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\begin{document}

\chapter*{october 24:}

\begin{center}
{\Huge \bf Math 431 Class Lecture Notes}
\end{center}

\bigskip

\begin{itemize}
\item
Ideals in a Dedekind domain
\item 
Chinese Remainder Theorem
\item
An application
\end{itemize}

\section{Ideals in a Dedekind domain}
We reviewed and finished the proof that ideals in a Dedekind domain
factor uniquely intro prime ideals (described in the notes
for 10/12).  The main steps were

\begin{enumerate}
\item Showing that every ideal $I$ in a Dedekind domain~$R$ is invertible, i.e.,
that there is an ideal~$J$ such that $IJ$ is principal.
\item Using this to show that if $I \subset J$ is equivalent to $J|I$
(the latter meaning that there is an ideal~$K$ such that $IK = J$.
\item Showing that Cancellation holds for ideal multiplication, i.e.,
that $IJ = IK$ implies that $J = K$.
\item Showing that every ideal factors uniquely into a product of
 ideals.
\end{enumerate}

If two ideals $I$ and $J$ have factorizations
$$I = \prod_{i=1}^r P_i^{a_i}, \qquad
J = \prod_{i=1}^r P_i^{b_i}$$
then their greatest common divisor is
$$\gcd(I,J) := \prod_i P_i^{\mbox{min}(a_i,b_i)}.$$
By the divisibility/inclusion result, this means that $\gcd(I,J)$ is the
smallest ideal containing $I$ and~$J$, i.e., that
$$\gcd(I,J) = I+J.$$
Similarly, the least common multiple of $I$ and~$J$ is
$$\lcm(I,J) := \prod_i P_i^{\mbox{max}(a_i,b_i)}$$
and this turns out to be the largest ideal contained in both~$I$
and~$J$, i.e.,
$$\lcm(I,J) = I \cap J.$$

The reverse relationship between inclusion and divisibility takes a little
bit of time to become comfortable with, but a moment's thought shows that it
is consistent with our experience in, e.g., the ring of integers~$\Z$.

\section{Chinese Remainder Theorem}

For this section, we let the ring~$R$ be any ring (commutative, with identity).

\begin{lemma}
If $I+J=(1)=R$ then $IJ=I\bigcap J$.
\end{lemma}

\begin{proof}  Clearly, if $x\in IJ$, then $x\in I\bigcap J$, so 
$IJ\subset I\bigcap J$.  
Suppose $u\in I\bigcap J$.  There exist elements $i$ and $j$ of $I$ and
$J$, respectively, such that $i+j=1$.  Then $u=u(i+j)=ui+uj\in IJ$.
\end{proof}

The Chinese Remainder Theorem says that a ring element can be specified by
giving its congruence class modulo each ideal in a collection of pairwise
relatively prime ideals, and this element is unique modulo the product
(intersection) of the ideals.

\begin{example}
If you are give constants $a,b,c$ and told that
$x \equiv a \bmod 7$, 
$x \equiv b \bmod 11$, 
$x \equiv c \bmod 13$ then $x$ is uniquely determined modulo~1001.
This has been used as a basis for a magic trick.
\end{example}

\begin{theorem} (Chinese Remainder Theorem)
If $I_1, I_2, \cdots, I_n$ are pairwise 
relatively prime ideals (i.e., $I_i+I_j=R$, for 
$i\ne j$) then 
$$R/\bigcap I_i\simeq R/\prod I_i \simeq \bigoplus R/I_i.$$
\end{theorem}

\begin{proof}
Consider the case $n = 2$.  Let $\phi_1$ and $\phi_2$
be the natural maps from $R$ to $R/I_1$ and from $R$ to $R/I_2$, respectively,
and define $\phi:R \longrightarrow R/I_1\bigoplus R/I_2$ by 
$\phi (x)=(\phi_1(x),\phi_2(x))$.  
It is clear that $\ker(\phi)=I_1\cap I_2$, so that all that has to be done
is to show that $\phi$ is surjective.  

Let $y = (y_1 \bmod I_1 ,y_2 \bmod I_2)$ be an element of 
$R/I_1\bigoplus R/I_2$. 
Then since $I_1+I_2=1$, there are $i_1\in I_1,
i_2\in I_2$ such that $i_1+i_2=1$.  Let $s x=y_2 i_1+y_1 i_2$.  
A simple calculation shows that
$x\equiv y_1 \bmod I_1$ and $x\equiv y_1 \bmod I_1$, so $\phi$ is surjective.

The case of arbitrary~$n$ can be proved by an induction argument based on
the $n = 2$ case, or by generalizing the proof in that case in a straightforward
way  \end{proof}

\section{An application}

As an example application of this result, we will prove the following useful
result that says that ideals in a number ring can be generated by two elements,
and one of them can even be specified arbitrarily.

\begin{theorem} Let $R = \Z_F$ be the ring of integers of a number field~$F$.
If $I$ is an ideal and $\alpha$ is a nonzero element of~$I$, then
there is a $\beta$ such that $I = (\alpha,\beta)$.
\end{theorem}

\begin{proof}
Let $I = \prod_i P_i^{a_i}$ be the prime factorization of~$I$.  Then the 
principal ideal $(\alpha)$ can be factored in the form
$$(\alpha) = J \prod_i P_i^{b_i}$$
where $b_i \ge a_i$, and the prime ideals dividing $J$ are disjoint from the $P_i$.
Choose $\beta_i \in P_i^{a_i}-P_i^{a_i+1}$.  (The ideals $P^a$ and $P^{a+1}$ are
distinct, by unique factorization, so such elements exist.)  Now use the Chinese
Remainder Theorem to find a $\beta$ such that
$$\mbox{ for each } i, \beta \equiv \beta_i \bmod P_i^{a_i+1}$$
and $\beta \equiv 1 \bmod J$.  Then the ideal $(\beta)$ has a factorization of the form
$$(\beta) = K \prod_i P_i^{a_i}$$
where $K$ is relatively prime to~$J$.  It follows that
$$(\alpha,\beta) = (\alpha)+(\beta) = \gcd(\alpha,\beta) = \prod_i P_i^{a_i} = I$$
as desired.
\end{proof}

\end{document}
