M commutative-algebra-notes.pdf → commutative-algebra-notes.pdf

M commutative-algebra-notes.tex → commutative-algebra-notes.tex

@@ -58,6 +58,11 @@ \newenvironment{cor}[1]
 {\renewcommand\theinnercor{#1}\innercor}
 {\endinnercor}
 
+\newtheorem{innerremark}{Remark}
+\newenvironment{remark}[1]
+{\renewcommand\theinnerremark{#1}\innerremark}
+{\endinnerremark}
+
 \newtheorem{innereg}{Example}
 \newenvironment{eg}[1]
 {\renewcommand\theinnereg{#1}\innereg}
@@ -187,6 +192,39 @@ \hspace*{4em}$\Longleftrightarrow$ $A$ has only one maximal ideal.\\
     \hspace*{4em}$\Longleftrightarrow$ all the nonunits of $A$ form an ideal.
 \end{defn}
 
+\begin{prop}{(R/m=Field)}\label{maximal-quotient-field}
+  Let $R$ be a commutative ring with identity, and let $m \subsetneq R$ be a maximal ideal. Then the quotient ring $R/m$ is a field.
+\end{prop}
+\begin{proof}
+  Let $x + m$ be a nonzero element of $R/m$. Since
+  $$x + m \neq m$$
+  we have $x \not\in m$.
+
+  Consider the ideal $m + (x)$, where $(x)$ is the principal ideal generated by $x$.
+  Since $x \not\in m$, this ideal strictly contains $m$:
+  $$m \subsetneq m + (x)$$
+
+  Because $m$ is maximal, there are no proper ideals strictly between $m$ and $R$. Therefore,
+  $m + (x) = R$
+
+  In particular, $1 \in m + (x)$. Hence, there exist
+  $a \in m$ and $y \in R$ such that $a + yx = 1$.
+
+  Passing to the quotient modulo $m$, we obtain
+  $$
+    (y+m)(x+m)
+    = yx+m
+    = 1+m,
+  $$
+  because $a \in m$ implies $a+m=m$.
+
+  Thus, $y+m$ is the multiplicative inverse of
+  $x+m$. Therefore, every nonzero element of
+  $R/m$ is invertible, and hence $R/m$ is a field.
+\end{proof}
+
+
+
 \subsection{Z and K[X], two Principal Ideal Domains}
 
 
@@ -323,6 +361,10 @@ \begin{itemize}
 \item since $\psi$ is $A$-linear, $A[\psi]$ is a commutative ring.
 \item $M$ is a module over $A[\psi]$, so $\psi$ beomes multiplication by a ring element.
 \end{itemize}
+
+\begin{remark}{}
+  Vector spaces are defined over fields, a Module is the generalization of a vector space where the scalars come from a ring instead of a field.
+\end{remark}
 
 \subsection{Cayley-Hamilton theorem, Nakayama lemma, and corollaries}
 
@@ -1153,8 +1195,8 @@
   This extends on point (b) from the previous proposition \ref{R.4.3}.
 \end{lemma}
 \begin{proof}
-  Suppose $y$ is integral over $A$. By definition $\exists~~ f \in A[T]$, with
-  $f$ monic, such that $f(y)=0$.
+  Suppose $y$ is integral over $A$. By definition $\exists~ f \in A[T]$, with
+  $f$ monic, such that $f(y)=0$. (here $T$ is a formal variable).
 
   Let $deg(f)=d$, so that for $f(y)=0$ we have
   $$y^d + a_{d-1} y^{d-1} + \ldots + a_1 y + a_0 = 0 ~~~~ a_i \in A$$
@@ -1392,7 +1434,7 @@ Equivalently, $V(I) = \emptyset ~~\Longleftrightarrow  1 \in I$.
 
 \vspace{0.4cm}
 
-\begin{thm}{R.4.10}[Weak Nullstellensatz - Zariski's lemma] \label{zariski}
+\begin{thm}{R.4.10}[Weak Nullstellensatz - Zariski's lemma] \label{zariski2}
   let $k$ a field, $K$ a $k$-algebra which
   \begin{enumerate}
     \item is finitely generated as a $k$-algebra
@@ -1762,7 +1804,7 @@ $$\mathfrak{P} \in Spec~ A \longleftrightarrow X \subseteq V(J)$$
   where $X = V(\mathfrak{P})$.
 \end{proof}
 
-\subsection{Zariski, Weak Nullstellensatz, Nullstllensatz}
+\subsection{Again: Zariski, Weak Nullstellensatz, Nullstllensatz}
 This subsection proves again Zariski's lemma, Weak Nullstellensatz and Nullstellensatz, in a more interconnected way than the proofs described previously.
 
 \begin{lemma}{}[Zariski's lemma]
@@ -1777,10 +1819,10 @@ $K$ a fingen $k$-algebra and a field $\Longleftrightarrow$ $K=k[X_1, \ldots, X_n]$ a field.\\
   Want to show that $K$ is algebraic over $k$.
 
   Since $K$ is a fingen $k$-algebra, by Noether normalization \ref{noether-normalization},\\
-  $\exists~ \{y_1, ldots, y_r\} \subset K$ such that
+  $\exists~ \{y_1, \ldots, y_r\} \subset K$ such that
   \begin{itemize}
     \item the set $\{y_1, \ldots, y_r\}$ is algebraically independent over $k$
-    \item $K$ is integral over the polynomial ring $R=k[y_1, \ldots, y_r] \sbuset K$.
+    \item $K$ is integral over the polynomial ring $R=k[y_1, \ldots, y_r] \subset K$.
   \end{itemize}
 
   By \ref{R.4.9}:\\
@@ -1825,10 +1867,11 @@ proven at previous proof.
 \end{proof}
 
 
+
 \vspace{0.8cm}
 Now, let $m\subset k[X_1, \ldots, X_n]$ a maximal ideal. Let $K=k[X_1, \ldots, X_n]$.
 
-Then $\frac{k[X_1, \ldots, X_n]}{m} = \frac{K}{m}$ is a field (by basic ideal quotient theory).
+Then $\frac{k[X_1, \ldots, X_n]}{m} = \frac{K}{m}$ is a field (by basic ideal quotient theory \ref{maximal-quotient-field}).
 
 If it's a fingen $k$-algebra $\Longrightarrow$ by Zariski's lemma, $\frac{K}{m}$ is algebraic over $k$.
 
@@ -1840,7 +1883,7 @@
 Thus, $\forall x_i \in k,~ \psi(x_i)=a_i \in k$.
 And by definition of a quotient ring,
 
-$psi(x_i)$ is the coset
+$\psi(x_i)$ is the coset
 \begin{align*}
   \psi(x_i)&=x_i+m\\
            &\Longrightarrow a_i = x_i +m\\
@@ -1895,7 +1938,7 @@ \begin{enumerate}[a.]
   \item already proven in the previous steps.
     
   \item $I(V(J)) = rad J$\\
-  ie. vanishing ideal of a variety = radical of the ideal definingh the variety
+  ie. vanishing ideal of a variety = radical of the ideal defining the variety
 
   where $rad~J = \{ f \in R | f^n \in J ~\text{for some}~ n>0 \}$.
 
@@ -2852,7 +2895,7 @@ General rule of saturation:\\
   two multiplicative sets $S,~T$ yield the same localization, $S^{-1}A = T^{-1}A$, iff the have the same saturation.
 
   Saturation of $S$: $\hat{S}$, set of all elements in $A$ that divide some element of $S$:
-  $$\hat{S} = \{ a \in A ~|~ \exist~ b \in A ~\text{s.th.}~ ab \in S \}$$
+  $$\hat{S} = \{ a \in A ~|~ \exists~ b \in A ~\text{s.th.}~ ab \in S \}$$
 
   \vspace{0.4cm}
   Example 1: $A=\mathbb{Z}$,\\
@@ -2894,8 +2937,8 @@
     Let $u \in U$; the image of $u$ in $S^{-1}A$ must be a unit.
 
     Units in $S^{-1}A$ are the fractions $\frac{s}{a}$ such that their inverse exists.\\
-    \hspace*{2em} $\exist~ x=\frac{a}{s}$ such that $u \frac{a}{s} = \frac{1}{1}$\\
-    \hspace*{2em} $\Longrightarrow~ \frac{ua}{s} = \frac{1}{1} ~\Longrightarrow~ \exist s' \in S$ such that $s'(ua-s)=0$\\
+    \hspace*{2em} $\exists~ x=\frac{a}{s}$ such that $u \frac{a}{s} = \frac{1}{1}$\\
+    \hspace*{2em} $\Longrightarrow~ \frac{ua}{s} = \frac{1}{1} ~\Longrightarrow~ \exists s' \in S$ such that $s'(ua-s)=0$\\
     $\Longrightarrow~ s'ua=s's \in S$
 
     since $s,s' \in S$, their product is also in $S$.