New results and more detailed proofs

algebra.tex

@@ -7589,20 +7589,28 @@ \section{Finite and integral ring extensions}
 Let $R \to S$ be a ring map.
 Let $f_1, \ldots, f_n \in R$ generate the unit ideal.
 \begin{enumerate}
-\item If each $R_{f_i} \to S_{f_i}$ is integral, so is $R \to S$.
-\item If each $R_{f_i} \to S_{f_i}$ is finite, so is $R \to S$.
+\item $R \to S$ is integral if and only if
+$R_{f_i} \to S_{f_i}$ is integral, for all $i=1,\dots,n$.
+\item $R \to S$ is finite if and only if
+$R_{f_i} \to S_{f_i}$ is finite, for all $i=1,\dots,n$.
 \end{enumerate}
 \end{lemma}
 
 \begin{proof}
-Proof of (1).
+Suppose that $R\to S$ is integral (resp., finite).
+By Lemma \ref{lemma-base-change-integral},
+each $R_{f_i} \to S_{f_i}$ is integral
+(resp., finite), using Lemma \ref{lemma-tensor-localization}).
+The converse of part (2) is a particular case of Lemma \ref{lemma-cover}.
+We prove the converse of part (1).
+Suppose then $R_{f_i} \to S_{f_i}$ is integral, for all $i=1,\dots,n$.
 Let $s \in S$. Consider the ideal $I \subset R[x]$ of
 polynomials $P$ such that $P(s) = 0$. Let $J \subset R$
 denote the ideal (!) of leading coefficients of elements of $I$.
 By assumption and clearing denominators
 we see that $f_i^{n_i} \in J$ for all $i$
 and certain $n_i \geq 0$. Hence $J$ contains $1$ and we see
-$s$ is integral over $R$. Proof of (2) omitted.
+$s$ is integral over $R$.
 \end{proof}
 
 \begin{lemma}

categories.tex

@@ -755,6 +755,130 @@ \section{Fibre products}
 and $g\in \Mor_{\mathcal C}(z, y)$.
 \end{definition}
 
+\begin{proof}
+\label{definition-diagonal-morphism}
+Let $\mathcal{C}$ be a category, let $f:x\to y$ be a
+morphism in $\mathcal{C}$, and suppose that $x\times_y x$ exists.
+The {\it diagonal morphism}, denoted $\Delta_f$ or $\Delta_{x/y}$,
+is the unique morphism $x\to x\times_y x$ coming from the identity $x\to x$.
+\end{proof}
+
+\begin{lemma}
+\begin{slogan}
+The horizontal fibre product of the vertical fibre products
+equals the vertical fibre product of the horizontal fibre products.
+\end{slogan}
+\label{lemma-3x3-pullback}
+Let $\mathcal{C}$ be a category that has fibre products and suppose that
+$$
+\xymatrix{
+X_1 \ar[r]\ar[d] & X_0\ar[d] & X_2 \ar[l]\ar[d]\\
+S_1 \ar[r] & S_0 & S_2 \ar[l] \\
+Y_1 \ar[r]\ar[u] & Y_0\ar[u] & Y_2 \ar[l]\ar[u]
+}
+$$
+is a commutative diagram in $\mathcal{C}$. Then we have
+$$
+(X_1 \times_{S_1} Y_1) \times_{X_0 \times_{S_0} Y_0} (X_2 \times_{S_2} Y_2)
+= (X_1 \times_{X_0} X_2) \times_{S_1 \times_{S_0} S_2} (Y_1 \times_{Y_0} Y_2).
+$$
+\end{lemma}
+
+\begin{proof}
+First proof: Use the fact that the Yoneda embedding
+$\mathcal{C}\to\text{Fun}(\mathcal{C}^\text{opp},Sets)$
+preserves limits to reduce it to the case of $Sets$,
+where it is easier to check the two sets are isomorphic.
+
+\medskip\noindent
+Second proof: Define a pair of anti-parallel canonical maps
+between the two sides of the equality from the statement.
+Do so leveraging the universal properties of all
+the involved fibre products.
+Verify that these maps are mutually inverse.
+\end{proof}
+
+\begin{lemma}
+\label{lemma-magic-square}
+Let $\mathcal{C}$ be a category that has fibre products.
+Let $X\to S$, $Y\to S$, $S\to T$ be morphisms in $\mathcal{C}$.
+There is a cartesian diagram
+$$
+\xymatrix{
+X \times_T Y \ar[r] \ar[d] & X \times_S Y \ar[d] \\
+T \ar[r]^(.4){\Delta_{T/S}} \ar[r] & T \times_S T
+}
+$$
+\end{lemma}
+
+\begin{proof}
+Apply Lemma \ref{lemma-3x3-pullback} to the diagram
+$$
+\xymatrix{
+S \ar[r]\ar[d] & S \ar[d] & X \ar[l]\ar[d]\\
+S \ar[r]       & T        & T \ar[l] \\
+S \ar[r]\ar[u] & S \ar[u] & Y \ar[l]\ar[u]
+}
+$$
+\end{proof}
+
+\begin{lemma}
+\label{lemma-base-change-diagonal-1-cat}
+\begin{slogan}
+	The base change of the diagonal is the diagonal of the base change.
+\end{slogan}
+Let $\mathcal{C}$ be a category with fibre products.
+Let $X\to S$, $S'\to S$ be morphisms in $\mathcal{C}$,
+and let $X'=S'\times_SX$. There is a cartesian square
+$$
+\xymatrix{
+X' \ar[r]^(.37){\Delta_{X'/S'}}\ar[d] & X'\times_{S'}X'\ar[d]\\
+X \ar[r]^(.35){\Delta_{X/S}} & X\times_{S}X
+}
+$$
+\end{lemma}
+
+\begin{proof}
+Apply Lemma \ref{lemma-3x3-pullback} to the diagram
+$$
+\xymatrix{
+X \ar[r]\ar[d] & X \ar[d] & X' \ar[l]\ar[d]\\
+X \ar[r]       & S        & S' \ar[l] \\
+X \ar[r]\ar[u] & X \ar[u] & X' \ar[l]\ar[u]
+}
+$$
+\end{proof}
+
+\begin{lemma}
+\label{lemma-pasting-law-pullbacks}
+Let
+$$
+\xymatrix{
+	X' \ar@{->}[r] \ar@{->}[d] & X \ar@{->}[d] \\
+	Y' \ar@{->}[r] \ar@{->}[d] & T \ar@{->}[d] \\
+	Y \ar@{->}[r] & S
+}
+$$
+be a commutative diagram in a category,
+and suppose that the lower square is cartesian.
+Then the outer rectangle is cartesian
+if and only if the upper square is cartesian.
+\end{lemma}
+
+\begin{proof}
+Suppose each inner square is cartesian. Then the outer rectangle is cartesian by Lemma \ref{lemma-3x3-pullback} applied to the diagram
+$$
+\xymatrix{
+	X \ar[r]\ar[d] & T \ar[d] & T \ar[l]\ar[d]\\
+	S \ar[r]       & S        & S \ar[l] \\
+	S \ar[r]\ar[u] & S \ar[u] & Y \ar[l]\ar[u]
+}
+$$
+
+\medskip\noindent
+We omit the proof of the other implication.
+\end{proof}
+
 \begin{definition}
 \label{definition-representable-morphism}
 A morphism $f : x \to y$ of a category $\mathcal{C}$ is said to be

morphisms.tex

@@ -216,6 +216,43 @@ \section{Immersions}
 \noindent
 In this section we collect some facts on immersions.
 
+\begin{lemma}
+\label{lemma-immersion}
+Let $f:X\to Y$ be a morphism of schemes.
+Then $f$ is an immersion if and only if it is a homeomorphism
+onto a locally closed subset of $Y$ and $f^\sharp:f^{-1}\mathcal{O}_Y\to\mathcal{O}_X$
+is surjective.
+\end{lemma}
+
+\begin{proof}
+Suppose $f$ is an immersion. There is an open subset $U\subset Y$
+such that $f$ equals the composite $X\xrightarrow{i}U\xrightarrow{j}Y$,
+where $i$ is a closed immersion. On the level of sheaves, equation $f=j\circ i$
+means that the composite
+$i^{-1}j^{-1}\mathcal{O}_Y
+\xrightarrow{i^{-1}(j^\sharp)}i^{-1}\mathcal{O}_U
+\xrightarrow{i^\sharp}\mathcal{O}_X$ equals $f^\sharp$.
+Since $j^\sharp$ is an isomorphism, we get that $i^\sharp$ is onto.
+Now apply Lemma \ref{lemma-closed-immersion}.
+
+\medskip\noindent
+Conversely, suppose that $f$ is a homeomorphism onto a locally closed
+subset of $Y$ and that $f^{-1}\mathcal{O}_Y\to\mathcal{O}_X$ is surjective.
+There is an open subset $U\subset Y$ such that $f(X)\subset U$ and $f(X)$
+is closed in $U$. By Schemes, Lemma \ref{lemma-restrict-map-to-opens},
+the commutative diagram of topological spaces
+$$
+\xymatrix{
+	X \ar@{->}[r]^{i} \ar@/_/@{->}[rr]_{f} & U \ar@{->}[r]^{j} & Y
+}
+$$
+promotes to a commutative diagram of schemes,
+where the morphism $i^\sharp:i^{-1}\mathcal{O}_U\to\mathcal{O}_X$
+is given by $f^\sharp:f^{-1}\mathcal{O}_Y\cong i^{-1}j^{-1}\mathcal{O}_Y
+\cong i^{-1}\mathcal{O}_U\to\mathcal{O}_X$.
+Apply again Lemma \ref{lemma-closed-immersion}.
+\end{proof}
+
 \begin{lemma}
 \label{lemma-immersion-permanence}
 Let $Z \to Y \to X$ be morphisms of schemes.
@@ -10663,8 +10700,16 @@ \section{Integral and finite morphisms}
 \end{lemma}
 
 \begin{proof}
-See Algebra, Lemma \ref{algebra-lemma-integral-local}.
-Some details omitted.
+Implications (1)$\Rightarrow$(3)$\Rightarrow$(2) are clear.
+We see (2)$\Rightarrow$(1).
+Let $V=\Spec A\subset S$ be open affine,
+By Lemma \ref{lemma-characterize-affine}, $f^{-1}(V)=\Spec B$ is affine.
+Since any localization of an integral ring morphism is again integral
+(Algebra, Lemma \ref{algebra-lemma-integral-local}), we may
+(using Schemes, Lemma \ref{lemma-standard-open-two-affines})
+cover $\Spec A$ by affine open subschemes of the form $D(f_i)$,
+$f_i\in A$, $i=1,\dots,n$, such that $B_{f_i}$ is an integral $A_{f_i}$-algebra.
+By Algebra, Lemma \ref{algebra-lemma-integral-local}, $A\to B$ is integral.
 \end{proof}
 
 \begin{lemma}
@@ -10684,8 +10729,8 @@ \section{Integral and finite morphisms}
 \end{lemma}
 
 \begin{proof}
-See Algebra, Lemma \ref{algebra-lemma-integral-local}.
-Some details omitted.
+The proof is exactly the same as the proof of Lemma \ref{lemma-integral-local},
+but substituting ``integral'' by ``finite.''
 \end{proof}
 
 \begin{lemma}
@@ -10717,7 +10762,30 @@ \section{Integral and finite morphisms}
 \end{lemma}
 
 \begin{proof}
-See Algebra, Lemma \ref{algebra-lemma-base-change-integral}.
+Let
+$$
+\xymatrix{
+	X' \ar@{->}[r]^{f'} \ar@{->}[d] & S' \ar@{->}[d] \\
+	X \ar@{->}[r]^{f} & S
+}
+$$
+be a cartesian diagram of schemes, and suppose that $f$ is finite
+(resp., integral). Lemma \ref{lemma-base-change-affine} tells us that
+$f'$ is affine. Each point $s'\in S$ has an affine open neighborhood
+$V\subset S'$ that maps to an open affine $U\subset S$.
+In the proof of the last invoked lemma one sees that the restricted diagram
+$$
+\xymatrix{
+	(f')^{-1}(V) \ar@{->}[d] \ar@{->}[r]^{f'} & V \ar@{->}[d] \\
+	f^{-1}(U) \ar@{->}[r]^{f} & U
+}
+$$
+is a cartesian square of affine schemes.
+Hence, using Algebra, Lemma \ref{algebra-lemma-base-change-integral}, one sees that
+$f'$ is finite, by Lemma \ref{lemma-finite-local}
+(resp., that $f'$ is integral, by Lemma \ref{lemma-integral-local}).
+Since these $V$ cover $S'$, we finish by Lemma \ref{lemma-finite-local}
+(resp., by Lemma \ref{lemma-integral-local}).
 \end{proof}
 
 \begin{lemma}
@@ -13688,12 +13756,21 @@ \section{Relative normalization}
 \end{lemma}
 
 \begin{proof}
-By Lemma \ref{lemma-normalization-localization} we may assume $X = \Spec(A)$
+For continuous maps of sober spaces, the following property
+is local on the target: ``all generic points of the irreducible
+components of the target are hit by generic points of irreducible
+components of the source'' (use Topology, Lemma
+\ref{topology-lemma-correspondence-sober-generic-points-open}).
+Hence, by Lemma \ref{lemma-normalization-localization} we may assume $X = \Spec(A)$
 is affine. Choose a finite affine open covering $Y = \bigcup \Spec(B_i)$.
 Then $X' = \Spec(A')$ and the morphisms $\Spec(B_i) \to Y \to X'$
 jointly define an injective $A$-algebra map $A' \to \prod B_i$.
 Thus the lemma follows from
-Algebra, Lemma \ref{algebra-lemma-injective-minimal-primes-in-image}.
+Algebra, Lemma \ref{algebra-lemma-injective-minimal-primes-in-image}
+(and also that by Topology, Lemma
+\ref{topology-lemma-correspondence-sober-generic-points-open},
+every generic point of an irreducible component in $Y$ is a generic
+point in $\Spec B_i$ of an irreducible component, for some $i$).
 \end{proof}
 
 \begin{lemma}
@@ -13776,7 +13853,9 @@ \section{Relative normalization}
 of schemes. Let $X'$ be the normalization of $X$ in $Y$. Assume
 \begin{enumerate}
 \item $Y$ is a normal scheme,
-\item quasi-compact opens of $Y$ have finitely many irreducible components.
+\item $Y$ has locally finitely many irreducible components
+(see Properties, Section
+\ref{properties-section-locally-finitely-many-irred-comp}).
 \end{enumerate}
 Then $X'$ is a disjoint union of integral normal schemes. Moreover,
 the morphism $Y \to X'$ is dominant and induces a bijection of
@@ -13811,14 +13890,18 @@ \section{Relative normalization}
 Properties, Lemma \ref{properties-lemma-locally-normal} that $X'$ is normal.
 On the other hand, each $U'$ is a finite disjoint union of irreducible
 schemes, hence every quasi-compact open of $X'$ has finitely many irreducible
-components (by a topological argument which we omit). Thus $X'$
+components (Properties, Lemma
+\ref{properties-lemma-characterize-locally-finitely-many-irred-comp}).
+Thus $X'$
 is a disjoint union of normal integral schemes by
 Properties, Lemma \ref{properties-lemma-normal-locally-finite-nr-irreducibles}.
-It is clear from the description of $X'$ above that $Y \to X'$
-is dominant and induces a bijection on irreducible components
-$V \to U'$ for every affine open $U \subset X$. The bijection of
+It is clear that $V \to U'$ induces a bijection on irreducible components
+for every affine open $U \subset X$. The bijection of
 irreducible components for the morphism $Y \to X'$
-follows from this by a topological argument (omitted).
+follows from this plus Topology, Lemma
+\ref{topology-lemma-bijection-irreducible-components-local-target}.
+The morphism $Y\to X'$ is dominant by Lemma
+\ref{lemma-characterize-normalization}.
 \end{proof}
 
 \begin{lemma}
@@ -13977,7 +14060,13 @@ \section{Normalization}
 canonical maps of Schemes, Section \ref{schemes-section-points}.
 Note that this morphism is quasi-compact by assumption and
 quasi-separated as $Y$ is separated (see
-Schemes, Section \ref{schemes-section-separation-axioms}).
+Schemes, Lemmas \ref{lemma-compose-after-separated},
+\ref{schemes-lemma-quasi-separated-coproduct}, and
+\ref{schemes-lemma-affine-separated}).
+This morphism induces a bijection on irreducible components
+almost by definition
+(Topology, Lemma
+\ref{topology-lemma-characterize-bijection-irreducible-components-sober}).
 
 \begin{definition}
 \label{definition-normalization}
@@ -14057,13 +14146,17 @@ \section{Normalization}
 This proves (2). Part (3) follows from
 Algebra, Lemma \ref{algebra-lemma-characterize-reduced-ring-normal},
 or Lemma \ref{lemma-normalization-in-disjoint-union}.
-Part (4) holds because it is clear that $f^{-1}(U) \to U$ is the morphism
+Finally, we prove part (4).
+By Lemma \ref{lemma-normalization-localization} and Topology,
+Lemma \ref{topology-lemma-correspondence-sober-generic-points-open},
+the morphism $f^{-1}(U) \to U$ is the morphism
 $$
 \Spec\left(\prod \kappa(\mathfrak q_i)\right)
 \longrightarrow
 \Spec(A)
 $$
 where $f : Y \to X$ is the morphism (\ref{equation-generic-points}).
+Hence (4) follows from (2) and (3).
 \end{proof}
 
 \begin{lemma}
@@ -14142,7 +14235,9 @@ \section{Normalization}
 The morphism $\nu$ is integral by Lemma \ref{lemma-characterize-normalization}.
 By Lemma \ref{lemma-normal-normalization} the
 morphism $Y \to X^\nu$ induces a bijection on irreducible components,
-and by construction of $Y$ this implies that $X^\nu \to X$ induces
+and by Topology, Lemma
+\ref{topology-lemma-bijection-irreducible-components-2-out-of-3},
+this implies that $X^\nu \to X$ induces
 a bijection on irreducible components. By construction $f : Y \to X$
 is dominant, hence also $\nu$ is dominant. Since an integral morphism is
 closed (Lemma \ref{lemma-integral-universally-closed}) this implies that