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\title{Lecture 5-13: Localization of commutative rings}
\date{May 13, 2024}
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Picking up from where we left off last time, we continue to continue to study arbitrary commutative rings rather than quotients of polynomial rings over fields, developing a basic technique that will prove quite useful for the quotients as well.  In particular we use it to show that there is a finite-to-one covering map from any algebraic set $V$ of dimension $d$ to affine space $\mathbf A^d$. 
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Let $R$ be a commutative ring.  You know that there is a bijection between ideals of a quotient $R/I$ and ideals of $R$ containing $I$.  What if one wanted to study ideals of $R$ contained in $I$ instead?  We will see that there is indeed a way to focus attention at least on prime ideals contained in $I$.  To do this, we generalize the construction of the field of quotients of an integral domain.  Let $D$ be a \alert{multiplicatively closed} subset of $R$, so that (by definition) $1\in D$ and $ab\in D$ whenever $a,b\in D$.  
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\begin{alertblock}{Definition of localization $D^{-1}R$; DF, p. 707}
The ring $D^{-1}R$, the localization of $R$ by $D$, consists of all equivalence classes of ordered pairs $(d,r)\in D\times R$, subject to the relation $(d,r)\sim(e,s)$ if there is $x\in D$ with $x(er-ds)=0$.  The equivalence class of $(d,r)$ is denoted $\frac r d$.  We make $D^{-1}R$ into a ring by the usual rules for adding and multiplying fractions:  $\frac r d+\frac s e=\frac {re+ds} {de},\frac r e \frac s d= \frac {rs} {de}$.  There is a natural map $R\rightarrow D^{-1}R$ sending $r$ to $\frac r 1$.
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In constructing $D^{-1}R$ we are adjoining multiplicative inverses of the elements of $D$ to $R$.  If $R$ is an integral domain and $D$ is the nonzero elements of $R$, then the construction reduces to that of the field of quotients of $R$.
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One checks immediately that the relation $\sim$ is indeed reflexive, symmetric, and transitive; note that the $x$ appearing in the definition of $\sim$ is crucial to proving its transitivity.  One also checks that the ring operations are well defined on equivalence classes.  The map from $R$ to $D^{-1}R$ is injective if and only if $D$ contains no zero divisors; in general its kernel is the set of $r\in R$ for which there is $d\in D$ with $dr=0$.
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For our purposes the most important example of this construction occurs when $D$ is the complement of a prime ideal $P$ of $R$; then primeness of $P$ guarantees that $D$ is multiplicatively closed.  In this case we denote $D^{-1}R$ by $R_P$ and call it the \alert{localization of $R$ at $P$}.  Another important example has $D=\{f^n:n\in\mathbb N\}$, the powers of a non-nilpotent element $f$ of $R$; here $D^{-1}R$ is denoted $R_f$.
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Ideals in a localization $D^{-1}R$ are closely related to ideals in $R$.  More precisely, we have
\begin{alertblock}{Proposition; DF, p. 709}
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\item For any ideal of $J$ of $D^{-1}R$ we have $J=J^{ce}$; in particular, every ideal of $D^{-1}R$ is the extension of an ideal of $R$ and distinct ideals in $D^{-1}R$ have distinct contractions in $R$.
\item For any ideal $I$ of $R$ we have $I^{ec}=\{r\in R: dr\in I\text{ for some }d\in D\}$; also $I^e = D^{-1}R$ if and only if $I$ intersects $D$.
\item Extension and contraction give order-preserving bijections between prime ideals of $R$ not intersecting $D$ and prime ideals of $D^{-1}R$.
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\begin{proof}
We already know that $J^{ce}\subseteq J$; conversely, if $\frac a d\in J$, then $a=d\frac a d\in J$, so $a\in J^c$, showing that $J^{ce}=J$.  If $I$ is an ideal of $R, r\in R$ and $d\in D$ has $dr=a\in I$ then $\frac r 1=\frac a d\in I^e$, so $r\in I^{ec}$.  Conversely, if $r\in I^{ec}$, then $\frac r 1=\frac a d$ for some $a\in I,d\in D$, so that $x(dr-a)=0,xdr=xa\in I$ for some $x\in D$, whence the second assertion holds.  In particular, we have $I^e=D^{-1}R$ if and only if $1\in I^{ec}$, so that $I$ intersects $D$.  If $Q$ is prime in $D^{-1}R$ then we have already observed that its contraction $Q^c$ is prime in $R$.  Conversely, if $P$ is prime in $R$ and $\frac a {d_1}\frac b {d_2}\in Q=P^e$, then $\frac {ab} {d_1d_2}=\frac c d$ for some $c\in P,d\in D$, whence $x(dab-d_1d_2c)=0$ for some $x\in D$, forcing $xdab\in P$ and then $ab\in P$ since $P$ is prime and disjoint from $D$.  Then either $a\in P$ or $b\in P$, forcing $\frac a {d_1}\in Q$ or $\frac b {d_2}\in Q$, as desired.  Since $P^{ec}=P$ we get the bijection of the third assertion.
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Thus we have achieved the goal set out at the beginning:  if $P$ is prime in $R$, then the localization $R_P$ is such that its prime ideals correspond bijectively to prime ideals of $R$ lying in $P$.  In particular, $R_P$ has a unique maximal ideal, namely the extension $P^e=PR_P$.  We call a ring $R$ with a unique maximal ideal $M$ \alert{local} (DF, p. 717). In this case, $M$ consists precisely of the nonunits in $R$, since any nonunit $x$ lies in a proper principal ideal $(x)$, which can be enlarged to a maximal ideal necessarily coinciding with $M$.  
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Now we can complete the proof of Corollary 27 (DF, p. 695), as promised last time.
\begin{alertblock}{Corollary 50, DF, p. 720}
Given a ring extension $R\subseteq S$ with $S$ integral over $R$ and $P$, a prime ideal of $R$, there is a prime ideal $Q$ of $S$ with $Q^c=Q\cap R=P$.  
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Let $D$ be the complement of $P$ in $R$.  It is easy to check that $D^{-1}S$ is integral over $D^{-1}R=R_P$; let $\mathfrak m$ be a maximal ideal of $D^{-1}S$ (any ring has at least one maximal ideal, by something called Zorn's Lemma (see DF, p. 909, and Proposition 11 on p. 254).  We saw last time that the contraction $\mathfrak m^c=\mathfrak m\cap R_P$ of $\mathfrak m$ is maximal in $R_P$ and thus equal to $PR_P$.  Taking the contraction $\mathfrak m\cap S$ of $\mathfrak m$ in $S$, we get a prime ideal $Q$ with $Q\cap R=P$, as desired.
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Now let $k$ be an algebraically closed field and $V\subset\mathbf A^n$ an algebraic set.  We have seen by Noether normalization that the coordinate ring $k[V]$ is a finitely generated integral extension of $P_d=k[x_1,\ldots,x_d]$, where $d=\dim V$, whence there is a morphism $\pi$ from $V$ to affine $d$-space $\mathbf A^d$.  Since points of either $V$ or $\mathbf A^d$ correspond bijectively to maximal ideals of their coordinate rings, we see from the aforementioned Corollary 27 that \alert{this morphism is surjective and has finite fibers}.  As mentioned last time, however, It is not a topological covering map, however, since the fibers need not have the same size in general.  The fibers of the map from $V$ to $\mathbf A^d$ do have constant size on a Zariski open subset of $\mathbf A^d$, but not on the entire space.
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 Now let $V$ be a variety, $v$ a point of $V$.  We have the maximal ideal $M=M_v$ of the coordinate ring $k[V]$ consisting of all functions vanishing at $v$.  The localization $k[V]_M$ of $k[V]$ at $M$ may be viewed as a subring of the function field $k(V)$; it consists of all quotients $\frac f g$ of polynomials  $f,g\in k[V]$ with $g(v)\ne0$.  We call such a quotient \alert{regular at $v$}; the ring or all such quotients is called the \alert{local ring of $V$ at $v$} and is denoted $\mathcal O_{v,V}$ (DF, p. 722).  More generally, any quotient of polynomials $\frac f g$ defined on an open subset of $V$ (but not necessarily on all of $V$) and equal in some neighborhood of a point $v$ at which it is defined to a function in $\mathcal O_{v,V}$ is called \alert{regular on $V$}; the ring of all such is denoted $\mathcal O_V$.  The collection of regular functions defined at all points of an open subset $U$ of $V$ is denoted $\mathcal O_{U,V}$.
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\begin{example}
Let $V$ be the zero locus of $xz-yw$ in $\mathbf A^4$.  This polynomial is easily seen to be irreducible in $k[x,y,z,w]$; since the latter ring is a UFD, the principal ideal $(xy-zw)$ is prime, so that $V$ is a variety.  The function $f=\frac x y$ is defined at all points $(x,y,z,w)\in V$ with $y\ne0$ and is regular at such points; since $\frac x y=\frac w z$ at any point in $V$ at which both quotients are defined, we have $f=\frac w z$ at any point in $V$ where $z\ne0$.  Thus $f\in\mathcal O_V$; the domain $D$ of definition of $f$ consists of all points $(x,y,z,w)\in V$ at which at least one of $y,z$ is not 0.  It is easy to check that there is no single quotient $\frac p q$ of polynomials which equals $f$ at all points of $D$.  See DF, p. 721.
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It is not difficult to see that \alert{if $R$ is an integral domain with field of fractions $K$, then the intersection $\cap_M R_M$ of all localizations of $R$ at maximal ideals $M$, regarded as a subring of $K$, is just $R$} (Proposition 48, DF, p. 720).  Indeed, suppose that $a\in K$ lies in the intersection.  Then $I_a=\{d\in R: da\in R\}$ is an ideal of $R$; if it is proper then it lies in a maximal ideal $M$.  Writing $a=\frac r d$ for some $r\in R,d\notin M$ we then get $d\in I_a$, a contradiction; so $I_a=R,a\in R$, as claimed.  As a consequence, \alert{the ring $\mathcal O_{V,V}\cong R$} (Proposition 51, DF, p. 722).
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