ring.tex

\chapter{Ring}
\section{Basic Concepts}
\begin{definition}{Ring}{}
    A \textbf{ring} is a set $R$ together with two binary operations $+$ and $\cdot$ on $R$ such that
    \begin{enumerate}[(i)]
        \item $(R,+)$ is an abelian group.
        \item $(R,\cdot)$ is a monoid.
        \item $\cdot$ is distributive over $+$,
        \begin{align*}
            a\cdot(b+c)=a\cdot b+a\cdot c\\
            (a+b)\cdot c=a\cdot c+b\cdot c
        \end{align*}
    \end{enumerate}
\end{definition}

A ring is a monoid object in the category $\mathsf{Ab}$. In other words, a ring is an $\mathsf{Ab}$-enriched category with only one object.\\
A ring $R$ is an $R$-module over itself.\\
A ring $R$ is a $Z(R)$-algebra and also a $\mathbb{Z}$-algebra. In fact, we have the following category isomorphism
\[
    \mathsf{Ring}\cong \mathbb{Z}\text{-}\mathsf{Alg}.
\]
\begin{definition}{Unit Group of a Ring}{}
    Let $R$ be a ring. The \textbf{unit group} of $R$ is the group of invertible elements of $R$ under multiplication, denoted by $R^\times$. We can define a functor $(-)^\times:\mathsf{Ring}\to\mathsf{Grp}$ that sends a ring to its unit group
    \[
        \begin{tikzcd}[ampersand replacement=\&]
            \mathsf{Ring}\ \&[-25pt]\&[+10pt]\&[-30pt]\mathsf{Grp}\&[-30pt]\&[-30pt] \\ [-15pt] 
            R \arrow[dd, "f"{name=L, left}] 
            \&[-25pt] \& [+10pt] 
            \& [-30pt]R^{\times}\arrow[dd, "f|_{R^\times}"{name=R}] \\ [-10pt] 
            \&  \phantom{.}\arrow[r, "(-)^\times", squigarrow]\&\phantom{.}  \&   \\[-10pt] 
            S  \& \& \& S^{\times}
        \end{tikzcd}
    \]
\end{definition}


\begin{proposition}{Adjunction $\mathbb{Z}\left[-\right]\dashv \left(-\right)^\times$}{}
    $(-)^\times:\mathsf{Ring}\to\mathsf{Grp}$ has a left adjoint which sends each group $G$ to the group ring $\mathbb{Z}\left[G\right]$.
\end{proposition}


Next we define the morphisms in the category $\mathsf{Ring}$.
\begin{definition}{Ring Homomorphism}{}
    Let $R$, $S$ be rings. A \textbf{ring homomorphism} from $R$ to $S$ is a map $f:R\to S$ such that
    \begin{enumerate}[(i)]
        \item $f(a+b)=f(a)+f(b)$ for all $a,b\in R$.
        \item $f(a\cdot b)=f(a)\cdot f(b)$ for all $a,b\in R$.
        \item $f(1_R)=1_S$.
    \end{enumerate}
\end{definition}




\begin{definition}{Zero Divisor}{zero_divisor}
    Assume that $a$ is an element of a ring $R$. 
    \begin{itemize}
        \item $a$ is called a \textbf{left zero divisor} if there exists a nonzero $x$ in $R$ such that $ax = 0$.
        \item $a$ is called a \textbf{right zero divisor} if there exists a nonzero $x$ in $R$ such that $xa = 0$.
        \item $a$ is called a \textbf{zero divisor} if there exists a nonzero $x$ in $R$ such that $ax = xa = 0$.
        \item $a$ is called a \textbf{regular element} if it is neither a left zero divisor nor a right zero divisor.
    \end{itemize}
\end{definition}

$0$ is never a regular element.

\begin{proposition}{Set of Regular Elements is a Multiplicative Set}{set_of_regular_elements_is_a_multiplicative_set}
    The set of regular elements in a ring $R$ is a multiplicative set.
\end{proposition}
\begin{prf}
    Let $a,b$ be regular elements in $R$. We are to show that $ab$ is also a regular element. Suppose that there exists a nonzero $x\in R$ such that $(ab)x=0$. Then $a(bx)=0$. Since $a$ is a regular element, we have $bx=0$. Since $b$ is also a regular element, we have $x=0$. This is a contradiction. 
    
    Similarly, if there exists a nonzero $y\in R$ such that $y(ab)=0$, then we can deduce that $(ya)b=0$. Since $b$ is a regular element, we have $ya=0$. Since $a$ is also a regular element, we have $y=0$. This is also a contradiction. Therefore, $ab$ is a regular element.
\end{prf}

\begin{definition}{Domain}{domain}
    A nonzero ring $R$ is called a \textbf{domain} if one of the following equivalent conditions holds:
    \begin{enumerate}[(i)]
        \item 0 is the only left zero divisor in $R$.
        \item 0 is the only right zero divisor in $R$.
    \end{enumerate}
\end{definition}


\begin{definition}{Ideal}{}
    Let \( R \) be a ring and \( I \subseteq R \) be an additive subgroup.
    \begin{itemize}
        \item \textbf{Left ideal}: If for every \( r \in R \), \( rI \subseteq I \), then \( I \) is called a left ideal of \( R \).
        \item \textbf{Right ideal}: If for every \( r \in R \), \( Ir \subseteq I \), then \( I \) is called a right ideal of \( R \).
        \item \textbf{Two-sided ideal}: If \( I \) is both a left and a right ideal, then it is called a two-sided ideal.
    \end{itemize}

A left, right, or two-sided ideal \( I \) that satisfies \( I \ne R \) is called a \textbf{proper ideal}. In commutative rings, left and right ideals are the same and are simply called ideals. 
\end{definition}
Sometimes we just say ideal to mean two-sided ideal.

\begin{proposition}{Equivalent Characterizations of Ideals}{equivalent_characterizations_of_ideals}
    Let \( R \) be a ring. Then
    \begin{enumerate}[(i)]
        \item Submodules of the left \( R \)-module \( R \) are precisely the left ideals of \( R \).
        \item Submodules of the right \( R \)-module \( R \) are precisely the right ideals of \( R \).
        \item Sub-bimodules of the \( R \)-\( R \) bimodule \( R \) are precisely the two-sided ideals of \( R \).
    \end{enumerate}
    Especially, if \( R \) is commutative, then the ideals of \( R \) are precisely the submodules of the \( R \)-module \( R \).
\end{proposition}

\begin{definition}{Kernel of a Ring Homomorphism}{}
    Let $f:R\to S$ be a ring homomorphism. The \textbf{kernel} of $f$ is the set
    \[
        \ker f=f^{-1}(0_S)=\{r\in R\mid f(r)=0_S\}.
    \]
\end{definition}

\begin{proposition}{Kernel of a Ring Homomorphism is an Ideal}{kernel_of_ring_homomorphism_is_an_ideal}
    Let $f:R\to S$ be a ring homomorphism. Then $\ker f$ is a two-sided ideal of $R$. If $S$ is not a zero ring, then $\ker f$ is a proper ideal of $R$.
\end{proposition}
\begin{prf}
    Let $a,b\in\ker f$ and $r\in R$. Then $f(a)=f(b)=0_S$. Hence $f(a+b)=f(a)+f(b)=0_S$, $f(ra)=f(r)f(a)=0_S$ and $f(ar)=f(a)f(r)=0_S$. Therefore, $a+b,ra,,ar\in\ker f$.\\
    If $S$ is not a zero ring, then $1_S\ne 0_S$. Hence $f(1_R)=1_S\ne 0_S$, which implies $1_R\notin\ker f$. Therefore, $\ker f$ is a proper ideal of $R$.
\end{prf}

\begin{proposition}{Surjective Ring Homomorphisms Map Ideals to Ideals}{}
    A ring homomorphism $f:R\to S$ is surjective if and only if maps two-sided ideals to two-sided ideals.
\end{proposition}
\begin{prf}
    First, assume that \( f \) is surjective. Let \( I \) be any two-sided ideal of \( R \). Since $f$ is a homomorphism between addition group $(R,+)$ and $(S,+)$, $(f(I),+)$ is a subgroup of $(S,+)$. Now take any element \( y \in f(I) \) and any \( s \in S \), we are to show $sy\in f(I)$ and $ys\in I$. Since \( f \) is surjective, there exists an \( r \in R \) such that \( s = f(r) \). Also, since \( y \in f(I) \), there exists an \( x \in I \) with \( y = f(x) \). Then
\[
s y = f(r) f(x) = f(rx).
\]
Because \( I \) is a two-sided ideal in \( R \), we have \( rx \in I \), hence \( f(rx) \in f(I) \). Similarly,
\[
y s = f(x) f(r) = f(xr) \in f(I).
\]
Thus, \( f(I) \) is a two-sided ideal of \( S \).

Conversely, suppose that for every two-sided ideal \( I \) of \( R \), the set \( f(I) \) is a two-sided ideal in \( S \). In particular, consider the two-sided ideal \( I = R \). Then \( f(R) \) is a two-sided ideal of \( S \). Note that \(1_S\in f(R) \). We have \( f(R) = S \), so \( f \) is surjective.

\end{prf}



\begin{definition}{Reduced Ring}{}
    A ring $R$ is called \textbf{reduced} if it has no nonzero nilpotent elements, or equivalently, if for any $x\in R$, $x^2=0\implies x=0$.
\end{definition}


\begin{proposition}{Examples of Reduced Ring}{}
    \begin{enumerate}[(i)]
        \item Subrings, products, and localizations of reduced commutative rings are again reduced rings.
        \item Every integral domain is reduced.
        \item $\mathbb{Z}/n\mathbb{Z}$ is reduced if and only if $n=0$ or $n$ is square-free.
    \end{enumerate}
\end{proposition}
\begin{proof}
    \begin{enumerate}[(i)]
        \item Let $R$ be a reduced ring and $S$ be a  multiplicative subset of $R$. For any $\frac{f}{s}\in S^{-1}R$, if $\left(\frac{f}{s}\right)^n=\frac{f^n}{s^n}=0$, then there exists $t\in S$ such that $tf^n=0$, which implies $(tf)^n=0$. Since $R$ is reduced, we have $tf=0$, which means $\frac{f}{s}=0$. Hence $S^{-1}R$ is reduced.
    \end{enumerate}
\end{proof}


\begin{definition}{Local Ring}{}
    A ring $R$ is called \textbf{local} if it has a unique maximal ideal.
\end{definition}

\begin{definition}{Local Ring Homomorphism}{}
    Let $f:R\to S$ be a ring homomorphism and $\mathfrak{m}_R$ and $\mathfrak{m}_S$ be the unique maximal ideals of $R$ and $S$ respectively. Then $f$ is called a \textbf{local ring homomorphism} if $f(\mathfrak{m}_R)\subseteq \mathfrak{m}_S$ or equivalently $f^{-1}(\mathfrak{m}_S)=\mathfrak{m}_R$.
\end{definition}

\section{Construction}
\subsection{Initial Object and Terminal Object}
\begin{proposition}{Initial Object in $\mathsf{Ring}$}{}
    The ring $\mathbb{Z}$ is an initial object in $\mathsf{Ring}$. That is, for any ring $R$, there exists a unique ring homomorphism
    \begin{align*}
        \varphi:\mathbb{Z}&\longrightarrow R\\
        n&\longmapsto n\cdot 1_R
    \end{align*}
\end{proposition}
\begin{proof}
    If $\psi: \mathbb{Z} \to R$ is a ring homomorphism, then $\psi(1) = 1_R$ implies 
    \[
        \psi(n) = \psi(1+\cdots+ 1) = \psi(1)+\cdots+ \psi(1) = n \cdot 1_R.
    \]
    Therefore, $\psi = \varphi$.
\end{proof}

\begin{definition}{Characteristic of a Ring}{characteristic_of_a_ring}
    Let $R$ be a ring and $\varphi:\mathbb{Z}\to R$ be the unique ring homomorphism. Then $\ker\mathrm{\varphi}= n\mathbb{Z}$ and $\mathrm{im}\varphi=\mathbb{Z}/n\mathbb{Z}$, where $n\in\mathbb{Z}_{\ge 0}$.
    The \textbf{characteristic} of $R$ is defined to be $n$, denoted by $\mathrm{char}(R)$.

    Equivalently, $\mathrm{char}(R)$ is the smallest positive integer $n$ such that $n\cdot 1_R=0_R$ if such an integer exists. Otherwise, the characteristic of $R$ is $0$.
\end{definition}

\begin{proposition}{Terminal Object in $\mathsf{Ring}$}{}
    The ring $\mathbb{Z}$ is an initial object in $\{0\}$.
\end{proposition}

Since the forgetful functor $\mathsf{Ring}\to\mathsf{Set}$ is a right adjoint, it preserves all limits. Hence the underlying set of the terminal object in $\mathsf{Ring}$ is the terminal object in $\mathsf{Set}$, which is the singleton set $\{*\}$.



\subsection{Quotient Object}
\begin{definition}{Quotient Ring}{}
    Let $R$ be a ring and $I$ be a two-sided ideal of $R$. Equip the additive group \( R / I \) with the following multiplication operation:
\[
(r+I) \cdot (s+I) := (rs + I), \quad r, s \in R .
\]
Then \( R / I \) forms a ring, which is called the \textbf{quotient ring of \( R \) modulo \( I \)}. The quotient map \( R \rightarrow R / I \) is called the quotient homomorphism.
\end{definition}

\begin{proposition}{Universal Property of Quotient Rings}{}
    Let $R$ be a ring and $I$ be a two-sided ideal of $R$. Then the quotient map $\pi:R\to R/I$ is a surjective ring homomorphism with kernel $I$. Moreover, for any ring $S$ and any ring homomorphism $f:R\to S$ such that $I\subseteq\ker f$, or equivalently $f(I)=\{0\}$, there exists a unique ring homomorphism $\bar{f}:R/I\to S$ such that the following diagram commutes
    \[
    \begin{tikzcd}[ampersand replacement=\&]
        R \arrow[r, "f"] \arrow[d, "\pi"'] \& S \\
        R/I \arrow[ru, "\exists!\bar{f}"', dashed] \&  
    \end{tikzcd}
    \] 
    Moreover, $\bar{f}$ is injective if and only if $\ker f=I$. $\bar{f}$ is surjective if and only if $f$ is surjective.
\end{proposition}

\begin{proposition}{Kernel of a Ring Homomorphism is a Two-sided Ideal}{}
    Let $f:R\to S$ be a ring homomorphism. Then $\ker f$ is an two-sided ideal of $R$.
\end{proposition}

\begin{proposition}{Image of a Ring Homomorphism is a Subring}{}
    Let $f:R\to S$ be a ring homomorphism. Then $\mathrm{im}f$ is a subring of $S$.
\end{proposition}

\begin{theorem}{The Fundamental Theorem of Ring Homomorphisms}{}
    Let $f:R\to S$ be a ring homomorphism. Then $R/\ker f\cong \mathrm{im}f$.
\end{theorem}

\begin{proposition}{}{}
    Let $f:R\to S$ be a ring homomorphism. Then we have the following map
    \begin{align*}
       \left\{\text{two-sided ideals of }S\right\}&\longrightarrow\left\{\text{two-sided ideals of }R\text{ containing }\ker f\right\}\\
        I_2&\longmapsto f^{-1}(I_2)
    \end{align*}
    Furthermore, if $f$ is surjective, then we obtain a bijection
    \begin{align*}
        \left\{\text{two-sided ideals of }S\right\}&\xlongrightarrow{\sim}\left\{\text{two-sided ideals of }R\text{ containing }\ker f\right\}\\
        I_2&\longmapsto f^{-1}(I_2)\\
        f(I_1)&\longmapsfrom I_1
    \end{align*}
\end{proposition}


\begin{corollary}{}{}
    Let $R$ be a ring and $I$ be a two-sided ideal of $R$. Then we have the following bijection
    \begin{align*}
        \left\{\text{two-sided ideals of }R/I\right\}&\xlongrightarrow{\sim}\left\{\text{two-sided ideals of }R\text{ containing }I\right\}\\
        J/I:=\left\{j+I\mid j\in J\right\} &\longmapsfrom J
    \end{align*}
\end{corollary}


\begin{example}{}{coset_of_generated_ideals_quotient_generates_ideals}
    Let $R$ be a commutative ring and $a_1,\cdots,a_n\in R$ be elements in $R$. Then 
    \[
    \left(a_1,\cdots,a_n\right)/(a_1) = \left(a_2+(a_1),\cdots,a_n+(a_1)\right)
    \]
    is a ideal of $R/(a_1)$.
\end{example}
\begin{prf}
    For any $x=r_1a_1+\cdots+r_na_n+(a_1)\in \left(a_1,\cdots,a_n\right)/(a_1)$, we have
    \[
    x=r_1a_1+\cdots+r_na_n+(a_1)=r_2(a_2+(a_1))+\cdots+r_n(a_n+(a_1))\in \left(a_2+(a_1),\cdots,a_n+(a_1)\right).
    \]
    This implies $\left(a_1,\cdots,a_n\right)/(a_1)\subseteq \left(a_2+(a_1),\cdots,a_n+(a_1)\right)$.\\
    Conversely, for any $y=r_2(a_2+(a_1))+\cdots+r_n(a_n+(a_1))\in \left(a_2+(a_1),\cdots,a_n+(a_1)\right)$, we have
    \[
    y=r_2a_2+\cdots+r_na_n+(a_1)\in \left(a_1,\cdots,a_n\right)/(a_1).
    \]
    This implies $\left(a_2+(a_1),\cdots,a_n+(a_1)\right)\subseteq \left(a_1,\cdots,a_n\right)/(a_1)$. Therefore, we prove the equality.
\end{prf}

\begin{theorem}{Third Isomorphism Theorem}{}
    Let $R$ be a ring and $I$ and $J$ be two-sided ideals of $R$ such that $I\subseteq J$. Then $J/I$ is a two-sided ideal of $R/I$ and we have the following isomorphism
    \[
        \frac{R/I}{J/I}\cong R/J.
    \]    
\end{theorem}


\subsection{Free Object}
\begin{definition}{Free Ring}{}
    Let $S$ be a set. The \textbf{free ring} on $S$, denoted by $\mathrm{Free}_{\mathsf{Ring}}(S)$, together with a function $\iota:S\to \mathrm{Free}_{\mathsf{Ring}}(S)$, is defined by the following universal property: for any ring $R$ and any function $f:S\to R$, there exists a unique ring homomorphism $\widetilde{f}:\mathrm{Free}_{\mathsf{Ring}}(S)\to R$ such that the following diagram commutes
    \begin{center}
        \begin{tikzcd}[ampersand replacement=\&]
            \mathrm{Free}_{\mathsf{Ring}}(S)\arrow[r, dashed, "\exists !\,\widetilde{f}"]  \& R \\[0.3cm]
            S\arrow[u, "\iota"] \arrow[ru, "f"'] \&  
        \end{tikzcd}
    \end{center}
    The free ring $\mathrm{Free}_{\mathsf{Ring}}(S)$ can be contructed as the free $\mathbb{Z}$-algebra on $\mathrm{Free}_{\mathsf{Mon}}(S)$
    \[
        \mathrm{Free}_{\mathsf{Ring}}(S)\cong\bigoplus_{w\in\mathrm{Free}_{\mathsf{Mon}}(S)}\mathbb{Z}w.  
    \]
\end{definition}


\begin{example}{Forgetful Functor $U:\mathsf{Ring}\to\mathsf{Set}$}{}
    The forgetful functor $U:\mathsf{Ring}\to\mathsf{Set}$ forgets the ring structure and retains only the underlying set.
    \begin{enumerate}[(i)]
        \item $U$ is representable by $\left(\mathbb{Z}[x],x\right)$.
        \item $U$ is faithful but not full.
    \end{enumerate}
\end{example}


\begin{proposition}{Free-Forgetful Adjunction $\mathrm{Free}_{\mathsf{Ring}}\dashv U$}{}
    The free ring functor $\mathrm{Free}_{\mathsf{Ring}}$ is left adjoint to the forgetful functor $U:\mathsf{Ring}\to\mathsf{Set}$.
\end{proposition}




\subsection{Graded Object}
\begin{definition}{$I$-Graded Ring (Internal Definition)}{I_graded_ring}
    Let $(I,+)$ be a monoid. An \textbf{$I$-graded ring} is a ring $(R,+,\cdot)$ together with a family of subgroups $(R_i)_{i\in I}$ of $(R,+)$ such that
    \begin{enumerate}[(i)]
        \item $R=\bigoplus_{i\in I}R_i$.
        \item $R_iR_j\subseteq R_{i+j}$ for all $i,j\in I$.
    \end{enumerate}
    Elements in $R_i-\{0\}$ are called \textbf{homogeneous elements of degree $i$}.
\end{definition}

\begin{definition}{Graded Ideal}{graded_ideal}
    Let $R$ be an $I$-graded ring with grading $(R_i)_{i\in I}$. An ideal $J$ of $R$ is called \textbf{graded} if
    \[
    J=\bigoplus_{i\in I}J\cap R_i.
    \]
\end{definition}

\begin{proposition}{Homogeneous Elements Generate Graded Ideal}{}
    Let $R$ be a $I$-graded ring with grading $(R_i)_{i\in I}$ and $\mathfrak{a}$ be a two-sided ideal of $R$. Then $\mathfrak{a}$ is a graded ideal if and only if $\mathfrak{a}$ is generated by homogeneous elements. That is,
    \[
    \mathfrak{a}\text{ is a graded ideal }\iff \mathfrak{a}=\left\langle \bigcup_{i\in I}H_i\right\rangle,\quad H_i\subseteq R_i.
    \]
\end{proposition}

\begin{prf}
If $\mathfrak{a}$ is a graded ideal, then 
\[
    \mathfrak{a}=\bigoplus_{i\in I}\mathfrak{a}\cap R_i=\left\langle \bigcup_{i\in I}\left( \mathfrak{a}\cap R_i\right)\right\rangle,
\]
which means $\mathfrak{a}$ is generated by homogeneous elements.\\

Assume $\mathfrak a$ is generated by homogeneous elements, say $\mathfrak a = \left\langle\, \bigcup_{i\in I}H_i \right\rangle$, where $H_i\subseteq R_i$. Let $H=\bigcup_{i\in I}H_i $. Then for any $a\in\mathfrak a$, it can be written as
\[
a=\sum_{k=1}^nr_k h_ks_k.
\]
where $r_k,s_k\in R$, $h_k\in H$. Assume that $r_k,s_k$ have the following decomposition
\begin{align*}
    r_k=\sum_{i\in I}r_{k,i},\quad r_{k,i}\in R_i,\\
    s_k=\sum_{j\in I}s_{k,j},\quad s_{k,j}\in R_j,
\end{align*}
Then we have
\[
a=\sum_{k=1}^n\left(\sum_{i\in I}r_{k,i}\right)h_k\left(\sum_{j\in I}s_{k,j}\right)=\sum_{k=1}^n\sum_{i\in I}\sum_{j\in I} r_{k,i}h_ks_{k,j}.
\]
Suppose $h_k\in R_m$, then $r_{k,i}h_ks_{k,j}\in R_{i+m+j}$. Also we note $h_k\in \mathfrak{a}$ implies $r_{k,i}h_ks_{k,j}\in \mathfrak{a}$. Hence 
\[
a=\sum_{k=1}^n\sum_{i\in I}\sum_{j\in I} r_{k,i}h_ks_{k,j}\in \sum_{i\in I} \mathfrak{a} \cap R_i=\bigoplus_{i\in I}\mathfrak{a}\cap R_i.
\]
It is clear that $\bigoplus\limits_{i\geq 0}(\mathfrak a\cap R_i) \subseteq \mathfrak a $. Therefore, we show that
\[
    \mathfrak a=\bigoplus_{i\geq 0}(\mathfrak a\cap R_i) , 
\]
which means $\mathfrak{a}$ is a graded ideal.
\end{prf}


\begin{proposition}{Membership Criterion for Graded Ideals}{membership_criterion_for_graded_ideals}
    Let $R$ be a $I$-graded ring with grading $(R_i)_{i\in I}$ and $\mathfrak{a}$ be a graded ideal of $R$. Assume $x=\sum_{i\in I}x_i\in R$ with $x_i\in R_i$. Then 
    \[
    x\in \mathfrak{a}\iff x_i\in \mathfrak{a}\text{ for all }i\in I.
    \]
\end{proposition}
\begin{prf}
    Suppose $x\in \mathfrak{a}$. Since $\mathfrak{a}=\bigoplus_{i\in I}\mathfrak{a}\cap R_i$, there exists $y_i\in \mathfrak{a}\cap R_i$ such that
    \[
    x=\sum_{i\in I}y_i.
    \]
    Since $R=\bigoplus_{i\in I}R_i$, the decomposition of $x$ is unique. Hence $x_i=y_i\in \mathfrak{a}$ for all $i\in I$. The converse is clear.
\end{prf}

\section{Category Properties}
The category $\mathsf{Ring}$ is both complete and cocomplete.

\begin{proposition}{Equivalence Chracaterization of Monomorphisms in $\mathsf{Ring}$}{}
    Let $f:R\to S$ be a ring homomorphism. Then the following are equivalent:
    \begin{enumerate}[(i)]
        \item $f$ is a monomorphism.
        \item $f$ is injective.
        \item $\ker f=\{0_R\}$.
    \end{enumerate}
\end{proposition}


\begin{proposition}{Sujective Ring Homomorphisms are Epimorphisms}{}
    Every surjective homomorphism of rings is an epimorphism. However, the converse is not true in general.
\end{proposition}


\begin{proposition}{Equivalence Chracaterization of Isomorphisms in $\mathsf{Ring}$}{}
    Let $f:R\to S$ be a ring homomorphism. Then the following are equivalent:
    \begin{enumerate}[(i)]
        \item $f$ is an isomorphism.
        \item $f$ is bijective.
        \item $\ker f=\{0_R\}$ and $\mathrm{im}f=S$. 
    \end{enumerate}
\end{proposition}