\classheader{2013-01-11}


The number $x=\frac{1+\sqrt{5}}{2}$ is integral over $\Z$ because it is a root of the polynomial $x^2-x-1=0$. However, the number $x=\frac{\sqrt{5}}{2}$ is not integral over $\Z$ because it is a root of $4x^2-5=0$ and not of any monic integer polynomial.

\begin{proposition-N}
If $B$ is a ring and $A\subset B$ is a subring, and $x\in B$, then $x$ is integral over $A$ if and only if $A[x]\subset B$ is a finitely-generated $A$-module.
\end{proposition-N}

Recall that we say an abelian group $M$ is an $A$-module when for any $a\in A$ and $x\in M$, we define their product $ax\in M$, and this satisfies
\begin{align*}
a(x+y)&=ax+ay\\
(a+b)x&=ax+bx\\
(ab)x&=a(bx)\\
1x&=x
\end{align*}
An $A$-module $M$ is finitely generated when there exist $x_1,\ldots,x_n\in M$ such that
\[M=\{a_1x_1+\cdots+a_nx_n\mid a_1,\ldots,a_n\in A\}.\]
\begin{proof}
If $x$ is integral over $A$, then it satisfies some monic polynomial over $A$, say
\[x^n+a_1x^{n-1}+\cdots+a_n=0.\]
Then $A[x]$ is generated by $1,x,\ldots,x^{n-1}$ as an $A$-module, because
\begin{align*}
x^n&=-(a_n+a_{n-1}x+\cdots+a_1x^{n-1})\\
x^{n+1}&=-x(a_n+a_{n-1}x+\cdots+a_1x^{n-1})=-(a_nx+\cdots+a_2x^{n-1})-a_1x^n\\
x^{n+2}&=\cdots
\end{align*}
Conversely, if $A[x]$ is generated by $f_1,\ldots,f_m\in A[x]$, then there is some $r\in\N$ such that
\[f_1,\ldots,f_m\in A+Ax+\cdots+Ax^r\]
(for example, $r=$ the largest $\deg(f_i)$). We have $A[x]\subset A+Ax+\cdots Ax^r$, so $x^{r+1}=a_0+a_1x+\cdots+a_rx^r$ for some $a_i\in A$, and therefore $x$ is integral.
\end{proof}

As an illustration of this proposition, note that $\frac{2}{3}\in\Q$ is not integral over $\Z$ because 
\[\textstyle\Z[\frac{2}{3}]=\bigcup_{n\geq 1}(\Z+\Z\frac{2}{3}+\cdots+\Z\frac{2^n}{3^n})\]
cannot be finitely generated as a $\Z$-module.


\subsection*{Noetherian rings and modules}

\begin{definition}
A commutative ring $A$ is noetherian when any ideal of $A$ is finitely generated.
\end{definition}

\begin{theorem}[Hilbert, 1888]
Let $A$ be a commutative ring and $B$ is a finitely generated ring over $A$. If $A$ is noetherian, then $B$ is noetherian.
\end{theorem}

When we say that $B$ is a ring over $A$, what we really mean is that we have fixed a ring homomorphism $\phi:A\to B$. Then we say that $B$ is finitely generated (as a ring) over $A$ when there exist $b_1,\ldots,b_n\in B$ such that $B=\phi(A)[b_1,\ldots,b_n]$.

Hilbert's theorem demonstrates that most of the rings that come up in algebraic geometry or number theory, which are finitely generated rings over either a field $k$ or over $\Z$, are noetherian.


\begin{definition}
Let $A$ be a commutative ring and $M$ is an $A$-module. Then we say that $M$ is a noetherian $A$-module when all $A$-submodules of $M$ are finitely generated as $A$-modules.
\end{definition}

\begin{remark}
A commutative ring $A$ can be regarded as an $A$-module. An ideal of a commutative ring $A$ is precisely an $A$-submodule of $A$. Thus, $A$ is a noetherian ring if and only if it is a noetherian module over itself.
\end{remark}


\begin{proposition-N}
Let $A$ be a commutative ring, let $M$ be an $A$-module, and let $N$ a submodule of $M$.
\[N\text{ and }M/N\text{ are f.g. }A\text{-modules}\implies M\text{ is an f.g. }A\text{-module} \implies M/N\text{ is an f.g. }A\text{-module}\]
and
\[N\text{ and }M/N\text{ are noetherian }A\text{-modules}\iff M\text{ is a noetherian }A\text{-module}.\]
\end{proposition-N}

\begin{proposition-N}
If $A$ is a commutative ring and $M$ is a finitely generated $A$-module, then $M$ is a noetherian $A$-module.
\end{proposition-N}
\begin{proof}[Proof of Prop. 3]
Because $M$ is finitely generated, we have $M=Ax_1+\cdots+Ax_n$ for some $x_i\in M$. Then we have a surjection $A^n\to M$, defined by mapping $(a_1,\ldots,a_n)$ to $a_1x_1+\cdots+a_nx_n$. Letting $N=\ker(h)$, we then have $A^n/N\cong M$. An easy induction argument shows that $A^n$ is a noetherian $A$-module for any $n$ (note that $A^n/(A^{n-1}\oplus 0)\cong A$). Thus, using Proposition 2, $M\cong A^n/N$ is noetherian.
\end{proof}

Now let's prove Proposition 1 from last time, i.e. that given a ring $B$ and a subring $A$, the integral closure of $A$ in $B$ is a subring of $B$.

\begin{proof}
First, let's do the case that $A$ is noetherian. Assume that $x,y\in B$ are integral over $A$. Then $A[x,y]$ is finitely generated as an $A$-module, because $x^n+a_1x^{n-1}+\cdots+a_n=0$ and $y^m+c_1y^{m-1}+\cdots+c_m=0$ implies that $A[x,y]$ is generated by $x^iy^j$ for $0\leq i<n$ and $0\leq j<m$.

Clearly, $x+y,xy\in A[x,y]$. Equivalently, we know that $A[x+y]\subset A[x,y]$ and $A[xy]\subset A[x,y]$. By Proposition 3, $A[x,y]$ is a noetherian $A$-module, so both $A[x+y]$ and $A[xy]$ are finitely generated as $A$-modules, and therefore $x+y$ and $xy$ are integral over $A$ by Proposition 1.

Now let's do the general case. Any ring $A$ whatsoever is a union of subrings of $A$ each of which is finitely generated over $\Z$. This is clear because any finite subset $\{a_1,\ldots,a_n\}$ of $A$ is contained in a subring of $A$ which is finitely generated over $\Z$, specifically, $\phi(\Z)[a_1,\ldots,a_n]$ where $\phi:Z\to A$ is the unique ring homomorphism (specifically $\phi(n)=n\cdot 1$).

Thus, if $x,y\in B$ are integral over $A$, say $x^n+a_1x^{n-1}+\cdots+a_n=0$ and $y^m+c_1y^{m-1}+\cdots+c_m=0$ for some $a_i,c_i\in A$, then not only are $x$ and $y$ integral over $A$, they are integral over the subring $A'=\phi(A)[a_1,\ldots,a_n,c_1,\ldots,c_m]$ of $A$. Because $A'$ is finitely generated over $\Z$, it is noetherian. By the noetherian case, we have that $x+y$ and $xy$ are integral over $A'$, and therefore they are clearly also integral over $A$.
\end{proof}