\chapter{Ring homomorphisms}

\section{Ring homomorphism}

\begin{definition}[Ring homomorphism]
Let $(R,+_R,\cdot_R)$ and $(S,+_S,\cdot_S)$ be two rings. A \emph{ring homomorphism} $f : R \to S $ is a function between rings that 'preserves' the ring's operation in the following sense; if $r_1,r_2$ are elements of $R$, we have the following.
\[ f(r_1 +_R r_2) = f(r_1) +_S f(r_2) \]
\[ f(r_1 \cdot_R r_2) = f(r_1) \cdot_S f(r_2) \]
\[ f(1_R) = 1_S \]
\end{definition}

\begin{definition}[Alternative definition for a ring homomorphism]
Let $(R,+_R,\cdot_R)$ and $(S,+_S,\cdot_S)$ be two rings. A function $f : R \to S $  is a ring homomorphism iff $f$ is a group homomorphism from $(R,+_R)$ to $(S,+_S)$ and the following 'monoid homomorphism' rule holds.
\[ f(r_1 \cdot_R r_2) = f(r_1) \cdot_S f(r_2) \]
\end{definition}

We may also want to consider  a variant where we drop the requirement of $f(1_R)=1_S$; rngs don't assume the existence of such an object, hence this condition doesn't make sense in that context. Even if it does exist, we may want to consider this weaker definition of a homomorphism in certain contexts anyway. Since this definition is the best that can be done for a rng, we call it a rng homomorphism.
 
\begin{definition}[Rng homomorphism]
Let $(R,+_R,\cdot_R)$ and $(S,+_S,\cdot_S)$ be two rngs. A \emph{rng homomorphism} $f : R \to S $ is a function between rings that 'preserves' the rng's operation in the following sense; if $r_1,r_2$ are elements of $R$, we have the following.
\[ f(r_1 +_R r_2) = f(r_1) +_S f(r_2) \]
\[ f(r_1 \cdot_R r_2) = f(r_1) \cdot_S f(r_2) \]
\end{definition}


\section{Properties of ring homomorphism}
\section{Examples of ring homomorphism}

\section{Ring monomorphism}
\section{Ring epimorphism}
\section{Ring endomorphism}

These types of ring homorphisms have particularly interesting properties, but ring isomorphisms are perhaps the most important class of homomorphisms.

\chapter{Ring isomorphism}
\section{Properties of Ring isomorphism}
\section{Examples of Ring isomorphism}
\section{Ring automorphism}



Some Theorems

A \emph{Canonical map} is a function between two objects that arises from their definitions. It is a function used to define the behaviour of some object.

\begin{definition}[Kernel (ring homomorphism)]
\[ \mathrm{ker}(f)= \{r \in R : f(r)=0_S \}\]
\end{definition}




\begin{definition}
\[ \mathrm{Im}(f) = \{ f(r) \in H : r \in R \]
\end{definition}


\begin{proposition}
	\[ f : R \to S \text{ is a ring homomorphism } \implies \mathrm{Im}(f) \leq S \]
\end{proposition}


Image

Isometry


\section{First isomorphism theorem}

The first isomorphism theorem generalizes to rings.

\begin{lemma}
	\[ f : R \to S \text{ is a ring homomorphism } \implies \mathrm{ker}(f) \text{ is an ideal of } R \]
\end{lemma}





\begin{lemma}
\[ f : R \to S \text{ is a ring homomorphism } \implies [ r_1 + \mathrm{ker}(f) = r_2 + \mathrm{ker}(f) \iff f(r_1)=f(r_2) ] \]
\end{lemma}


\begin{theorem}[First isomorphism theorem for rings]
Let $f : R \to S$ be a ring homomorphism . Then $\mathrm{Im}(f) \cong R / \mathrm{ker}(f)$ the following isomorphism $k$.
\[ k : r +\mathrm{ker}(f) \mapsto f(r)\]
\begin{tikzcd}[column sep=small]
	R \arrow{r}{f} \arrow{d}{\pi} & \mathrm{Im}(f) \\
R/\mathrm{ker}(f) \arrow{ru}{k}
\end{tikzcd}
\end{theorem}

\begin{proposition}
For any ring $R$, the following $ f : \mathbb{Z} \to R$ is a ring homomorphism.
\[f(n) = \sum^{n}_{i=1} 1_R \]
\end{proposition}




Though we've proven this ring version of the theorem by expressing our ideas with the objects of ring theory rathee than group theory, the machinery of category theory provides a natural way to extend certain "theorems" to a plethora of structures; such results are called universal properties in category theory. 

Mathematical ranting aside, we can now prove some new results.

Ring homomorphisms between 2 fields are injective.


let f F \to K be a ring homomorphism
For any subfield F' \leq F, f(F') is a subfield of K
For any subfield K' \leq K, f^{-1}(K') is a subfield of F


Let S be a set of ring homomorphisms F \to K, then Eq(S) is a subfield of F







\section{Ring characteristic}



Imagine we want to make ring homomorphisms from $\mathbb{Z}$, one can show that there is a unique homomorphism from $\mathbb{Z}$ to any ring.
\[f(n) = f(\sum^{n}_{k=1} 1) = \sum^{n}_{k=1} f(1) = \sum^{n}_{k=1} 1_R \]


We will now introduce the idea of a ring characteristic; a value that characterizes when this ring wraps back to $0_r$, essentially determining the nature of how this homomorphism acts on a ring.

\begin{definition}[Ring characteristic]
The \emph{ring characteristic} is the order of the cyclic subgroup generated by the multiplicative identity.
\[\mathrm{char}(R) = \mathrm{ord}_{+}(1_R)\]
If the subgroup has infinite order, we say it has characteristic $0$.
\end{definition}



This homomorphism is special; it essentially means that $\mathbb{Z}$ lives inside every ring, integers being interpreted as $n$ repeated additions of the multiplicative identity. This can go for groups too; take the cyclic subgroup generated by some element $g$ and define the group homomorphism $f(n)=g^n$. By the first isomorphism theorem for groups, this is isomorphic to some $\mathbb{Z}/n \mathbb{Z}$; the concept is similar, though in rings this exists in a more canonical way.

Back to rings, let's see how this homomorphism interacts with 'modular' rings (rings with non-zero characteristic).

For rings with characteristic $n$, the kernel of $f$ will be $n\mathbb{Z}$, so by the first isomorphism theorem for rings $f : \mathbb{Z} /n\mathbb{Z} \to R$ is a ring isomorphism on $f(\mathbb{Z} / n \mathbb{Z})$, but since the image of this function is merely a subgroup of $R$, it is a monomorphism on $R$.

\begin{proposition}
For any ring $R$, the following $ f : \mathbb{Z} / \mathrm{char}(R) \mathbb{Z} \to R$ is a ring monomorphism.
\[f(n) = \sum^{n}_{i=1} 1_R \]
\end{proposition}

Notice that if $R$ is isomorphic to $\mathbb{Z}$ (repeated addition of multiplicative identity reaches all elements and never 'wraps' back to $0_R$), the kernel is $0\mathbb{Z}$; this is why we use the conventionn of giving infinite order subgroups characteristic $0$.

\subsection{Theorems related to this special homomorphism}

By noticing that some $\mathbb{Z} / n \mathbb{Z}$ lives in every ring, we can prove some interesting results.


\begin{proposition}
Let $D$ be a domain, then $\mathrm{char}(D)$ is either prime or 0.
\end{proposition}

If $D$ is a finite domain, then the $f(\mathbb{Z} / \mathrm{char}(R) \mathbb{Z})$ living inside must be a domain too since it is a subring. From our study in group theory however $\mathbb{Z} / n\mathbb{Z}$ with nonprime $n$ permit zero divizors, hence $\mathrm{char}(R)$ must be prime.


\begin{theorem}[Freshman's dream]
	\[(x+y)^{p^r} = x^{p^r}+y^{p^r}\]
\end{theorem}


\subsection{Frobenius map}


\[f(r)=r^p\]
let p be prime, R be a commutarive ring of characteristic p
then the Fobenius map is a homomorphism

If F is a finite of characteristic p then the frobenius map is an automorphism