M .github/workflows/typos.toml → .github/workflows/typos.toml

@@ -1,5 +1,10 @@
 [default.extend-words]
 thm = "thm"
+
+# equations stuff
 ba = "ba"
-Strang = "Strang" # name
-Bootle = "Bootle" # name
+nd = "nd"
+
+# names
+Strang = "Strang"
+Bootle = "Bootle"

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

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

@@ -178,6 +178,61 @@ Notation: local ring $A$, its maximal ideal $\mM$, residue field $K=A/\mM$:
   $$A \supset \mM ~\text{or}~ (A, \mM) ~\text{or}~ (A, \mM, K)$$
 \end{defn}
 
+\subsection{$\mathbb{Z}$ and $K[X]$, two Principal Ideal Domains}
+
+
+\begin{lemma}{}
+  $\mathbb{Z}$ is a PID.
+\end{lemma}
+\begin{proof}
+  Let $I$ a nonzero ideal of $\mathbb{Z}$.
+
+  Since $I \neq \{0\}$, there is at least one nonzero integer in $I$. Choose the smallest element of $I$, namely $d$.
+
+  Observe that $(d) \subseteq I$, since $d \in I$. Then, every multiple $nd \in I$, since $I$ is an ideal.
+
+  Take $a \in I$. By the Euclidean division algorithm in $\mathbb{Z}$, $a=qd+r$, with $q,r \in \mathbb{Z}$ and $0 \leq r \leq d$.
+
+  Then $r = a - qd \in I$, but $d$ was chosen to be the smallest positive element of $I$, so the only possibility is $r=0$.
+
+  Hence, $a=qd$, so $a \in (d)$, giving $I \subseteq (d)$.
+
+  Since we had $(d) \subseteq I$ and now we got $I \subseteq (d)$, we have $I = (d)$, so every ideal of $\mathbb{Z}$ is principal. Thus $\mathbb{Z}$ is a Principal Ideal Domain(PID).
+\end{proof}
+
+\begin{lemma}{}
+  $K[X]$ is a PID.
+\end{lemma}
+\begin{proof}
+  This proof follows very similarly to the previous proof.\\
+
+  Let $K$ be a field, $K[X]$ a polynomial ring.
+
+  Take $\{0\} \neq I \subseteq K[X]$.
+
+  Since $I \neq \{0\}$, there is at least one non-zero polynomial in $I$.
+
+  Let $p(X) \in I$ be of minimal degree among nonzero elements of $I$.
+
+  Observe that $(p(X)) \subseteq I$, because $p(X) \in I$ and $I$ is an ideal.
+
+  Let $f(X) \in I$. By Euclidean division algorithm in $K[X]$, $\exists q, r \in K[X]$ such that $f(X) = q(X) \cdot p(X) + r(X)$ with eithr $r(X)=0$ or $deg(r) < deg(p)$.
+
+  Since $f,p \in I$, then $r(X) = f(X) - q(X)\cdot p(X) \in I$
+
+  If $r(X) \neq 0$, then $deg(r) < deg(p)$, which contradicts the minimality of $deg(p)$ in $I$.
+
+  Therefore, $r(X)=0$, thus $f(X)=q(X)\cdot p(X)$, hence $f(X) \in (p(X))$.
+  Henceforth, $I \subseteq (p(X))$.
+
+  Then, since $(p(X)) \subseteq I$ and $I \subseteq (p(X))$, we have that $I = (p(X))$.
+
+  So every ideal of $K[X]$ is principal; thus $K[X]$ is a PID.
+
+\end{proof}
+
+
+
 \subsection{Lemmas, propositions and corollaries}
 
 Let $\Sigma$ be a partially orddered set. Given subset $S \subset \Sigma$, an \emph{upper bound} of $S$ is an element $u \in \Sigma$ such that $s<u \forall s \in S$.
@@ -237,7 +292,7 @@ \end{proof}
 
 \section{Modules}
 
-\subsection{Modules}
+\subsection{Modules concepts}
 
 Let $A$ be a ring. An $A$-module is an Abelian group $M$ with a multiplication
 map
@@ -792,6 +847,8 @@ \begin{enumerate}[a.]
     \item the intersection of two prime ideals is prime
     \item the ideal $P_1+P_2$ generated by $2$ prime ideals $P_1,P_2$ is prime
     \item if $\psi: A \longrightarrow B$ ring homomorphism, then $\psi^{-1}$ takes maximal ideals of $B$ to maximal ideals of $A$
+    \item the map $\psi^{-1}$ of Proposition 1.2 takes maximal ideals of $A/I$
+      to maximal ideals of $A$
   \end{enumerate}
 \end{ex}
 \begin{proof}
@@ -832,7 +889,114 @@
       But $(0)$ is not maximal in $\mathbb{Z}$, because $\mathbb{Z}/(0) \cong \mathbb{Z}$ is not a field.
 
       Thus the preimages of maximal ideals under arbitrary ring homomorphisms need not be maximal.
+
+    \item $\psi: A \longrightarrow A/I$ quotient homomorphism, $I \subseteq A$ an ideal.
+
+      Let $M$ a maximal ideal of $A/I$, then $\frac{(A/I)}{M}$ is a field (Proposition 1.3).
+
+      By the isomorphism theorems,
+      $$\frac{(A/I)}{M} \cong \frac{A}{\psi^{-1}(M)}$$
+
+
+      Since $\frac{(A/I)}{M}$ is a field, the quotient $\frac{A}{\psi^{-1}(M)}$ is a field, so $\psi^{-1}(M)$ is a maximal ideal of $A$.
+
+      $\Longrightarrow~$ under $\psi$, preimages of maximal ideals are maximal.
   \end{enumerate}
+\end{proof}
+
+\begin{ex}{R.1.12.a}
+  if $I,J$ ideals and $P$ prime ideal, prove that
+  $$IJ \subset P ~\Longleftrightarrow~ I \cap J \subset P ~\Longleftrightarrow~ I ~\text{or}~ J \subset P$$
+\end{ex}
+\begin{proof}
+  assume $I \subseteq P$ (for $J \subseteq P$ will be the same, symmetric), take $x \in IJ$,
+
+  then
+  $$x = \sum_{k=1}^n a_k b_k$$
+  with $a_k \in I,~ b_k \in J$.
+
+  Each $a_k \in I \subseteq P$. Since $P$ an ideal,
+  $$\sum_{k=1}^n a_k b_k \in P$$
+
+  thus $x \in P$, hence $IJ \subseteq P$.
+
+  So $I \subseteq P$ or $J \subseteq P$ $~\Longrightarrow IJ \subseteq P$.
+
+  \vspace{0.5cm}
+  Conversely,\\
+  assume $P$ prime and $IJ \subseteq P$.
+
+  Suppose by contradiction that $I \not\subseteq P$ and $J \not\subseteq P$.
+  \begin{itemize}
+    \item[-] since $I \not\subseteq P,~ \exists a \in I$ with $a \not\in P$
+    \item[-] since $J \not\subseteq P,~ \exists b \in J$ with $b \not\in P$
+  \end{itemize}
+  Since $a \in I,~ b \in J,~~ ab \in IJ \subseteq P$, but $P$ is prime, so $ab \in P$ implies that $a \in P$ or $b \in P$. This contradicts $a,b$ being taken outside of $P$.
+
+  Thus $I \not\subseteq P$ and $J \not\subseteq P$ are false.
+
+  \vspace{0.3cm}
+  So both directions are proven, hence
+  $$IJ \subseteq P ~\Longrightarrow~ I \subseteq P ~\text{or}~ J \subseteq P$$
+
+\end{proof}
+
+\begin{ex}{R.1.18}
+  Use Zorn's lemma to prove that any prime ideal $P$ contains a minimal prime ideal.
+\end{ex}
+\begin{proof}
+  Let $P$ prime ideal of $R$.
+
+  $$S = \{ Q \subseteq R ~|~ Q ~\text{a prime ideal AND}~ Q \subseteq P \}$$
+
+  Goal: show that $S$ has a minimal element, the minimal ideal contained in $P$.
+
+  $P \subset S$, so $S$ is nonempty.
+
+  Let $C \subseteq S$ be a chain (= totally ordered subset) with respect to inclusion.
+
+  Define
+  $$Q_C = \bigcap_{Q \in C} Q$$
+
+  Clearly $Q_C \subseteq P$, since each $Q \in C$ is $Q \subseteq P$.
+
+  Since $C$ is ordered by inclusion, it is a decreasing chain of prime ideals.
+
+  Intersection of a decreasing chain of prime ideals is again a prime ideal:
+  \begin{itemize}
+    \item[-] if $ab \in Q_C$, then $ab \in Q ~\forall Q \in C$
+    \item[-] since $Q$ prime, $\forall Q \in C$ either $a \in Q$ or $b \in Q$
+  \end{itemize}
+
+  If there were some $Q_1,~ Q_2 \in C$ with $a \in Q_1$ and $b \not\in Q_2$, then by total ordering, either $Q_1 \subseteq Q_2$ or $Q_2 \subseteq Q_1$.
+
+  In either case: contradiction, since the smaller one would have to contain the element that was assumed to be excluded.
+
+  Thus $\forall Q \in C$ the same element $a, b$ must lie in all $Q$. $\Longrightarrow~$ lies in the intersection of them, $Q_C$.
+
+  Henceforth, $Q_C$ is a prime ideal and lies in $S$, and its a lower bound of $C$ in $S$.
+
+  Now, $S$ is nonempty, and every chain in $S$ has a lower bound in $S$ (its intersection).\\
+  Therefore, $S$ has a minimal element $P_{min}$.
+
+  By construction, $P_{min}$ is a prime ideal $P_{min} \subseteq P$, and by minimality there are no strictly smaller prime ideals inside $P$.
+
+  So $P_{min}$ is a minimal prime ideal, contained in $P$.
+\end{proof}
+
+\begin{ex}{R.1.10}
+\end{ex}
+\begin{proof}
+\end{proof}
+
+\begin{ex}{R.1.11}
+\end{ex}
+\begin{proof}
+\end{proof}
+
+\begin{ex}{R.1.4}
+\end{ex}
+\begin{proof}
 \end{proof}
 
 \subsection{Exercises Chapter 2}