of algebras.
+        Now #{\sh{B}} is locally free, and further, by hypothesis, of rank 1 everywhere, and so #{X=Y}.
+      }
+
+      \li{
+        We can suppose that #{Y} and #{X} are \em{affine}.
+        We can further easily reduce to proving the following:
+        if #{Y=\Spec(A)}, with #{A} local, and if #{f^{-1}(y)} is non-empty (where #{y} is the closed point of #{Y}), then #{X=Y} (indeed, this would imply that every #{y\in f(X)} has an open neighbourhood #{U} such that #{X|U=U}).
+        We will then have that #{X=\Spec(B)}, and wish to prove that #{A=B}.
+        But for this we can reduce to proving the analogous claim where we replace #{A} by #{\widehat{A}}, and #{B} by #{B\otimes_A\widehat{A}}
+        (taking into account the fact that #{\widehat{A}} is faithfully flat over #{A}).
+        We can thus suppose that #{A} is \em{complete}.
+        Let #{x} be the point over #{y}.
+        By \ref{sga1-i.2.2}, #{\sh{O}_x} is finite over #{A}, and is thus (being flat and radicial over #{A}) identical to #{A}.
+        So #{X=Y\coprod X'} (disjoint sum).
+        But since #{X} is radicial over #{Y}, #{X'} is empty.
+      }
+    }
+  }
+}
+
+\subtree[sga1-i.5.2]{
+  \taxon{corollary}
+  \number{I.5.2}
+
+  \p{
+    Let #{f\colon X\to Y} be a morphism that is both a \em{closed immersion} and \em{étale}.
+    If #{X} is connected, then #{f} is an isomorphism from #{X} to a connected component of #{Y}.
+  }
+
+  \proof{
+    Indeed, #{f} is also an open immersion.
+  }
+}
+
+\p{
+  We thus deduce:
+}

trees/sga1/sga1-i/sga1-i.tree

@@ -13,3 +13,4 @@
 \transclude{sga1-i.2}
 \transclude{sga1-i.3}
 \transclude{sga1-i.4}
+\transclude{sga1-i.5}