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\begin{document}
\centerline{Math 361 lecture for Wednesday, Week 9}
\bigskip

\centerline{Generators for ideals in a number ring}
\bigskip

Let~$K$ be a number field with ring of integers $\mathfrak{O}_{K}$.  We have
seen that every ideal in $\mathfrak{O}_{K}$ may be uniquely factored into a
product of prime ideals.  Given nonzero ideals $\mathfrak{a}$ and~$\mathfrak{b}$ of
$\mathfrak{O}_{K}$, define their \emph{greatest common divisor} to be the
ideal~$\mathfrak{c}$ such that (1) $\mathfrak{c}|\mathfrak{a}$
and~$\mathfrak{c}|\mathfrak{b}$ and (2) if $\mathfrak{d}$ is any ideal dividing
$\mathfrak{a}$ and $\mathfrak{b}$, then $\mathfrak{d}|\mathfrak{c}$.  Since
division of ideals in~$\mathfrak{O}_{K}$ is the same as containment, the
condition for $\mathfrak{c}$ is that (1) $\mathfrak{c}\supseteq\mathfrak{a}$
and~$\mathfrak{c}\supseteq\mathfrak{b}$ and (2) if
$\mathfrak{d}\supseteq\mathfrak{a}$ and $\mathfrak{d}\supseteq\mathfrak{b}$,
then $\mathfrak{d}\supseteq\mathfrak{c}$.    Note
that~$\mathfrak{c}\supseteq\mathfrak{a}$ and~$\mathfrak{c}\supseteq\mathfrak{b}$
if and only if
\[
  \mathfrak{c}\supseteq\mathfrak{a}+\mathfrak{b}.
\]
It follows that
\[
  \gcd(\mathfrak{a},\mathfrak{b})=\mathfrak{a}+\mathfrak{b}.
\]
Similarly, we define the \emph{least common multiple} to be ideal~$\mathfrak{c}$
such that $\mathfrak{c}$ is the multiple of~$\mathfrak{a}$ and of~$\mathfrak{b}$
such that if $\mathfrak{d}$ is any other common multiple, then~$\mathfrak{c}$
divides~$\mathfrak{d}$.  So $\mathfrak{c}\subseteq\mathfrak{a}$
and~$\mathfrak{c}\subseteq\mathfrak{b}$ and if the same containments hold
for~$\mathfrak{d}$, then $\mathfrak{d}\subseteq\mathfrak{c}$.  It follows that
\[
  \lcm(\mathfrak{a},\mathfrak{b})=\mathfrak{a}\cap\mathfrak{b}.
\]
If we have factorizations into primes
\[
  \mathfrak{a} =\prod_{i=1}^{k} \mathfrak{p}_{i}^{e_{i}}\quad\text{and}\quad
  \mathfrak{b} =\prod_{i=1}^{k} \mathfrak{p}_{i}^{\ell_{i}},
\]
(taking some~$e_{i}=0$ or~$\ell_{i}=0$, if necessary), then
\[
  \gcd(\mathfrak{a},\mathfrak{b})=\prod_{i=1}^{k} \mathfrak{p}_{i}^{\min\{e_{i},\ell_{i}\}}
  \quad\text{and}\quad
  \lcm(\mathfrak{a},\mathfrak{b})=\prod_{i=1}^{k}
  \mathfrak{p}_{i}^{\max\{e_{i},\ell_{i}\}}.
\]
In particular, if~$\mathfrak{a}$ and~$\mathfrak{b}$ relatively prime,
then~$\mathfrak{a}+\mathfrak{b}=\gcd(\mathfrak{a},\mathfrak{b})=(1)=\mathfrak{O}_{K}$.

We know that not every number ring is a PID.  However, our main result shows
that~$\mathfrak{O}_{K}$ is not far from being a PID: every ideal can be generated by two elements.

\begin{thm}\label{thm:09wed1}
  Let~$\mathfrak{a}$ be a nonzero ideal of~$\mathfrak{O}_{K}$, and
  let~$0\neq\beta\in\mathfrak{a}$.  Then there
  exists~$\alpha\in\mathfrak{O}_{K}$ such that
  \[
    \mathfrak{a}=(\alpha,\beta).
  \]
\end{thm}

To prove this theorem, we use the following:

\begin{lem}\label{lemma:09wed1}
If $\mathfrak{a}$ and $\mathfrak{b}$ are nonzero ideals of~$\mathfrak{O}_{K}$,
then there exists $\alpha\in\mathfrak{a}$ such that
\[
  \alpha\mathfrak{a}^{-1}+\mathfrak{b}=\mathfrak{O}_{K}.
\]
\end{lem}
\begin{proof} (We are following the proof of Lemma~5.19 in the text.)
First note that if~$\alpha\in\mathfrak{a}$, then we can multiply
$(\alpha)\subseteq\mathfrak{a}$ through by $\mathfrak{a}^{-1}$ to
get~$\alpha\mathfrak{a}^{-1}\subseteq\mathfrak{O}_{K}$.  Hence,
$\alpha\mathfrak{a}^{-1}$ will be an ideal not just a fractional ideal. 

Suppose that~$\mathfrak{b}=\prod_{i=1}^{k} \mathfrak{p}_{i}^{e_i}$ is the
factorization of~$\mathfrak{b}$ into prime ideals, and suppose that we can
find~$\alpha\in\mathfrak{a}$ such that $\alpha\mathfrak{a}^{-1}$ is relatively
prime to $\mathfrak{p}_{i}$ for $i=1,\ldots,k$.  Then~$\alpha\mathfrak{a}^{-1}$
will be relatively prime to $\mathfrak{b}$, and the result will follow:
$\alpha\mathfrak{a}^{-1}+\mathfrak{b}=\mathfrak{O}_{K}$.

Now
\[
  \mathfrak{p}_{i}|(\alpha\mathfrak{a}^{-1})
  \Leftrightarrow
  \alpha\mathfrak{a}^{-1}\subseteq \mathfrak{p}_{i}
  \Leftrightarrow
  (\alpha)\subseteq \mathfrak{a}\mathfrak{p}_{i}
  \Leftrightarrow
  \alpha\in\mathfrak{a}\mathfrak{p}_{i}.
\]
So it suffices to find~$\alpha\in\mathfrak{a}$ such
that~$\alpha\in\mathfrak{a}\setminus\mathfrak{a}\mathfrak{p}_{i}$ for all~$i$.
If~$k=1$, there is no problem.  So suppose that~$k>1$.

For each~$i=1,\ldots,k$, define
\[
  \mathfrak{a}_{i} = \mathfrak{a}\mathfrak{p}_{1}\cdots \widehat{\mathfrak{p}_{i}}\cdots \mathfrak{p}_{k}
\]
where the hat over~$\mathfrak{p}_{i}$ means to omit that factor.
Since~$\mathfrak{p}_{i}\subsetneq\mathfrak{O}_{K}$ it follows
that~$\mathfrak{a}_{i}\mathfrak{p}_{i}\subsetneq\mathfrak{a}_{i}$.
Let~$\alpha_{i}\in \mathfrak{a}_{i}\setminus
(\mathfrak{a}_{i}\mathfrak{p}_{i})$.  Then define 
\[
  \alpha = \alpha_{1}+\cdots+\alpha_{k}.
\]
Since~$\alpha_{i}=\mathfrak{a}_{i}\subseteq\mathfrak{a}$ for all~$i$, it follows
that~$\alpha\in\mathfrak{a}$.

It remains to be shown that for each~$i$, we have
$\alpha\not\in\mathfrak{a}\mathfrak{p}_{i}$.  For sake of contradiction, suppose
that~$\alpha\in\mathfrak{a}\mathfrak{p}_{i}$ for some~$i$.
For each~$j\neq i$, from the definition of
$\mathfrak{a}_{j}$, it follows that~$\mathfrak{a}\mathfrak{p}_{i}$
divides~$\mathfrak{a}_{j}$.  Hence,
\[
  \alpha_{j}\in\mathfrak{a}_{j}\subseteq \mathfrak{a}\mathfrak{p}_{i}.
\]
It follows that
\[
  \alpha_{i}=\alpha-\alpha_{1}-\cdots-\widehat{\alpha_{i}}-\cdots-\alpha_{k}\in\mathfrak{a}\mathfrak{p}_{i}.
\]
Therefore, we have~$\mathfrak{a}_{i}|(\alpha_{i})$ and
$(\mathfrak{a}\mathfrak{p}_{i})|(\alpha_{i})$.  So $\lcm(\mathfrak{a}_{i},\mathfrak{a}\mathfrak{p}_{i})|(\alpha_{i}).$ 
However,
\[
  \lcm(\mathfrak{a}_{i},\mathfrak{a}\mathfrak{p}_{i})
  =
  \lcm(\mathfrak{a}\mathfrak{p}_{1}\cdots \widehat{\mathfrak{p}_{i}}\cdots \mathfrak{p}_{k},\mathfrak{a}\mathfrak{p}_{i})
  =
  \mathfrak{a}\lcm(\mathfrak{p}_{1}\cdots \widehat{\mathfrak{p}_{i}}\cdots \mathfrak{p}_{k},\mathfrak{p}_{i})
  =
  \mathfrak{a}\mathfrak{p}_{1}\cdots\mathfrak{p}_{i}\cdots
  \mathfrak{p}_{k}=\mathfrak{a}_{i}\mathfrak{p}_{i}.
\]
So $(\mathfrak{a}_{i}\mathfrak{p}_{i})|(\alpha_{i})$, from which we get the
contradiciton~$\alpha_{i}\in\mathfrak{a}_{i}\mathfrak{p}_{i}$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:09wed1}]
  Define~$\mathfrak{b}=\beta\mathfrak{a}^{-1}$.  By Lemma~\ref{lemma:09wed1},
  there exists $\alpha\in\mathfrak{a}$ such that
  \[
    \alpha\mathfrak{a}^{-1}+\beta\mathfrak{a}^{-1}=\mathfrak{O}_{K}.
  \]
  Multiplying through by~$\mathfrak{a}$, we get
  \[
    (\alpha,\beta)=(\alpha)+(\beta)=\mathfrak{a}.
  \]
\end{proof}
\medskip

\textbf{Factorization of rational integers in number rings.} Give~$a\in\Z$, how
does the principal ideal~$a\mathfrak{O}_{K}=(a)\subseteq\mathfrak{O}_{K}$ factor into primes?
By factoring~$a$ into primes in~$\Z$, we see that it suffices determine, for each rational prime~$p$,
how the principal ideal~$p\mathfrak{O}_{K}=(p)$ factors in~$\mathfrak{O}_{K}$.

\begin{defn}
 Let~$\mathfrak{p}$ be a prime ideal of~$\mathfrak{O}_{K}$. Last time, we saw
 that there exists a unique rational prime~$p$ such that~$N(\mathfrak{p})=p^{f}$
 where~$1\leq f\leq n$. The integer~$f$ is called the \emph{inertial degree}
 of~$\mathfrak{p}$.
 \end{defn}

 \begin{thm}\label{thm:09wedfactor} (The $e_{i}$-$f_{i}$ theorem.)
    Let $p$ be a rational prime, and say $(p)=\prod_{i=1}^{k}
    \mathfrak{p}_{i}^{e_{i}}$ is the prime factorization of the ideal~$(p)$
    in~$\mathfrak{O}_{K}$.  Then
    \[
      \sum_{i=1}^{k} e_{i}f_{i}=n.
    \]
    were~$f_{i}$ is the inertial degree of~$\mathfrak{p}_{i}$ for each~$i$
    and~$n=[K\mathbin{:}\Q]$.
 \end{thm}
 \begin{proof}
  For each~$\mathfrak{p}_{i}$, we know $N(\mathfrak{p}_{i})=p_{i}^{f_{i}}$ for
  some rational prime $p_{i}$.  We have seen that the norm of an ideal is
  element of the ideal.  Hence, $p_{i}^{f_{i}}\in\mathfrak{p}_{i}$.
  Since~$\mathfrak{p}_{i}$ is prime, we have~$p_{i}\in\mathfrak{p}_{i}$.
  However, we also know that~$p\in\mathfrak{p}_{i}$.  So $p_{i}=p$.  (If~$p_{i}$
  and $p$ were distinct primes both in~$\mathfrak{p}_{i}$, then we get that
  $1\in\mathfrak{p}_{i}$.  However, then~$\mathfrak{p}_{i}=\mathfrak{O}_{K}$,
  contradicting the fact that~$\mathfrak{p}_{i}$ is prime.)
  Take norms 
  \[
    p^{n}=N((p))=\prod_{i=1}^{k} N(\mathfrak{p}_{i})^{e_{i}} =\prod_{i=1}^{k}
    p^{f_{i}e_{i}}=p^{\sum_{i=1}^{k} e_{i}f_{i}}.
  \]
  The result follows from equating exponents.
 \end{proof}

The next result allows us to factor rational integers in number fields with a
\emph{power basis}, i.e., whose ring of integers has the form $\Z[\theta]$ for
some algebraic integer~$\theta$. 

\begin{thm}
  Let~$K=\Q(\theta)$ be a number field of degree~$n$ such
  that~$\mathfrak{O}_{K}=\Z[\theta]$.  Suppose that~$p$ is a rational prime, and
  let~$f$ be the minimal polynomial for $\theta$ over~$\Q$.
  Let~$\FF_{p}=\Z/p\Z$,  the field with~$p$ elements.
  If $g\in\Z[x]$, we let $\overline{g}$ denote the image of~$g$ in~$\FF_{p}[x]$
  under the quotient mapping $\Z[x]\to \Z[x]/(p)=\FF_{p}[x]$.

  Suppose that
  \[
    \overline{f}=\prod_{i=1}^{k} \overline{f}_{i}^{\,e_{i}}
  \]
  is the factorization of~$f$ as an element of~$\FF_{p}[x]$ into monic
  irreducibles $\overline{f}_{i}$.
  Let~$\mathfrak{p}_{i}=(p,f_{i}(\theta))$ for $i=1,\ldots,k$.  Then each
  $\mathfrak{p}_{i}$ is prime and
  \[
    (p)=\prod_{i=1}^{k} \mathfrak{p}_{i}^{e_{i}}=\prod_{i=1}^{k}
    (p,f_{i}(\theta))^{e_{i}}
  \]
  is the prime factorization of~$(p)$ in~$\mathfrak{O}_{K}$.
\end{thm}

\begin{eg}
   Let $K=\Q(\sqrt{-6})$.  Since~$-6\neq 1\bmod 4$, we know
   that~$\mathfrak{O}_{K}=\Z[\sqrt{-6}]=\Span_{\Z}\{1,\sqrt{-6}\}$.  One of the
   homework problems gives a method for factoring the principal ideal~$(6)$
   in~$\mathfrak{O}_{K}$.  Alternatively, we may use Theorem~\ref{thm:09wedfactor}.  
   
   The minimal polynomial for~$\sqrt{-6}$ over~$\Q$ is $f=x^2+6$.  Modulo $2$,
   we have $f = x^{2}\bmod 2$.  Hence,
   $(2)=(2,\sqrt{-6})^{2}\subset\mathfrak{O}_{K}$.  Similarly,
   $(3)=(3,\sqrt{-6})\subset\mathfrak{O}_{K}$.  So the factorization of~$(2)$
   into primes in~$\mathfrak{O}_{K}$ is 
   \[
     (2)=(2,\sqrt{-6})^{2}(3,\sqrt{-6})^{2}.
   \]
   On the other hand, we have
   \[
     f = x^2+1=(x-1)(x+1)\bmod 5.
   \]
   Hence, the prime factorization of~$(5)$ in~$\mathfrak{O}_{K}$ is
   \[
     (5)=(5,-1+\sqrt{-6})(5,1+\sqrt{-6}).
   \]
   Exercise: what about $(7)$?
\end{eg}

\begin{proof}[Proof of Theorem~\ref{thm:09wedfactor}.]
\textbf{Step 1.} First, we show that sending~$x$ to~$\theta$ induces an
isomorphism
\[
 \overline{\phi}\colon \Z[x]/(f)\xrightarrow{\sim} \Z[\theta].
\]
To see this consider the mapping of rings $\phi\colon\Z[x]\to\Z[\theta]$
determined $\phi(x)=\theta$.  It is clearly surjective, and since~$f(\theta)=0$,
we have $f\in\ker(\phi)$.  Hence, $\overline{\phi}$ is well-defined.  Suppose
that~$g\in\ker \overline{\phi}$, i.e., $g(\theta)=0$.  Since~$f$ is the minimal
polynomial of~$\theta$, we have~$g = fh$ for some $h\in\Q[x]$.  It is easy to
check that since~$f$ is monic, we must have $h\in\Z[x]$ (or one may use Gauss's
lemma and the fact that~$f$ is monic).  Hence, $g=0\in \Z[x]/(f)$.
\medskip

\textbf{Step 2.} For each~$i$, we have a natural sequence of surjections
\[
  \Z[x]\to \FF_{p}[x]\to \FF_{p}[x]/(f_{i})=\Z[x]/(p,f_{i})
\]
Since~$f_{i}|f$, we see~$f$ is in the kernel.  So we get a surjection
\[
  \phi_{i}\colon\Z[\theta]\simeq\Z[x]/(f)\to \Z[x]/(p,f_{i}).
\]
where
\[
  \phi_{i}(g(\theta)) = \overline{g(x)} \in \Z[x]/(p,f_{i})
\]
for all~$g\in\Z[x]$.

\textbf{Step 3.} We claim~$(p,f_{i}(\theta))$ is a prime ideal in~$\Z[\theta]$.
We have the isomorphism
\[
  \Z[\theta]/\ker(\phi_{i}) \xrightarrow{\sim} \FF_{p}[x]/(f_{i}).
\]
Since~$\FF_{p}=\Z/p\Z$ is a field, $\FF_{p}[x]$ is a PID.  Then, since~$f_{i}$
is irreducible, $(f_{i})$ is a maximal ideal (if $(f_{i})\subset(g)$, then $f=gh$ for
some $h$.  So either~$g$ is a unit and~$(g)=\FF_{p}[x]$ or $h$ is a unit
and~$(f)=(g)$.    Therefore, $\FF_{p}[x]/(f_{i})$ is a field, which implies
$\ker(\phi_{i})$ is a maximal, hence, prime ideal.  So to show
$(p,f_{i}(\theta))$ is prime, it suffices to show that
\[
  \ker(\phi_{i}) =(p,f_{i}(\theta)).
\]
It is clear that~$(p,f_{i}(\theta))\subseteq\ker(\phi_{i})$. For the opposite inclusion,
let $g(\theta)\in\ker(\phi_{i})$.  Then $g(x)=p\ell(x)+h(x)f_{i}(x)$.  Then, 
\[
  g(\theta)=p\ell(\theta)+h(\theta)f_{i}(\theta)\in (p,f_{i}(\theta)).
\]

\textbf{Step 4.} We claim that if $i\neq j$, then $(p,f_{i}(\theta))\neq
(p,f_{j}(\theta))$.  To see this, suppose that these ideals are equal.  Then, 
\[
  (p,f_{i}(\theta)) = (p,f_{j}(\theta)) 
  \Rightarrow
  f_{j}(\theta)\in (p,f_{i}(\theta))\in\ker(\phi_{i})
  \Rightarrow
  \phi_{i}(f_{j}) = 0
  \Rightarrow
  f_{j}(x) \in (p,f_{i}).
\]
Therefore, $f_{j}=hf_{i}\bmod p$, i.e., $f_{j}=hf_{i}$ in $\FF_{p}[x]$.
However, $f_{j}$ is irreducible in~$\FF_{p}[x]$.  Hence, $h$ is a unit, i.e.,
$h\in\{1,2,\ldots,p-1\}$ in~$\FF_{p}$.  Since~$f_{i}$ and $f_{j}$ are both
monic, it follows that~$f_{i}=f_{j}$. 
\medskip

\textbf{Step 5.} We now show that
\[
  (p) = \prod_{i=1}^{k} (p,f_{i}(\theta))^{e_{i}}.
\]
We use the fact that for ideals in~$\mathfrak{O}_{K}$, 
\[
  (\mathfrak{a}+\mathfrak{b})(\mathfrak{a}+\mathfrak{c})\subseteq
  \mathfrak{a}+\mathfrak{b}\mathfrak{c}.
\]
We have
\begin{align*}
  \prod_{i=1}^{k} (p,f_{i}(\theta))^{e_{i}}
  &=\prod_{i=1}^{k} \left((p)+(f_{i}(\theta))\right)^{e_{i}}\\
  &\subseteq \prod_{i=1}^{k} ( (p)+(f_{i}(\theta)^{e_{i}}))\\
  &\subseteq (p)+\left(\prod_{i=1}^{k}f_{i}(\theta)^{e_{i}} \right)\\
  &=(p)+(f(\theta))\\
  &=(p).
\end{align*}
since~$f(\theta)=0$.  Thus, $(p)$ divides $\prod_{i=1}^{k} (p,f_{i}(\theta))^{e_{i}}=\prod_{i=1}^{k} \mathfrak{p}_{i}^{e_{i}}$.
It follows that
\[
  (p)=\prod_{i=1}^{k} \mathfrak{p}_{i}^{\ell_{i}}
\]
for some $0\leq \ell_{i}\leq e_{i}$.

Take norms:
\[
  \mathfrak{p}^{n} = \prod_{i=1}^{k} \mathfrak{p}_{i}^{\ell_{i}}.
\]
Now, 
\[
  N(\mathfrak{p}_{i})=\left|\Z[\theta]/\mathfrak{p}_{i}\right|
\]
and
\[
  \Z[\theta]/\mathfrak{p}_{i} =\Z[\theta]/\ker(\phi_{i})\simeq
  \FF_{p}[x]/(f_{i}).
\]
The elements of $\FF_{p}[x]/(f_{i})$ are exactly
$a_{0}+a_{1}x+\cdots+a_{d_{i}-1}x^{d_{i}-1}$
where the $a_{i}$ are $\FF_{p}=\Z/p\Z$ and $d_{i}=\deg(f_{i})$.  Therefore
\[
  \left|\FF_{p}[x]/(f_{i})\right| = p^{\deg(f_{i})}.
\]
It follows that
\[
  p^{n}=\prod_{i=1}^{k} N(\mathfrak{p}_{i})^{\ell_{i}}=\prod_{i=1}^{k}
  p^{\deg(f_{i})\ell_{i}}=p^{\sum_{i=1}^{k} \deg(f_{i})\ell_{i}},
\]
and hence
\[
  n = \sum_{i=1}^{k} \deg(f_{i})\ell_{i}.
\]
On the other hand, $f=\prod_{i=1}^{k} f_{i}^{e_{i}}$ implies
\[
  n = \deg(f)=\sum_{i=1}^{k} \deg(f_{i})e_{i}.
\]
Since $0\leq \ell_{i}\leq e_{i}$, we must have~$\ell_{i}=e_{i}$ for all~$i$.
\end{proof}

\begin{remark}
  It would be nice to have criteria for deciding when a number field has a power
  basis.
\end{remark}
\end{document}
