add R.4.3's proof & 4.4

.github/workflows/typos.yml

@@ -1,7 +1,7 @@
 name: typos
 on:
   pull_request:
-    branches: [ main ]
+    branches: [ master ]
     types: [ready_for_review, opened, synchronize, reopened]
   push:
     branches:

commutative-algebra-notes.tex

@@ -1023,7 +1023,7 @@ As in with rings, it is equivalent to say that
     \item $y$ is integral over $A$
     \item subring $A'[y] \subset B$ generated by $A' = \psi(A)$ and $y$ is finite over $A$
     \item $\exists$ an $A$-subalgebra $C \subset B$ such that $A'[y] \subset C$ and $C$ is finite over $A$
-  \end{ennumerate}
+  \end{enumerate}
 
   Notes: $A'$ is the image of $A$ in $B$, ie. $A' = \psi(A)$.\\
   $A'[y]$ is the smallest subring of $B$ containing both coefficients from $A$ and the element $y$.
@@ -1069,9 +1069,78 @@ As in with rings, it is equivalent to say that
       Moreover, if $y \in B$ is integral over $\tilde{A}$ then $y \in \tilde{A}$, so that $\tilde{\tilde{A}} = \tilde{A}$.
   \end{enumerate}
 \end{prop}
-\begin{proof}
+\begin{proof}.\\
+  \begin{enumerate}[a.]
+    \item if $\{ \beta_1, \ldots, \beta_n \}$ generate $B$ as an $A$-module and $\{ \gamma_1, \ldots, \gamma_n \}$ generate $C$ as an $B$-module,\\
+      then the set of products $\{ \betA_i \gamma_j \}$ generates $C$ as an $A$-module.
+
+      Since there are $n \times m$ generators (ie. finite), $C$ is finite over $A$.
+
+    \item proof by induction:
+
+      base case: if $y_1$ integral over $A ~\Longrightarrow~$ it satisfies a monic polynomial.
+
+      Thus $A[y_1]$ is generated as an $A$-module by $\{1, y_1, y_1^2, \ldots, y_1^{n-1} \}$, making it a finite $A$-algebra.
+
+      inductive step: let $R_k = A[y_1, \ldots, y_k]$. Assume $R_k$ is finite over $A$.
+
+      Since $y_{k+1}$ is integral over $A~~ \Longrightarrow~$ it is also integral over $R_k$.
+
+      Thus $R_{k+1} = R_k[y_{k+1}]$ is finite over $R_k$.
+
+      Applying part (a) (transitivity of finiteness), if $R_{k+1}$ is finite over $R_k$ and $R_k$ finite over $A$, then $R_{k+1}$ is finite over $A$.
+
+      Consequence: since any $f \in A[y_1, \ldots, y_m]$ belongs to a finite $A$-algebra, $f$ must be integral over $A$ (since an element is integral iff it is contained in a finite extension).
+
+    \item let $x \in C$, since $x$ integral over $B$, it satisfies:
+      $$x^n + b_{n-1} x^{n-1} + \ldots + b_1 x + b_0 = 0,~~~~~b_i \in B$$
+
+      Let $B''=A[b_0, b_1, \ldots, b_{n-1}]$. Since each $b_i \in B$ and $B$ is integral over $A$\\
+      \hspace*{2em} $\Longrightarrow~$ each $b_i$ is integral over $A$.
+
+      Since all $b_i$ are integral over $B'$ $~~\Longrightarrow~ B'[x]$ is a finite $B'$-algebra.
+
+      By part (a) (transitivity of finiteness), $B'[x]$ is a finite $A$-algebra.
+
+      Therefore, $x$ is integral over $A$.
+
+    \item
+      \begin{enumerate}[I.]
+        \item subring:\\
+          let $x,y \in \tilde{A}$. Want to show $x+y, xy \in \tilde{A}$:
+
+          by part (b), the algebra $A[x,y]$ is finite over $A$.
+
+          Since $x+y,xy \in A[x,y]$, they are integral over $A$.
+
+          Thus $x+y, xy \in \tilde{A}$, since $\tilde{A} = \{ b \in B ~|~ b ~\text{integral over}~ A \}$.
+
+        \item idempotence\\
+          let $z \in B$ be integral over $\tilde{A}$
+          
+          we have a chain $A \subseteq \tilde{A} \subseteq \tilde{A}[x]$.
+
+          By definition, $\tilde{A}$ is integral over $A$, and $z$ is integral over $\tilde{A}$\\
+          thus by part (c), $z$ is integral over $A$.
+
+          Therefore, $z \in \tilde{A}$.
+      \end{enumerate}
+  \end{enumerate}
 \end{proof}
 
+
+\begin{defn}{4.4}{Integral closure.}
+  Given the ring $\tilde{A}$ from \ref{R.4.3}.(d), ie. $\tilde{A} = \{ y \in B ~|~ y ~\text{integral over}~ A \} \subset B$,
+  $\tilde{A}$ is the \emph{integral closure} of $A$ in $B$.
+
+  If $A=\tilde{A}$, then $A$ is \emph{integrally closed} in $B$.
+
+  An integral domain $A$ is \emph{normal} if it is \emph{integrally closed in its field of fractions}, that is if
+  $$A = \tilde{A} \subset K = Frac(A)$$
+
+  For any integral domain $A$, the integral closure of $A$ in its field of fractions $K=Frac(A)$ is also called the \emph{normalization} of $A$.
+\end{defn}
+
 \newpage
 
 \section{Exercises}