Last update: 3 June 2013
Let $A$ be a commutative ring with 1. Let $X=\text{Spec}\left(A\right)$ denote the set of all prime ideals of $A\text{.}$ $\text{(}p$ is a prime ideal $\iff A/p$ is an integral domain; thus $A$ itself is not a prime ideal.) If $x\in X$ it is sometimes convenient to write ${j}_{x}$ for the ideal $x\text{.}$ For each subset $E$ of $A,$ let $V\left(E\right)=\{x\in X:{j}_{x}\supseteq E\}\text{.}$ If $E$ consists of a single element $f,$ we write $V\left(f\right)$ in place of $V\left(\left\{f\right\}\right)\text{.}$
Lemma (3.1).
(i) | $V\left(0\right)=X\text{;}$ $V\left(1\right)=\varnothing \text{.}$ |
(ii) | If $E\subseteq E\prime ,$ then $V\left(E\right)\supseteq V\left(E\prime \right)\text{.}$ |
(iii) | $V\left(\underset{\lambda}{\cup}{E}_{\lambda}\right)=\underset{\lambda}{\cap}V\left({E}_{\lambda}\right)\text{.}$ |
(iv) | $V\left(EE\prime \right)=V\left(E\right)\cup V\left(E\prime \right)\text{.}$ |
Proof. | |
Only (iv) is not entirely trivial. Clearly $V\left(EE\prime \right)\supseteq V\left(E\right)\cup V\left(E\prime \right)\text{.}$ Conversely, if $x\notin V\left(E\right)\cup V\left(E\prime \right)$ then there exist $f\in E$ and $f\prime \in E\prime $ such that $f\notin {j}_{x}$ and $f\prime \notin {j}_{x}\text{;}$ since ${j}_{x}$ is prime, we have $ff\prime \notin {j}_{x},$ hence $x\notin V\left(EE\prime \right)\text{.}$ $\square $ |
It follows from (3. 1) that the sets $V\left(E\right)$ satisfy the axioms for closed setsJn a topology on $X\text{.}$ This topology is called the Zariski topology or spectral topology on $X,$ and it is the only one we shall use.
If $a$ is an ideal in $A,$ the radical $r\left(a\right)$ of $a$ is the set of all $f\in A$ such that some power of $f$ lies in $a\text{;}$ it is also the intersection of all the prime ideals of $A$ which contain $a\text{.}$ In particular, the radical $r\left(0\right)$ of the zero ideal is the set $N$ of all nilpotent elements of $A\text{;}$ this ideal is called the nilradical of $A\text{.}$
If $E$ is a subset of $A$ and if $a$ is the ideal generated by $E,$ then $V\left(E\right)=V\left(a\right)=V\left(r\left(a\right)\right)\text{.}$
We need some more notation:
${A}_{x}={A}_{{j}_{x}}=$ local ring of $A$ with respect to the prime ideal ${j}_{x}\text{;}$
${m}_{x}={j}_{x}{A}_{x}=$ maximal ideal of ${A}_{x}\text{:}$
$k\left(x\right)={A}_{x}/{m}_{x}=$ residue field of ${A}_{x}\cong $ field of fractions of $A/{j}_{x}\text{.}$
If $f\in A,$ $f\left(x\right)$ denotes the class of $f$ mod. ${j}_{x}$ in $A/{j}_{x}\subseteq k\left(x\right)\text{.}$ Thus $f\left(x\right)=0$ if and only if $f\in {j}_{x}\text{.}$
$D\left(f\right)=X-V\left(f\right)=\{x\in X:f\left(x\right)\ne 0\}=$ 'support' of $f\in A\text{;}$ it is an open set.
Finally, if $Y\subseteq X,$ $j\left(Y\right)$ denotes $\underset{y\in Y}{\cap}{j}_{y}\text{.}$ Thus $j\left(\left\{x\right\}\right)={j}_{x}\text{.}$ Then we have the following formulas:
Lemma (3.2).
(i) | $j\left(\varnothing \right)=A,$ $j\left(X\right)=N$ (the nilradical of $A\text{).}$ |
(ii) | If $Y\subseteq Y\prime ,$ then $j\left(Y\right)\supseteq j\left(Y\prime \right)\text{.}$ |
(iii) | $j\left(\underset{\lambda}{\cup}{Y}_{\lambda}\right)=\underset{\lambda}{\cap}j\left({Y}_{\lambda}\right)\text{.}$ |
(iv) | $j\left(V\left(E\right)\right)=$ radical of the ideal generated by $E\text{.}$ |
(v) | $V\left(j\left(Y\right)\right)=\stackrel{\u203e}{Y}\text{.}$ |
It follows from (iv) and (v) that $a\mapsto V\left(a\right),$ $Y\mapsto j\left(Y\right)$ gives an order-reversing one-one correspondence between closed subsets of $X$ and ideals $a$ in $A$ such that $a=r\left(a\right)\text{.}$ Hence, if the ring $A$ is Noetherian, $X=\text{Spec}\left(A\right)$ is a Noetherian space. (The converse of this is false: $X$ can be Noetherian and $A$ not Noetherian. For example, let $B$ be a polynomial ring $k[{x}_{1},{x}_{2},\dots ]$ over a field in a countable infinity of indeterminuates, let $b$ be the ideal generated by ${x}_{1},{x}_{2}^{2},\dots ,{x}_{n}^{n},\dots ,$ and let $A=B/b\text{.}$ Then $A$ is not Noetherian but has exactly one prime ideal.)
If $x,y\in X$ then $y\in \left\{\stackrel{\u203e}{x}\right\}$ (i.e., $y$ is a specialization of $x\text{)}$ if and only if ${j}_{x}\subseteq {j}_{y}\text{.}$ Hence $\left\{x\right\}$ is a closed set (by abuse of language, $x$ is a closed point of $X\text{)}$ if and only if ${j}_{x}$ is a maximal ideal of $A\text{.}$ Thus $X$ is a ${T}_{1}$ space (every point is closed) if and only if every ideal of $A$ is maximal, i.e., $\text{dim}\hspace{0.17em}A=0\text{.}$ However, $X$ is always a ${T}_{0}\text{-space}$ (this means that, given any two distinct points $x,y$ in $X,$ then either there is a neighbourhood of $y$ which does not contain $x,$ or else a neighbourhood of $x$ which does not contain $y\text{).}$
Next, let us look at the open sets $D\left(f\right),$ $f\in A\text{.}$ First, from (3.1) (iv) we have
$$D\left(fg\right)=D\left(f\right)\cap D\left(g\right)\phantom{\rule{1em}{0ex}}(f,g\in A)\text{.}$$Proposition (3.3).
(i) | The open sets $D\left(f\right)$ form a base of open sets for the topology of $X\text{.}$ |
(ii) | Each $D\left(f\right)$ is quasi-compact. In particular $X=D\left(1\right)$ is quasi-compact. |
Proof. | |
(i) If $U$ is an open set in $X,$ then $U=X-V\left(E\right)$ for some $E\subseteq A\text{;}$ we have $V\left(E\right)=\underset{f\in E}{\cap}{\mathrm{???}}y$ (3.1) (iii), hence $U=\underset{f\in E}{\cup}D\left(f\right)\text{.}$ (ii) By virtue of (i) it is enough to show that every covering of a set $D\left(f\right)$ by open sets $D\left({f}_{\lambda}\right)$ has a finite subcovering. Suppose then that $D\left(f\right)\subseteq \underset{\lambda \in L}{\cup}D\left({f}_{\lambda}\right)\text{;}$ let $a$ be the ideal of $A$ generated by the ${f}_{\lambda},$ then $V\left(f\right)\supseteq \cap V\left({f}_{\lambda}\right)=V\left(a\right),$ hence $V\left(r\left(f\right)\right)\supseteq V\left(r\left(a\right)\right)$ and therefore $r\left(f\right)\subseteq r\left(a\right),$ so that $f\in r\left(a\right)$ and therefore ${f}^{n}\in a$ for some $n>0\text{.}$ Say ${f}^{n}=\sum _{\lambda \in J}{a}_{\lambda}{f}_{\lambda},$ where $J$ is some finite subset of $L\text{.}$ Then ${f}^{n}\in b,$ where $b$ is the ideal generated by the ${f}_{\lambda},$ $\lambda \in J\text{;}$ hence $V\left(f\right)=V\left({f}^{n}\right)\supseteq V\left(b\right)=\underset{\lambda \in J}{\cap}V\left({f}_{\lambda}\right)\text{.}$ Taking complements, we have $D\left(f\right)\subseteq \underset{\lambda \in J}{\cup}D\left({f}_{\lambda}\right),$ as required. $\square $ |
The open sets $D\left(f\right)$ $(f\in A)$ will be called basic open sets.
Let $a$ be an ideal of $A\text{.}$ Then the ideals of $A/a$ correspond one-to-one to the ideals of $A$ which contain $a,$ and therefore $\text{Spec}(A/a)$ is canonically homeomorphic to the closed subspace $V\left(a\right)$ of $\text{Spec}\left(A\right)\text{.}$ In particular, $\text{Spec}\left(A\right)$ and $\text{Spec}(A/N)$ are canonically homeomorphic $\text{(}N=$ nilradical of $A\text{).}$
Proposition (3.4). $X=\text{Spec}\left(A\right)$ is irreducible $\iff A/N$ is an integral domain.
Proof. | |
From what has just been said, we may as well take $N=0\text{.}$ Suppose $X$ is reducible; then there exist proper closed subsets ${Y}_{1},{Y}_{2}$ in $X$ such that ${Y}_{1}\cup {Y}_{2}=X,$ and therefore $j\left({Y}_{1}\right)\cap j\left({Y}_{2}\right)=j\left(X\right)=N=0$ (by (3.2)). But $j\left({Y}_{1}\right)$ and $j\left({Y}_{2}\right)$ are $\ne 0,$ hence there exist ${f}_{i}\in j\left({Y}_{i}\right)$ such that ${f}_{i}\ne 0,$ and ${f}_{1}{f}_{2}\in j\left({Y}_{1}\right)\cap j\left({Y}_{2}\right)=0\text{.}$ Hence $A$ is not an integral domain. Conversely, if $A$ is not an integral domain we have $f,g$ in $A$ such that $f\ne 0,$ $g\ne 0$ and $fg=0\text{.}$ Hence $V\left(f\right)\ne X,$ $V\left(g\right)\ne X$ (since $N=0\text{);}$ but $X=V\left(fg\right)=V\left(f\right)\cup V\left(g\right)\text{.}$ Consequently $X$ is reducible. $\square $ |
In the correspondence between closed subsets of $X$ and ideals of $A$ which are equal to their radicals, the irreducible closed subsets correspond to the prime ideals. In particular the irreducible components of $X$ correspond to the minimal prime ideals of $A\text{.}$ Furthermore, $x\mapsto \left\{\stackrel{\u203e}{x}\right\}$ gives a one-to-one correspondence between the points of $X$ and the irreducible closed subsets of $X,$ i.e., every irreducible closed subset of $X$ has exactly one generic point. For if $x\in X,$ then $\left\{x\right\}$ is irreducible by (2.1) (iv). If $\left\{\stackrel{\u203e}{x}\right\}=\left\{\stackrel{\u203e}{y}\right\},$ then each of $x$ and $y$ is a specialization of the other, so that ${j}_{x}={j}_{y},$ i.e. $x=y\text{.}$ Conversely, if $Y$ is an irreducible subset of $X,Y$ corresponds to a prime ideal ${j}_{x}$ of $X,$ i.e. $Y=V\left({j}_{x}\right)=\left\{\stackrel{\u203e}{x}\right\}\text{.}$
Let $k$ be a field, $K$ an algebraically closed extension of $k,$ and let $V$ be a $(k,K)\text{-affine}$ variety as in Chapter 1; let $A$ be the coordinate ring of $V$ (a $k\text{-algebra,}$ finitely generated with no nilpotent elements), and let $X=\text{Spec}\left(A\right)\text{.}$ What is the relationship between $V$ and $X\text{?}$ Let us assume that $K$ is a universal domain in the sense of Weil, i.e. that $K$ has infinite transcendence degree over $k\text{;}$ this is just to give us plenty of elbow room. Let $x\in V,$ then $x$ determines a homomorphism $A\mapsto K,$ whose kernel is a prime ideal of $A,$ i.e. an element $x\prime $ of $X\text{.}$ Conversely, if $p$ is any prime ideal of $A,$ we can embed $A/p$ in $K$ (for the field of fractions of $A/p$ is a finitely generated field extension of $k,$ hence is an algebraic extension of a pure transcendental extension of $k\text{)}$ and thus we have a homomorphism $A\mapsto K$ with kernel $p\text{.}$ Hence $x\mapsto x\prime $ is a map of $V$ onto $X,$ and $X$ is obtained from $V$ by identifying 'equivalent' points in $V,$ i.e. points which are generic specializations of each other.
At the other extreme, if $k=K,$ then $V$ may be identified with the set of maximal ideals of $A,$ i.e. with the set of closed points of $X\text{:}$ so in this case the map $V\to X$ described above is injective (and not in general surjective).
Let $A,A\prime $ be two rings and let $\phi :A\prime \to A$ be a ring homomorphism (which is always assumed to map identity element to identity element). If $x\in X=\text{Spec}\left(A\right),$ then ${\phi}^{-1}\left({j}_{x}\right)$ is a prime ideal in $A\prime ,$ hence a point of $X\prime =\text{Spec}\left(A\prime \right)\text{.}$ Thus we have a mapping
$$\text{Spec}\left(\phi \right)={}^{a}\phi :X\u27f6X\prime ,$$said to be associated with $\phi \text{.}$ Let ${\phi}^{x}$ denote the embedding of $A\prime /{\phi}^{-1}\left({j}_{x}\right)$ in $A/{j}_{x}$ induced by $\phi \text{;}$ then ${\phi}^{x}$ extends to a field monomorphism
$${\phi}^{x}:k\left({}^{a}\phi \left(x\right)\right)\u27f6k\left(x\right)\text{.}$$Lemma (3.5).
(i) | ${}^{a}{\phi}^{-1}\left(V\left(E\prime \right)\right)=V\left(\phi \left(E\prime \right)\right),$ for any subset $E\prime $ of $A\prime \text{.}$ In particular: |
(ii) | ${}^{a}{\phi}^{-1}\left(D\left(f\prime \right)\right)=D\left(\phi \left(f\prime \right)\right)\phantom{\rule{1em}{0ex}}(f\prime \in A\prime )\text{.}$ |
(iii) | ${}^{a}\phi \left(V\left(a\right)\right)=V\left({\phi}^{-1}\left(a\right)\right)\phantom{\rule{1em}{0ex}}\left(a\hspace{0.17em}\text{any ideal of}\hspace{0.17em}A\right)\text{.}$ |
Proof. | |
(i) is straightforward and (ii) follows from (i). To prove (iii) we may assume that $a=r\left(a\right),$ since $V\left(r\left(a\right)\right)=V\left(a\right)$ and $r\left({\phi}^{-1}\left(a\right)\right)={\phi}^{-1}\left(r\left(a\right)\right)\text{.}$ Put $Y=V\left(a\right),$ and let $a\prime =j\left({}^{a}\phi \left(Y\right)\right)\text{;}$ then $V\left(a\prime \right)=\stackrel{\u203e}{{}^{a}\phi \left(Y\right)}$ by (3.2) (v). Also: $$\begin{array}{ccc}f\prime \in a\prime & \u27fa& f\prime \left(x\prime \right)=0\hspace{0.17em}\text{for all}\hspace{0.17em}x\prime \in {}^{a}\phi \left(Y\right)\\ & \u27fa& f\prime \in {\phi}^{-1}\left({j}_{x}\right)\hspace{0.17em}\text{for all}\hspace{0.17em}x\in Y\\ & \u27fa& \phi \left(f\prime \right)\in j\left(Y\right)=j\left(V\left(a\right)\right)=a\\ & \u27fa& f\prime \in {\phi}^{-1}\left(a\right)\text{.}\end{array}$$Hence ${}^{a}\phi \left(V\left(a\right)\right)={}^{a}\phi \left(Y\right)=V\left(a\prime \right)=V\left({\phi}^{-1}\left(a\right)\right)\text{.}$ $\square $ |
From (i) or (ii) above it follows that ${}^{a}\phi $ is continuous. Clearly, if ${A}^{\prime \prime}$ is another ring, $\phi \prime :{A}^{\prime \prime}\to A\prime $ another ring homomorphism, then $a^(\phi \circ \phi \prime )=a^\phi \prime \circ {}^{a}\phi \text{;}$ so that Spec is a contravariant functor from the, category of rings and ring homomorphisms to the category of topological spaces and continuous maps.
Examples.
(1) If $a$ is an ideal in $A$ and $\phi :A\to A/a$ the projection, then ${}^{a}\phi :\text{Spec}(A/a)\to \text{Spec}\left(A\right)$ is a homeomorphism of $\text{Spec}(A/a)$ onto $V\left(a\right)\text{.}$
(2) Let $S$ be a multiplicatively closed subset of $A$ (i.e. $S$ is closed under finite products, so that in particular $1\in S$ (take the empty product!)). Then we can form the ring of fractions ${S}^{-1}A,$ and we have a canonical mapping $\phi :A\to {S}^{-1}A,$ hence ${}^{a}\phi :\text{Spec}\left({S}^{-1}A\right)\to \text{Spec}\left(A\right)\text{.}$ It is a well-known and not difficult fact of commutative algebra that the prime ideals of ${S}^{-1}A$ are in one-one correspondence (under ${}^{a}\phi \text{)}$ with the prime ideals of $A$ which don't meet $S,$ and consequently ${}^{a}\phi $ is a homeomorphism of $\text{Spec}\left({S}^{-1}A\right)$ onto the set of all $x\in X$ such that ${j}_{x}\cap S=\varnothing \text{.}$ (In general this subset of $X$ is neither open nor closed, nor even locally closed.)
(3) In particular, $\text{Spec}\left({A}_{x}\right)$ may be canonically identified with the subspace of $X$ consisting of all generizations of $x,$ i.e. all $y$ such that $x\in \left\{\stackrel{\u203e}{y}\right\}\text{.}$
(4) As another example, let $f\in A$ and let $S$ be the set of all ${f}^{n}$ $(n\ge 0)\text{.}$ In this case ${S}^{-1}A$ is usually denoted by ${A}_{f}\text{.}$ Then $\text{Spec}\left({A}_{f}\right)$ is identified with the set of all $x\in X$ such that ${j}_{x}$ contains no power of $f,$ i.e. such that $f\notin {j}_{x}\text{.}$ Hence
Proposition (3.6). If $\phi :A\to {A}_{f}$ is the canonical homomorphism $(f\in A),$ then ${}^{a}\phi $ is a homeomorphism of $\text{Spec}\left({A}_{f}\right)$ onto the open set $D\left(f\right)\text{.}$
(5) The 'characteristic morphism'. Since $A$ has an identity element, there is a canonical mapping $\phi :\mathbb{Z}\to A,$ where $\mathbb{Z}$ is the ring of integers; hence ${}^{a}\phi :X\to \text{Spec}\left(\mathbb{Z}\right)\text{.}$ Now the points of $\text{Spec}\left(\mathbb{Z}\right)$ are $\left(0\right)$ and the prime ideals $\left(p\right)$ $\text{(}p$ a positive prime number), and ${}^{a}\phi \left(x\right)$ is just the ideal generated by the characteristic of the residue field $k\left(x\right)$ of $x\text{.}$
Proposition (3.7). Let $\phi :A\prime \to A$ be a ring homomorphism, ${}^{a}\phi :X\to X\prime $ the associated map.
(i) | If $\phi $ is surjective, ${}^{a}\phi $ is a closed embedding (i.e. a homeomorphism of $X$ onto a closed subset of $X\prime \text{).}$ |
(ii) | If $\phi $ is injective, ${}^{a}\phi $ is dominant (i.e. ${}^{a}\phi \left(X\right)$ is dense in $X\prime \text{).}$ |
Proof. | |
(i) is just Example 1 above. (ii) follows from (3.5) (iii): $\stackrel{\u203e}{{}^{a}\phi \left(X\right)}=\stackrel{\u203e}{{}^{a}\phi \left(V\left(0\right)\right)}=V\left({\phi}^{-1}\left(0\right)\right)=V\left(0\right)$ (since $\phi $ is injective) $=X\prime \text{.}$ $\square $ |
This is a typed excerpt of the book "Algebraic Geometry: Introduction to Schemes - I.G. Macdonald".