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\title{Lecture 5-20: The prime spectrum of a ring}
\date{May 20, 2024}
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We generalize the Zariski topology to an arbitrary Noetherian ring $R$ in such a way that we add more points to affine space $\mathbf A^n$ and give it a richer structure.
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In an attempt to generalize the Zariski topology from coordinate rings $k[V]$ of varieties to general Noetherian rings we might start with the set of maximal ideals of $R$.  This set however does not have good functorial properties:  the inverse image $f^{-1}(M)$ of a maximal ideal $M\subset S$ under a ring homomorphism $f:R\rightarrow S$ need not be maximal in $R$ (think of the inclusion of $\mathbb Z$ in $\mathbb Q$ and the 0 ideal).  The inverse image $P=f^{-1}(Q)$ of a {\sl prime} ideal $Q$ of $S$ is prime in $R$, however, since the quotient $R/P$ is a subring of $S/Q$ and so cannot have zero divisors if $S/Q$ does not.  Accordingly, we start with the set Spec~$R$ of prime ideals in $R$, calling this the \alert{prime spectrum of $R$} (DF, p. 731).  
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We define the \alert{Zariski topology} on Spec~$R$ by decreeing that the closed sets are the sets $\mathcal Z(I)$ of prime ideals containing a fixed ideal $I$ of $R$ (DF, p. 733).  Clearly any prime ideal $P$ containing an ideal $I$ also contains its radical $\sqrt I$, since a power $x^n\in P$ if and only if $x\in P$.  Conversely, \alert{the radical $\sqrt I$ of $I$ is exactly the intersection $\mathcal I(I)=\cap\mathcal Z(I)$ of all prime ideals $P$ containing $I$} (DF, Proposition 12, p. 674).  To prove this, let $x\in R$ be such that $x^n\notin I$ for all $n$ and let $\mathcal S$ be the collection of ideals in $R$ excluding all powers of $x$.  Then $\mathcal S$ is not empty since $I\in\mathcal S$; let $P$ be a maximal element of $\mathcal S$ under inclusion.  If $y,z\notin P$ then $P+(y),P+(z)$ are strictly larger than $P$ and so they must contain powers $x^r,x^s$ of $x$; but then $P+(yz)$ contains $x^{r+s}$, forcing $yz\notin P$, whence $P$ is prime.
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Hence any radical ideal $I$ is the intersection $\mathcal I(\mathcal P)$ of the set $\mathcal P$ of prime ideals containing $I$, and conversely any intersection of prime ideals is radical.  We deduce
\begin{alertblock}{Proposition 53, DF, p. 732} 
There are inclusion-revsersing bijections $\mathcal Z(I)\mapsto \mathcal I(I),I\mapsto\mathcal Z(I)$ between the sets of Zariski closed subsets of Spec~$R$ and of radical ideals of $R$.
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We also see that finite unions and arbitrary intersections of Zariski closed sets are closed, since $\mathcal Z(IJ) = \mathcal Z(I)\cup\mathcal Z(J)$ and $\mathcal Z(\sum_i I_i) = \cap_i\mathcal Z(I_i)$.
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As with algebraic sets, one also readily checks that a Zariski closed subset $\mathcal Z(I)$ of Spec~$R$ is irreducible (not the union of two proper closed subsets) if and only if its radical ideal $\mathcal I(I)$ is prime.
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If $R=k[V]$ is the coordinate ring of an algebraic set $V$, then the prime spectrum $X=$ Spec~$R$ contains the set mSpec $k[V]$ of maximal ideals of $k[V]$, which corresponds bijectively to $V$.  For every maximal ideal $M$, the singleton set $\{M\}=\mathcal Z(M)$ and so is closed; by abuse of language we call $M$ a \alert{closed point} in $X$ (DF, p. 733).  Now however we have additional points in $X$ that are not closed.  For example, if $V$ is a variety, then the closure of $\{0\}$ is all of $X$.  We call this ideal a \alert{generic point}.
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Returning now to a general Noetherian ring $R$, let $f:R\rightarrow S$ be a homomorphism of rings.  Then $f$ induces a map $f^*:$ Spec $S\rightarrow$ Spec $R$ sending a prime ideal $Q$ of $S$ to the prime ideal $f^{-1}(Q)$ of $R$.  It is easy to check that the inverse image under $f^*$ of the Zariski closed set $\mathcal Z(I)$ for an ideal $I$ of $R$ is the closed set $\mathcal Z(J), J$, the ideal of $S$ generated by $f(I)$, whence we get
\begin{alertblock}{Proposition 35, DF, p. 734}
With notation as above the induced map $f^*$ from Spec~$S$ to Spec~$R$ is continuous with respect to the Zariski topology.
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Just as an arbitrary Zariski-continuous map from one variety $V$ to another one $W$ need not be a morphism and so need not correspond to a ring homomorphism, there is no reason to expect an arbitrary continuous map from Spec~$S$ to Spec~$R$ to arise from a ring homomorphism from $R$ to $S$.
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Now let $f\in R$.  Corresponding to the Zariski principal open set $D_f\subset\mathbf A^m$ of nonzeroes of a polynomial $f\in P_m$ we have the \alert{basic open set} $X_f$ of ideals in $X=$~Spec~$R$ not containing $f$.  Since a Zariski closed set $\mathcal Z(I)$ correpsonds to a finitely generated ideal $I=(f_1,\ldots,f_m)$, we see that \alert{the $X_f$ form a basis for the Zariski topology} and in fact \alert{every Zariski open set is a finite union of $X_f$} (DF, Proposition 56, p. 738).  
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We now define the analogue of regular function for a Noetherian ring.  
\begin{alertblock}{Definition; DF, p. 739}
Let $U$ be a nonempty open subset of Spec~$R$.  Define $\mathcal O(U)$ to be the set of functions $s:U\rightarrow\sqcup_{Q\in U} R_Q$ from $U$ to the disjoint union of the localizations $R_Q$ for $Q\in U$ such that 
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\item $s(Q)\in R_Q$ for $Q\in U$, and
\item for every $P\in U$ there is a basic open neighborhood $X_f\subseteq U$ of $P$ in $U$ and an element $\frac a {f^n}$ in the localization $R_f$ defining $s$ on $X_f$, so that $s(Q)=\frac a {f^n}$ for $Q\in X_f$.
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It is easy to check that each $\mathcal O(U)$ is closed under addition and multiplication and that there is a natural restriction map from $\mathcal O(U)$ to $\mathcal O(U')$ whenever $U'$ is a nonempty open subset of $U$. 
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\begin{alertblock}{Definition; DF, p. 740}
With notation as above, the collection of rings $\mathcal O(U)$ as $U$ runs over the nonempty Zariski open subsets of $X=$~Spec~$R$ together with the restriction maps $\mathcal O(U)\rightarrow\mathcal O(U')$ for $U'\subseteq U$ is called the {\sl structure sheaf} of $X$ and is denoted $\mathcal O$ or $\mathcal O_X$.  The elements $s$ of $\mathcal O(U)$ are called {\sl sections of $\mathcal O$ over $U$}.  The elements of $\mathcal O(X)$ are called {\sl global sections}.
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A straightforward argument then proves
\begin{alertblock}{Proposition 57, DF, p. 740}
If $X=$~Spec~$R$ has structure sheaf $\mathcal O$, then the global sections of $\mathcal O$ identify naturally with the elements of $R$.  More generally, if $X_f$ is a basic open set, then $\mathcal O(X_f)$ identifies with the localization $R_f$.
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\begin{alertblock}{Definition; DF, p. 741}
If $P\in X$ then the localization $R_P$ of $R$ at $P$ is called the {\sl stalk of $\mathcal O$ at $P$}.
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There is a nice picture of this on p. 742.  The pair $(X,\mathcal O_X)$ with $X=$~Spec~$R$ is called an \alert{affine scheme}.  The stalk $\mathcal O_P=R_P$ of $\mathcal O$ at $P$ may be viewed as a direct limit of rings $\mathcal O(U)$ as $U$ runs through the open sets containing $P$.
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We have mentioned that continuous maps from $X=$~Spec~$S$ to $X'=$~Spec~$R$ do not generally correspond to ring homomorphisms from $R$ to $S$; but if the continuous map is paired with compatible ring homomorphisms from sections over open subsets for $X$ to corresponding sections for $X'$ commuting with restriction maps and taking stalks to stalks, then it does correspond to a ring homomorphism.  In this way we get a bijection between ring homomorphisms and what are called morphisms of affine schemes.
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Finally, a \alert{locally ringed space} is a topological space $X$ equipped with a ring $\mathcal O(U)$ attached to every nonempty open subset $U$ with a compatible set of homomorphisms from $\mathcal O(U)$ to $\mathcal O(U')$ if $U'\subseteq U$ and with local conditions on the sections, such that the stalks (direct limits of $\mathcal O(U)$ as $U$ runs through the open sets containing a fixed point $P\in X$) are local rings.  A general \alert{scheme} is a locally ringed space in which each point lies in a neighborhood isomorphic to an affine scheme (with some compatibility conditions between such neighborhoods).  These are the algebro-geometric analogues of differentiable manifolds and play a fundamental role in modern algebraic geometry.
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