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\begin{document}
\pagestyle{plain}
\title{Commutative algebra used in Intro to A.G. course (up to Class 16)}
\author{Ravi Vakil}
\date{Friday, November 5, 1999.}
\maketitle

{\parskip=12pt % closing bracket is just before the bibliography 

Topics:  Rings, localization, primes, completion, dimension theory.

Class 3:
If $k$ is a field, and $B$ is
an integral domain which is a finitely-generated $k$-algebra.
Then $\dim B = tr.deg_k K(B)$ (where $K(B)$ is the quotient field of
$B$), and for any prime ideal $\fp$ in $B$,
$$
ht \fp + \dim B/\fp = \dim B.$$  
(I can't remember if this was central to anything I said.

Class 11:  The tensor product of two domains is also a domain.
(In case of finitely-generated domain over $\kb$, and the tensor
is over $\kb$.)

Class 13:
Transcendence degree.  Krull's Hauptidealsatz.

Class 15:
If $A$ is a noetherian local ring with maximal ideal
$\fm$ and residue field $\kb$, then $\dim_{\kb} \fm/\fm^2 \geq \dim A$.  


Class 16:
Let $(R,\fm)$ be a noetherian local domain of dimension one.
Then the following are equivalent.
\begin{enumerate}
\item[(i)] $R$ is a discrete valuation ring;
\item[(ii)] $R$ is integrally closed (I'll speak about integral closures next day);
\item[(iii)] $R$ is a regular local ring;
\item[(iv)] $\fm$ is a principal ideal.
\end{enumerate}

Class 17:  
{\bf Theorem (Integral closure is a local property).} 
If $R$ is integrally closed if and only if the localization of $R$ at
each of its maximal ideals is integrally closed.

Theorem on finiteness of integral closure (in finite field extension).

Take any Dedekind domain $R$, and let $K$ be its field of fractions.
Let $L$ be a finite extension of $K$, and let $S$ be the integral
closure of $R$ in $L$.  Then $S$ is also a Dedekind domain.  (Not used in this
generality, but stated for the benefit of the subset of students who are thinking about schemes.)  From Hartshorne I.6.3.

Class 18:  I may use the following fact (but this weekend I might try to
come up with a proof of the special case I'll need).
 {\bf Maximality of valuation rings.} If $(A, \fm_A)$, $(B,
\fm_B)$ are local rings contained in a field $K$, we say that $B$ {\em
dominates} $A$ if $A \subset B$ and $\fm_B \cap A = \fm_A$) Let $K$ be
a field.  A local ring $R$ contained in $K$ is a valuation ring of $K$
iff it is a maximal element of the set of local rings contained in
$K$, with respect to domination.  

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iamhere2

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