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\newenvironment{proof}{\noindent{\sc Proof:~}}{\hfill$\Box$\par\medskip}

\begin{document}

\section{Rings}

Algebra is the study of mathematical {\em structures} and the
possible, structure respecting {\em maps} that can be
constructed between them.
One of the most important of the structures in modern algebra is
called {\em the ring}. It was defined in the late 1800's by Kummer
while working on the problem of unique factorization.
The name comes from his early examples, which were number
systems including elements, such as the imaginary quantity $i$, which
when taken to successive powers, eventually circle around onto themselves.

A set of elements is a ring if there are two functions on the
set, called addition and multiplication, which abide by a
long list of properties.
\begin{enumerate}
\item
Both addition and multiplication must be associative, binary operations.
\item 
Addition is an abelian group. That is, for every element $a$ in
the group there is a unique inverse $-a$, and there is a unique
{\em zero element $0$} for which $a+0=0+a=a$ for any $a$ in the
group. Addition is also required to be commutative, $a+b=b+a$ for
any $a$ and $b$ in the group.
\item 
Multiplication left and right distributes over addition:
$a(b+c)=ab+ac$ and $(b+c)a=ba+ca$, any $a, b$ and $c$.
\item
Typically by ring one means {\em a ring with unit}. That is, there
is a unique {\em unit element}, aptly denoted $1$, for which
$a1=1a=a$ for any element $a$. 
\end{enumerate}
A {\em commutative ring} is a ring in which multiplication
is commutative. 
The most common rings in everyday life are commutative with unit.

The definition of an abelian group might be better understood by
the alternative definition.
A group (abelian or not) is a set of elements for which the
equations $a + x =b $ and $x + a = b $ always have a unique solution
for $x$ given any $a$ and $b$. One sees here the need for the
definition of a group --- it corresponds to the need to solve
simple equations. If one writes $a+x=a$, then it implies the
existence of a zero element. And if one writes $a+x=0$, it implies
for every $a$ the existence of a unique inverse.

\medskip
\noindent{\em Examples:\/}
\begin{itemize}
\item
The integers $\{\, \ldots, -1, 0, 1, 2, \ldots\,\}$ with
the standard addition and multiplication operations form a ring.
\item
The integers modulo some number $\{\, 0, 1, 2, \ldots
N-1\,\}$ with modular arithmetic, form a ring.
\item
Polynomials with coefficients in a ring, form a ring.
For instance, polynomials in $x$ and $y$ with integer coefficients,
denoted ${\bf Z}[x,y]$. As example elements we
can give $1, 5x+2$ or $7x^3y+9xy+2x+1$.
\end{itemize}

\section{Map of rings}

As important and structures are the maps between structures.
In this way we can say that two obects are the same or delineate
how they differ. Well chosen maps can make problems easy to solve
by highlighting the most important structural imperatives.
Suppose $R$ and $R'$ are two rings.
A map of these rings can be more than a passive pairing of
items in $R$ with items in $R'$.
It might {\em respect} or {\em preserve} the ring structures.
Often a map between two rings is tacitly assumed to preserve the
ring structure, or one can be explicit by saying that is is
a {\em map of rings.}

Concentrating our attention on rings, a map $f:R\rightarrow R'$
preserves addition if for any $r_1$ and $r_2$ of $R$,
\[ f(r_1+r_2) = f(r_1) + f(r_2).\]
What this is saying that the same element of $R'$ results either
if one first adds then maps or first maps then adds.
Another way of looking at it is to see ``phantom'' hands
adding the ``shadows'' of $r_1$ and $r_2$ in $R'$ as you add
them in $R$. Your result and the phantom's result would agree.
\begin{figure}
\begin{picture}(380,170)(0,0)
% number lines
\put(0,57){$R'$}
\put(0,107){$R$}
\put(0,50){\line(1,0){300}}
\put(0,100){\line(1,0){300}}

% points on the line
\put(60,57){$f(r_1)$}
\put(50,50){\circle*{3}}
\put(60,107){$r_1$}
\put(50,100){\circle*{3}}
\put(110,57){$f(r_2)$}
\put(100,50){\circle*{3}}
\put(110,107){$r_2$}
\put(100,100){\circle*{3}}
\put(240,57){$f(r_1+r_2)=f(r_1)+f(r_2)$}
\put(230,50){\circle*{3}}
\put(240,107){$r_1+r_2$}
\put(230,100){\circle*{3}}

% lower addition arrows and box
\put(50,50){\line(0,-1){40}}
\put(50,10){\vector(1,0){100}}
\put(100,50){\line(0,-1){20}}
\put(100,30){\vector(1,0){50}}
\put(150,0){\framebox(40,40){$+$}}
\put(190,20){\line(1,0){40}}
\put(230,20){\vector(0,1){30}}

% upper addition arrows and box
\put(50,100){\line(0,1){40}}
\put(50,140){\vector(1,0){100}}
\put(100,100){\line(0,1){20}}
\put(100,120){\vector(1,0){50}}
\put(150,110){\framebox(40,40){$+$}}
\put(190,130){\line(1,0){40}}
\put(230,130){\vector(0,-1){30}}

% connecting map
\put(50,90){\vector(0,-1){30}}
\put(100,90){\vector(0,-1){30}}
\put(230,90){\vector(0,-1){30}}

\end{picture}
\caption{A map of rings preserves addition.}
\end{figure}


When an correspondence, or {\em map}, preserves an operation,
we can expect it to preserve properties which are a necessary result
of the operation.
For instance, it the above correspondence preserves addition,
and $r_1$ happens to be the inverse of $r_2$, then it must
be that $r_1{}'$ is the inverse of $r_2{}'$. We leave a proof
of this to the reader.

A map $f:R\rightarrow R'$ between two rings
$R$ and $R'$ is a {\em map of rings} if it preserves addition,
multiplication and the unit element. That is,
\begin{enumerate}
\item For any $x, y \in R$, $f(x+y)=f(x)+f(y)$. It follows logically
that $f(0)=0'$, where $0$ and $0'$ are the zero elements of
$R$ and $R'$, respectively, and that $f(-x) = -f(x)$, for any $x$.
\item
For any $x, y \in R$, $f(xy)=f(x)f(y)$.
\item If $1$ and $1'$ are the unit elements in $R$ and $R'$,
respectively, $f(1)=1'$.
\end{enumerate}
In the theory of rings without unit, the third condition would not
be included.

\section{Ideals and Quotient Rings}

The rings Kummer worked with often had difficiencies in their
structure. Namely, the principle of unique factorization into
irreducibles failed. He perceived that the principle could be
recovered if elements were introduced into to ring as
new factors for many elements. He considered these elements
{\em ideal} since they existed in order to fulfill the
ideal of unique factorization.
It was later discovered that such elements needn't be
created from nothing, they could be found by taking subsets
of existing elements. And it was soon discovered that these
subsets, called {\em ideals}, were the basic building blocks
of any map of rings.

Given a ring $R$, an ideal $I$ in $R$ is a non-empty subset of $R$ which
is closed by addition and additive inverses and by multiplication
by any element of $R$. That is,
\begin{enumerate}
\item The set $I$ is closed by addition:  
if $i$ and $i'$ is in $I$, then $i+i'$ is in $I$.
\item The set $I$ is closed by additive inverses: if $i$ is
in $I$, then $-i$ is in $I$.
\item The set $I$ is closed by multiplication by any element
of $R$: if $i$ is in $I$ and $r$ is in $R$, then $ir$ and $ri$
must be in $I$.
\end{enumerate}
Given an ideal $I$ in $R$, and an element $r$ of $R$, the
{\em translate of $I$ by $r$} is the set:
\[ r+I = \set{r+i}{i\in I}.\]
The collection of translates of an ideal of a ring can itself
be given the structure of a ring. It is called the {\em quotient
ring} of the ring by the ideal.
The current goal is to describe more fully the creation of
the quotient ring.
\begin{theorem}
Given $I$ an ideal of $R$ and $r_1$ and $r_2$ elements of $R$,
if $r_1-r_2\in I$ then $r_1+I=r_2+I$, else these two translates
have no elements in common.
\end{theorem}
\begin{proof}
If $r_1-r_2\in I$, then let this difference be $i$. So $r_1=r_2+i$
and $r_1$ is in $r_2+I$ as well as in $r_1+I$. Hence the
there is a common element.
Conversly, if $r_1+I$ and $r_2+I$ share any element in common, then there
is an $i\in I$ such that $r_1=r_2+i$, which implies that
$r_1-r_2\in I$.
We have shown that the two translates intersect if and only if the
difference of the translations is in $I$.

We go on to show that if the translate intersect, they are identical.
Once again, let $r_1=r_2+i$ with $i\in I$, by supposition,
and for any $r_1+i'$ in $r_1+I$, we set let $i''=i+i'$.
Then $r_1+i'$ is also in $r_2+I$, because $r_2+i''$ is in
$r_2+I$ and $r_2+i''=r_2+i+i'=r_1+i'$.
We also show likewise that any $r_2+i$ in $r_2+I$ is in $r_1+I$.
So these sets are exactly the same.
\end{proof}

We can define addition and multiplication of translates
of $I$ so that the set of all translates becomes a ring,
called the quotient ring of $R$ modulo $I$, $R/I$.
Consider the map $\phi:R\rightarrow R/I$ which takes
$r\in R$ to the translate $r+I$.
We define addition and multiplication in $R/I$ so that
$\phi$ is a map of rings.
Consider first addition. For $\phi$ to respect addition,
then for any $r_1$ and $r_2$ in $R$,
\[ \phi(r_1+r_2) = \phi(r_1) + \phi(r_2).\]
Hence adding $r_1+I$ and $r_2+I$ should result in $(r_1+r_2)+I$.
We can make this the definition of addition in $R/I$ provided
all elements of the quotient ring have at least on representation
as $r+I$ for some $r$ in $R$, and if an element of the quotient
ring can be represented in two ways, $r_1+I = r_2+I$ for $r_1$
and $r_2$ two different elements of $R$, either representation can
be used to determine the result of the addition.

The definition for multiplication is similar. The equation,
\[\phi(r_1r_2)=\phi(r_1)\phi(r_2)\]
must hold for any $r_1$ and
$r_2$ in $R$, so we define,
\[ (r_1+I)(r_2+I)= (r_1r_2)+I.\]
Again, we have the burden of showing that translations equivalent
to $r_1+I$ and $r_2+I$ multiply to an translation equivalent to
$r_1r_2+I$.

\begin{theorem}
With addition and multiplication as defined, the collection of
translations of the ideal $I$ form a ring.
\end{theorem}
\begin{proof}
We first show that addition and multiplication are defined consistantly,
That is, different choices for the representation of
a translation don't matter.

Suppose $r_1+I=r_1{}'+I$ and $r_2+I=r_2{}'+I$. Then both
$r_1-r_1{}'$ and $r_2-r_2{}'$ are in $I$, and so is
\[ (r_1-r_1{}') + (r_2-r_2{}') = (r_1+r_2)-(r_1{}'+r_2{}').\]
Thus $(r_1+r_2)+I = (r_1{}'+r_2{}')+I$. This shows that
any representation of a translation gives the same result for
addition as defined.

If $r_2+I=r_2{}'+I$ then $r_2-r_2{}'\in I$, and by the rules of
an ideal,
\[ r_1(r_2-r_2{}') = r_1 r_2-r_1 r_2{}'\in I,\]
for any $r_1\in R$.
Likewise, if $r_1+I=r_1{}'+I$, then $r_1{}-r_1{}'\in I$, and
by the rules of an ideal,
\[ (r_1-r_1{}')r_2{}'= r_1 r_2{}' - r_1{}' r_2{}' \in I,\]
for any $r_2{}'\in R$.
The sum of two elements in an ideal rests inside the ideal, so,
\[ (r_1 r_2-r_1 r_2{}') + (r_1 r_2{}' - r_1{}' r_2{}') =
    r_1 r_2 - r_1{}' r_2{}'\in I.\]
Therefore $r_1r_2+I=r_1{}'r_2{}'+I$, as was to be shown.

Now that addition and multiplication are defined, we must show
that they fulfil the properties of a ring.
All the properties enjoyed by $R$ are passed on to $R/I$, since
we can compute in $R/I$ with representatives taken from $R$.
This shows that $R$ being commutative with unit implies that
$R/I$ will also be commutative with unit.
\end{proof}



\section{Applications}

We can now see that the integers mod $n$, for any integer
$n$, is also understood as ${\bf Z}/n{\bf Z}$. That is,
the integers modulo the ideal of all multiples of $n$.


\subsection{Ideals in the ring of integers}

In the ring of integers $\bf Z$ there is one ideal for
every integer $m$ in $\bf Z$ and it is the set of all multiples of
$m$,
\[ m{\bf Z} = \set{mi}{i\in{\bf Z}}.\]
Unless $m$ is $1$ or $-1$, the ideal $m{\bf Z}$ does not include
all integers. We can ``translate'' the ideal by some integer $k$,
\[ k + m {\bf Z} = \set{k+i}{i\in m{\bf Z}}.\]
Try to imagine what these translates look like on the number line.
You have an infinitely long ruler with inches marked off
in the positive and negative directions from some arbitrary
zero point.
The ideal $m{\bf Z}$ is the set taking every $m$-th
mark, starting from zero. Keeping this configuration rigid
but moving it entirely left or right, you have a picture of the
different translates of the ideal.

Given any integer $i$, it is contained in 
the translate $i + m{\bf Z}$. If one were to
take $i$ modulo $m$ in the fashion of our childhood,
(where $r$ is equivalent to $i$ mod $m$, for the 
$r$ safisfying,
\[ 0 \le r < m \mbox{  and  } i=mq+r\;\;\;)\]
then,
\[ i + m{\bf Z} = r + m{\bf Z},\]
where by ``equal'' we mean that they are exactly the same
sets.
This is the foundation of another way of looking at
the integers modulo $m$. We can either say that
$i$ modulo $m$ is the remainder when $i$ is divided
by $m$, or it is the translate of the ideal $m{\bf Z}$ in
which $i$ lies.

Addition modulo $m$ also has a reinterpretation as translations
of the ideal $m{\bf Z}$.
The sum of $i$ and $j$ modulo $m$ is the translate
$(i+j) + m{\bf Z}$. In fact, if $i'$ is any integer
in $i+ m{\bf Z}$, and $j'$ is any integer in $j + m{\bf Z}$,
\[ (i' + j') + m{\bf Z} = (i+j) + m {\bf Z}.\]
In the children's approach to modular arithmetic,
we might say this in the form:
\[ ( i \pmod {m}) + (j \pmod {m}) \pmod{m} = (i+j) \pmod{m}.\]
That is, it doesn't matter when or how often we decide to
go modulo $m$, the result will end up the same.
The same is true of multiplication. The product of $i$ and
$j$ modulo $m$ is the translate $i'j' + m{\bf Z}$ where
$i'$ is any integer in $i + m{\bf Z}$ and $j'$ is
any integer in $j+m{\bf Z}$.

\subsection{Gaussian Integers}

Another informative example is the ring of {\em Gaussian
integers.} All of us are familiar with $i$, the imaginary
quantity whose sqaure is minus one. 
In a similar way that the complex numbers are the reals with
the adjunction of $i$, the Gaussian integers are the integers
with the adjunction of $i$.

Each Gaussian integer can be written as.
\[ a + b\,i, \quad a, b \in {\bf Z}.\]
To calculate with Guassian integers, we follow the rules,
\[ ( a_1 + b_1\, i ) + (a_2 + b_2\, i ) = (a_1+a_2) + (b_1+b_2)\,i\]
and,
\[ ( a_1 + b_1\, i ) (a_2 + b_2\, i ) = (a_1 a_2 - b_1 b_2) +
  (a_1 b_2 + b_1 a_2)\, i . \]
We demonstrate the theory of ideals can be used to ``construct''
the Gaussian integers as a quotient ring of the ring of 
polynomials with integer coefficients.

Starting from ${\bf Z}[x]$, the ring of polynomials in $x$ with
integers coefficients, we ``force'' $x$ to behave as does $i$.
That is, we proclaim that $x^2+1=0$. Hence every polynomial in the
ideal,
\[ {\cal I} = (x^2+1)\,{\bf Z}[x],\]
becomes zero. We therefore look at the quotient ring
${\bf Z}[x]/{\cal I}$
and show that this is exactly the same as the Gaussian integers.

The formal method of doing this might at first make one's head
spin, but it is less awesome than it first appears. We define
a map $\phi$ from ${\bf Z}[x]/{\cal I}$ to the Gaussian integers.
We then demonstrate that this map sets up a perfect correspondence
between the elements and operations in ${\bf Z}[x]/{\cal I}$
and the elements and operations of Gaussian integers.
That is, the two rings have completely parallel
structures, and this is mathematically sufficient to
claim that the rings are the same.

Here is the map.
Given any element of the quotient ring ${\bf Z}[x]/{\cal I}$,
take any polynomial $f$ in the residue class and consider
its remainder when divided by $x^2+1$,
\[  f= q(x^2+1) + r, \]
where $r$ will be a polynomial of the form $bx+a$.
Then $\phi(f) = a+b\,i.$
Any other $f'$ from the same residue class will give
the same result, since $f'-f$ is a multiple of $x^2+1$.

We claim that $\phi$ is a map of rings.
\begin{eqnarray*}
 \phi(f+g) &=& \phi\bigl( q_f(x^2+1)+r_f + q_g(x^2+1)+r_g\bigr)\\
  &=& \phi\bigl((q_f+q_g)(x^2+1)+r_f+r_g\bigr) \\
  &=& r_f + r_g = \phi(f)+\phi(g).
\end{eqnarray*}
The computation for multiplication is more involved:
\begin{eqnarray*}
 \phi(fg) &=& \phi\bigl((q_f(x^2+1)+r_f)(q_g(x^2+1)+r_g)\bigr)\\
 &=& \phi\bigl((q_fq_g(x^2+1)+q_fr_g + q_gr_f)(x^2+1)+r_fr_g\bigr),
\end{eqnarray*}
but,
\begin{eqnarray*}
 r_fr+g &=& a_fa_g + (a_fb_g+a_gb_f)x + b_fb_gx^2 \\
 &=& a_fa_g-b_fb_g + (a_fb_g+a_gb_f)x + b_fb_g(x^2+1),
\end{eqnarray*}
setting
\[ q = q_fq_g(x^2+1)+q_fr_g+q_gr_f+b_fb_g,\]
we have.
\begin{eqnarray*}
 \phi(fg) &=& \phi\bigl(q(x^2+1) + a_fa_g-b_fb_g + (a_fb_g+a_gb_f)x\bigr) \\
  &=& a_fa_g-b_fb_g + (a_fb_g+a_gb_f)\,i \\
  &=& (a_f+b_f\,i)(a_g+b_g\,i) = \phi(f)\phi(g).
\end{eqnarray*}


\section{The Lattice Theory of Ideals}

Ideals can be investigated as mathematical entities onto themselves. 
Then lend themselves to more than one structure, our interest
shall be in their structure as a {\em lattice}.
Given a set with a partial order, a lattice has for
every two elements $a$ and $b$ in the lattice, a least element
greater than the both, called the {\em join} of $a$ and
$b$, and a greatest element less than the both, called
the {\em meet} of $a$ and $b$.
The set of ideals of a ring can be given a partial order
by set inclusion: $A\ge B$ if and only if
$A\supseteq B$.
This done, simple interpretations of the join and meet are
available. 
If $A$ and $B$ be two ideals of a ring $R$, then the set of
all sums,
\[ \set{a+b}{a\in A, \; b\in B},\]
can be shown to be the join of $A$ and $B$. That is, the
sum is an ideal and it is the smallest ideal in $R$ containing
$A$ and $B$. When speaking generally about lattices,
we can use the traditional notation for a join, $A\vee B$.
However, in the particular case at hand, this join can be
denoted $A+B$, since it is truely the sum of the two ideals.
Likewise, the set-theoretic intersection of $A$ and $B$,
\[ \set{c}{c\in A \mbox{ and } c\in B}, \]
is the meet of $A$ and $B$, it is the
largest ideal in $R$ which is contained in both $A$ and $B$.
In general, the meet is denoted $A\wedge B$, but in this
case we also use the obvious notation $A\cap B$.
Note also that there is an smallest ideal in any ring,
the ideal containing only $0$, and a largest ideal, the
ring itself.

A {\em Hasse Diagram} gives a picture of the
arrangement of elements in a lattice.
Suppose $a$ and $b$ are two elements of a lattice. Then $a$ is said
to {\em cover} $b$ if it is the smallest element 
of the lattice strictly larger than $b$.
That is, $a>b$ and for no $c$ do we have $a>c>b$.
One draws a Hasse Diagram by placing a point for each
element of the lattice, and if $a$ covers $b$ then $a$
is placed higher on the page than $b$ with a line connecting
the two points.

A lattice can also be specified without reference to a partial
order. In which case we just assume there to be a set of elements
and two operations on the set, which we shall call the sum
and the intersection, because we will be only concerned with
the lattice of the ideals of a ring, which are binary,
associative, commutative, idempotent operations, that is,
\begin{enumerate}
\item $A\cdot(B\cdot C) = (A\cdot B)\cdot C,$
\item $A\cdot B = B \cdot A$,
\item $A\cdot A = A$,
\end{enumerate}
where $A, B$ and $C$ are any ideals in the ring and
$\cdot$ can be either intersection or addition; and
the {\em absorption identies} hold,
\begin{enumerate}
\item $A = A \cap ( A + B ) $,
\item $A = A + ( A \cap B ),$
\end{enumerate}
for any ideals $A$ and $B$.

For ideals, an addition law holds, the {\em modular
identity},
\[ (A\cap B)+(A\cap C) = A \cap ( B + (A\cap C)),\]
and any ideals $A, B$ and $C$ in the ring.
A lattice with the modular identity is called a {\em modular
lattice}.
In a modular lattice the following
diagram can not appear embedded in the it's Hasse diagram,
\[  \begin{array}{ccccc}
  & & x & & \\
  & \swarrow & & \searrow & \\
  a & & & & \\
  \downarrow & & & & b \\
  c & & & & \\
  & \searrow & & \swarrow & \\
  & & y & & \\
\end{array} \]
Which is to say that in the lattice there
are not five elements $x, y, a, b $ and $c$ such that,
\begin{enumerate}
\item The element $x$ is larger than any other and $y$ is smaller
than any other.
\item The elements $a$ and $c$ are comparable and $a$ is the larger.
\item The element $b$ is incomparable with both $a$ and $c$.
\item The join of $b$ and either $a$ or $c$ is $x$ and the
meet of $b$ with either $a$ or $c$ is $y$.
\end{enumerate}
In fact, the absence of this diagram embedded in a Hasse diagram
is both necessary and sufficient for the lattice to be modular.

Suppose the lattice of ideals of two rings $R$ and $S$ are
${\cal R}$ and ${\cal S}$, respectively.
Any ring map $\phi:R\rightarrow S$ gives a map
in the opposite direction, $\phi^*:{\cal S}\rightarrow{\cal R}$.
Given an ideal $s$ of $S$, define,
\[ \phi^*(s) = \set{r\in R}{\phi(r)\in s}.\]
First, we show that $\phi^*(s)$ is an ideal.
It is never empty, since at least $0\in\phi^*(s)$.
If $r_1$ and $r_2$ are in $\phi^*(s)$, then,
\[ \phi(r_1+r_2)=\phi(r_1)+\phi(r_2),\]
which is an element of $s$, hence $r_1+r_2$ is
an element of $\phi^*(s)$.
If $r_1$ is in $\phi^*(s)$, and $r_2$ is any element of $R$, then,
\[ \phi(r_1 r_2) = \phi(r_1)\phi(r_2),\]
which is an element of $s$, hence $r_1 r_2$ is an element
of $\phi^*(s)$. This gives that $\phi^*(s)$ is an ideal of $R$.

Under the additional hypotheses that $\phi$ is surjective,
we will show that $\phi^*$ is a {\em lattice map}, that is,
\begin{enumerate}
\item For any $s_1, s_2\in {\cal S},$
\[ \phi^*(s_1 + s_2) = \phi^*(s_1) + \phi^*(s_2).\]
\item For any $s_1, s_2\in {\cal S},$
\[ \phi^*( s_1 \cap s_2 ) = \phi^*(s_1) \cap \phi^*(s_2).\]
\end{enumerate}
Since $s_1+s_2$ contains both $s_1$ and $s_2$, then
$\phi^*(s_1+s_2)$ contains both $\phi^*(s_1)$ and
$\phi^*(s_2)$. Hence,
\[ \phi^*(s_1+s_2) \supseteq \phi^*(s_1) + \phi^*(s_2),\]
the sum on the right being defined as the smallest such ring
in $R$.
Consider now any $r\in \phi^*(s_1+s_2)$.
By definition, there exists $\sigma_1\in s_1$ and $\sigma_2\in
s_2$ such that $\phi(r)=\sigma_1+\sigma_2$. Because $\phi$ is
surjective, there exists $\rho_1\in R$ such
that $\phi(\rho_1)=\sigma_1$.
From the simple calculation,
\[ \phi(r-\rho_1) = \phi(r)-\phi(\rho_1) = \sigma_1+\sigma_2 -\sigma_1
= \sigma_2,\]
we see that,
\[ (r-\rho_1)\in\phi^{-1}(\sigma_2).\]
Hence we can chose a $\rho_2\in\phi^{-1}(\sigma_2)$ such that
$r=\rho_1+\rho_2$. That is, $r\in\phi^*(s_1)+\phi^*(s_2)$.

The second result is purely set-theoretic.
Since both $s_1$ and $s_2$ contain $s_1\cap s_2$,
both $\phi^*(s_1)$ and $\phi^*(s_2)$ contain $\phi^*(s_1\cap s_2),$
and so by definition,
\[ \phi^*(s_1)\cap\phi^*(s_2)\supseteq\phi^*(s_1\cap s_2),\]
the ideal on the left being the largest such ideal in $R$.
Taking any $r\in\phi^*(s_1)\cap\phi^*(s_2),$ let $\sigma=\phi(r)$.
Since $r\in\phi^*(s_1)$, then $\sigma\in s_1$; likewise,
$\sigma\in s_2$; hence $\sigma\in s_1\cap s_2$. Therefore
$r\in\phi^*(s_1\cap s_2)$.

For a surjective mapping $\phi$, with kernel $K\in{\cal R}$,
we then have for any $s\in{\cal S}$, 
\[ \phi^*(s) = \phi^*(s+0)=\phi^*(s) + K.\]
Hence, the quotient of a ring distroys all ideals contained
in or incompatible with the kernel.
The reader is encouraged to form a more exact theorem.

\end{document}

