Ring homomorphisms are for comparing rings.
Let $R$ and $S$ be rings with identities
${1}_{R}$ and ${1}_{S}$,
respectively.

A ring homomorphism from $R$ to $S$
is a function $f:R\to S$ such that
 (a)
If $r,s\in R$
then $f(r+s)=f\left(r\right)+f\left(s\right)$,
 (b)
If $r,s\in R$
then $f\left(rs\right)=f\left(r\right)f\left(s\right)$,
 (c)
$f\left({1}_{R}\right)={1}_{S}$.

A ring isomorphism is a bijective ring homomorphism.

Two rings $R$ and $S$ are isomorphic,
$R\simeq S$, if there exists a ring isomorphism
$f:R\to S$ between them.
Two rings are isomorphic if the elements of the rings and their operations match up exactly.
Think of rings that are isomorphic as being "the same".
HW: Give an example of two rings $R$ and $S$ that are isomorphic
as groups but not as rings.
In the case of group, condition (b) in the definition of ring homomorphism forced condition (c). (See
Proposition 1.1.11). This does not happen here since rings don't necessarily have multiplicative inverses.
Let
$f:R\to S$ be a ring homomorphism.
Let ${0}_{R}$ and ${0}_{\mathrm{S}}$
be the zeros for $R$ and $S$ respectively. Then
 (a)
$f\left({0}_{R}\right)={0}_{S}$.
 (b)
If $r\in R$ then
$f(r)=f\left(r\right)$.
 A subring $S$ of a ring $R$
is a subset $S\subseteq R$ such that
 (a)
If ${s}_{1},{s}_{2}\in S$ then
${s}_{1}+{s}_{2}\in S$,
 (b)
$0\in S$,
 (c)
If $s\in S$ then
$s\in S$,
 (d)
If ${s}_{1},{s}_{2}\in S$ then
${s}_{1}{s}_{2}\in S$,
 (e)
$1\in S$,
 The zero ring, $\left(0\right)$, is the set containing
only $0$ with operations $0+0=0$ and $0\cdot 0=0$.
Cosets
 An additive subgroup of a ring $R$
is a subset $I\subseteq R$ of $R$
such that
 (a)
If ${h}_{1},{h}_{2}\in I$ then
${h}_{1}+{h}_{2}\in I$,
 (b)
$0\in I$,
 (c)
If $h\in I$ then
$h\in I$,
Let $R$ be a ring and let
$I$ be an additive subgroup of $R$. We will use the
subgroup $I$ to divide up the ring $R$.

A coset of $I$ in $R$
is a set $r+I=\{r+h\phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}h\in I\}$, where
$r\in R$.
 $R/I$ (pronounced "$R$ mod
$I$") is the set of cosets of $I$ in $R$.
Let $R$ be a ring and let
$I$ be an additive subgroup of $R$. Then the cosets of
$I$ in $R$ partition $R$.
Notice that the proofs of Proposition (rgptn) and Proposition
(gpptn) are essentially the same.
HW: Write a very short proof of Proposition (rgptn) by using (gpptn).
Quotient rings $\leftrightarrow $ Ideals
Let $R$ be a ring and let
$I$ be an additive subgroup of $R$. We can try to make the set
$R/I$ of cosets of $I$ in $R$
into a ring by defining an addition operation and a multiplication operation on cosets.
The only problem is that this doesn't work for the cosets of just any additive subgroup, the subgroup
has to have special properties.
HW: Let $R$ be a ring and let $I$ be an additive subgroup of
$R$. Show that $I$ is a normal subgroup of $R$.
 An ideal $I$ is a subset of a ring $R$ such that
 (a)
If $a,b\in I$ then $a+b\in I$,
 (b)
If $i\in I$ and $r\in R$
then $ir\in R$ and
$ri\in R$.
 The zero ideal $\left(0\right)$ of $R$
is the ideal containing only the zero element of $R$.
HW: Show that if $I$ is an ideal of a ring $R$ then
$0\in I$ and if $a\in I$
then $a\in I$.
HW: Show that an ideal $I$ of a ring $R$ is an additive subgroup of
$R$.
Let $I$ be an additive subgroup of a ring $R$.
Then $I$ is an ideal of $R$
if and only if $R/I$ with operations given by
$$({r}_{1}+I)+({r}_{2}+I)=({r}_{1}+{r}_{2})+I\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}{r}_{1}({r}_{2}+I)={r}_{1}{r}_{2}+I$$
is a ring.
Notice that the proofs of Proposition 2.1.6 and proposition 1.1.8 are essentially the same.
HW: Write a shorter proof of Proposition 2.1.6 by using Proposition 1.1.8.
 The quotient ring $R/I$ is the ring
of cosets of an ideal $I$ of a ring $R$ with operations
given by
$({r}_{1}+I)+({r}_{2}+I)=({r}_{1}+{r}_{2})+I$
and
$({r}_{1}+I)({r}_{2}+I)={r}_{1}{r}_{2}+I$.
So we have successfully made $R/I$ into a ring when
$I$
is an ideal of $R$.
Kernel and image of a homomorphism
 The kernel of a ring homomorphism
$f:R\to S$ is the set
$$\mathrm{ker}f=\{r\in R\phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}f\left(r\right)={0}_{S}\},$$
where ${0}_{S}$ is the zero element of $S$.
 The image of a ring homomorphism
$f:R\to S$ is the set
$$\mathrm{im}f=\left\{f\right(r\left)\phantom{\rule{0.5em}{0ex}}\right\phantom{\rule{0.5em}{0ex}}r\in R\}.$$
Note that $\mathrm{ker}f=\{r\in R\phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}f\left(r\right)={0}_{S}\}$ not
$\{r\in R\phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}f\left(r\right)={1}_{S}\}$. If $\mathrm{ker}f$ was $\{r\in R\phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}f\left(r\right)={1}_{S}\}$ then $\mathrm{ker}f$ would not necessarily be a subgroup of $R$
(not to mention an ideal) and we couldn't even hope to get homomorphism theorems like we did for groups.
Let $f:R\to S$ be a ring
homomorphism. Then
 (a)
$\mathrm{ker}f$ is an ideal of $R$.
 (b)
$\mathrm{im}f$ is a subring of $S$.
Let $f:R\to S$ be a ring homomorphism.
Let ${0}_{R}$ be the zero element of $R$.
Then
 (a)
$\mathrm{ker}f=\left\{{0}_{R}\right\}$
if and only if
$f$ is injective.
 (b)
$\mathrm{im}f=S$ if and only if
$f$ is surjective.
Notice that the proof of Proposition (rginjsur)(b) does not use the fact that
$f:R\to S$ is a homomorphism,
only the fact that
$f:R\to S$ is a function.
 (a)
Let $f:R\to S$ be a ring homomorphism
and let $K=\mathrm{ker}f$. Define
$$\begin{array}{cccc}\hat{f}:& R/\mathrm{ker}f& \u27f6& S\\ & r+K& \u27fc& f\left(r\right)\end{array}$$
Then $\hat{f}$ is a well defined injective ring homomorphism.
 (b)
Let $f:R\to S$ be a ring homomorphism
and define
$$\begin{array}{cccc}f\prime :& R& \u27f6& \mathrm{im}f\\ & r& \u27fc& f\left(r\right)\end{array}$$
Then $f\prime $ is a well defined surjective ring homomorphism.
 (c)
If
$f:R\to S$ is a ring homomorphism
then
$$R/\mathrm{ker}f\simeq \mathrm{im}f,$$
where the isomorphism is a ring isomorphism.
Direct sums
Suppose $S$ and $T$ are rings.
The idea is to make $S\times T$
into a ring.
 The direct sum $S\oplus T$ of two rings
$S$ and $T$
is the set $S\times T$ with operations given by
$$({s}_{1},{t}_{1})+({s}_{2},{t}_{2})=({s}_{1}+{s}_{2},{t}_{1}+{t}_{2})\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}({s}_{1},{t}_{1})({s}_{2},{t}_{2})=({s}_{1}{s}_{2},{t}_{1}{t}_{2}),$$
for ${s}_{1},{s}_{2}\in S$,
${t}_{1},{t}_{2}\in T$.
 More generally, given rings ${R}_{1},{R}_{2},\dots ,{R}_{\mathrm{n}}$,
the direct sum
${R}_{1}\oplus {R}_{2}\oplus \cdots \oplus {R}_{\mathrm{n}}$
is the set
${R}_{1}\times {R}_{2}\times \cdots \times {R}_{\mathrm{n}}$
with the operations given by
$$({s}_{1},\dots ,{s}_{i},\dots ,{s}_{n})+({t}_{1},\dots ,{t}_{i},\dots ,{t}_{n})=({s}_{1}+{t}_{1},\dots ,{s}_{i}+{t}_{i},\dots ,{s}_{n}+{t}_{n})$$
$$\text{and}\phantom{\rule{2em}{0ex}}({s}_{1},\dots ,{s}_{i},\dots ,{s}_{n})({t}_{1},\dots ,{t}_{i},\dots ,{t}_{n})=({s}_{1}{t}_{1},\dots ,{s}_{i}{t}_{i},\dots ,{s}_{n}{t}_{n})$$
where ${s}_{i},{t}_{i}\in {R}_{i}$,
and
${s}_{i}+{t}_{i}$
and
${s}_{i}{t}_{i}$
are given by the operations for the ring
${R}_{i}$. The operations in the direct sum are just
the operations from the original rings acting componentwise.
HW: Show that these are good definitions, i.e. that, as defined above,
$S\oplus T$ and
${R}_{1}\oplus {R}_{2}\oplus \cdots \oplus {R}_{\mathrm{n}}$
are rings with zeros given by
$({0}_{S},{0}_{T}$
and
$({0}_{{R}_{1}},\dots ,{0}_{{R}_{n}}$, respectively,
and identities given by
$({1}_{S},{1}_{T}$
and
$({1}_{{R}_{1}},\dots ,{1}_{{R}_{n}}$, respectively.
Further definitions
There are many things which help to characterize a ring; some of these will be studied in depth
in later chapters. Some definitions are given here for reference.
 A commutative ring is a ring $R$ such that if
$a,b\in R$ then
$ab=ba$.
 The center of a ring $R$ is the set
$$Z\left(R\right)=\{z\in R\phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}\text{if}\phantom{\rule{0.5em}{0ex}}r\in R\phantom{\rule{0.5em}{0ex}}\text{then}\phantom{\rule{0.5em}{0ex}}zr=rz\}.$$
HW: Give an example of a noncommutative ring.
HW:Prove that $Z\left(R\right)$ is a subring of $R$.
HW: Give an example to show that $Z\left(R\right)$ is not
necessarily an ideal of $R$.
HW: What two elements are always in the center of $R$?
 The characteristic, $\mathrm{char}\left(R\right)$, of a
ring $R$ is the smallest positive integer $n$ such that
$1+1+\cdots +1$
($n$ times) is 0. If such an integer does not exist,
$\mathrm{char}\left(R\right)$ is 0.
Let $R$ be a ring. Let ${0}_{R}$ and
${1}_{R}$ be the zero and the identity in $R$,
respectively.
 (a)
The function $\varphi :\mathbb{Z}\to R$
given by
$$\varphi \left(0\right)={0}_{R},\phantom{\rule{2em}{0ex}}\varphi \left(m\right)=\underset{\underset{m\phantom{\rule{0.5em}{0ex}}\text{times}}{\u23df}}{{1}_{R}+\cdots +{1}_{R}},\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\varphi (m)=\varphi \left(m\right),\phantom{\rule{2em}{0ex}}\text{for}\phantom{\rule{0.5em}{0ex}}m\in {\mathbb{Z}}_{>0},$$
is the unique ring homomorphism from $\mathbb{Z}$ to $R$.
 (b)
$\mathrm{ker}\varphi =n\mathbb{Z}=\left\{nk\phantom{\rule{0.5em}{0ex}}\right\phantom{\rule{0.5em}{0ex}}k\in \mathbb{Z}\}$,
where $n=\mathrm{char}\left(R\right)$
is the characteristic of the ring $R$.
HW: Show that if $\mathrm{char}\left(R\right)=2$ then
$1=1$ in $R$.
 A left inverse of an element $b$ of a ring $R$
is an element $c\in R$ such that
$cb=1$.
 A right inverse of an element $b$ of a ring $R$
is an element $c\in R$ such that
$bc=1$.
 An inverse of an element $b$ of a ring $R$
is an element $c\in R$ such that
$cb=bc=1$.
 A unit is an element of a ring that has an inverse.
 If $R$ is a ring, ${R}^{\times}$
is the set of units of $R$.
HW: Show that if $b\in R$ has both a left inverse and a right
inverse then they must be equal.
HW: Give an example of a ring $R$ and an element of $R$ that has a left
inverse but not a right inverse.
HW: What element of a ring is always a unit?
HW: Give an example of a ring such that ${R}^{\times}=R\left\{0\right\}$.
 Let $R$ be a ring and let $S$ be a subset of
$R$. The ideal generated by $S$,
$\left(S\right)$, is the ideal of $R$ such that
 (a)
$S\subseteq \left(S\right)$,
 (b)
If $J$ is an ideal of $R$ and $S\subseteq J$ then $\left(S\right)\subseteq J$.
 An ideal is principal if it is generated by one element.
The ideal $\left(S\right)$ is the smallest ideal of
$R$ containing $S$.
Think of $\left(S\right)$ as gotten by adding to $S$
exactly those elements of $R$
that are needed to make an ideal.
 A proper ideal of a ring $R$ is an ideal that is not
equal to $R$.
 A maximal ideal of a ring $R$ is a proper ideal of
$R$ that is not contained in any other proper ideal of $R$.
HW: Show that a proper ideal does not contain any units.
Every proper ideal $I$ of a ring $R$
is contained in a maximal ideal of $R$.
 A local ring is a commutative ring with only one maximal ideal.
 A simple ring is a ring with no ideals except $\left(0\right)$ and $R$.
 A ring $R$ is a division ring if every nonzero element of
$R$ has an inverse in $R$.
 A field is a commutative ring $\mathbb{F}$ such that every
nonzero element of $\mathbb{F}$ has an inverse in $\mathbb{F}$.
Notes and References
These notes are written to highlight the analogy between groups and group actions,
rings and modules, and fields and vector spaces.
References
[Ram]
A. Ram,
Notes in abstract algebra,
University of Wisconsin, Madison 19931994.
[Bou]
N. Bourbaki,
Algèbre, Chapitre 9: Formes sesquilinéaires et formes quadratiques,
Actualités Sci. Ind. no. 1272 Hermann, Paris, 1959, 211 pp.
MR0107661.
[Ru]
W. Rudin,
Real and complex analysis, Third edition, McGrawHill, 1987.
MR0924157.
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