\chapter{Quotient rings}

\section{Ideals}



Consider a subgroup $N$. In group theory we based quotient groups on the following equivalence relation.

\[g \sim h \iff h^{-1}g \in N\]

For the quotient map to be well defined however, requiring $N$ to be a normal subgroup is necessary and sufficient.


In ring theory, we take left cosets with respect to subgroups of the additive group, so we have the following.

\[r \sim s \iff r-s \in I\]

By the definition of a ring, $I$ is Abelian and therefore normal, however we require an extra condition for the quotient map to be well defined with multipliction.
\[ (r+i)(s+j) = rs + rj + si + ij \]
We want some expression $rs + i_{*}$, by assuming that $ri \in I$ for any ring element $r$ and any $i\in I$, then $rj + si + ij  \in I$ and hence $rs + (rj + si + ij) \in rs+I$

Such an $I$ is called an \emph{right-sided ideal}.

Remember quotient groups? It's quite nice that cosets partition a group (thanks Lagrange); this along with the ability of normal subgroups to make our quotient map a well defined group homomorphism allowed the creation of a group on these 'divided' equivalence classes.
In group theory, quotient groups must be formed from normal subgroups to ensure that the quotient group operation functions as desired. One may ask; what's the ideal (pun intended) subring that can be used to make a notion of a 'quotient ring'?

Unfortunately, subrings won't do the the trick, but if we want to emulate the idea of left multiplication cosets, we can form some kind of set that would ideally (this joke is going to get seriously overused in this part) do the trick.



\begin{definition}[Two sided ideal]
Let $(R,+,\cdot)$ be a ring. A \emph{two-sided ideal of $R$} is a subgroup $I$ of $(R,+)$ closed under multiplication from any element in $R$. For brevity, two-sided ideals will just be called ideals.
\[(I,+) \leq (R,+) \land \forall r \in R \forall i \in I [ri \in I \land ir \in I]\]
\end{definition}

\begin{proposition}[Alternative definition for an ideal]
Let $(R,+,\cdot)$ be a ring. A \emph{two-sided ideal of $R$} is a subgroup $I$ of $(R,+)$ closed under addition, multiplication, and multiplication from any element in $R$. For brevity, two-sided ideals will just be called ideals.
\[r \in R,i,j \in I\]
\[i+j \in I\]
\[ ij \in I \]
\[ ri \in I\]
\[ ir \in I\]
\end{proposition}

One can define specific ideals where arbitrary ring multiplication is allowed only from the left or right.

\begin{definition}[Left ideal]
Let $(R,+,\cdot)$ be a ring. A \emph{left ideal of $R$} is a subgroup $I$ of $(R,+)$ closed under addition, multiplication, and left multiplication from any element in $R$.
\[r \in R,i,j \in I\]
\[i+j \in I\]
\[ ij \in I \]
\[ ir \in I\]
\end{definition}

We'll focus our attention on two-sided ideals from now on.

\begin{proposition}
If $1_R \in I$, then $I=R$.
\end{proposition}

\begin{proposition}
Let $I,J \subseteq R$ be ideals, then $I \cap J$ is an ideal of $R$.
\end{proposition}

We can also define the sum of ideals, which is an ideal itself.

\begin{definition}[Sum of ideals]
\[I + J = \{ i+j : i \in I \land j \in J \}\]
\end{definition}






\section{Ideal generating notation}

Consider $R$ as a commutative ring, since $(R,+)$ is an Abelian group, any element $r \in R$ generates its own cyclic subgroup of $(R,+)$, to make this set a (double-sided) ideal we just need to close this with multiplication over any ring element.

\[ \langle x \rangle = \{ \lambda x  : \lambda \in R \} \]


Why does this form an ideal? The elements of the cyclic subgroup of $x$ are representable as $\sum^{n}_{i=1} x$. Since $\lambda$ is an arbitrary ring element, we can consider lambdas of the form  $\lambda=\sum^{n}_{i=1} 1_R$ so that we have $(\sum^{n}_{i=1} 1_R)x = \sum^{n}_{i=1} x$ which is in the cyclic subgroup. Since $\lambda$ is again an arbitrary ring element, letting it be of the form $\lambda=r(\sum^{n}_{i=1} 1_R)$ where $r \in R$ satisfies all properties of an ideal.

Since
\[ \langle x_1 ,x_2 , \cdot x_n \rangle = \{ \sum^{n}_{k=1} \lambda_k x_k  : \lambda_k \in R \} \]
\[ \langle X \rangle = \{ \sum^{n}_{k=1} \lambda_k x_k  : \lambda_k \in R \land x_k \in X \} \]
Note that when using this notation.
We require topological structure to have a notion of convergence, therefore we can only consider finite sums when generating ideals (if you're familiar with linear algebra, perhaps one could compare $X$ generating the ideal in a way similar to a Hamel basis).


\[ \langle ij : i\in I , j \in J \rangle \]

\[ \langle r_1 r_2 , \hdots , r_n \rangle = \{ \sum^{n}_{k=1} \lambda_k r_k  : \lambda_k \in R \} \]

\[ \langle f : f \in M \rangle = \{ \sum_{f_k \in N} \lambda_k f_k  : \lambda_k \in R, N \subseteq M, |N|< \infty \} \]
Ideals equal to some are 'finitely generated'.


With ideal generating notation, we can now produce a sleek definition for the product of ideals.

\begin{definition}[Product of ideals]
\[ IJ = \langle  \{ ij : i \in I \land j \in J \} \rangle \]
\end{definition}




\section{Basic types of ideals}
\subsection{Zero ideals}
\begin{definition}
Let $(R,+,\cdot)$ be a ring, then $\{0_R\}$ is the \emph{zero ideal}.
\end{definition}

We should check that this definition is well defined, that is, this set is indeed always an ideal. When describing the ideals over a certain ring, We'll often need to tag along the fact that the zero ideal is an ideal over said ring; otherwise there's not too much to say about it.

\begin{proposition}
	Let $I$ be a ideal of the field $(F,+,\cdot)$, then either $I$ is the zero ideal or $I=F$.
\end{proposition}


\subsection{Unit ideals}
We've proven that when the identity is in our ideal, the ideal must be the whole ring. This is another trivial ring that we may as well mention for completeness.

\begin{definition}
Let $(R,+,\cdot)$ be a ring, then $R$ is the \emph{unit ideal}.
\end{definition}

\subsection{Principle ideal}
\begin{definition}
A \emph{principal ideal} is an ideal of the form $\langle d \rangle$. In other words, it is generated by one element.
\end{definition}

If D is a domain where all its ideals are principal ideals, PID (principal ideal domain).


\begin{example}
$(\mathbb{Z},+,\cdot)$ is a PID and all its ideals are $d\mathbb{Z}$. This is because the only subgroups of $(\mathbb{Z},+)$ are of the form $d\mathbb{Z}$, and these can be verified to be principal ideals since addition is obviously satisfied (being a subgroup), multiplication in and out of the ideal is satisfied since $r(dn)=d(rn)\in d\mathbb{Z}$.
Note that $0\mathbb{Z}$ corresponds to the zero ideal.
\begin{example}


\section{Quotient rings}

\begin{definition}
A \emph{quotient ring on $R$} is a ring of unique left 'addition' cosets on the ideal $I$. Addition and multiplication are defined in the following way.
\[ ( R / I , \oplus, \otimes )\]
\[ (r+I) \oplus (s+I) = (r+s)+I\]
\[ (r+I) \otimes (s+I) = rs+I\]
\end{definition}

We should make sure that $\oplus, \otimes$ are well defined operations; we didn't introduce ideals for nothing!


For the sake of simplicity, we'll often use the equivalence class notation $[r]=r+I$ with the original ring's  addition and multiplication symbols $+,\times$ for elements, addition, and multiplication of a quotient ring. Here's an example below.

\[ (r+I) \otimes ((s+I) \oplus (t+I)) = [r]([s]+[t]) \]

\subsection{Examples of quotient rings}

\section{Prime ideal}

For quotient groups, there aren't many general properties of the normal subgroup that give interesting properties to the quotient group. For quotient rings, the properties of the ideal used in the quotient ring can say a great deal regarding the quotient ring's properties.

For instance, One may want to know when a quotient ring is a domain; this question happens to depend greatly on the nature of the ideal.

Recall that domains are categorized by the fact that $xy=0_D$ iff either $x,y$ equals $0_D$. $[0_D]$ is the left coset representing the domain itself, so assuming our quotient ring is a domain, when $[x][y]=[0_R]$ (in other words, when $xy \in I$), either $[x]$ or $[y]$ equals $[0_R]$ (in other words, $x \in I$ or $y \in I$); meaning that one of them is in the same coset (the ideal) as $0_R$.

This allows us to forge precisely the right ideal that makes a quotient ring a domain.

\begin{definition}
A \emph{prime ideal $I$} is an ideal such that for any ring elements $r,s$, if $rs$ is in the ideal, then either $r$ or $s$ is in the ideal.
\end{definition}

Note that this looks slightly similar to Euclid's lemma! The difference being that it is inclusion of an ideal rather than divisibility by a prime.


\begin{proposition}
$R / I$ is a domain iff $I$ is a prime ideal.
\end{proposition}

\section{Maximal ideal}

Similarly, one wants to know which type of ideal forms a field.
Assuming that the original ring is a commutative ring and we've made a quotient ring that is a field, we have

	$[r][s]=[rs]=[1_R]$
	so for any $r \notin I$, there is some $s \in R$ such that $rs-1_R \in I$.

Now imagine a strictly larger ideal $J$ such that $r \in J$. Since $rs \in J$ and $rs-1_R \in J$, we can prove that $1_R \in J$ which means $J=R$. Since $r$ was chosen arbitrarily, this means quotient rings that are fields have the largest non-unit ideal possible on the commutative ring.

\begin{definition}[Maximal ideal]
A \emph{maximal ideal $I$} is an ideal such that any ideal $J$ where $I \subset J$ equals the entire ring.
\end{definition}


\begin{proposition}
$R / I$ is a field iff $I$ is a maximal ideal.
\end{proposition}

since for a field $R / I$ with an ideal $J$ with $I \subset J$, we have both $rs-1_R \in J$ and $rs \in J$ for some $r \in J,r \notin I$, and therefore$1_R \in J$, meaning that $R=J$.
Here's a proof sketch for the other direction: if $I$ is maximal then for any $j \notin I$ we have $I \oplus \langle j \rangle = R$. Since $1_R \in R$, we have the expression $i +rz=1_R$, where $i \in I, z \in \langle j \rangle, r \in R$, so then by the properties of ideals $rz \in \langle j \rangle$. We then have $rz-1_R = -i$ so $rz-1_R \in I$. Since $r$ is an arbitrary ring element, this holds for any element in the commutative ring.


Since prime ideals and maximal ideals have a direct correspondence to domains and fields respectively, we can prove the following.


\begin{corollary}
Maximal ideals are prime ideals.
\end{corollary}





\section{Generalized number theory in integral domains}

Let's generalize number theory in an integral domain.

\begin{definition}[Divisibility in an integral domain]
\[a|r \iff \exists k \in R [ak=r] \]
\end{definition}



\begin{definition}[Reducible element]
$r$ is irreducible iff it is not a unit nor 0 and the following holds.
\[r=ab \implies a \in R^{*} \lor b \in R^{*}\]
\end{definition}

\begin{definition}[GCD in an integral ring]
\[\mathrm{gcd}(a,b)=n \iff \forall d \in R [d|a \land d|b \implies d|n ] \]
\end{definition}

A ring such that any pair of elements have a GCD is called a GCD domain

\begin{proposition}
PIDs are GCD domains.
\end{proposition}


\begin{lemma}[Euclid's lemma (ring theory)]
Let $R$ be a GCD domain
\[ n|ab \land \gcd(n,a)=1_R \implies n|b \]
\end{lemma}



\begin{lemma}[Bézout's lemma (ring theory)]
Let $R$ be a PID
\[ \exists x,y \in R [  ax+by = \mathrm{gcd}(a,b) ] \]
\end{lemma}