\classheader{2013-01-09}

\subsection*{Noetherian rings and integral closure; intro. to algebraic number theory}

Let $A$ be a commutative ring with unity $1\in A$. A subring of $A$ must contain 1; thus, for example, $2\Z$ is not a subring of $\Z$. We also require that a ring homomorphism respects the unities of the rings.

\subsubsection*{Integral elements, algebraic integers}

The spirit of algebraic number theory:

\begin{quote}
To study $\Z$, it is better not only to consider $\Z$, but also to consider the friends $\Z[i]$, $\Z[\sqrt{2}]$, $\ldots$ of $\Z$, for they are happy to help $\Z$.
\end{quote}

Numbers like $i$, $\sqrt{2}$, $\sqrt{5}$, $\ldots$ are algebraic integers. However, $\frac{\sqrt{5}}{2}$ is not an algebraic integer. But even though the shape of $\frac{1+\sqrt{5}}{2}$ is a fraction, it is an algebraic integer, and similarly $\frac{-1+\sqrt{-3}}{2}$ is an algebraic integer. Now we need a precise definition of algebraic integer.

\begin{definition}
Let $A$ be a subring of a ring $B$. Then we say an element $x\in B$ is integral over $A$ when there is an $n\geq 1$ and $a_1,\ldots,a_n\in A$ such that $x^n+a_1x^{n-1}+\cdots+a_n=0$.
\end{definition}

For example, if we take $A=\Z$ and $B=\Q(\sqrt{5})$, the element $x=\frac{1+\sqrt{5}}{2}$ satisfies $x^2-x-1=0$. Thus, $\frac{1+\sqrt{5}}{2}$ is integral over $\Z$.

\begin{definition}
Given a field $K\supseteq \mathbb{Q}$, we say that an element $x\in K$ is an algebraic integer when $x$ is integral over $\Z$.
\end{definition}

\begin{definition}
Let $A$ be a subring of a ring $B$. The integral closure of $A$ in $B$ is the set \[\{x\in B\mid x\text{ is integral over }A\}.\] When we have a field $K$ that is a finite extension of $\Q$, we denote the integral closure of $\Z$ in $K$ by $\mathcal{O}_K$. Sometimes, it is called the ring of integers of $K$, or the integer ring of $K$.
\end{definition}

When $K=\Q$, we have $\mathcal{O}_K=\Z$. When $K=\Q(i)$, we have $\mathcal{O}_K=\Z[i]$.

\begin{proposition-N}
The integral closure of $A$ in $B$ is a subring of $B$.
\end{proposition-N}

We will prove this claim later.


\begin{proposition-N}
Let $A\subseteq B\subseteq C$ be rings. Let $B'$ be the integral closure of $A$ in $B$, and let $C'$ be the integral closure of $A$ in $C$. Then $C'$ is the integral closure of $B'$ in $C$.
\end{proposition-N}

We will prove this claim later. When I say that we will prove something later, it either means I will give it later in class, or on the homework, or I will just forget to prove it.

\begin{corollary}
$\mathcal{O}_K$ is a ring.
\end{corollary}

Let $K\supseteq \Q$ be a quadratic extension, so that $K=\Q(\sqrt{m})$ for a squarefree $m\in\Z$. Then
\[\mathcal{O}_K=\begin{cases}
\Z[\sqrt{m}] & \text{ if }m\equiv 2,3\bmod 4,\\
\Z[\frac{1+\sqrt{m}}{2}] & \text{ if }m\equiv 1\bmod 4.
\end{cases}\]
Thus, for example, if $K=\Q(\sqrt{5})$, we have $\mathcal{O}_K=\Z[\frac{1+\sqrt{5}}{2}]$.

Now consider $A=k[T]$ and $B=k(T)(\sqrt[\leftroot{-2}\uproot{2}n]{f(T)})$ where $f\in k[T]$ is a polynomial with no multiple factors. Then the integral closure of $k[T]$ in $B$ is $k[T,\sqrt[\leftroot{-2}\uproot{2}n]{f(T)}]$. For example, the integral closure of $\C[T]$ in $\C(T,\sqrt{T^3+1})$ is $\C[T,\sqrt{T^3+1}]$.

\begin{definition}
Let $A$ be an integral domain wth field of fractions $K$. We say that $A$ is integrally closed if the integral closure of $A$ in $K$ is just $A$. Sometimes we instead say that $A$ is normal.
\end{definition}

For example, for any finite extension $K\supseteq\Q$, the ring $\cal{O}_K$ is normal.

However, $\Z[\sqrt{5}]$ is not normal, because $\frac{1+\sqrt{5}}{2}$ is in the field of fractions $\Q(\sqrt{5})$ and is integral over $\Z[\sqrt{5}]$, so it is in the integral closure of $\Z[\sqrt{5}]$ in $\Q(\sqrt{5})$, but it is not itself in $\Z[\sqrt{5}]$.

Here is another example of a non-normal ring: $\C[T,\sqrt{T^3}]$. The field of fractions of this ring is $\C(\sqrt{T})$, and $\sqrt{T}\notin \C[T,\sqrt{T^3}]$ even though it is integral over $\C[T,\sqrt{T^3}]$. Note that $\C[T,\sqrt{T^3}]\cong \C[X,Y]/(Y^2-X^3)$, where $X\mapsto T$ and $Y\mapsto \sqrt{T^3}$.

The ring $\C[T,T\sqrt{T^3+1}]$ is not normal, because $\sqrt{T^3+1}\notin \C[T,T\sqrt{T^3+1}]$. Note that $\C[T,T\sqrt{T^3+1}]\cong \C[X,Y]/(Y^2-X^2(X^3+1))$, where $X\mapsto T$ and $Y\mapsto T\sqrt{T^3+1}$.

\begin{center}
\begin{tikzpicture}
\begin{axis}[title={$y^2=x^3$}, no markers, axis lines=middle,axis y line=center,axis equal,axis y line=middle, axis x line=middle, xmin=-3.5, xmax=3.5, ymin=-3.5, ymax=3.5,xtick={-3,-2,...,3},ytick={-3,-2,...,3},axis equal,width=7cm, height=7cm,tick label style={color=gray}]
\addplot[variable=\t, samples=100, domain=0:3,color=blue,thick] ({t},{t*sqrt(t)});
\addplot[variable=\t, samples=100, domain=0:3,color=blue,thick] ({t},{-t*sqrt(t)});
\end{axis}
\end{tikzpicture}\qquad\qquad\begin{tikzpicture}
\begin{axis}[title={$y^2=x^2(x^3+1)$}, no markers, axis lines=middle,axis y line=center,axis equal,axis y line=middle, axis x line=middle, xmin=-3.5, xmax=3.5, ymin=-3.5, ymax=3.5,xtick={-3,-2,...,3},ytick={-3,-2,...,3},axis equal,width=7cm, height=7cm,tick label style={color=gray}]
\addplot +[no markers,raw gnuplot,thick] gnuplot {
set contour base;
set cntrparam levels discrete 0.00;
unset surface;
set view map;
set isosamples 1000;
set xrange[-4:1.5];
set yrange[-3:3];
splot y^2-x^2*(x^3+1);
};
\end{axis}
\end{tikzpicture}
\end{center}
The property of not being normal can be thought of as as a singularity in the corresponding graph, i.e. the set of maximal ideals has a singularity.

The integral closure of $A=\C[T,T\sqrt{T^3-1}]$ in its field of fractions $\C(T,\sqrt{T^3+1})$ is $\C[T,\sqrt{T^3+1}]\cong\C[X,Y]/(Y^2-(X^3+1))$. The intersection of a maximal ideal of $\C[T,\sqrt{T^3+1}]$ with $A$ is a maximal ideal of $A$. This map of maximal ideals corresponds to the map of points
\[\{(x,y)\mid y^2=x^3+1\}\xrightarrow{\;\;(x,y)\,\mapsto\,(x,xy)\;\;}\{(x,y)\mid y^2=x^2(x^3+1)\}.\]
\begin{center}
\begin{tikzpicture}
\begin{axis}[title={$y^2=x^3+1$}, no markers, axis lines=middle,axis y line=center,axis equal,axis y line=middle, axis x line=middle, xmin=-3.5, xmax=3.5, ymin=-3.5, ymax=3.5,xtick={-3,-2,...,3},ytick={-3,-2,...,3},axis equal,width=7cm, height=7cm,tick label style={color=gray}]
\addplot +[no markers,raw gnuplot,thick] gnuplot {
set contour base;
set cntrparam levels discrete 0.00;
unset surface;
set view map;
set isosamples 1000;
set xrange[-4:3];
set yrange[-3:3];
splot y^2-x^3-1;
};
\end{axis}
\end{tikzpicture}\quad\raisebox{1.2in}{$\xrightarrow{\;\;\;\;\text{surjection}\;\;\;\;}$}\quad\begin{tikzpicture}
\begin{axis}[title={$y^2=x^2(x^3+1)$}, no markers, axis lines=middle,axis y line=center,axis equal,axis y line=middle, axis x line=middle, xmin=-3.5, xmax=3.5, ymin=-3.5, ymax=3.5,xtick={-3,-2,...,3},ytick={-3,-2,...,3},axis equal,width=7cm, height=7cm,tick label style={color=gray}]
\addplot +[no markers,raw gnuplot,thick] gnuplot {
set contour base;
set cntrparam levels discrete 0.00;
unset surface;
set view map;
set isosamples 1000;
set xrange[-4:1.5];
set yrange[-3:3];
splot y^2-x^2*(x^3+1);
};
\end{axis}
\end{tikzpicture}
\end{center}
Taking the integral closure is like dissolving the singularity. The same thing is occurring when we take $\Z[\frac{1+\sqrt{5}}{2}]$ instead of $\Z[\sqrt{5}]$.

\begin{proposition}
$\C[T,\sqrt{T^3+1}]$ is the integral closure of $\C[T]$ in $\C(T)(\sqrt{T^3+1})$
\end{proposition}
\begin{proof}
An element $f\in\C(T)(\sqrt{T^3+1})$ is of the form $f=g+h\sqrt{T^3+1}$, where $g,h\in\C(T)$. If $f$ is integral over $\C[T]$, then the conjugate $\overline{f}=g-h\sqrt{T^3+1}$ is also integral over $\C[T]$, so $f+\overline{f}=2g\in\C(T)$ and $f\overline{f}=g^2-(T^3+1)h^2\in\C(T)$ are integral over $\C[T]$. Now using a lemma,
\begin{lemma}
PID's are normal.
\end{lemma}
we see that we must have $2g\in\C[T]$ and $g^2-(T^3+1)h^2\in\C[T]$. Therefore $g\in\C[T]$, hence $(T^3+1)h^2\in\C[T]$, and because $T^3+1$ has no repeated factors, this implies that $h\in\C[T]$. Hence $f\in\C[T,\sqrt{T^3+1}]$.
\end{proof}





