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

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

@@ -2060,6 +2060,34 @@
 
 \section{Rings of fractions $S^{-1}A$ and localization}
 
+\subsection{Small recap}
+\paragraph{principal ideal:} generated by a single element, ie. for $a \in A$, $(a)=aA=\{ra ~|~ r\in A\}$.
+
+\paragraph{unit:} $u \in A$ is a unit if $\exists v \in A$ with $uv=1$.
+
+\paragraph{proper ideal:} $I \subsetneq A$, equivalently $1 \not\in I$. Therefore, a proper ideal does not have units.
+
+\paragraph{maximal ideal:}
+$\mM \subset A$ ($A$ ring) with $m \neq A$ and there is no ideal $I$ strictly between $\mM$ and $A$. ie. if $\mM$ maximal and $\mM \subseteq I \subseteq A$, either $\mM=I$ or $I=A$.
+\begin{itemize}
+  \item is proper
+  \item for $m$ maximal ideal, $\frac{A}{m}$ is a field.
+  \item every maximal ideal is prime
+  \item elements of a maximal ideal are always non-units\\
+    \hspace*{2em} since if $x \in m$ were a unit, then $1=x^{-1}x \in m$, a contradiction since if $1 \in m$ then $m=A$, which would mean that it is not proper; and it is maximal, hence proper.
+\end{itemize}
+
+\paragraph{prime ideal:} proper ideal $P$ such that
+$$ab \in P ~~\Longrightarrow~ a \in P ~\text{or}~ b \in P$$
+for $P$ prime, $\frac{A}{P}$ is an integral domain.
+
+\paragraph{Spec(A):} prime spectrum, $Spec(A)=\{ P \subset A ~|~ P ~\text{is prime}\}$
+
+
+\vspace{0.6cm}
+
+
+
 \subsection{Rings of fractions $S^{-1}A$}
 
 \begin{defn}{6.1}[ring of fractions] \label{def.6.1}
@@ -2091,10 +2119,6 @@ is a ring homomorphism.
   \end{enumerate}
 \end{prop}
 
-\begin{eg}{ }
-  TODO
-\end{eg}
-
 \begin{lemma}{6.2} \label{6.2}
   For $f \in A$, write $S = \{ 1, f, f^2, \ldots \}$, and $A_f \cong S^{-1}A$.
 
@@ -2102,6 +2126,7 @@ Then
   $$A_f \cong \frac{A[X]}{(Xf-1)}$$
 \end{lemma}
 \begin{proof}
+  (adjoining a variable $X$ and imposing $Xf=1$ is the algebraic way of saying "add $f^{-1}$".)\\
   Define the homomorphism
   \begin{align*}
     \psi: A[X] &\longrightarrow A_f\\
@@ -2320,6 +2345,44 @@ $$ T^{-1}(A/I) = S^{-1}(A/I) \cong \frac{S^{-1}A}{S^{-1}I}$$
   it's easy to see that this is a ring homomorphism.
 \end{proof}
 % TODO 6.6 (proposition and corollary)
+
+
+\vspace{1.0cm}
+\subsection{Key ideas}
+Main idea: make some choosen elements invertible.
+
+For localization, it is important to not get confused between "localization at an element" and "localization at a prime ideal".
+
+\paragraph{Localization at an element $f$.}
+
+Let $S=\{1, f, f^2, f^3, \ldots \}$ a multiplicative set. Then $S^{-1}A$ is the ring of fractions, and
+$$A_f= S^{-1}A = A \big[ \frac{1}{f} \big] = \{\frac{a}{f^n} ~|~ a \in A, n \geq 0\}$$
+
+ie. "make $f$ invertible", then $\frac{1}{f} \in A_f$.\\
+$\Longrightarrow~ A_f$ is the ring obtained from $A$ by making every $f \in S$ invertible.
+
+Lemma \ref{6.2} will show that $A_f \cong \frac{A[X]}{(Xf-1)}$.
+
+\paragraph{Localization at a prime ideal P.}
+
+For $P \in Spec(A)$ (ie. $P$ is a prime ideal),
+
+$$A_P = ( \underbrace{A \setminus P}_{S} )^{-1} A = \{ \frac{a}{s} ~|~ a \in A, s \not\in P \}$$
+
+ie. "make every element of $A$ not in $P$ invertible". Since $p \in P$, then $\frac{1}{p} \not\in A_P$.\\
+
+Given the ideal $I \subseteq A$, its extension to the localized ring is
+$$I A_P = \{ \frac{a}{s} ~|~ a \in I, s \not\in P \}$$
+
+\vspace{0.6cm}
+
+The only nonunits are fractions whose numerator lies in $P$: $\{ \frac{a}{s} ~|~ a \in P, s \not\in P \} = P A_P$
+
+Hence, $A_P$ has one maximal ideal: $P A_P$.
+
+$(A_P, P A_P)$ denotes the local ring $A_P$ together with $P A_P$, often called the localization of $A$ at $P$ as a local ring.
+
+\vspace{1.0cm}