Expand algebraic and tensor product content in notes

Added detailed explanations and new propositions on tensor products, algebraic elements, minimal polynomials, and coproducts in commutative algebra and field theory. Improved definitions and universal properties, clarified relationships between irreducible and prime elements in PIDs, and expanded the treatment of normal extensions and splitting fields. Various minor corrections and enhancements to mathematical rigor and cross-referencing were also made.

Author: hooyuser

algebraic_construction.tex

@@ -1,5 +1,6 @@
 \documentclass{report}
 
+
 % \usepackage[extract=python]{memoize}
 % \mmzset{%
 %   prefix=memos/,
@@ -11,17 +12,9 @@
 \input{preamble}
 
 
-\usepackage[page,toc,titletoc,title]{appendix}
-\usepackage{chemfig}
-\usepackage{relsize}
-\usepackage{multirow}
-\usepackage{colortbl}
-\usepackage{mathrsfs}
-\usepackage{stmaryrd}
-\usepackage{bbm}
-\usepackage{graphicx}
 
-\usepackage{leftindex}
+% make index
+\makeindex[columns=2, title=Index, intoc]
 
 \newcommand{\divides}{\mid}
 
@@ -161,6 +154,8 @@
   \csname qendmydef\endcsname
 }
 
+
+
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 % Start document
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
@@ -250,6 +245,8 @@ \chapter*{Notation Conventions}
 
 \include{appendices}
 
+\printindex
+
 % \begin{Theorem}{Title}{label}
 %     This is the statement of the theorem.
 % \end{Theorem}

associative_algebra.tex

@@ -3537,7 +3537,7 @@ \subsection{Fibered Product and Fibered Coproduct}
 
 \end{example}
 
-\begin{proposition}{}{}
+\begin{proposition}{Diagonal Base Change for Fibered Products}{diagonal_base_change_fibered_product}
     Let $\mathsf{C}$ be a category which admits fibered products. Let $\sigma: S\to T$, $f: X\to S$, and $g:Y\to S$ be morphisms in $\mathsf{C}$. Then there is an  isomorphism
     \[
       X \times_{S} Y \cong \left(X \times_{T} Y\right) \times_{(S \times_{T} S)} S.

category_theory.tex

@@ -318,10 +318,10 @@ \subsection{Prime Elements}
 \end{definition}
 
 
-\begin{proposition}{}{}
+\begin{proposition}{}{divisibility_ideal_characterization}
     Let $R$ be a commutative ring.
     \begin{enumerate}[(i)]
-        \item $a \mid b\iff(b) \subseteq (a)$.
+        \item $a \divides b\iff(b) \subseteq (a)$.
         \item $u\in R^\times \iff (u) = R  \iff \forall r\in R,\,u\mid r$.
     \end{enumerate}
 \end{proposition}
@@ -337,7 +337,7 @@ \subsection{Prime Elements}
 
 
 
-\begin{proposition}{Prime Element and Prime Ideal}{equivalent_characterization_of_prime_element}
+\begin{proposition}{Prime Element and Nonzero Prime Ideal}{equivalent_characterization_of_prime_element}
     Suppose $R$ is a commutative ring and $a\in R$. Then
     \[
         a\text{ is prime }\iff (a)\text{ is a nonzero prime ideal}.
@@ -433,14 +433,28 @@ \section{Integral Domain}
 
 Associatedness can also be described in terms of the action of $R^\times$ on $R$ via multiplication: two elements of $R$ are associates if they are in the same $R^\times$-orbit.
 \begin{definition}{Irreducible Element}{}
-    Let $R$ be an integral domain. An element $a\in R$ is called \textbf{irreducible} if
+    Let $R$ be an integral domain. An non-zero element $a\in R$ is called \textbf{irreducible} if
     \begin{enumerate}[(i)]
         \item $a\notin R^\times$, i.e. $a$ is not a unit.
         \item $a=bc\implies b\in R^\times\text{ or }c\in R^\times$.
     \end{enumerate}    
 \end{definition}
 
 0 is never an irreducible element.
+
+\begin{proposition}{Dividing Irreducible Element in PID Implies Associate or Unit}{dividing_irreducible_element_in_PID_implies_associate_or_unit}
+    Let $R$ be an integral domain and $f,g\in R$. If $f$ is irreducible and $f \in (g)$, then one of the following holds:
+    \begin{enumerate}[(i)]
+        \item $g \in R^\times$, i.e. $(g) = R$.
+        \item $g$ is an associate of $f$, i.e. $(g) = (f)$.
+    \end{enumerate}
+\end{proposition}
+\begin{prf}
+    Since $f \in (g)$, there exists $h \in R$ such that $f = gh$. Since $f$ is irreducible, we have $g \in R^\times$ or $h \in R^\times$. If $g \in R^\times$, then $(g) = R$. If $h \in R^\times$, then $g$ is an associate of $f$ and $(g) = (f)$.
+\end{prf}
+
+
+
 \begin{proposition}{Prime Element $\implies$ Irreducible Element in Integral Domain}{prime_element_implies_irreducible_element_in_integral_domain}
     Let $R$ be an integral domain. Then every prime element in $R$ is irreducible.
 \end{proposition}
@@ -505,16 +519,30 @@ \section{Principal Ideal Domain}
 
 
 \begin{proposition}{Nonzero Prime Ideal $\iff$ Maximal Ideal in PID}{nonzero_prime_ideal_iff_maximal_ideal_in_PID}
-    Let $R$ be a PID and $(p)$ be any nonzero prime ideal in $R$. Then $(p)$ is a maximal ideal and $p$ is an irreducible element.
+    Let $R$ be a PID and $p\in R$. Then the following are equivalent:
+    \begin{enumerate}[(i)]
+        \item $p$ is a prime element.
+        \item $p$ is an irreducible element.
+        \item $(p)$ is a nonzero prime ideal.
+        \item $(p)$ is a maximal ideal.
+    \end{enumerate}
 \end{proposition}
 
 \begin{prf}
-    Let $I=(p)\subseteq R$ be a prime ideal. We only need to show $R/I$ is a field. Let $\overline{a}\in R/I$ be a nonzero element. Then $a\notin I$. Since $I$ is prime, $a$ is not a multiple of any prime element in $R$. Thus $a$ is irreducible. Since $R$ is a PID, $a$ is prime. Thus $\overline{a}$ is prime in $R/I$. Since $R/I$ is an integral domain, $\overline{a}$ is a maximal ideal in $R/I$. That implies $R/I$ is a field. Hence $I$ is a maximal ideal.
-
-    By \Cref{th:equivalent_characterization_of_prime_element}, $p$ is a prime element. Since \hyperref[th:irreducible_element_iff_prime_element_in_UFD]{ in UFD, prime elements are irreducible elements}, we conclude $p$ is irreducible.
-\end{prf}
+    By \Cref{th:equivalent_characterization_of_prime_element}, (i)$\iff$(iii). Since \hyperref[th:irreducible_element_iff_prime_element_in_UFD]{ in UFD, prime elements are irreducible elements}, (i)$\iff$(ii). Since maximal ideals are prime, (iv)$\implies$(iii).
 
 
+    To show (ii)$\implies$(iv), suppose $p$ is irreducible. Let $I$ be an ideal of $R$ such that 
+    \[
+    (p)\subseteq I \subseteq R.
+    \]
+    Since $R$ is a PID, there exists some $a\in R$ such that $I=(a)$. Then we have $(p)\subseteq (a)$, which implies there exists $r\in R$ such that $p=ra$ by \Cref{th:divisibility_ideal_characterization}. Because $p$ is irreducible, by definition we have $a\in R^\times$ or $r\in R^\times$. 
+    \begin{itemize}
+        \item If $a\in R^\times$, then $I=(a)=R$.
+        \item If $r\in R^\times$, then $a=r^{-1}p$, which implies $(a)\subseteq (p)$ and thus $I=(a)=(p)$.
+    \end{itemize}
+    This shows that there are no ideals strictly between $(p)$ and $R$. Hence $(p)$ is maximal.
+\end{prf}
 
 
 \section{Construction}
@@ -547,6 +575,20 @@ \subsection{Product}
     This implies $I_1\times I_2$ is an ideal of $R_1\times R_2$.
 \end{prf}
 
+\subsection{Coproduct}
+Since we have the isomorphisms of categories
+$$
+\mathsf{CRing}\cong (\mathsf{CRing}/\mathbb{Z})\cong \mathbb{Z}\hspace{0.5pt}{\text -}\mathsf{CAlg},
+$$
+the coproduct in the category of commutative rings is given by the tensor product of commutative $\mathbb{Z}$-algebras.
+\begin{definition}{Coproduct of Commutative Rings}{}
+    Let $R$, $S$ be commutative rings. The \textbf{coproduct} of $R$ and $S$, denoted by $R\otimes_{\mathbb{Z}} S$, is the tensor product of $R$ and $S$ over $\mathbb{Z}$. The addition and multiplication are defined as follows:
+    \begin{align*}
+        (r_1\otimes s_1)+(r_2\otimes s_2)&=(r_1+r_2)\otimes (s_1+s_2)-r_1\otimes s_2 - r_2\otimes s_1,\\
+        (r_1\otimes s_1)\cdot(r_2\otimes s_2)&=(r_1r_2)\otimes (s_1s_2).
+    \end{align*}
+\end{definition}
+
 
 \subsection{Quotient Ring}
 \begin{definition}{Quotient Ring}{}
@@ -902,12 +944,12 @@ \subsection{Localization}
 \begin{prf}
     Note
     \[
-        \{I \in \operatorname{Spec} R: I \cap S=\varnothing\}=\{I \in \operatorname{Spec} R: I \cap (R-\frak{p})=\varnothing\}=\{I \in \operatorname{Spec} R: I\subseteq \frak{p}\}.
+        \{I \in \operatorname{Spec} R: I \cap S=\varnothing\}=\{I \in \operatorname{Spec} R: I \cap (R-\mathfrak{p})=\varnothing\}=\{I \in \operatorname{Spec} R: I\subseteq \mathfrak{p}\}.
     \]
     For any ideal $S^{-1}I \in \spec S^{-1}R$, where
     $I\in  \{I \in \operatorname{Spec} R: I \cap S=\varnothing\}$, we have $I\subseteq \mathfrak{p}$, which implies $S^{-1}I\subseteq  S^{-1}\mathfrak{p}$. Thus we see $S^{-1}\mathfrak{p}$ is the unique maximal ideal of $S^{-1}R$. 
 
-    According to \Cref{th:localization_at_ideal_respects_quotients}, we have an isomorphism $(R/\frak{p})_{\frak{p}}\cong R_\frak{p}/\frak{p}R_\frak{p}$ and the following commutative diagram in $R\text{-}\mathsf{CAlg}$ 
+    According to \Cref{th:localization_at_ideal_respects_quotients}, we have an isomorphism $(R/\mathfrak{p})_{\mathfrak{p}}\cong R_\mathfrak{p}/\mathfrak{p}R_\mathfrak{p}$ and the following commutative diagram in $R\text{-}\mathsf{CAlg}$ 
     \[
         \begin{tikzcd}
             R \arrow[r, "\pi_\mathfrak{p}"] \arrow[d, "l_{R-\mathfrak{p}}"'] &[+5em] R/\mathfrak{p} \arrow[d, "l_{R-\mathfrak{p}}"]              \\[+2em]