From 5b492925db7c4dc4fc28196e957dca637c00dda1 Mon Sep 17 00:00:00 2001 From: Derived Cat Date: Sun, 22 Dec 2024 00:41:49 -0500 Subject: [PATCH] update --- associative_algebra.tex | 40 ++-- category_theory.tex | 113 ++++++++++- commutative_ring.tex | 437 +++++++++++++++++++++++++++++++++------- field.tex | 26 +++ module.tex | 55 ++++- set_theory.tex | 30 ++- valuation_theory.tex | 181 +++++++++++++++++ 7 files changed, 777 insertions(+), 105 deletions(-) diff --git a/associative_algebra.tex b/associative_algebra.tex index 14f5ff5..daca3be 100644 --- a/associative_algebra.tex +++ b/associative_algebra.tex @@ -212,6 +212,19 @@ \section{Determinant and Trace} \end{enumerate} \end{definition} +\begin{example}{Left Multiplication Endomorphism}{} + Let $R$ be a commutative ring and $A$ be an $R$-Algebra. Suppose $A$ is a free $R$-module of finite rank $n$. Given any $a\in A$, the left multiplication endomorphism $l_a\in\mathrm{End}_{R\text{-}\mathsf{Mod}}(A)$ is defined by + \begin{align*} + l_a:A &\longrightarrow A\\ + x &\longmapsto ax. + \end{align*} +\end{example} + + +\begin{definition}{Trace of Elements in $R$-algebra}{} + Let $A$ be an $R$-algebra. The \textbf{trace} of an element $a\in A$ is defined as the trace of the left multiplication endomorphism $l_a\in\mathrm{End}_{R\text{-}\mathsf{Mod}}(A)$. +\end{definition} + @@ -230,32 +243,7 @@ \subsection{Construction} X\arrow[u, "\iota"] \arrow[ru, "f"'] \& \end{tikzcd} \end{center} - The free $R$-module $\mathrm{Free}_{R\text{-}\mathsf{CAlg}}(X)$ can be contructed as the polynomial algebra $R[X]$. + The free commutative $R$-algebra $\mathrm{Free}_{R\text{-}\mathsf{CAlg}}(X)$ can be contructed as the polynomial algebra $R[X]$. \end{definition} -\begin{definition}{Finite-type Commutative Algebra}{} - Let $R\to A$ be a commutative ring homomorphism. We say $A$ is a \textbf{finite-type $R$-algebra}, or that $R\to A$ is \textbf{of finite type}, if one of the following equivalent conditions holds: - \begin{enumerate}[(i)] - \item there exists a finite set of elements $a_1,\cdots,a_n$ of A such that every element of $A$ can be expressed as a polynomial in $a1,\cdots,an$, with coefficients in $K$. - \item there exists a finite set $X$ such that $A\cong R[X]/I$ as $R$-algebra where $I$ is an ideal of $R[X]$. - \end{enumerate} - -\end{definition} - - -\begin{proposition}{Finite Generation Implies Finite Type}{} - Let $A$ be a $R$-algebra. If $A$ is finitely generated as an $R$-module, then $A$ is a finite-type $R$-algebra. -\end{proposition} -\begin{prf} - This holds because if each element of $A$ can be expressed as an $R$-linear combination of finitely many elements of $A$, then each element of $A$ can also be expressed as a polynomial in finitely many elements of $A$ with coefficients in $R$. - - An alternative proof can be given by utilizing the universal property of the free contruction. Suppose $A$ is finitely generated as an $R$-module. Then there exists some a finite set $X=\{x_1,\cdots,x_n\}$ and a surjective $R$-linear map $\varphi:R^{\oplus X}\to A$. Define $f=\varphi\circ \iota$, where $\iota:X\to R^{\oplus X}$ is the inclusion map. - \begin{center} - \begin{tikzcd}[ampersand replacement=\&] - R^{\oplus X}\arrow[r, dashed, "\exists !\,\widetilde{j}"] \&R[X]\arrow[r, dashed, "\exists !\,\widetilde{f}"] \& A\\[0.3cm] - \& X\arrow[ul, "\iota"] \arrow[u, "j"] \arrow[ru, "f:=\varphi\circ \iota"'] \& - \end{tikzcd} - \end{center} - The universal property of free $R$-module induces a unique $R$-linear map $\widetilde{j}:R^{\oplus}\to R[X]$ such that $j=\widetilde{j}\circ \iota$. And the universal property of free commutative $R$-algebra induces a unique $R$-algebra homomorphism $\widetilde{f}:R[X]\to A$ such that $f=\widetilde{f}\circ j$. Note $f=\varphi\circ \iota=\left(\widetilde{f}\circ \widetilde{j}\right)\circ \iota$. By the uniqueness of the universal property of $R^{\oplus}$, we have $\widetilde{f}\circ \widetilde{j}=\varphi$. Since $\widetilde{f}\circ \widetilde{j}$ is surjective, $\widetilde{f}$ must be surjective, which implies $A$ is a finite-type $R$-algebra. -\end{prf} diff --git a/category_theory.tex b/category_theory.tex index 533083f..c3604a3 100644 --- a/category_theory.tex +++ b/category_theory.tex @@ -2536,7 +2536,7 @@ \section{Limit and Colimit} \begin{proposition}{}{} - If $F:\mathsf{C}\to \mathsf{D}$ creates limits for $\mathcal{K}$ and $\varprojlim F\circ K$ exists for all $K\in\mathcal{K}$, then $\varprojlim K$ exists for all $K\in\mathcal{K}$ and $F$ preserves limits for $\mathcal{K}$. + Suppose $\mathcal{K}\subseteq\mathrm{Ob}\left([\mathsf{J}, \mathsf{C}]\right)$. If $F:\mathsf{C}\to \mathsf{D}$ creates limits for $\mathcal{K}$ and $\varprojlim F\circ K$ exists for all $K\in\mathcal{K}$, then $\varprojlim K$ exists for all $K\in\mathcal{K}$ and $F$ preserves limits for $\mathcal{K}$. \end{proposition} \begin{prf} @@ -2599,10 +2599,6 @@ \section{Limit and Colimit} -\begin{proposition}{Limits Commute with Limits}{} - Let $\mathsf{C}$ be a category and $F:\mathsf{I}\times \mathsf{J}\to\mathsf{C}$ be a diagram. If $\varprojlim\limits_{i\in \mathsf{I}}\varprojlim\limits_{j\in \mathsf{J}}F(i,j)$ and $\varprojlim\limits_{j\in \mathsf{J}}\varprojlim\limits_{i\in \mathsf{I}}F(i,j)$ exist, then they are naturally isomorphic. -\end{proposition} - \begin{definition}{Exact Functor}{exact_functor} Let $F$ be a functor between finitely complete categories categories $\mathsf{C}$ and $\mathsf{D}$. @@ -2778,6 +2774,113 @@ \section{Limit and Colimit} \end{definition} +\begin{proposition}{Limits Commute with Limits}{} + Let $F: \mathsf{I}\times \mathsf{J}\to \mathsf{C}$ be a functor. Suppose for each $i\in \mathrm{Ob}\left(\mathsf{I}\right)$, the limit of the diagram $F(i,-):\mathsf{J}\to \mathsf{C}$ exists and is denoted by $F_{i,\infty}:=\varprojlim F(i,-)$. Define a diagram $F_{-,\infty}:\mathsf{I}\to \mathsf{C}$ as follows + \[ + \begin{tikzcd}[ampersand replacement=\&] + \mathsf{I}\&[-25pt]\&[+10pt]\&[-30pt]\mathsf{C}\&[-30pt]\&[-30pt] \\ [-15pt] + i \arrow[dd, "\lambda"{name=L, left}] + \&[-25pt] \& [+10pt] + \& [-30pt] \varprojlim F(i,-) \arrow[dd, "F_{-,\infty}(\lambda)"{name=R}] \\ [-10pt] + \& \phantom{.}\arrow[r, "F_{-,\infty}", squigarrow]\&\phantom{.} \& \\[-10pt] + i' \& \& \& \varprojlim F(i',-) + \end{tikzcd} + \] + where $F_{-,\infty}(\lambda)$ is induced by the universal property of $\varprojlim F(i', -)$ + \[ + \begin{tikzcd}[ampersand replacement=\&, background color=mypropbg!10] + \& \varprojlim F(i,-) \arrow[ld] \arrow[rd] \& \\[+10pt] +F(i,j) \arrow[rr, black!35] \arrow[dd, "{F(\lambda, \mathrm{id}_j)}"'] \& \& F(i,j') \arrow[dd, "{F(\lambda, \mathrm{id}_{j'})}"] \\[+10pt] + \& \varprojlim F(i', -)\arrow[ld] \arrow[rd] \& \\[+10pt] +F(i',j) \arrow[rr] \& \& F(i', j') + % absulute arrow + \arrow[from=1-2, to=3-2, "F_{-,\infty}(\lambda)"', near end, crossing over] + \end{tikzcd} + \] + Then $\varprojlim F_{-,\infty}$ exists if and only if $\varprojlim F$ exists. Moreover, if $\varprojlim F$ exists, we have natural isomorphism + \[ + \varprojlim F\cong \varprojlim F_{-,\infty}. + \] +\end{proposition} +\begin{prf} + For any cone + $$ + h=\left(c\xlongrightarrow{h_i} F_{i,\infty}\right)_{i \in\mathrm{Ob}\left(\mathsf{I}\right)}\in \mathrm{Ob}\left(\mathsf{Cone}\left(\mathsf{C}, \varprojlim F_{-,\infty}\right)\right), + $$ + we can construct a cone + $$ + \phi(h):=\left( c\xlongrightarrow{h_i} F_{i,\infty}\xlongrightarrow{\pi_{i,j}} F(i,j)\right)_{(i,j) \in\mathrm{Ob}\left(\mathsf{I}\times \mathsf{J}\right)}\in \mathrm{Ob}\left(\mathsf{Cone}\left(\mathsf{C}, \varprojlim F\right)\right). + $$ + To verify it is a cone, we need to check for any morphism $(\lambda, \mu):(i,j)\to (i',j)$ in $\mathsf{I}\times \mathsf{J}$, the following diagram commutes + \[ + \begin{tikzcd} + & c \arrow[ld, "h_i"'] \arrow[rd, "h_{i'}"] & \\[+10pt] +{F_{i,\infty}} \arrow[d, "{\pi_{i,j}}"'] & & {F_{i',\infty}} \arrow[d, "{\pi_{i',j'}}"] \\[+10pt] +{F(i,j)} \arrow[rr, "{F(\lambda,\mu)}"'] & & {F(i',j')} +\end{tikzcd} +\] +Note that $F(\lambda,\mu)= F(\mathrm{id}_{i'},\mu)\circ F(\lambda,\mathrm{id}_{j'})$. It is sufficient to show the following diagram commutes + \[ + \begin{tikzcd} + {F(i,j)} \arrow[rrrr, "{F(\lambda,\mathrm{id}_{j'})}"] \arrow[dddd, "{F(\mathrm{id}_i,\mu)}"'] & & & & {F(i',j)} \arrow[dddd, "{F(\mathrm{id}_{i'},\mu)}"] \\ + & & c \arrow[ld, "h_i"'] \arrow[rd, "h_{i'}"] & & \\ + & {F_{i,\infty}} \arrow[rr, "{F_{-,\infty}(\lambda)}"'] \arrow[luu, "{\pi_{i,j}}"'] \arrow[ldd, "{\pi_{i,j'}}"] & & {F_{i',\infty}} \arrow[ruu, "{\pi_{i',j}}"] \arrow[rdd, "{\pi_{i',j'}}"'] & \\ + & & & & \\ + {F(i,j')} \arrow[rrrr, "{F(\lambda,\mathrm{id}_{j'})}"'] & & & & {F(i',j')} + \end{tikzcd} + \] + + \noindent Conversely, for any cone + $$ + g=\left( c \xlongrightarrow{g_{i,j}} F(i,j)\right)_{(i,j) \in\mathrm{Ob}\left(\mathsf{I}\times \mathsf{J}\right)}\in \mathrm{Ob}\left(\mathsf{Cone}\left(\mathsf{C}, \varprojlim F\right)\right), + $$ + we can construct a cone + $$ + \psi(g):=\left( c\xlongrightarrow{q_i} F_{i,\infty}\right)_{i \in\mathrm{Ob}\left(\mathsf{I}\right)}\in \mathrm{Ob}\left(\mathsf{Cone}\left(\mathsf{C}, \varprojlim F_{-,\infty}\right)\right). + $$ + through the universal property of $\varprojlim F(i,-)$ + \[ + \begin{tikzcd} + & c \arrow[d, "q_i"', dashed] \arrow[ldd, "{g_{i,j}}"'] \arrow[rdd, "{g_{i,j'}}"] & \\[+20pt] + & {\varprojlim F(i,-)} \arrow[ld, "{\pi_{i,j}}"] \arrow[rd, "{\pi_{i,j'}}"'] & \\ [+5pt] +{F(i,j)} \arrow[rr, "{F(\mathrm{id}_i,\mu)}"'] & & {F(i,j')} +\end{tikzcd} +\] + + + +\end{prf} + +\begin{proposition}{Colimits Commute with Colimits}{} + Let $F: \mathsf{I}\times \mathsf{J}\to \mathsf{C}$ be a functor. Suppose for each $i\in \mathrm{Ob}\left(\mathsf{I}\right)$, the colimit of the diagram $F(i,-):\mathsf{J}\to \mathsf{C}$ exists and is denoted by $F_{i,\infty}:=\varinjlim F(i,-)$. Define a diagram $F_{-,\infty}:\mathsf{I}\to \mathsf{C}$ as follows + \[ + \begin{tikzcd}[ampersand replacement=\&] + \mathsf{I}\&[-25pt]\&[+10pt]\&[-30pt]\mathsf{C}\&[-30pt]\&[-30pt] \\ [-15pt] + i \arrow[dd, "\lambda"{name=L, left}] + \&[-25pt] \& [+10pt] + \& [-30pt] \varinjlim F(i,-) \arrow[dd, "F_{-,\infty}(\lambda)"{name=R}] \\ [-10pt] + \& \phantom{.}\arrow[r, "F_{-,\infty}", squigarrow]\&\phantom{.} \& \\[-10pt] + i' \& \& \& \varinjlim F(i',-) + \end{tikzcd} + \] + where $F_{-,\infty}(\lambda)$ is induced by the universal property of $\varinjlim F(i,-)$ + \[ + \begin{tikzcd}[ampersand replacement=\&, background color=mypropbg!10] + \& \varinjlim F(i,-) \& \\[+10pt] +F(i,j) \arrow[ru]\arrow[rr, black!35] \arrow[dd, "{F(\lambda, \mathrm{id}_j)}"'] \& \& F(i,j') \arrow[lu]\arrow[dd, "{F(\lambda, \mathrm{id}_{j'})}"] \\[+10pt] + \& \varinjlim F(i', -) \& \\[+10pt] +F(i',j) \arrow[ru]\arrow[rr] \& \& F(i', j')\arrow[lu] + % absulute arrow + \arrow[from=1-2, to=3-2, "F_{-,\infty}(\lambda)"', near end, crossing over] + \end{tikzcd} + \] + Then $\varinjlim F_{-,\infty}$ exists if and only if $\varinjlim F$ exists. Moreover, if $\varinjlim F$ exists, we have natural isomorphism + \[ + \varinjlim F\cong \varinjlim F_{-,\infty}. + \] +\end{proposition} + + \subsection{Product and Coproduct} \begin{definition}{Binary Product}{} diff --git a/commutative_ring.tex b/commutative_ring.tex index 8f9187f..94c77d7 100644 --- a/commutative_ring.tex +++ b/commutative_ring.tex @@ -7,7 +7,7 @@ \section{Basic Concepts} -\begin{definition}{Noetherian Commutative Ring}{} +\begin{definition}{Noetherian Commutative Ring}{Noetherian_commutative_ring} Let $R$ be a commutative ring. We say $R$ is \textbf{Noetherian} if one of following conditions holds: \begin{enumerate}[(i)] \item $R$ as an $R$-module is Noetherian. @@ -209,6 +209,14 @@ \subsection{Ideals} The ideal $J\supseteq I$ is radical, prime, or maximal if and only if $J/I$ is radical, prime, or maximal respectively. \end{proposition} +\begin{definition}{Jacobson Radical}{} + Let $R$ be a commutative ring and $\mathfrak{m}$ be a maximal ideal of $\mathfrak{m}$. The \textbf{Jacobson radical} of $R$, denoted by $\mathfrak{J}_R$, is the intersection of all maximal ideals of $R$, denoted by + \[ + \mathfrak{J}_R=\bigcap_{\substack{\mathfrak{m} \in \mathrm{MaxSpec} R }} \mathfrak{m}. + \] + +\end{definition} + \subsection{Prime Elements} \begin{definition}{Divisibility}{} @@ -264,6 +272,37 @@ \subsection{Local Commutative Ring} \end{enumerate} \end{definition} +\begin{lemma}{Nakayama's Lemma}{Nakayama_lemma} + Let $R$ be a commutative ring and $M$ be a finitely generated $R$-module. If the image of $m_1,\cdots,m_n$ in $M/\mathfrak{J}_R M$ generates $M/\mathfrak{J}_R M$ as an $R/\mathfrak{J}_R$-module, then $m_1,\cdots,m_n$ generates $M$ as an $R$-module. + +\end{lemma} + +\begin{proposition}{}{} + Let $(R, \mathfrak{m})$ be a local commutative ring and $M$ be a finitely generated $R$-module. Let + \[ +\mathfrak{m}M = \left\{ \sum_{i=1}^n r_i m_i \;\middle|\; r_i \in \mathfrak{m}, m_i \in M, n \in \mathbb{N} \right\} +\] +denote the submodule of $M$ generated by $\mathfrak{m}$-action and $\pi: M \to M/\mathfrak{m}M$ be the natural projection. Then we have +\begin{itemize} + \item $M/\mathfrak{m}M$ is an $R/\mathfrak{m}$-vector space. + \item If $(v_1, \ldots, v_n)$ is a $R/\mathfrak{m}$-basis for $M/\mathfrak{m}M$, then $(\pi^{-1}(v_1), \ldots, \pi^{-1}(v_n))$ is a minimal generating set for $M$. + \item If $(m_1, \ldots, m_k)$ is a minimal generating set for $M$, then $(\pi(m_1), \ldots, \pi(m_k))$ is a $R/\mathfrak{m}$-basis for $M/\mathfrak{m}M$. +\end{itemize} +As a result, the minimal number of generators of $M$ is equal to $\dim_{R/\mathfrak{m}}(M/\mathfrak{m}M)$. +\end{proposition} +\begin{prf} + The action of $R/\mathfrak{m}$ on $M/\mathfrak{m}M$ is defined by + \[ + \overline{r}\cdot\pi(m)=\pi(rm). + \] + This is well-defined since if $\overline{r_1}=\overline{r_2}$, then $r_1-r_2=s \in \mathfrak{m}$. For any $m\in M$, we have + \[ + r_1m-r_2m=sm\in \mathfrak{m}M \implies \pi(r_1m)=\pi(r_2m). + \] + Suppose $(v_1, \ldots, v_n)$ is a $R/\mathfrak{m}$-basis for $M/\mathfrak{m}M$. The rest of the proof is a direct application of \hyperlink{th:Nakayama_lemma}{Nakayama's lemma}. +\end{prf} + + \section{Integral Domain} \begin{definition}{Associate}{} @@ -337,22 +376,9 @@ \section{Principal Ideal Domain} \end{prf} -\section{Krull Dimension} -\begin{definition}{Length of a Chain of Prime Ideals}{} - Let $R$ be a commutative ring and $\mathfrak{p}_0\subsetneq\mathfrak{p}_1\subsetneq\cdots\subsetneq\mathfrak{p}_n$ be a chain of prime ideals of $R$. The \textbf{length} of the chain is defined to be $n$. -\end{definition} -\begin{definition}{Height of a Prime Ideal}{} - Let $R$ be a commutative ring and $\mathfrak{p}$ be a prime ideal of $R$. The \textbf{height} of $\mathfrak{p}$ is defined to be the supremum of the lengths of all chains of prime ideals of $R$ contained in $\mathfrak{p}$ - \[ - \mathrm{ht}(\mathfrak{p})=\sup\left\{n\in\mathbb{N}\mid\exists\text{ a chain of prime ideals }\mathfrak{p}_0\subsetneq\mathfrak{p}_1\subsetneq\cdots\subsetneq\mathfrak{p}_n=\mathfrak{p}\right\}. - \] -\end{definition} -\begin{definition}{Height of an Ideal}{} - Let $R$ be a commutative ring and $I$ be an ideal of $R$. The \textbf{height} of $I$ is defined to be the height of the prime ideal $\mathfrak{p}$ generated by $I$, denoted by $\mathrm{ht}(I)$ -\end{definition} \section{Polynomial Ring} @@ -452,40 +478,6 @@ \subsection{Localization} Localization is the most economical way to make a multiplicative subset invertible. -\begin{proposition}{}{} - Let $R$ be a commutative ring and $S \subseteq R$ be a multiplicative subset. The category of $S^{-1} R$ modules is equivalent to the category of $R$-modules $M$ with the property that every $s \in S$ acts as an automorphism on $M$. The following functor $F$ gives a equivalence of categories: - \[ - \begin{tikzcd}[ampersand replacement=\&] - S^{-1} R\text{-}\mathsf{Mod}\&[-25pt]\&[+10pt]\&[-30pt] R\text{-}\mathsf{Mod}\text{ where }S\text{ act as automorphisms}\&[-30pt]\&[-30pt] \\ [-15pt] - M \arrow[dd, "f"{name=L, left}] - \&[-25pt] \& [+10pt] - \& [-30pt] M\arrow[dd, "f"{name=R}] \&[-30pt]\\ [-10pt] - \& \phantom{.}\arrow[r, "F", squigarrow]\&\phantom{.} \& \\[-10pt] - N \& \& \& N\& - \end{tikzcd} - \] -\end{proposition} - -\begin{prf} - Assume $S$ is a multiplicative subset of communitative ring $R$ and the localization map is $\varphi:R\to S^{-1}R$. Then $R$ can acts on $S^{-1}R$-module $M$ through - \[ - R\xrightarrow{\varphi}S^{-1}R\xrightarrow{\sigma_M'}\mathrm{End}_{\mathsf{Ab}}(M), - \] - which enables us to regard $M$ as an $R$-module. Furthermore, since - \[ - \sigma_M'(\varphi(S))\subseteq \sigma_M' \left(\left(S^{-1}R\right)^\times\right)\subseteq \left(\mathrm{End}_{\mathsf{Ab}}(M)\right)^\times=\mathrm{Aut}_{\mathsf{Ab}}(M), - \] - every $s \in S$ acts as an automorphism on $M$.\\ - Conversely, if $M$ is an $R$-module such that every $s\in S$ acts as an automorphism on $M$, i.e. $\sigma_M:R\to\mathrm{End}_{\mathsf{Ab}}(M)$ satisfies $\sigma_M(S)\subseteq \mathrm{Aut}_{\mathsf{Ab}}(M)$, then by unversal property - \begin{center} - \begin{tikzcd}[ampersand replacement=\&] - - S^{-1}R\arrow[rr, "\sigma_M'", dashed]\&\& \mathrm{End}_{\mathsf{Ab}}(M) \& \\ - \&R \arrow[ru, "\sigma_M"'] \arrow[lu, "\varphi"] \& - \end{tikzcd} - \end{center} - we can define a $S^{-1}R$-module structure on $M$ by lifting $\sigma_M$ to $\sigma_M'$. It is easy to check that these two functors are quasi-inverse to each other. -\end{prf} \begin{proposition}{Properties of Localization of Rings}{prop_of_localization_of_rings} Let $R$ be a commutative ring and $S\subseteq R$ be a multiplicative subset. Then @@ -493,7 +485,7 @@ \subsection{Localization} \item $S^{-1}R=0$ if and only if $0\in S$. \item If $0\notin S$, then $\frac{a}{s}$ is invertible in $S^{-1}R$ if and only if there exists $r\in R$ such that $ra\in S$. \item If $0\notin S$, the localization map $\varphi:R\to S^{-1}R$ is injective if and only if $S$ contains no zero divisors. - \item If $R$ is an integral domain, then $S^{-1}R$ is also an integral domain. + \item Localization preserves nilradical: $\mathfrak{N}_{S^{-1}R}=S^{-1}\mathfrak{N}_R$. Especially, $R$ is reduced $\implies$ $S^{-1}R$ is reduced. \end{enumerate} \end{proposition} @@ -511,6 +503,7 @@ \subsection{Localization} $$ \varphi\text{ is injective}\iff \ker \varphi=\{0\}\iff \forall s\in S,\forall r\in R-\{0\},sr\ne 0\iff S\text{ contains no zero divisors}. $$ + \item By \Cref{th:properties_of_localization_of_ideals}, localization commutes with taking radical. Thus we have $\mathfrak{N}_{S^{-1}R}=\sqrt{0(S^{-1}R)}=S^{-1}\sqrt{0R}=S^{-1}\mathfrak{N}_R$. If $R$ is reduced, then the nilradical of $R$ is $\mathfrak{N}_R=(0)$. Thus we have $\mathfrak{N}_{S^{-1}R}=S^{-1}\mathfrak{N}_R=S^{-1}(0)=(0)$, which implies $S^{-1}R$ is reduced. \end{enumerate} \end{prf} @@ -541,27 +534,28 @@ \subsection{Localization} \[ S^{-1}I=\left\{\frac{a}{s}\midv a\in I, s\in S\right\}. \] - $S^{-1}I$ is a $S^{-1}R$-submodule of $S^{-1}R$. Suppose the localization map is $\varphi:R\to S^{-1}R$, $S^{-1}I$ can also defined as the ideal generated by $\varphi(I)$ in $S^{-1}R$ + $S^{-1}I$ is a $S^{-1}R$-submodule of $S^{-1}R$. Suppose the localization map is $l_S:R\to S^{-1}R$, $S^{-1}I$ can also defined as the ideal generated by $l_S(I)$ in $S^{-1}R$ \[ - S^{-1}I=\langle \varphi(I)\rangle=\left\{\frac{r}{s}\frac{a}{1}\midv a\in I, \frac{r}{s}\in S^{-1}R\right\}. + S^{-1}I=\langle l_S(I)\rangle=\left\{\frac{r}{s}\frac{a}{1}\midv a\in I, \frac{r}{s}\in S^{-1}R\right\}. \] \end{definition} -\begin{proposition}{Properties of Localization of Ideals}{} - Let $R$ be a commutative ring, $S$ be a multiplicative set in $R$, and $0\notin S$. Suppose the localization map is $\varphi:R\to S^{-1}R$. Then we have maps between the sets of ideals of $R$ and $S^{-1}R$: +\begin{proposition}{Properties of Localization of Ideals}{properties_of_localization_of_ideals} + Let $R$ be a commutative ring, $S$ be a multiplicative set in $R$, and $0\notin S$. Suppose the localization map is $l_S:R\to S^{-1}R$. Then we have maps between the sets of ideals of $R$ and $S^{-1}R$: \begin{align*} - \mathcal{I}(R)=\left\{\text{ideals of }R\right\}\xrightleftarrows[\varphi^{-1}]{\quad S^{-1}\quad} + \mathcal{I}(R)=\left\{\text{ideals of }R\right\}\xrightleftarrows[l_S^{-1}]{\quad S^{-1}\quad} \left\{\text{ideals of }S^{-1}R\right\}=\mathcal{I}(S^{-1}R) \end{align*} \begin{enumerate}[(i)] - \item $S^{-1}\circ \varphi^{-1}=\mathrm{id}_{\mathcal{I}(S^{-1}R)}$. As a result, $S^{-1}$ is surjective and $\varphi^{-1}$ is injective. + \item $S^{-1}\circ l_S^{-1}=\mathrm{id}_{\mathcal{I}(S^{-1}R)}$. As a result, $S^{-1}$ is surjective and $l_S^{-1}$ is injective. \item For any ideal $J$ of $S^{-1}R$, there exists an ideal $I$ of $R$ such that $S^{-1}I=J$. \item If $I$ is a ideal of $R$, then $S^{-1}I=S^{-1}R\iff I\cap S\ne\varnothing$. - \item $\varphi$ induces a bijection between the set of prime ideals of $R$ that do not intersect $S$ and the set of prime ideals of $S^{-1}R$. That is, the following restriction of $S^{-1}$ and $\varphi^{-1}$ are bijections: + \item $l_S$ induces a bijection between the set of prime ideals of $R$ that do not intersect $S$ and the set of prime ideals of $S^{-1}R$. That is, the following restriction of $S^{-1}$ and $l_S^{-1}$ are bijections: \begin{align*} - \{I \in \operatorname{Spec} R: I \cap S=\varnothing\} \xrightleftarrows[\varphi^{-1}]{\quad S^{-1}\quad} - \spec S^{-1}R + \left\{I \in \operatorname{Spec}\left(R\right): I \cap S=\varnothing\right\} \xrightleftarrows[l_S^{-1}]{\quad S^{-1}\quad} + \spec\left(S^{-1}R\right) \end{align*} + \item If $I$ is an ideal of $R$, then $S^{-1}\sqrt{I} = \sqrt{S^{-1}I}$. \end{enumerate} \end{proposition} @@ -569,17 +563,37 @@ \subsection{Localization} \begin{enumerate}[(i)] \item Let $J$ be an ideal of $S^{-1}R$. We have \[ - S^{-1}\varphi^{-1}(J)=\left\{\frac{x}{s}\midv x\in \varphi^{-1}(J),s\in S\right\}=\left\{\frac{x}{s}\midv \frac{x}{1}\in J,s\in S\right\}=\left\{\frac{1}{s}\frac{x}{1}\midv \frac{x}{1}\in J,s\in S\right\}=J. + S^{-1}l_S^{-1}(J)=\left\{\frac{x}{s}\midv x\in l_S^{-1}(J),s\in S\right\}=\left\{\frac{x}{s}\midv \frac{x}{1}\in J,s\in S\right\}=\left\{\frac{1}{s}\frac{x}{1}\midv \frac{x}{1}\in J,s\in S\right\}=J. \] \item It is a direct consequence of the surjectivity of $S^{-1}$. \item Let $I$ be an ideal of $R$. We have \[ S^{-1}I=S^{-1}R\iff \frac{1}{1} \in S^{-1}I \iff \exists t,s\in S,a\in I, t(a-s)=0\iff ta=ts\in I\cap S\ne\varnothing \iff I\cap S\ne\varnothing. \] + \item Omitted. + \item For any $\frac{a}{s} \in S^{-1}\sqrt{I}$, there exists $n \in \mathbb{N}$ such that $a^n \in I$. Since $s^n \in S$, we have $\left(\frac{a}{s}\right)^n \in S^{-1}I$, which implies $\frac{a}{s} \in \sqrt{S^{-1}I}$. Hence $S^{-1}\sqrt{I} \subseteq \sqrt{S^{-1}I}$. + + Conversely, for any $x \in \sqrt{S^{-1}I}$, since $\sqrt{S^{-1}I}$ is an ideal of $S^{-1}R$, there exists $a\in r$ and $s\in S$ such that $x=\frac{a}{s}$. $\frac{a}{s}\in \sqrt{S^{-1}I}$ means there exists $n \in \mathbb{N}$ such that $\left(\frac{a}{s}\right)^n \in S^{-1}I$. Thus there exists $t \in S$ and $b\in I$ such that $\left(\frac{a}{s}\right)^n =\frac{b}{t}$. And this is equivalent to $uta^n =ubs^n$ for some $u\in S$. Note $(uta)^n=u^nt^{n-1}s^nb\in I$, we have $uta\in\sqrt{I}$. Now we get $x=\frac{a}{s} =\frac{uta}{uts}\in S^{-1}\sqrt{I}$. Hence $\sqrt{S^{-1}I} \subseteq S^{-1}\sqrt{I}$. \end{enumerate} \end{prf} +\begin{proposition}{Localization Respects Quotients}{localization_at_ideal_respects_quotients} + Let $R$ be a commutative ring, $S$ be a multiplicative set in $R$, and $I$ be an ideal of $R$. Then we have an $R$-algebra isomorphism $S^{-1}(R/I)\cong (S^{-1}R)/(S^{-1}I)$ and the following commutative diagram in $R\text{-}\mathsf{CAlg}$ (and accordingly in $\mathsf{CRing}$) + \[ + \begin{tikzcd} + R \arrow[r, "\pi_I"] \arrow[d, "l_S"'] &[+5em] R/I \arrow[d, "l_S"] \\[+2em] + S^{-1}R \arrow[r, "\pi_{S^{-1}I}"'] & S^{-1}(R/I)\cong (S^{-1}R)/(S^{-1}I) + \end{tikzcd} + \] +\end{proposition} +\begin{prf} + From \Cref{th:localization_respects_quotients} we get the commutative diagram in $R\text{-}\mathsf{Mod}$. Since localization map and quotient map are both ring homomorphisms, the commutative diagram holds in $R\text{-}\mathsf{CAlg}$. +\end{prf} + + + + \begin{example}{Localization at a Prime Ideal}{} Let $R$ be a commutative ring and $\mathfrak{p}$ be a prime ideal of $R$. Then $S=R-\mathfrak{p}$ is a multiplicative set. The localization $S^{-1}R$ is called the \textbf{localization of $R$ at $\mathfrak{p}$}, denoted by $R_\mathfrak{p}$. $R_\mathfrak{p}$ is a local ring with unique maximal ideal \[ @@ -589,11 +603,31 @@ \subsection{Localization} \end{example} \begin{prf} - Since for any ideal $I\in \{I \in \operatorname{Spec} R: I \cap S=\varnothing\}$, we have + Note + \[ + \{I \in \operatorname{Spec} R: I \cap S=\varnothing\}=\{I \in \operatorname{Spec} R: I \cap (R-\frak{p})=\varnothing\}=\{I \in \operatorname{Spec} R: I\subseteq \frak{p}\}. + \] + 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 $R\text{-}\mathsf{CAlg}$ + \[ + \begin{tikzcd} + R \arrow[r, "\pi_\mathfrak{p}"] \arrow[d, "l_{R-\mathfrak{p}}"'] &[+5em] R/\mathfrak{p} \arrow[d, "l_{R-\mathfrak{p}}"] \\[+2em] + R_\mathfrak{p} \arrow[r, "\pi_{\mathfrak{p}\hspace{-1pt}R_\mathfrak{p}}"'] & (R/\mathfrak{p})_\mathfrak{p}\cong R_\mathfrak{p}/\mathfrak{p} R_\mathfrak{p} + \end{tikzcd} + \] +\end{prf} + +\begin{proposition}{Localization of Integral Domain at a Prime Ideal}{} + Let $R$ be an integral domain and $\mathfrak{p}$ be a prime ideal of $R$. Then $R_\mathfrak{p}$ is an integral domain and we have \[ - I\subseteq \mathfrak{p}\implies S^{-1}I\subseteq S^{-1}\mathfrak{p}. + R = \bigcap_{\mathfrak{p}\in \spec R} R_\mathfrak{p}= \bigcap_{\mathfrak{m}\in \mathrm{MaxSpec} R} R_\mathfrak{m}. \] - Thus we see $S^{-1}\mathfrak{p}$ is the unique maximal ideal of $S^{-1}R$. + +\end{proposition} +\begin{prf} + First we need to show $R_\mathfrak{p}$ is an integral domain. Suppose $\frac{a}{s},\frac{b}{t}\in R_\mathfrak{p}$ such that $\frac{a}{s}\cdot\frac{b}{t}=\frac{ab}{st}=\frac{0}{1}$. Then there exists $u\in R-\mathfrak{p}$ such that $uab=0$. Since $R$ is an integral domain and $u\ne0$, either $a=0$ or $b=0$, which implies $R_\mathfrak{p}$ is an integral domain.\\ \end{prf} @@ -606,6 +640,279 @@ \subsection{Localization} $R_f=\{0\}\iff 0\in S\iff \exists n\in\mathbb{Z}_{\ge0},\;f^n=0$. \end{prf} +\section{Commutative Ring Homomorphism} + +\subsection{Commutative Ring Homomorphism of Finite Type} + +\begin{definition}{Finite-type Commutative Algebra}{} + Let $R\to A$ be a commutative ring homomorphism. We say $A$ is a \textbf{finite-type $R$-algebra}, or that $R\to A$ is \textbf{of finite type}, if one of the following equivalent conditions holds: + \begin{enumerate}[(i)] + \item there exists a finite set of elements $a_1,\cdots,a_n$ of $A$ such that every element of $A$ can be expressed as a polynomial in $a_1,\cdots,a_n$, with coefficients in $R$. + \item there exists a finite set $X$ such that $A\cong R[X]/I$ as $R$-algebra where $I$ is an ideal of $R[X]$. + \end{enumerate} + +\end{definition} + + +\subsection{Finite Commutative Ring Homomorphism} + +\begin{definition}{Finite Commutative Ring Homomorphism}{} + Let $\varphi:R\to S$ be a homomorphism between two commutative rings. We say $\varphi$ is \textbf{finite} if $S$ as an $R$-module is finitely generated. +\end{definition} + +\begin{definition}{Finitely Generated Commutative Algebra}{} + Let $R\to A$ be a commutative ring homomorphism. We say $A$ is a \textbf{finitely generated $R$-algebra} or $A$ is \textbf{finite} over $R$ if one of the following equivalent conditions holds: + \begin{enumerate}[(i)] + \item $R\to A$ is finite. + \item $A$ as an $R$-module is finitely generated. + \end{enumerate} +\end{definition} + +\begin{proposition}{}{} + Let $\varphi:R\to S$ be a finite homomorphism between two commutative rings. Let $M$ be an $S$-module. Then $M$ is a finitely generated $R$-module if and only if $M$ is a finitely generated $S$-module. +\end{proposition} + +\begin{proposition}{Composition of Finite Ring Homomorphisms is Finite}{} + Let $\varphi:R\to S$ and $\psi:S\to T$ be two finite homomorphisms between commutative rings. Then $\psi\circ\varphi:R\to T$ is also finite. +\end{proposition} + + + + + +\begin{proposition}{Finite Generation Implies Finite Type}{finite_generation_implies_finite_type} + Let $A$ be a $R$-algebra. If $A$ is finitely generated as an $R$-module, then $A$ is a finite-type $R$-algebra. +\end{proposition} + +\begin{prf} + This holds because if each element of $A$ can be expressed as an $R$-linear combination of finitely many elements of $A$, then each element of $A$ can also be expressed as a polynomial in finitely many elements of $A$ with coefficients in $R$. + + An alternative proof can be given by utilizing the universal property of the free contruction. Suppose $A$ is finitely generated as an $R$-module. Then there exists some a finite set $X=\{x_1,\cdots,x_n\}$ and a surjective $R$-linear map $\varphi:R^{\oplus X}\to A$. Define $f=\varphi\circ \iota$, where $\iota:X\to R^{\oplus X}$ is the inclusion map. + \begin{center} + \begin{tikzcd}[ampersand replacement=\&] + R^{\oplus X}\arrow[r, dashed, "\exists !\,\widetilde{j}"] \&R[X]\arrow[r, dashed, "\exists !\,\widetilde{f}"] \& A\\[0.3cm] + \& X\arrow[ul, "\iota"] \arrow[u, "j"] \arrow[ru, "f:=\varphi\circ \iota"'] \& + \end{tikzcd} + \end{center} + The universal property of free $R$-module induces a unique $R$-linear map $\widetilde{j}:R^{\oplus}\to R[X]$ such that $j=\widetilde{j}\circ \iota$. And the universal property of free commutative $R$-algebra induces a unique $R$-algebra homomorphism $\widetilde{f}:R[X]\to A$ such that $f=\widetilde{f}\circ j$. Note $f=\varphi\circ \iota=\left(\widetilde{f}\circ \widetilde{j}\right)\circ \iota$. By the uniqueness of the universal property of $R^{\oplus}$, we have $\widetilde{f}\circ \widetilde{j}=\varphi$. Since $\widetilde{f}\circ \widetilde{j}$ is surjective, $\widetilde{f}$ must be surjective, which implies $A$ is a finite-type $R$-algebra. +\end{prf} + +\begin{corollary}{}{} + Let $\varphi:R\to A$ be a finite homomorphism between two commutative rings. Then $\varphi$ is of finite type. +\end{corollary} +\begin{prf} + This is a reformulation of \Cref{th:finite_generation_implies_finite_type}. +\end{prf} + +\subsection{Integral Commutative Ring Homomorphism} + +\begin{definition}{Integral Element} + Let $\varphi:R\to S$ be a ring homomorphism between two commutative rings. An element $x\in S$ is called \textbf{integral} over $R$ if there exists a monic polynomial $f\in R[T]$ such that $f(x)=0$. +\end{definition} + +\begin{definition}{Generated Subalgebra}{} + Let $\varphi:R\to A$ be a ring homomorphism between two commutative rings and $x\in A$. By the universal property of polynomial ring, there exists a unique ring homomorphism $\psi:R[T]\to A$ such that + $\psi(T)=x$. + \begin{center} + \begin{tikzcd}[ampersand replacement=\&] + R[T]\arrow[r, dashed, "\exists !\,\psi"] \& A\\[0.3cm] + \{T\}\arrow[u, "\iota"] \arrow[ru, "\mathrm{const}_x"'] \& + \end{tikzcd} + \end{center} + The \textbf{$R$-subalgebra of $A$ generated by $x$} is defined as + \[ + R[x]:=\psi(R[T])=\left\{\sum_{i=0}^n r_i x^i\;\middle|\; r_i\in R\right\}. + \] +\end{definition} + +\begin{proposition}{Equivalent Definition of Integral Element}{} + Let $\varphi:R\to A$ be a ring homomorphism between two commutative rings and $x\in S$. Let $R[x]$ be the $R$-subalgebra of $A$ generated by $x$. Then $A$ is an $R[x]$-module. And the following statements are equivalent: + \begin{enumerate}[(i)] + \item $x$ is integral over $R$. + \item $R[x]$ is a finitely generated $R$-module. + \item There exists a faithful $R[x]$-submodule of $A$ that is finitely generated as an $R$-module and contains $x$. + \end{enumerate} + +\end{proposition} + +\begin{definition}{Integral Extension}{} + Let $\varphi:R\to S$ be a ring homomorphism between two commutative rings. If every element of $S$ is integral over $R$, then we say $\varphi$ is \textbf{integral} and $S$ is an \textbf{integral extension} of $R$. + +\end{definition} + +\begin{definition}{Integral Closure}{integral_closure} + Let $\varphi:R\to S$ be a ring homomorphism between two commutative rings. The set of all elements in $S$ that are integral over $R$ is called the \textbf{integral closure of $R$ in $S$}. The integral closure of $R$ in $S$ is an $R$-subalgebra of $S$, denoted by $\overline{R}^S$. +\end{definition} + + + +\begin{definition}{Integral Closure in Field of Fractions}{} + Let $R$ be an integral domain. The \textbf{integral closure of $R$} is defined to be the integral closure of $R$ in $\mathrm{Frac}(R)$. +\end{definition} + + + +\begin{lemma}{Finite $\implies$ Integral}{} + Let $\varphi:R\to S$ be a finite homomorphism between two commutative rings. Then $\varphi$ is integral. +\end{lemma} + +\begin{proposition}{Equivalent Definition of Integral Ring Homomorphism}{} + Let $\varphi:R\to S$ be a ring homomorphism between two commutative rings. The following are equivalent: + \begin{enumerate}[(i)] + \item $\varphi$ is finite. + \item $\varphi$ is integral and of finite type. + \item there exist $x_1,\cdots ,x_n\in S$ such that $S=R[x_1,\cdots ,x_n]$ and each $x_i$ is integral over $R$. + \end{enumerate} +\end{proposition} + + +\begin{proposition}{Composition of Integral Ring Homomorphisms is Integral}{} + Let $\varphi:R\to S$ and $\psi:S\to T$ be two integral homomorphisms between commutative rings. Then $\psi\circ\varphi:R\to T$ is also integral. +\end{proposition} + +\subsection{Normal Ring} +\begin{definition}{Normal Domain}{} + An integral domain $R$ is called \textbf{normal} or \textbf{integrally closed} if $R$ equals its \hyperref[th:integral_closure]{integral closure} in its field of fractions. +\end{definition} + +\begin{proposition}{Equivalent Definition of Normal Domain}{} + Let $R$ be an integral domain. The following are equivalent: + \begin{enumerate}[(i)] + \item $R$ is normal. + \item For every prime ideal $\mathfrak{p}$ of $R$, the localization $R_\mathfrak{p}$ is normal. + \item For every maximal ideal $\mathfrak{m}$ of $R$, the localization $R_\mathfrak{m}$ is normal. + \end{enumerate} +\end{proposition} + + +\begin{definition}{Normal Ring}{} + An commutative $R$ is called \textbf{normal} if for every prime ideal $\mathfrak{p}$ of $R$, the localization $R_\mathfrak{p}$ is a normal domain. +\end{definition} +\begin{proposition}{}{} +A normal ring is integrally closed in its total ring of fractions. +\end{proposition} + +\subsection{Japanese Rings} +\begin{definition}{Japanese Ring}{} + Let $R$ be an integral domain with field of fractions $K=\mathrm{Frac}(R)$. + \begin{enumerate}[(i)] + \item We say $R$ is \textbf{N-1} if the integral closure of $R$ in $K$ is a finitely generated $R$-module. + \item We say $R$ is \textbf{N-2} or \textbf{Japanese} if for any finite extension $L/K$, the \hyperref[th:integral_closure]{integral closure} of $R$ in $L$ is finite over $R$. + \end{enumerate} +\end{definition} + +\begin{proposition}{}{} + Let $R$ be a Noetherian normal domain with fraction field $K=\mathrm{Frac}(R)$. Let $L/K$ be a finite separable field extension. Then the integral closure of $R$ in $L$ is finite over $R$. +\end{proposition} + +\section{Krull Dimension} +\begin{definition}{Length of a Chain of Prime Ideals}{} + Let $R$ be a commutative ring and $\mathfrak{p}_0\subsetneq\mathfrak{p}_1\subsetneq\cdots\subsetneq\mathfrak{p}_n$ be a chain of prime ideals of $R$. The \textbf{length} of the chain is defined to be $n$. +\end{definition} + + +\begin{definition}{Height of a Prime Ideal}{} + Let $R$ be a commutative ring and $\mathfrak{p}$ be a prime ideal of $R$. The \textbf{height} of $\mathfrak{p}$ is defined to be the supremum of the lengths of all chains of prime ideals of $R$ contained in $\mathfrak{p}$ + \[ + \mathrm{ht}(\mathfrak{p})=\sup\left\{n\in\mathbb{N}\mid\exists\text{ a chain of prime ideals }\mathfrak{p}_0\subsetneq\mathfrak{p}_1\subsetneq\cdots\subsetneq\mathfrak{p}_n=\mathfrak{p}\right\}. + \] +\end{definition} + +\begin{definition}{Krull Dimension}{} + Let $R$ be a commutative ring. The \textbf{Krull dimension} of $R$, denoted by $\dim R$, is defined to be the supremum of the heights of all prime ideals of $R$ + \[ + \dim R=\sup\left\{\mathrm{ht}(\mathfrak{p})\mid\mathfrak{p}\in \mathrm{Spec}(R)\right\}= \sup\left\{\mathrm{ht}(\mathfrak{m})\mid\mathfrak{m}\text{ is a maximal ideal of }R\right\}. + \] +\end{definition} + +\begin{proposition}{}{} + Let $R$ be a commutative ring and $\mathfrak{p}$ be a prime ideal of $R$. Then + \[ + \operatorname{ht}\left(\mathfrak{p}\right) = \dim R_{\mathfrak{p}}. + \] +\end{proposition} +\begin{proof} + Let $\mathfrak{p}_0\subsetneq\mathfrak{p}_1\subsetneq\cdots\subsetneq\mathfrak{p}_n=\mathfrak{p}$ be a chain of prime ideals of $R$. Since $R_{\mathfrak{p}}$ is a local ring, $\mathfrak{p}R_{\mathfrak{p}}$ is the unique maximal ideal of $R_{\mathfrak{p}}$. Then we have a chain of prime ideals of $R_{\mathfrak{p}}$ + \[ + \mathfrak{p}_0R_{\mathfrak{p}}\subsetneq\mathfrak{p}_1R_{\mathfrak{p}}\subsetneq\cdots\subsetneq\mathfrak{p}_nR_{\mathfrak{p}}=\mathfrak{p}R_{\mathfrak{p}}. + \] + That implies $\dim R_{\mathfrak{p}}\geq n$. Since the chain is arbitrary, we have $\dim R_{\mathfrak{p}}\geq \mathrm{ht}(\mathfrak{p})$. + + On the other hand, any prime ideal of $R_{\mathfrak{p}}$ is of the form $\mathfrak{q}R_{\mathfrak{p}}$ for some prime ideal $\mathfrak{q}$ of $R$ such that $\mathfrak{q}\subseteq\mathfrak{p}\ne \varnothing$. Suppose $\dim R_{\mathfrak{p}}=m$ and + \[ + \mathfrak{q}_0R_{\mathfrak{p}}\subsetneq\mathfrak{q}_1R_{\mathfrak{p}}\subsetneq\cdots\subsetneq\mathfrak{q}_{m-1}R_{\mathfrak{p}}\subsetneq\mathfrak{p}R_{\mathfrak{p}} + \] + is a chain of prime ideals of $R_{\mathfrak{p}}$. Then we have a chain of prime ideals of $R$ + \[ + \mathfrak{q}_0\subsetneq\mathfrak{q}_1\subsetneq\cdots\subsetneq\mathfrak{q}_{m-1}\subsetneq\mathfrak{p} + \] + That implies $\mathrm{ht}(\mathfrak{p})\geq m = \dim R_{\mathfrak{p}}$. Thus we have $\dim R_{\mathfrak{p}}= \mathrm{ht}(\mathfrak{p})$. +\end{proof} + +\subsection{Noetherian Local Rings} + +\begin{definition}{Ideal of Definition}{} +Let $(R,\mathfrak{m})$ be a \hyperlink{th:Noetherian_commutative_ring}{Noetherian} local commutative ring. An \textbf{ideal of definition} of $R$ is an ideal $\mathfrak{a}$ of $R$ such that $\sqrt{\mathfrak{a}}=\mathfrak{m}$. +\end{definition} + + +\begin{proposition}{Krull Dimension of Noetherian Local Rings}{} + Let $R$ be a Noetherian local commutative ring and $d\ge 0$ be an integer. Then the following statements are equivalent: + \begin{enumerate}[(i)] + \item $\dim R=d$. + \item There exists an ideal of definition $\mathfrak{a}=(a_1,\cdots,a_d)$ of $R$, and no ideal of definition of $R$ is generated by fewer than $d$ elements. + \end{enumerate} +\end{proposition} + +\begin{definition}{System of Parameters}{} + Let $(R,\mathfrak{m})$ be a Noetherian local commutative ring. A \textbf{system of parameters} of $R$ is a sequence of elements $a_1,\cdots,a_d\in \mathfrak{m}$ such that + \[ + \sqrt{(a_1,\cdots,a_d)}=\mathfrak{m}, + \] + that is, $(a_1,\cdots,a_d)$ is an ideal of definition of $R$. +\end{definition} + +\begin{definition}{Regular Local Ring}{} + Let $(R,\mathfrak{m})$ be a Noetherian local commutative ring of dimension $d$. $R$ is called a \textbf{regular local ring} if there $a_1,\cdots,a_d\in \mathfrak{m}$ such that $(a_1,\cdots,a_d)=\mathfrak{m}$. In this case, $a_1,\cdots,a_d$ is called a \textbf{regular system of parameters} of $R$. + +\end{definition} + +A regular local ring is a field if and only if it has Krull dimension $0$. A regular local ring is a DVR if and only if it has Krull dimension $1$. + +\begin{definition}{Regular Ring}{} + Let $R$ be a commutative ring. $R$ is called a \textbf{regular ring} if $R$ is a Noetherian ring and for every prime ideal $\mathfrak{p}$ of $R$, the localization $R_{\mathfrak{p}}$ is a regular local ring. + +\end{definition} + + + + +\section{Dedekind Domain} + +\begin{definition}{Fractional Ideal}{} + Let $R$ be an integral domain. A \textbf{fractional ideal} of $R$ is an $R$-submodule $I$ of $\mathrm{Frac}(R)$ such that there exists a nonzero $r\in R$ such that $rI\subseteq R$. +\end{definition} + +\begin{definition}{Dedekind Domain}{} + An integral domain $R$ is called a \textbf{Dedekind domain} if every nonzero ideal $I$ of $R$ can be written as a product of prime ideals of $R$ + \[ + I=\mathfrak{p}_1\cdots\mathfrak{p}_r + \] + uniquely up to permutation of the $\mathfrak{p}_i$. +\end{definition} + +\begin{proposition}{Equivalent Definition of Dedekind Domain}{} + Let $R$ be a commutative ring. The following are equivalent: + \begin{enumerate}[(i)] + \item $R$ is a Dedekind domain. + \item $R$ is Noetherian domain and for every nonzero maximal ideal $\mathfrak{m}$ of $R$, the localization $R_{\mathfrak{m}}$ is a DVR. + \item $R$ a Noetherian, normal domain, and $\dim R\le 1$. + \item Every non-zero fractional ideal of $R$ is invertible. + \end{enumerate} + +\end{proposition} + + \section{Absolute Value} \begin{definition}{Absolute Value on an Integral Ring}{} Let $R$ be an integral ring. An \textbf{absolute value} on $R$ is a function $|\cdot|:R\to \mathbb{R}_{\ge0}$ satisfying the following properties: @@ -757,9 +1064,3 @@ \section{Valuation Ring} \end{definition} -\section{Integral Element} -\begin{definition}{Integral Element}{} - Let $R$ be an integral domain and $K$ be its field of fractions. An element $x\in K$ is called \textbf{integral} over $R$ if there exists a monic polynomial $f\in R[x]$ such that $f(x)=0$. The set of all elements in $K$ that are integral over $R$ is called the \textbf{integral closure} of $R$ in $K$, denoted by $\overline{R}$. -\end{definition} - - diff --git a/field.tex b/field.tex index 5b95ddd..c2652c4 100644 --- a/field.tex +++ b/field.tex @@ -27,3 +27,29 @@ \section{Field Extension} \begin{theorem}{Hilbert's Nullstellensatz}{} If $\Bbbk$ is any field and $\mathfrak{m}$ is a maximal ideal of $\Bbbk\left[x_1, \ldots, x_n\right]$, then $\Bbbk\left[x_1, \ldots, x_n\right]/\mathfrak{m}$ is a finite extension of $\Bbbk$. \end{theorem} + + +\begin{definition}{Perfect Field}{perfect_field} + A field $K$ is \textbf{perfect} if every finite extension of $K$ is separable. +\end{definition} + +\begin{example}{Examples of Perfect Fields}{} + Examples of perfect fields include + \begin{itemize} + \item Field of characteristic $0$. + \item Finite field. + \item Algebraically closed field. + \item Field which is algebraic over a perfect field. + \end{itemize} +\end{example} + +\begin{definition}{Degree of Field Extension}{} + Let $L/K$ be a field extension. The \textbf{degree} of $L/K$ is the dimension of $L$ as a $K$-vector space, denoted by $[L:K]=\dim_K L$. +\end{definition} + +\begin{definition}{Finite Extension}{} + A field extension $L/K$ is \textbf{finite} if $[L:K]<\infty$. +\end{definition} + +\section{Trace and Norm of Field Extension} + diff --git a/module.tex b/module.tex index daa0a53..0dbe3de 100644 --- a/module.tex +++ b/module.tex @@ -96,6 +96,10 @@ \section{Basic Concepts} \end{enumerate} \end{proposition} +\begin{definition}{Faithful Module}{} + Let $R$ be a ring, and let $M$ be a left $R$-module. We say $M$ is \textbf{faithful} if $\operatorname{Ann}_R(M)=0$, or equivalently, the map $R\to \mathrm{End}_{\mathsf{Ab}}(M)$ is injective. +\end{definition} + \section{Construction} \subsection{Free Object} @@ -174,6 +178,43 @@ \subsection{Free Object} \subsection{Localization} + +\begin{proposition}{}{} + Let $R$ be a commutative ring and $S \subseteq R$ be a multiplicative subset. The category of $S^{-1} R$ modules is equivalent to the category of $R$-modules $M$ with the property that every $s \in S$ acts as an automorphism on $M$. The following functor $F$ gives a equivalence of categories: + \[ + \begin{tikzcd}[ampersand replacement=\&] + S^{-1} R\text{-}\mathsf{Mod}\&[-25pt]\&[+10pt]\&[-30pt] R\text{-}\mathsf{Mod}\text{ where }S\text{ act as automorphisms}\&[-30pt]\&[-30pt] \\ [-15pt] + M \arrow[dd, "f"{name=L, left}] + \&[-25pt] \& [+10pt] + \& [-30pt] M\arrow[dd, "f"{name=R}] \&[-30pt]\\ [-10pt] + \& \phantom{.}\arrow[r, "F", squigarrow]\&\phantom{.} \& \\[-10pt] + N \& \& \& N\& + \end{tikzcd} + \] +\end{proposition} + +\begin{prf} + Assume $S$ is a multiplicative subset of communitative ring $R$ and the localization map is $\varphi:R\to S^{-1}R$. Then $R$ can acts on $S^{-1}R$-module $M$ through + \[ + R\xrightarrow{\varphi}S^{-1}R\xrightarrow{\sigma_M'}\mathrm{End}_{\mathsf{Ab}}(M), + \] + which enables us to regard $M$ as an $R$-module. Furthermore, since + \[ + \sigma_M'(\varphi(S))\subseteq \sigma_M' \left(\left(S^{-1}R\right)^\times\right)\subseteq \left(\mathrm{End}_{\mathsf{Ab}}(M)\right)^\times=\mathrm{Aut}_{\mathsf{Ab}}(M), + \] + every $s \in S$ acts as an automorphism on $M$.\\ + Conversely, if $M$ is an $R$-module such that every $s\in S$ acts as an automorphism on $M$, i.e. $\sigma_M:R\to\mathrm{End}_{\mathsf{Ab}}(M)$ satisfies $\sigma_M(S)\subseteq \mathrm{Aut}_{\mathsf{Ab}}(M)$, then by unversal property + \begin{center} + \begin{tikzcd}[ampersand replacement=\&] + + S^{-1}R\arrow[rr, "\sigma_M'", dashed]\&\& \mathrm{End}_{\mathsf{Ab}}(M) \& \\ + \&R \arrow[ru, "\sigma_M"'] \arrow[lu, "\varphi"] \& + \end{tikzcd} + \end{center} + we can define a $S^{-1}R$-module structure on $M$ by lifting $\sigma_M$ to $\sigma_M'$. It is easy to check that these two functors are quasi-inverse to each other. +\end{prf} + + \begin{definition}{Localization of a Module}{} Let $R$ be a commutative ring, $S$ be a multiplicative set in $R$, and $M$ be an $R$-module. The \textbf{localization of the module} $M$ by $S$, denoted $S^{-1}M$, is an $S^{-1}R$-module that is constructed exactly as the localization of $R$, except that the numerators of the fractions belong to $M$. That is, as a set, it consists of equivalence classes, denoted $\frac{m}{s}$, of pairs $(m, s)$, where $m\in M$ and $s\in S$, and two pairs $(m, s)$ and $(n, t)$ are equivalent if there is an element $u$ in $S$ such that \[u(sn-tm)=0.\] @@ -271,16 +312,22 @@ \subsection{Localization} we see that $\frac{m}{s}\in \operatorname{im}S^{-1}(u)$, which means $\operatorname{im}S^{-1}(u)=\ker S^{-1}(v)$. Hence $S^{-1}$ is exact. \end{prf} -\begin{proposition}{Localization Respects Quotients}{} - Let $M$ be an $R$-module and $N$ be a submodule of $M$. Then we have an isomorphism $S^{-1}(M/N)\cong (S^{-1}M)/(S^{-1}N)$. +\begin{proposition}{Localization Respects Quotients}{localization_respects_quotients} + Let $M$ be an $R$-module and $N$ be a submodule of $M$. Then we have an isomorphism $S^{-1}(M/N)\cong (S^{-1}M)/(S^{-1}N)$ and the following commutative diagram + \[ + \begin{tikzcd} + M \arrow[r, "\pi_M"] \arrow[d, "S^{-1}"'] &[+5em] M/N \arrow[d, "S^{-1}"] \\[+2em] + S^{-1}M \arrow[r, "\pi_{S^{-1}N}"'] & S^{-1}(M/N)\cong (S^{-1}M)/(S^{-1}N) + \end{tikzcd} + \] \end{proposition} \begin{prf} - From the exact sequence + Since localization is exact, from the exact sequence \[ 0\longrightarrow N\longrightarrow M\longrightarrow M/N\longrightarrow 0, \] - we have the exact sequence + we obtain the following exact sequence \[ 0\longrightarrow S^{-1}N\longrightarrow S^{-1}M\longrightarrow S^{-1}(M/N)\longrightarrow 0. \] diff --git a/set_theory.tex b/set_theory.tex index b52a0dd..6ef0514 100644 --- a/set_theory.tex +++ b/set_theory.tex @@ -2,7 +2,9 @@ \chapter{Set Theory} \thispagestyle{empty} \setcounter{page}{1} -\section{Set} + + +\section{Construction} \subsection{Basic Operations} \begin{definition}{Family of Sets}{} Let $I$ be some index set. A \textbf{family of sets index by $I$} is a function that maps each index $i\in I$ to a set $A_i$, denoted by $\left(A_i\right)_{i\in I}$. @@ -16,6 +18,21 @@ \subsection{Basic Operations} \end{align*} \end{definition} +\begin{definition}{Difference and Complement}{} + Let $A,B$ be subsets of $X$. The \textbf{complement} of $A$ is defined as follows + \begin{align*} + A^{\complement}&:=\left\{x\in X:x\notin A \right\}. + \end{align*} + The \textbf{difference} of $A$ and $B$ is defined as follows + \begin{align*} + A-B&:=\left\{x:x\in A\text{ and }x\notin B \right\}. + \end{align*} + We have + \[ + A^{\complement}=X-A,\qquad A-B =A\cap B^{\complement}. + \] +\end{definition} + \begin{proposition}{Distribution Law}{} Suppose that $A_\alpha,A,B_\alpha,B$ are sets and $I$ is some index set. We have the following distribution law: \begin{itemize} @@ -83,7 +100,16 @@ \subsection{Limit of Sequence of Sets} \end{proposition} -\subsection{Relation} +\subsection{Cartesian Product} + +\begin{proposition}{Cartesian Product of Intersections of Sets}{} + \[ + \left(\prod_{i \in I} A_i\right) \cap \left(\prod_{i \in I} B_i\right) = \prod_{i \in I} \left(A_i \cap B_i\right). + \] +\end{proposition} + + +\section{Relation} \begin{definition}{Relation}{} An \textbf{$n$-ary relation} $R$ over sets $X_1, \cdots, X_n$ is a subset of the Cartesian product $X_1 \times \cdots \times X_n$. A \textbf{binary relation} $R$ over sets $X$ and $Y$ is a subset of $X \times Y$. We write $xRy$ to denote that $(x, y) \in R$. \end{definition} diff --git a/valuation_theory.tex b/valuation_theory.tex index f7a47b1..15624ac 100644 --- a/valuation_theory.tex +++ b/valuation_theory.tex @@ -174,3 +174,184 @@ \section{Valuation of Field} \] \end{prf} +\section{Absolute Value of Field} + +\begin{definition}{Absolute Value}{} + An \textbf{absolute value} on a field \( K \) is a function \( |\cdot| : K \to \mathbb{R}_{\geq 0} \) satisfying the following conditions: + +\begin{enumerate}[(i)] + \item For any \( x \in K \), \( |x| = 0 \iff x = 0 \). + \item For any \( x, y \in K \), \( |xy| = |x| \cdot |y| \). + \item For any \( x, y \in K \), \( |x + y| \leq |x| + |y| \). +\end{enumerate} + +A field equipped with an absolute value is called a \textbf{normed field}, denoted by \( (K, |\cdot|) \). +\end{definition} + +A normed field $(K, |\cdot|)$ induces a metric $d(x, y) = |x - y|$ on $K$, making it a Hausdorff topological field. + +\begin{definition}{Trivial Absolute Value}{} + An absolute value $|\cdot|$ on a field $K$ is called \textbf{trivial} if + \[ +|x| = \begin{cases} + 0 & \text{if } x = 0, \\ + 1 & \text{if } x \neq 0. +\end{cases} +\] + +\end{definition} + +\begin{proposition}{Properties of Absolute Value}{} + Let \( (K, |\cdot|) \) be a normed field. Then we have the following properties: +\begin{enumerate}[(i)] + \item $|1| = \left|-1\right| = 1$. + \item $|x| = \left|-x\right|$ for all \( x \in K \). + \item $|n\cdot 1| \le n$ for all \( n \in \mathbb{Z} \). +\end{enumerate} +\end{proposition} + +\begin{definition}{Equivalent Absolute Value}{} + Two absolute values on a field are said to be \textbf{equivalent} if as topological spaces they are homeomorphic. +\end{definition} + +\begin{proposition}{Equivalent Characterization of Equivalent Absolute Values}{} + Let $K$ be a field and $|\cdot|$ and $ |\cdot|_\star $ are two nontrivial absolute values on $K$. Then the following are equivalent: +\begin{enumerate}[(i)] + \item $|\cdot|$ and $ |\cdot|_\star $ are equivalent absolute values. + \item For any $x\in K$, $|x| < 1 \implies |x|_\star < 1$. + \item There exists $s > 0$ such that $|\cdot| = |\cdot|_\star^s$. +\end{enumerate} + +\end{proposition} +\begin{proof} + (i) $\implies$ (ii): For any $x\in K$ such that $|x|<1$, we have + \[ + \lim_{n\to \infty} |x^n| =\lim_{n\to \infty} |x|^n= 0 , + \] + which implies the sequence $\left(x^n\right)_{n=1}^\infty$ converges to 0 in $(K, |\cdot|)$. Since $(K, |\cdot|)$ and $(K, |\cdot|_\star)$ as topological spaces are homeomorphic, the sequence $\left(x^n\right)_{n=1}^\infty$ also converges to 0 in $(K, |\cdot|_\star)$, which means + \[ + \lim_{n\to \infty} |x|_\star^n=\lim_{n\to \infty} |x^n|_\star = 0. +\] + Therefore, $|x|_\star<1$. + + (ii) $\implies$ (iii): Assume condition (ii): $|x| < 1 \implies |x|_\star < 1$ for all $x \in K$. By considering the inverse $x^{-1}$, we also conclude that $|x| > 1 \implies |x|_\star > 1$. + + Let $y \in K$ such that $|y| > 1$. For any $x \in K^\times-\{1\}$, there exists a real number $r(x)$ such that + \[ + |x| = |y|^{r(x)}, + \] + with $r(x) \neq 0$. + + Consider a sequence of rational numbers $\left(\frac{m_i}{n_i}\right)_{i=1}^\infty$ such that $\frac{m_i}{n_i} > r(x)$, $n_i > 0$, and + \[ + \lim_{i \to \infty} \frac{m_i}{n_i} = r(x). + \] + Then by assumption (ii), we have + \[ + |x| < |y|^{m_i / n_i}\implies + \left| \frac{x^{n_i}}{y^{m_i}} \right| < 1\implies \left| \frac{x^{n_i}}{y^{m_i}} \right|_\star < 1\implies + |x|_\star < |y|_\star^{m_i / n_i}. + \] + + + Taking the limit as $i \to \infty$, we obtain + \[ + |x|_\star \leq |y|_\star^{r(x)}. + \] + + Similarly, by considering a sequence of rational numbers $\left(\frac{m_i}{n_i}\right)_{i=1}^\infty$ such that $\frac{m_i}{n_i} < r(x)$ and $\lim_{i \to \infty} \frac{m_i}{n_i} = r(x)$, we can show that + \[ + |x|_\star \geq |y|_\star^{r(x)}. + \] + Thus, we conclude that + \[ + |x|_\star = |y|_\star^{r(x)}. + \] + Taking the logarithm of both sides, we find + \[ + \log |x| = r(x) \log |y| \quad \text{and} \quad \log |x|_\star = r(x) \log |y|_\star. + \] + Dividing these equations yields + \[ + \frac{\log |x|}{\log |x|_\star} = \frac{\log |y|}{\log |y|_\star}. + \] + Take $t=\frac{\log |y|}{\log |y|_\star} > 0$. We conclude that for all $x \in K$, $|x| = |x|_\star^t$. This completes the proof. + + (iii) $\implies$ (i): Since $|\cdot| = |\cdot|_\star^s$, we can prove that the identity map $f: (K, |\cdot|)\to (K, |\cdot|_\star)$ is a homeomorphism by checking that both $f$ and $f^{-1}$ are continuous. For any $\epsilon$-ball $B_{|\cdot|_\star}(x,\epsilon)$, we have + \[ + f^{-1}(B_{|\cdot|_\star}(x,\epsilon)) = B_{|\cdot|}(x,\epsilon^{1/s}), + \] + which is open in $(K, |\cdot|)$. Thus $f$ is continuous. Similarly, $f^{-1}$ is continuous. +\end{proof} + +\begin{definition}{Places of a Field}{} + Let $K$ be a field. A \textbf{place} on $K$ is an equivalence classes of non-trivial absolute values on $K$. The set of all places on a field $K$ is denoted by $\mathtt{pl}_K$. +\end{definition} + +\begin{definition}{Archimedean Absolute Value}{} + An absolute value is called \textbf{Archimedean} if the set + \[ + \{|n| : n \in \mathbb{Z}\} + \] + is unbounded in $\mathbb{R}$ equipped with the Euclidean topology. Otherwise, it is called \textbf{non-Archimedean}. +\end{definition} + + +\begin{example}{Absolute Value Induced by Harr Measure on Locally Compact Hausdorff Topological Field}{} + Let $K$ be a locally compact Hausdorff topological field. Since the additive group $(K, +)$ is a locally compact Hausdorff group, a Haar measure $\mu $ can be defined on it. Using the Haar measure, an absolute value + $$ + |\cdot| : a \longmapsto |a| = \frac{\mu(aX)}{\mu(X)} + $$ + can be defined on the field $K$. The topology induced by this absolute value coincides with the original topology of $K$. +\end{example} + +\begin{proposition}{Classification of Non-discrete, Locally Compact Hausdorff Topological Fields}{} + Let $K$ be a non-discrete, locally compact Hausdorff topological field. There are exactly three possibilities for $K$: +\begin{enumerate}[(i)] + \item Archimedean fields: $\mathbb{R}$ and $\mathbb{C}$. + \item $p$-adic number fields: finite extensions of $\mathbb{Q}_p$. These are non-Archimedean fields of characteristic 0. + \item Function fields over a finite field $\mathbb{F}_q$: finite extensions of the field of formal Laurent series $\mathbb{F}_q(\!(x)\!)$, where $q = p^n$ is a power of a prime. These are non-Archimedean fields of characteristic $p$. +\end{enumerate} +\end{proposition} + + +\begin{definition}{Local Field}{} + A \textbf{local field} is a non-discrete, locally compact Hausdorff topological field. +\end{definition} + + +\begin{definition}{Completion of a Normed Field}{} + Let \( (K, |\cdot|) \) be a normed field. The \textbf{completion} of \( K \) with respect to the absolute value \( |\cdot| \) is the Cauchy completion of the metric space \( (K, d) \), where \( d(x, y) = |x - y| \). +\end{definition} + +The completion of a normed field is functorial, i.e. it is a functor from the category of normed fields to the category of complete normed fields. + +\begin{definition}{Global Field}{} + Let $K$ be a field. $K$ is called a \textbf{global field} if it satisfies the following properties: + \begin{enumerate}[(i)] + \item the completion of $K$ with respect to every place on $K$ is a local field + \item product formula: + \[ + \prod_{v\in \mathtt{pl}_K} |x|_v = 1. + \] + \end{enumerate} + It can be proven that there are exactly two types of global fields: + +\begin{enumerate}[(i)] + \item Number fields: finite extensions of $\mathbb{Q}$. + \item Function fields: finite extensions of $\mathbb{F}_q(t)$. +\end{enumerate} +\end{definition} + +\begin{definition}{Finite Places of a Global Field}{} + Let $K$ be a global field. A place $v$ on $K$ is called \textbf{finite} if either + \begin{enumerate}[(i)] + \item $v$ restricts to a $p$-adic place on $\mathbb{Q}$ for some prime $p$, or + \item $v$ restricts to a $p(t)$-adic place on $\mathbb{F}_q(t)$ for some prime element $p(t) \in \mathbb{F}_q[t]$. + \end{enumerate} +\end{definition} + + + + +