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Add Glueing Functions
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hooyuser committed Mar 5, 2024
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55 changes: 44 additions & 11 deletions category_theory.tex
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Expand Up @@ -401,7 +401,7 @@ \section{Category}
\end{tikzcd}
\]
}
By viewing sets in $\mathsf{Set}$ as discrete categories, we have an inclusion $\mathsf{Set}\hookrightarrow\mathsf{Cat}$. Hence the category of elements of $F$ is a special case of the Grothendieck construction when the image of $F$ on $\mathrm{Ob}(\mathsf{C})$ on
By viewing sets in $\mathsf{Set}$ as discrete categories, we have an inclusion $\mathsf{Set}\hookrightarrow\mathsf{Cat}$. Hence the category of elements of $F$ is a special case of the Grothendieck construction.


\section{String Diagram}
Expand Down Expand Up @@ -3756,7 +3756,7 @@ \subsection{Abelian Category}
\end{tikzcd}
\]
}
\prop{Equivalent Characterizations of Split Exact Sequence}{
\prop[equivalent_split_exact_sequence]{Equivalent Characterizations of Split Exact Sequence}{
Let $\mathsf{A}$ be an abelian category and
\[
\begin{tikzcd}[ampersand replacement=\&]
Expand All @@ -3766,12 +3766,36 @@ \subsection{Abelian Category}
be an exact sequence in $\mathsf{A}$. Then the following statements are equivalent
\begin{enumerate}[(i)]
\item The sequence is split.
\item $f$ has right inverse, i.e. there exists a morphism $r:Y\to X$ such that $f\circ r=\mathrm{id}_X$.
\item $g$ has left inverse, i.e. there exists a morphism $s:Z\to Y$ such that $s\circ g=\mathrm{id}_Z$.
\item $f$ has left inverse, i.e. there exists a morphism $r:Y\to X$ such that $r\circ f=\mathrm{id}_X$.
\begin{center}
\begin{tikzcd}[ampersand replacement=\&, column sep={4em, between origins}, row sep={4em, between origins}]
0\arrow[r] \&X\arrow[d, "\mathrm{id}_X"']\arrow[r, "f"] \& Y\arrow[ld, "\exists r", dashed] \arrow[r, "g"] \& Z \arrow[r] \& 0\\
\&X \& \& \&
\end{tikzcd}
\end{center}
\item $g$ has right inverse, i.e. there exists a morphism $s:Z\to Y$ such that $g\circ s=\mathrm{id}_Z$.
\begin{center}
\begin{tikzcd}[ampersand replacement=\&, column sep={4em, between origins}, row sep={4em, between origins}]
\& \& \& Z \arrow[d, "\mathrm{id}_Z"] \arrow[ld, "\exists s"', dashed] \& \\
0\arrow[r] \&X\arrow[r, "f"'] \& Y\arrow[r, "g"'] \& Z \arrow[r] \& 0
\end{tikzcd}
\end{center}
\item The homomorphism $f^*: \mathrm{Hom}_{\mathsf{A}}(Y,A)\to \mathrm{Hom}_{\mathsf{A}}(X,A)$ is surjective for any $A\in \mathrm{Ob}(\mathsf{A})$.
\item The homomorphism $g_*: \mathrm{Hom}_{\mathsf{A}}(A,X)\to \mathrm{Hom}_{\mathsf{A}}(A,Y)$ is surjective for any $A\in \mathrm{Ob}(\mathsf{A})$.
\item The homomorphism $g_*: \mathrm{Hom}_{\mathsf{A}}(A,Y)\to \mathrm{Hom}_{\mathsf{A}}(A,Z)$ is surjective for any $A\in \mathrm{Ob}(\mathsf{A})$.
\end{enumerate}
}
\pf{
(ii) $\implies$ (iv). Suppose $r:Y\to X$ is a left inverse of $f$. Given any $h\in \mathrm{Hom}_{\mathsf{A}}(X,A)$, there exists $h\circ r\in \mathrm{Hom}_{\mathsf{A}}(Y,A)$ such that $f^*(h\circ r)=h\circ r\circ f=h$, which means that $f^*$ is surjective.
\begin{center}
\begin{tikzcd}[ampersand replacement=\&, column sep={4em, between origins}, row sep={4em, between origins}]
0\arrow[r] \&X\arrow[d, "\mathrm{id}_X"']\arrow[r, "f"] \& Y \arrow[ldd, "h\circ r"]\arrow[ld, "r"', dashed] \arrow[r, "g"] \& Z \arrow[r] \& 0\\
\&X \arrow[d, "h"'] \& \& \& \\
\&A \& \& \&
\end{tikzcd}
\end{center}

(iv) $\implies$ (ii). Suppose $f^*$ is surjective. Given any $h\in \mathrm{Hom}_{\mathsf{A}}(X,A)$, there exists $h'\in \mathrm{Hom}_{\mathsf{A}}(Y,A)$ such that $f^*(h')=h$. This means that $h'\circ f=h$. Let $r:=h'$, then $r\circ f=h$.
}


\section{Complex}
Expand Down Expand Up @@ -3960,12 +3984,12 @@ \section{Resolution}
\subsection{Projective and Injective Objects}
\dfn{Projective Object}{
An object $P$ in an abelian category $\mathsf{A}$ is \textbf{projective} if for any epimorphism $f: X \rightarrow Y$ and any morphism $g: P \rightarrow Y$, there exists a morphism $h: P \rightarrow X$ such that $f \circ h = g$.
\begin{center}
\begin{tikzcd}[ampersand replacement=\&, column sep={5em, between origins}, row sep={5em, between origins}]
\& P \arrow[d, "g"] \arrow[ld, "\exists h"', dashed] \& \\
X \arrow[r, "f"', two heads] \& Y \arrow[r] \& 0
\end{tikzcd}
\end{center}
\begin{center}
\begin{tikzcd}[ampersand replacement=\&, column sep={5em, between origins}, row sep={5em, between origins}]
\& P \arrow[d, "g"] \arrow[ld, "\exists h"', dashed] \& \\
X \arrow[r, "f"', two heads] \& Y \arrow[r] \& 0
\end{tikzcd}
\end{center}
}

\prop{Equivalent Characterizations of Projective Objects}{
Expand All @@ -3985,6 +4009,15 @@ \subsection{Projective and Injective Objects}
}
\pf{
(i)$\implies$(ii). Since $\mathrm{Hom}_{\mathsf{A}}(P,-)$ is a left exact functor, it suffices to show that $\mathrm{Hom}_{\mathsf{A}}(P,-)$ preserves epimorphisms. Let $f: X \rightarrow Y$ be an epimorphism in $\mathsf{A}$. We need to show that $f_*:\mathrm{Hom}_{\mathsf{A}}(P,X)\to \mathrm{Hom}_{\mathsf{A}}(P,Y)$ is surjective. Let $g: P \rightarrow Y$ be a morphism in $\mathsf{A}$. Since $P$ is projective, there exists a morphism $h: P \rightarrow X$ such that $f_*(h)=f \circ h = g$. This means that $f_*$ is surjective.

(i)$\implies$(iii). If $P$ is projective, then for any short exact sequence $0 \to X \xrightarrow{f} Y \xrightarrow{g} P \to 0$, there exists a morphism $h: P \to Y$ such that $h \circ g = \mathrm{id}_P$.
\begin{center}
\begin{tikzcd}[ampersand replacement=\&, column sep={4em, between origins}, row sep={4em, between origins}]
\& \& \& P \arrow[d, "\mathrm{id}_P"] \arrow[ld, "\exists h"', dashed] \& \\
0\arrow[r] \&X\arrow[r, "f"'] \& Y\arrow[r, "g"', two heads] \& P \arrow[r] \& 0
\end{tikzcd}
\end{center}
By \Cref{th:equivalent_split_exact_sequence}, this implies that the short exact sequence splits.
}

\dfn{Injective Object}{
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17 changes: 16 additions & 1 deletion commutative_ring.tex
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Expand Up @@ -16,7 +16,7 @@ \section{Basic Concepts}
\end{enumerate}
}

\dfn{Noetherian commutative ring}{
\dfn{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.
Expand Down Expand Up @@ -292,6 +292,21 @@ \section{Principal Ideal Domain}
Let $I\subseteq R$ 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.
}

\section{Krull Dimension}
\dfn{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 $n$.
}

\dfn{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 $\mathfrak{p}$
\[
\mathrm{ht}(\mathfrak{p})=\sup\left\{n\in\mathbb{N}\mid\exists\text{ a chain of prime ideals }\mathfrak{p}_0\subsetneq\mathfrak{p}_1\subsetneq\cdots\subsetneq\mathfrak{p}_n=\mathfrak{p}\right\}.
\]
}
\dfn{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 $\mathrm{ht}(I)$
}

\section{Polynomial Ring}
\dfn{Polynomial Ring}{
Let $R$ be a commutative ring. The \textbf{polynomial ring} in $n$ variables over $R$ is the ring $R[x_1,\cdots,x_n]$ defined as the set of all formal sums $$\sum_{\alpha\in\mathbb{N}^n}a_\alpha x^\alpha$$ where $a_\alpha\in R$ satisfies $a_\alpha=0$ for all but finitely many $\alpha\in\mathbb{N}^n$ and $x^\alpha=x_1^{\alpha_1}\cdots x_n^{\alpha_n}$ for $\alpha=(\alpha_1,\cdots,\alpha_n)\in\mathbb{N}^n$. The addition and multiplication are defined as follows: $$\sum_{\alpha\in\mathbb{N}^n}a_\alpha x^\alpha+\sum_{\alpha\in\mathbb{N}^n}b_\alpha x^\alpha=\sum_{\alpha\in\mathbb{N}^n}(a_\alpha+b_\alpha)x^\alpha$$ and $$\left(\sum_{\alpha\in\mathbb{N}^n}a_\alpha x^\alpha\right)\left(\sum_{\beta\in\mathbb{N}^n}b_\beta x^\beta\right)=\sum_{\gamma\in\mathbb{N}^n}\left(\sum_{\alpha+\beta=\gamma}a_\alpha b_\beta\right)x^\gamma.$$
Expand Down
73 changes: 59 additions & 14 deletions module.tex
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Expand Up @@ -233,7 +233,15 @@ \subsection{Localization}
\]
}
\prop{Localization is an Exact Functor}{
Let $R$ be a commutative ring, $S$ be a multiplicative set in $R$. If $L\xrightarrow {u} M\xrightarrow {v} N$ is an exact sequence of $R$-module, then $S^{-1}L\xrightarrow {S^{-1}(u)} S^{-1}M\xrightarrow {S^{-1}(v)} S^{-1}N$ is an exact sequence of $S^{-1}R$-module.
Let $R$ be a commutative ring, $S$ be a multiplicative set in $R$. If
\[
L\xlongrightarrow {u} M\xlongrightarrow {v} N
\]
is an exact sequence of $R$-module, then
\[
S^{-1}L\xlongrightarrow {S^{-1}(u)} S^{-1}M\xlongrightarrow {S^{-1}(v)} S^{-1}N
\]
is an exact sequence of $S^{-1}R$-module.
}
\pf{
Suppose $\frac{m}{s}\in \ker S^{-1}(v) $. Then we have
Expand Down Expand Up @@ -345,21 +353,32 @@ \subsection{Localization}
By the uniqueness of the universal property of $M_f$, we have $\mu_f=\nu\circ \psi_f$. If there exists another $\nu'$ such that $\mu_f=\nu'\circ \psi_f$, there must be $\nu'\circ l_S=\nu'\circ \psi_f \circ l_f=\mu_f \circ l_f=\nu\circ l_S$. The uniqueness of such $\nu$ forces $\nu=\nu'$.\\
Therefore we show that $ S^{-1}M\cong \varinjlim_{f\in S}M_f$.
}
\prop{}{
View $S^{\prime-1} M$ as an $R$-module, then $S^{-1}\left(S^{\prime-1} M\right)$ is isomorphic to $\left(S S^{\prime}\right)^{-1} M$.
\prop[associative_localization]{}{
Suppose $R$ is a commutative ring, $S,S'$ are multiplicative sets in $R$, and $M$ is an $R$-module. View $S^{\prime-1} M$ as an $R$-module, then $S^{-1}\left(S^{\prime-1} M\right)$ is isomorphic to $\left(S S^{\prime}\right)^{-1} M$ as $R$-modules.
}
\pf{
Note that given a $R$-module $M$, we have not proved any universal property for $S^{-1} M$. Hence we cannot reason as in the preceding proof; we have to construct the isomorphism explicitly.

We define the maps as follows
$$
\begin{aligned}
& f: S^{-1}\left(S^{\prime-1} M\right) \longrightarrow\left(S S^{\prime}\right)^{-1} M, \quad \frac{x / s^{\prime}}{s} \mapsto x / s s^{\prime} \\
& g:\left(S S^{\prime}\right)^{-1} M \longrightarrow S^{-1}\left(S^{\prime-1} M\right), \quad x / t \mapsto \frac{x / s^{\prime}}{s} \text { for some } s \in S, s^{\prime} \in S^{\prime}, \text { and } t=s s^{\prime}
\end{aligned}
$$
Define
$$
\begin{aligned}
f: S^{-1}\left(S^{\prime-1} M\right) &\longrightarrow\left(S S^{\prime}\right)^{-1} M\\
\quad \frac{x / s^{\prime}}{s} &\longmapsto\frac{x}{ss'}
\end{aligned}
$$
To show that $f$ is well-defined, suppose that $\frac{x/s^{\prime}}{s}=\frac{y/t^{\prime}}{t}$, which means there exists $v\in S$ such that
\[
v\left(t\frac{x}{s'} - s\frac{y}{t'}\right)=\frac{vtx}{s'}-\frac{vsy}{t'}=0.
\]
This further implies that there exists $w\in S'$ such that $w(vtt'x-vss'y)=0$.
Then we see there exists $vw\in S S^{\prime}$ such that $vw(tt'x-ss'y)=0$, which means $\frac{x}{ss'}=\frac{y}{tt'}$. Thus $f$ is well-defined.

We have to check that these homomorphisms are well-defined, that is, independent the choice of the fraction. This is easily checked and it is also straightforward to show that they are inverse to each other.
Define
\[
\begin{aligned}
g:\left(S S^{\prime}\right)^{-1} M& \longrightarrow S^{-1}\left(S^{\prime-1} M\right)\\
\frac{x}{ss'} &\longmapsto \frac{x / s^{\prime}}{s} \text { for some } s \in S, s^{\prime} \in S^{\prime}
\end{aligned}
\]
and we can check that $g$ is well-defined in a similar way. It is clear that $f$ and $g$ are linear maps inverse to each other.
}


Expand Down Expand Up @@ -394,7 +413,7 @@ \subsection{Localization}
\pf{
(iii)$\implies$(i). Suppose $M_{\mathfrak{m}}$ is zero for all $\mathfrak{m}\in \mathrm{Max}\left(R\right)$. Given any $x\in M$, since $x$ maps to $0$ in $M_{\mathfrak{m}}$ for all $\mathfrak{m}\in \mathrm{Max}\left(R\right)$, there must be $x=0$ by \Cref{th:module_localization_glueing_property}. Thus $M$ is zero.\\
}
\cor{}{
\cor[exactness_is_a_local_property]{Exactness is a Local Property}{
Given a sequence of $R$-modules $M\to M'\to M''$, the following are equivalent:
\begin{enumerate}[(i)]
\item $M\to M'\to M''$ is exact,
Expand All @@ -410,6 +429,32 @@ \subsection{Localization}
for all $\mathfrak{m}\in \mathrm{Max}\left(R\right)$. Thus by \Cref{th:module_localization_glueing_property_cor} we have $H=0$, which implies $M\to M'\to M''$ is exact.
}

\prop{Glueing Functions}{
Let $R$ be a ring. Let $f_1, \ldots, f_n$ be elements of $R$ generating the unit ideal. Let $M$ be an $R$-module. The sequence
$$
\begin{aligned}
& 0 \longrightarrow M \xlongrightarrow{\alpha} \bigoplus_{i=1}^n M_{f_i} \xlongrightarrow{\beta} \bigoplus_{i, j=1}^n M_{f_i f_j} \\
&
\end{aligned}
$$
is exact, where $\alpha(m)=\left(\frac{m}{1}, \cdots, \frac{m}{1}\right)$ and $\beta\left(\dfrac{m_1}{f_1^{e_1}}, \cdots, \dfrac{m_n}{f_n^{e_n}}\right)=\left(\dfrac{m_i}{f_i^{e_i}}-\dfrac{m_j}{f_j^{e_j}}\right)_{(i, j)}$.
}
\pf{
According to \Cref{th:exactness_is_a_local_property}, it suffices to show that the localization of the sequence at any maximal ideal $\mathfrak{m}$ is exact. Given any maximal ideal $\mathfrak{m}$ of $R$, since $f_1, \ldots, f_n$ generate the unit ideal, there is an $i$ such that $f_i \notin \mathfrak{m}$. Without loss of generality we may assume $f_1\notin \mathfrak{m}$. Note that \Cref{th:associative_localization} guarantees $\left(M_{f_i}\right)_{\mathfrak{m}}=\left(M_{\mathfrak{m}}\right)_{f_i}$ and $\left(M_{f_i f_j}\right)_{\mathfrak{m}}=\left(M_{\mathfrak{m}}\right)_{f_i f_{j}}$. In particular we have $\left(M_{f_1}\right)_{\mathfrak{m}}=M_{\mathfrak{m}}$ and $\left(M_{f_1 f_i}\right)_{\mathfrak{m}}=\left(M_{\mathfrak{m}}\right)_{f_i}$, because $f_1\in M_{\mathfrak{m}}^{\times}$. Thus it is suffices to show that the sequence
\[
0 \longrightarrow M_{\mathfrak{m}} \xlongrightarrow{\alpha_{\mathfrak{m}}} \bigoplus_{i=1}^n \left(M_{\mathfrak{m}}\right)_{f_i} \xlongrightarrow{\beta_{\mathfrak{m}}} \bigoplus_{i, j=1}^n \left(M_{\mathfrak{m}}\right)_{f_i f_j}
\]
is exact for $f_1=1$.

Injectivity of $\alpha_{\mathfrak{m}}$ is trivial because the first component of $\alpha_{\mathfrak{m}}$ is the identity map on $M_{\mathfrak{m}}$.

For any $\mathbf{x}=\left(x_1, \dfrac{x_2}{f_2^{e_n}},\cdots, \dfrac{x_n}{f_n^{e_n}}\right)\in\ker \beta_{\mathfrak{m}}$ we have $\beta_{\mathfrak{m}}(\mathbf{x})=0$. Consider the $(1,i)$-component of $\beta_{\mathfrak{m}}\left(\mathbf{x}\right)$ for $i=2, \cdots, n$. Then we get
\[
x_1-\frac{x_i}{f_i^{e_i}}=0\implies \mathbf{x}=\left(x_1,x_1,\cdots,x_1\right)=\alpha_{\mathfrak{m}}\left(x_1\right)\implies \ker \beta_{\mathfrak{m}}\subseteq \operatorname{im}\alpha_{\mathfrak{m}}.
\]
For any $\mathbf{y}=\left(y, y, \cdots, y\right)\in \operatorname{im}\alpha_{\mathfrak{m}}$, we have $\beta_{\mathfrak{m}}\left(\mathbf{y}\right)=0$, which means $\operatorname{im}\alpha_{\mathfrak{m}}\subseteq \ker \beta_{\mathfrak{m}}$. Thus the sequence is exact and we complete the proof.
}

\subsection{Graded Object}

\dfn{$I$-Graded Module (External Definition)}{
Expand Down

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