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Zariski topology

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topology (point-set topology, point-free topology)

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Idea

The Zariski topology is a topology on the prime spectrum of a commutative ring. It serves as the basis for much of algebraic geometry.

We consider the definition in increasing generality and sophistication:

  1. First we discuss the naive Zariski topology on affine spaces k nk^n, consider the classical proofs and discover thereby the special role of prime ideals and maximal ideals;

  2. then we turn to the modern definition of the Zariski topology on affine varieties Spec(R)Spec(R) which takes the concept of prime (and maximal) ideals as primary, and again we provide the classical arguments;

  3. finally we discuss the abstract category theoretic perspective on these matters in terms of Galois connections and obtain slick category theoretic proofs of all the previous statements.

Starting with affine space k nk^n, then the idea of the Zariski topology is to take as the closed subsets those defined by the vanishing of any set of polynomials over kk in nn variables, hence the solution sets to equations of the form

iI(f i(x 1,,x n)=0) \underset{i \in I}{\forall} \left( f_i(x_1, \cdots, x_n) = 0 \right)

for f ik[X 1,,X n]f_i \in k[X_1, \cdots, X_n] polynomials. The open subsets of the topology are the complements of these vanishing sets.

It is clear that the vanishing set such a set of polynomials depends only on the ideal in the polynomial ring which is generated by them. Under this translation then forming the intersection of closed subsets corresponds to forming the sum of these ideals, and forming the union of closed subsets corresponds to forming the product of the corresponding ideals. This way the Zariski topology establishes a dictionary between topological concepts of the affine space k nk^n, and algebra inside the polynomial ring.

In particular one finds that the irreducible closed subsets of the Zariski topology correspond to the prime ideals in the polynomial ring (prop. 3 and prop. 4 below), and that the closed points correspond to the maximal ideals among these (prop. 6).

This motivates the modern refinement of the concept of the Zariski topology, where one considers any commutative ring RR and equips its set of prime ideals with a topology, by direct analogy with the previously naive affine space k nk^n, which is recovered with RR a polynomial ring and restricting attention to the maximal ideals (example 2 below).

These sets of prime ideals of a ring RR equipped with the Zariski topology are called the (topological spaces underlying) the prime spectrum of a commutative ring, denoted Spec(R)Spec(R).

The Zariski topology is in general not Hausdorff (example 1 below) which makes it sometimes be regarded as an “exotic” type of topology. But it is in fact sober (prop. 8 below) and hence as well-behaved in this respect as general locales are.

On affine space

We consider here, for kk a field, the vector space k nk^n equipped with a Zariski topology. This is the original definition of Zariski topology, and serves well to motivate the concept, but eventually it was superceded by a more refined concept of Zariski topologies of prime spectra, discussed in the next subsection below. In example 2 below we reconsider the naive case of interest in this subsection here from that more refined perspective.

Definition

Definition

(Zariski topology on affine space)

Let kk be a field, let nn \in \mathbb{N}, and write k[X 1,,X n]k[X_1, \cdots, X_n] for the set of polynomials in nn variables over kk.

For k[X 1,,X n]\mathcal{F} \subset k[X_1, \cdots, X_n] a subset of polynomials, let the subset V()k nV(\mathcal{F}) \subset k^n of the nn-fold Cartesian product of the underlying set of kk (the vanishing set of \mathcal{F}) be the subset of points on which all these polynomials jointly vanish:

V(){(a 1,,a n)k n|ff(a 1,,a n)=0}. V(\mathcal{F}) \coloneqq \left\{ (a_1, \cdots, a_n) \in k^n \,\vert\, \underset{f \in \mathcal{F}}{\forall} f(a_1, \cdots, a_n) = 0 \right\} \,.

These subsets are called the Zariski closed subsets.

Write

τ 𝔸 k n{k n\V()k n|k[X 1,,X n]} \tau_{\mathbb{A}^n_k} \;\coloneqq\; \left\{ k^n \backslash V(\mathcal{F}) \subset k^n \,\vert\, \mathcal{F} \subset k[X_1, \cdots, X_n] \right\}

for the set of complements of the Zariski closed subsets. These are called the Zariski open subsets of k nk^n.

Proposition

(Zariski topology is well defined)

Assuming excluded middle, then:

For kk a field and nn \in \mathbb{N}, then the Zariski open subsets of k nk^n (def. 1) form a topology. The resulting topological space

𝔸 k n(k n,τ 𝔸 k n) \mathbb{A}^n_k \;\coloneqq\; \left( k^n, \tau_{\mathbb{A}^n_k} \right)

is also called the nn-dimensional affine space over kk.

Proof

We need to show for { ik[X 1,,X n]} iI\{\mathcal{F}_i \subset k[X_1, \cdots, X_n]\}_{i \in I} a set of subsets of polynomials that

  1. iI(k n\V( i))=k n\V( )\underset{i \in I}{\cup} \left(k^n \backslash V(\mathcal{F}_i)\right) = k^n \backslash V(\mathcal{F}_\cup) for some k[X 1,,X n]\mathcal{F}_\cup \subset k[X_1, \cdots, X_n];

  2. if II is finite then iI(k n\V( ))=k n\ \underset{i \in I}{\cap} \left( k^n \backslash V(\mathcal{F}_{\cap})\right) = k^n \backslash \mathcal{F}_\cap for some k[X 1,,X n]\mathcal{F}_{\cap} \subset k[X_1, \cdots, X_n].

By de Morgan's law for complements (and using excluded middle) this is equivalent to

  1. iIV( i)=V( )\underset{i \in I}{\cap} V(\mathcal{F}_i) = V(\mathcal{F}_\cup) for some k[X 1,,X n]\mathcal{F}_\cup \subset k[X_1, \cdots, X_n];

  2. if II is finite then iIV( i)=V( )\underset{i \in I}{\cup} V(\mathcal{F}_i) = V(\mathcal{F}_{\cap}) for some k[X 1,,X n]\mathcal{F}_{\cap} \subset k[X_1, \cdots, X_n].

We claim that we may take

  1. =iI i\mathcal{F}_\cup = \underset{i \in I}{\cup} \mathcal{F}_i

  2. =iI i{iIf i|f i i}\mathcal{F}_{\cap} = \underset{i \in I}{\prod} \mathcal{F}_i \coloneqq \left\{ \underset{i \in I}{\prod} f_i \,\vert\, f_i \in \mathcal{F}_i \right\}.

(In the second line we have the set of all those polynomials which arise as products of polynomials with one factor from each of the i\mathcal{F}_i.)

Regarding the first point:

(a 1,,a n)iIV( i) iI((a 1,,a n)V( i)) iI(f i(f(a 1,,a n)=0)) fiI i(f(a 1,,a n)=0) (a 1,,a n)V(iI i) \begin{aligned} & (a_1, \cdots, a_n) \in \underset{i \in I}{\cap} V(\mathcal{F}_i) \\ \Leftrightarrow\; & \underset{i \in I}{\forall} \left( (a_1, \cdots, a_n) \in V(\mathcal{F}_i) \right) \\ \Leftrightarrow\; & \underset{i \in I}{\forall} \left( \underset{f \in \mathcal{F}_i}{\forall} \left( f(a_1, \cdots, a_n) = 0 \right) \right) \\ \Leftrightarrow\; & \underset{f \in \underset{i \in I}{\cup} \mathcal{F}_i}{\forall} \left( f(a_1, \cdots, a_n) = 0 \right) \\ \Leftrightarrow\; & (a_1, \cdots, a_n) \in V\left( \underset{i \in I}{\cup} \mathcal{F}_i \right) \end{aligned}

Regarding the second point, in one direction we have the immediate implication

(a 1,,a n)iIV( i) iI(f i(f(a 1,,a n)=0)) (f i)iI i(iIf i(a 1,,a n)=0) (a 1,,a n)V(iI i). \begin{aligned} & (a_1, \cdots, a_n) \in \underset{i \in I}{\cup} V(\mathcal{F}_i) \\ \Leftrightarrow\; & \underset{i \in I}{\exists} \left( \underset{f \in \mathcal{F}_i}{\forall} \left( f(a_1, \cdots, a_n) = 0 \right) \right) \\ \Rightarrow \; & \underset{(f_i) \in \underset{i \in I}{\prod} \mathcal{F}_i}{\forall} \left( \underset{i \in I}{\prod} f_i(a_1, \cdots, a_n) = 0 \right) \\ \Leftrightarrow\; & (a_1, \cdots, a_n) \in V\left( \underset{i \in I}{\prod} \mathcal{F}_i \right) \,. \end{aligned}

For the converse direction we need to show that

((a 1,,a n)V(iI i))((a 1,,a n)iIV( 1)). \left( (a_1 , \cdots , a_n) \in V\left( \underset{i \in I}{\prod} \mathcal{F}_i \right) \right) \;\Rightarrow\; \left( (a_1, \cdots, a_n) \in \underset{i \in I}{\cup} V(\mathcal{F}_1) \right) \,.

hence that

((f i)iI i(iIf i(a 1,,a n)=0))(iI(f i i(f i(a 1,,a n)=0))). \left( \underset{(f_i) \in \underset{i \in I}{\prod} \mathcal{F}_i }{\forall} \left( \underset{i \in I}{\prod} f_i(a_1, \cdots, a_n) = 0 \right) \right) \;\Rightarrow\; \left( \underset{i \in I}{\exists} \left( \underset{f_i \in \mathcal{F}_i}{\forall} \left( f_i(a_1, \cdots, a_n) = 0 \right) \right) \right) \,.

By excluded middle, this is equivalent to its contraposition, which by de Morgan's law is

(iI(f i i(f i(a 1,,a n)0)))((f i)iI i(iIf i(a 1,,a n)0)). \left( \underset{i \in I}{\forall} \left( \underset{f_i \in \mathcal{F}_i}{\exists} \left( f_i(a_1, \cdots, a_n) \neq 0 \right) \right) \right) \;\Rightarrow\; \left( \underset{(f_i) \in \underset{i \in I}{\prod} \mathcal{F}_i }{\exists} \left( \underset{i \in I}{\prod} f_i(a_1, \cdots, a_n) \neq 0 \right) \right) \,.

This now is true by the assumption that kk is a field: If all factors f i(a 1,a n)kf_i(a_1, \dots a_n) \in k are non-zero, then their product iIf i(a 1,,a n)k\underset{i \in I}{\prod} f_i(a_1, \cdots, a_n) \in k is non-zero.

Properties

Topological closures

Proposition

For kk a field and nn \in \mathbb{N}, consider a subset

Sk n S \subset k^n

of the underlying set of the nn-fold Cartesian product of kk with itself. Then the topological closure Cl(S)Cl(S) of this subset with respect to the Zariski topology τ 𝔸 k n\tau_{\mathbb{A}^n_k} (def. 1) is the vanishing set of all those polynomials that vanish on SS:

Cl(S)={fk[X 1,,X n]|(a 1,,a n)Sf(a 1,,a n)=0}. Cl(S) \;=\; \left\{ f \in k[X_1, \cdots, X_n] \,\vert\, \underset{(a_1, \cdots, a_n) \in S}{\forall} f(a_1, \cdots, a_n) = 0 \right\} \,.
Proof

We compute as follows:

Cl(S) Ck nclosedCSC =k[X 1,,X n]SV()V() =V(k[X 1,,X n]SV()) =V({fk[X 1,,X n]|(a 1,,a n)Sf(a 1,,a n)=0}). \begin{aligned} Cl(S) & \coloneqq \underset{ {C \subset k^n \, \text{closed}} \atop {C \supset S} }{\cap} C \\ & = \underset{ { \mathcal{F} \subset k[X_1, \cdots, X_n] } \atop { S \subset V(\mathcal{F}) } }{\cap} V(\mathcal{F}) \\ & = V\left( \underset{ { \mathcal{F} \subset k[X_1, \cdots, X_n] } \atop { S \subset V(\mathcal{F}) } }{\cup} \mathcal{F} \right) \\ & = V \left( \left\{ f \in k[X_1, \cdots, X_n] \,\vert\, \underset{(a_1, \cdots, a_n) \in S}{\forall} f(a_1, \cdots, a_n) = 0 \right\} \right) \,. \end{aligned}

Here the first equality is the definition of topological closure, the second is the definition of closed subsets in the Zariski topology (def. 1), the third is the expression of intersections of these in terms of unions of polynomials as in the proof of prop. 1, and then the last one is immediate.

Irreducible closed subsets as prime ideals

In every topological space the irreducible closed subsets play a special role, as being precisely the points in the space as seen in its incarnation as a locale (this prop.). The following shows that in the Zariski topology the irreducible closed subsets all come from prime ideals in the corresponding polynomial ring, and that when the ground field is algebraically closed, then they are in fact in bijection to the prime ideals. See also at schemes are sober.

Definition

(vanishing ideal of Zariski closed subset)#

Let kk be a field, and let nn \in \mathbb{N}. Then for V()k nV(\mathcal{F}) \subset k^n a Zariski closed subset, according to def. 1, hence for k[X 1,,X n]\mathcal{F} \subset k[X_1, \cdots, X_n] a set of polynomials, write

I(V())k[X 1,,X n] I(V(\mathcal{F})) \subset k[X_1, \cdots, X_n]

for the maximal subset of polynomials that still has the same joint vanishing set:

I(V()){fk[X 1,,X n]|(a 1,,a n)V()f(a 1,,a n)=0}. I(V(\mathcal{F})) \;\coloneqq\; \left\{ f \in k[X_1, \cdots, X_n] \,\vert\, \underset{(a_1, \cdots, a_n) \in V(\mathcal{F})}{\forall} f(a_1, \cdots, a_n) = 0 \right\} \,.

This set is clearly an ideal in the polynomial ring k[X 1,,X n]k[X_1, \cdots, X_n], called the vanishing ideal of V()V(\mathcal{F}).

Proposition

With excluded middle then:

Let kk be a field, let nn \in \mathbb{N}, and let V()k nV(\mathcal{F}) \subset k^n be a Zariski closed subset (def. 1). Then the following are equivalent:

  1. V()V(\mathcal{F}) is an irreducible closed subset;

  2. The vanishing ideal I(V())I(V(\mathcal{F})) (def. 2) is a prime ideal.

Proof

In one direction, assume that V()V(\mathcal{F}) is irreducible and consider f,gk[X 1,,X n]f,g \in k[X_1, \cdots, X_n] with fgI(V())f \cdot g \in I(V(\mathcal{F})). We need to show that then already fI(V())f \in I(V(\mathcal{F})) or gI(V())g \in I(V(\mathcal{F})).

Now since kk is a field, we have

(f(a 1,a n)g(a 1,,a n)=0)((f(a 1,,a n)=0org(a 1,,a n)=0)). \left( f(a_1, \cdots a_n) \cdot g(a_1, \cdots, a_n) = 0 \right) \Rightarrow \left( \left( f(a_1, \cdots, a_n) = 0 \,\text{or}\, g(a_1, \cdots, a_n) = 0 \right) \right) \,.

This implies that

V()V({f})V({g}) V(\mathcal{F}) \subset V(\{f\}) \cup V(\{g\})

and hence that

V()=(V()F({f}))(V()F({g})). V(\mathcal{F}) = (V(\mathcal{F}) \cap F(\{f\})) \,\,\cup\,\, (V(\mathcal{F}) \cap F(\{g\}) ) \,.

But since V({f})V(\{f\}), V({g})V(\{g\}) and V()V(\mathcal{F}) are all closed, by construction, their intersections are closed and hence this is a decomposition of V()V(\mathcal{F}) as a union of closed subsets. Therefore now the assumption that V()V(\mathcal{F}) is irreducible implies that

(V()=V()V({f}))or(V()=V()V({g})) ((V()V({f}))or(V()V({g}))) ((fI(X))or(gI(X))). \begin{aligned} & \left( \, V(\mathcal{F}) = V(\mathcal{F}) \cap V(\{f\}) \, \right) \,\text{or}\, \left( \, V(\mathcal{F}) = V(\mathcal{F}) \cap V(\{g\}) \, \right) \\ \Leftrightarrow \; & \left( \left( \, V(\mathcal{F}) \subset V(\{f\}) \, \right) \,\text{or}\, \left( \, V(\mathcal{F}) \subset V(\{g\}) \, \right) \right) \\ \Leftrightarrow \, & \left( \left( \, f \in I(X) \, \right) \,\text{or}\, \left( \, g \in I(X) \, \right) \right) \end{aligned} \,.

Now for the converse, assume that I(V())I(V(\mathcal{F})) is a prime ideal, and that V()=V( 1)V( 2)V(\mathcal{F}) = V(\mathcal{F}_1) \cup V(\mathcal{F}_2). We need to show that V()=V( 1)V(\mathcal{F}) = V(\mathcal{F}_1) or that V()=V( 2)V(\mathcal{F}) = V(\mathcal{F}_2).

Assume on the contrary, that there existed elements

(a 1,,a n)V( 1)\V( 2)and(b 1,,b n)V( 2)\V( 1) (a_1, \cdots, a_n) \in V(\mathcal{F}_1) \backslash V(\mathcal{F}_2) \;\text{and}\; (b_1, \cdots, b_n) \in V(\mathcal{F}_2) \backslash V(\mathcal{F}_1) \,

Then in particular the vanishing ideals would not contain each other

¬(I(V( 1))I(V( 2)))and¬(I(V( 2))I(V( 1))) \not\left( I(V(\mathcal{F}_1)) \subset I(V(\mathcal{F}_2)) \right) \,\,\,\text{and}\,\,\, \not\left( I(V(\mathcal{F}_2)) \subset I(V(\mathcal{F}_1)) \right)

and hence there were polynomials

fI(V( 1))\I(V( 2))andgI(V( 2))\I(V( 1)). f\in I(V(\mathcal{F}_1)) \backslash I(V(\mathcal{F}_2)) \,\,\,\text{and}\,\,\, g \in I(V(\mathcal{F}_2)) \backslash I(V(\mathcal{F}_1)) \,.

But since a product of polynomials vanishes at some point once one of the factors vanishes at that point, it would follows that

fgI(V( 1))I(V( 2))=I(V()), f \cdot g \in I(V(\mathcal{F}_1)) \cap I(V(\mathcal{F}_2)) = I(V(\mathcal{F})) \,,

which were in contradiction to the assumption that I(V())I(V(\mathcal{F})) is a prime ideal. Hence we have a proof by contradiction.

Proposition 3 gives an injection

{irreducible Zariski closed V()k n}AAA{prime ideals Ik[X 1,,X n]}. \left\{ \array{ \text{irreducible Zariski closed} \\ V(\mathcal{F}) \subset k^n } \right\} \overset{\phantom{AAA}}{\hookrightarrow} \left\{ \array{ \text{prime ideals} \\ I \in k[X_1, \cdots, X_n] } \right\} \,.

The following says that for algebraically closed fields then this is in fact a bijection:

Proposition

Let k=k¯k = \overline{k} be an algebraically closed field and let nn \in \mathbb{N}. Then the function

IrrClSub(𝔸 k n) PrimeIdl(k[X 1,,X n]) V() AAA I(V()) \array{ IrrClSub(\mathbb{A}^n_k) &\overset{}{\longrightarrow}& PrimeIdl(k[X_1, \cdots, X_n]) \\ V(\mathcal{F}) &\overset{\phantom{AAA}}{\mapsto}& I(V(\mathcal{F})) }

from prop. 3 is a bijection.

The proof uses Hilbert's Nullstellensatz.

Remark

(generalization to affine varieties)

Prop 3 suggests to consider the set of prime ideals of a polynomial ring k[X 1,,X n]k[X_1, \cdots, X_n] for general kk as more fundamental, in some sense, than the set k nk^n. Morover, the set of prime ideals makes sense for every commutative ring RR, not just R=k[X 1,,X n]R = k[X_1, \cdots, X_n], and hence this suggests to consider a Zariski topology on sets of prime ideals. This leads to the more general concept of Zariski topologies for affine varieties, def. 3 below.

Examples

Example

If the field kk is not a finite field, then the Zariski topology on the affine space (def. 1) is not Hausdorff.

This is because the solution set to a system of polynomials over an infinite polynomial is always a finite set. This means that in this case all the Zariski closed subsets V()V(\mathcal{F}) are finite sets. This in turn implies that the intersection of every pair of non-empty Zariski open subsets is non-empty.

But the Zariski topology is always sober, see prop. 8 below.

On affine varieties

Definition

Definition

(Zariski topology on set of prime ideals)

Let RR be a commutative ring. Write PrimeIdl(R)PrimeIdl(R) for its set of prime ideals. For R\mathcal{F} \subset R any subset of elements of the ring, consider the subsets of those prime ideals that contain \mathcal{F}:

V(){pPrimeIdl(R)|p}. V(\mathcal{F}) \;\coloneqq\; \left\{ p \in PrimeIdl(R) \,\vert\, \mathcal{F} \subset p \right\} \,.

These are called the Zariski closed subsets of PrimeIdl(R)PrimeIdl(R). Their complements are called the Zariski open subsets.

Proposition

(Zariski topology well defined)

Assuming excluded middle, then:

Let RR be a commutative ring. Then the collection of Zariski open subsets (def. 3) in its set of prime ideals

τ Spec(R)P(PrimeIdl(R)) \tau_{Spec(R)} \subset P(PrimeIdl(R))

satisfies the axioms of a topology, the Zariski topology.

This topological space

Spec(R)(PrimeIdl(R),τ Spec(R)) Spec(R) \coloneqq (PrimeIdl(R), \tau_{Spec(R)})

is called (the space underlying) the prime spectrum of the commutative ring.

Proof

For R\mathcal{F} \subset R write I()I(\mathcal{F}) \subset \mathcal{F} for the ideal which is generated by \mathcal{F}. Evidently the Zariski closed subsets depend only on this ideal

V(I())=V() V(I(\mathcal{F})) = V(\mathcal{F})

and therefore it is sufficient to consider the V()V(\mathcal{F}) for the case that R\mathcal{F} \subset R is not just a subset, but an ideal.

So let {F iIdl(R)} iI\{F_i \in Idl(R)\}_{i \in I} be a set of ideals in RR and let {V( i)PrimeIdl(R)} iI\{V(\mathcal{F}_i) \subset PrimeIdl(R)\}_{i \in I} be the corresponding set of Zariski closed subsets. We need to show that there exists , R\mathcal{F}_\cup, \mathcal{F}_\cap \subset R such that

  1. iIV( i)=V( )\underset{i \in I}{\cap} V(\mathcal{F}_i) = V(\mathcal{F}_\cup);

  2. if II is finite set then iIV( i)=V( )\underset{i \in I}{\cup} V(\mathcal{F}_i) = V(\mathcal{F}_\cap).

We claim that

  • =iI i{iIf i,R|f i i}\mathcal{F}_{\cup} = \underset{i \in I}{\sum} \mathcal{F}_i \coloneqq \left\{ \underset{i \in I}{\sum} f_i \,, \in R\;\vert\; f_i \in \mathcal{F}_i \right\}

  • =iI i{iIf iR|f i i}\mathcal{F}_{\cap} = \underset{i \in I}{\prod} \mathcal{F}_i \coloneqq \left\{ \underset{i \in I}{\prod} f_i \, \in R \;\vert\; f_i \in \mathcal{F}_i \right\},

Regarding the first point:

By using the various definitions, we get the following chain of logical equivalences:

piIV( i) iI(pV( i)) iI( ip) (iI i)p pV(iI i). \begin{aligned} & p \in \underset{i \in I}{\cap} V(\mathcal{F}_i) \\ \Leftrightarrow\; & \underset{i \in I}{\forall} \left( p \in V(\mathcal{F}_i) \right) \\ \Leftrightarrow\; & \underset{i \in I}{\forall} \left( \mathcal{F}_i \subset p \right) \\ \Leftrightarrow\; & \left(\underset{i \in I}{\sum} \mathcal{F}_i\right) \subset p \\ \Leftrightarrow\; & p \in V\left( \underset{i \in I}{\sum} \mathcal{F}_i \right) \,. \end{aligned}

Regarding the second point, in one direction we have the immediate implication

piIV( i) iI( ip) (f i)iI i(iIf ip) pV(iI i). \begin{aligned} & p \in \underset{i \in I}{\cup} V(\mathcal{F}_i) \\ \Leftrightarrow\; & \underset{i \in I}{\exists} \left( \mathcal{F}_i \subset p \right) \\ \Rightarrow \; & \underset{(f_i) \in \underset{i \in I}{\prod} \mathcal{F}_i}{\forall} \left( \underset{i \in I}{\prod} f_i \in p \right) \\ \Leftrightarrow\; & p \in V\left( \underset{i \in I}{\prod} \mathcal{F}_i \right) \,. \end{aligned}

For the converse direction we need to show that

(pV(iI i))(piIV( 1)). \left( p \in V\left( \underset{i \in I}{\prod} \mathcal{F}_i \right) \right) \;\Rightarrow\; \left( p \in \underset{i \in I}{\cup} V(\mathcal{F}_1) \right) \,.

hence that

((f i)iI i(iIf ip))(iI(f i i(f ip))). \left( \underset{(f_i) \in \underset{i \in I}{\prod} \mathcal{F}_i }{\forall} \left( \underset{i \in I}{\prod} f_i \in p \right) \right) \;\Rightarrow\; \left( \underset{i \in I}{\exists} \left( \underset{f_i \in \mathcal{F}_i}{\forall} \left( f_i \in p \right) \right) \right) \,.

By excluded middle, this is equivalent to its contraposition, which by de Morgan's law is

(iI(f i i¬(f ip)))((f i)iI i¬(iIf ip))is. \left( \underset{i \in I}{\forall} \left( \underset{f_i \in \mathcal{F}_i}{\exists} \not \left( f_i \in p \right) \right) \right) \;\Rightarrow\; \left( \underset{(f_i) \in \underset{i \in I}{\prod} \mathcal{F}_i }{\exists} \not \left( \underset{i \in I}{\prod} f_i \in p \right) \right)is \,.

This holds by the assumption that pp is a prime ideal.

Properties

We discuss some properties of the Zariski topology on prime spectra of commutative rings.

Topological closures

Lemma

(topological closure of points)

Let RR be a commutative ring and consider Spec(R)=(PrimeIdl(R),τ Spec(R))Spec(R) = (PrimeIdl(R), \tau_{Spec(R)}) its prime spectrum equipped with the Zariski topology (def. 3).

Then the topological closure of a point pPrimeIdl(R)p \in PrimeIdl(R) is V(p)PrimeIdl(R)V(p) \subset PrimeIdl(R) (def. 3).

Proof

By definition the topological closure of {p}\{p\} is

Cl({p})IIdl(R)pV(I)V(I). Cl(\{p\}) \underset{ {I \in Idl(R) } \atop { p \in V(I) } }{\cap} V(I) \,.

Hence unwinding the definitions, we have the following sequence of logical equivalences:

qCl({q}) qIIdl(R)pV(I)V(I) IIdl(R)Ip(qV(I)) IIdl(R)Ip(Iq) pq qV(p) \begin{aligned} & q \in Cl(\{q\}) \\ \Leftrightarrow\; & q \in \underset{ {I \in Idl(R)} \atop { p \in V(I) } }{\cap} V(I) \\ \Leftrightarrow\; & \underset{ { I \in Idl(R) } \atop { I \subset p } }{\forall} (q \in V(I)) \\ \Leftrightarrow\; & \underset{ { I \in Idl(R) } \atop { I \subset p } }{\forall} (I \subset q) \\ \Leftrightarrow\; & p \subset q \\ \Leftrightarrow\; & q \in V(p) \end{aligned}

Recall:

Lemma

(prime ideal theorem)

Assuming the axiom of choice or at least the ultrafilter principle then:

For RR a commutative ring and IRI \subset R a proper ideal, then II is contained in some prime ideal.

The axiom of choice even implies that every proper ideal is contained in a maximal ideal (by this prop.).

Proposition

(maximal ideals are closed points)

Let RR be a commutative ring, consider the topological space Spec(R)=(PrimeIdl(R),τ Spec(R))Spec(R) = (PrimeIdl(R),\tau_{Spec(R)}), i.e. its prime spectrum equipped with the Zariski topology from def. 3.

Then the maximal ideals inside the prime ideals constitute closed points.

Assuming the axiom of choice or at least the ultrafilter principle then also the converse is true:

Then the inclusion of maximal ideals 𝔪MaxIdl(R)PrimeIdl(R)\mathfrak{m} \in MaxIdl(R) \subset PrimeIdl(R) into all prime ideals is precisely the inclusion of the subset of closed points into all points of Spec(R)Spec(R).

ClosedPoints(Spec(R))MaxIdl(R)PrimeIdl(R). ClosedPoints(Spec(R)) \simeq MaxIdl(R) \subset PrimeIdl(R) \,.
Proof

By lemma 1 we have

Cl({p})=V(p) Cl(\{p\}) = V(p)

and hence we need to show that

𝔪=V(𝔪) \mathfrak{m} = V(\mathfrak{m})

precisely if 𝔪\mathfrak{m} is maximal.

In one direction, assume that 𝔪\mathfrak{m} is maximal. By definition V(𝔪)V(\mathfrak{m}) contains all the prime ideals pp such that 𝔪p\mathfrak{m} \subset p. That 𝔪\mathfrak{m} is maximal means that it is not contained in a larger proper ideal, in particular not in any larger prime ideal, and hence V(𝔪)={𝔪}V(\mathfrak{m}) = \{\mathfrak{m}\}.

In the other direction, assume that 𝔪\mathfrak{m} is a prime ideal such that V(𝔪)={𝔪}V(\mathfrak{m}) = \{\mathfrak{m}\}. By definition this means equivalently that the only prime ideal pp with 𝔪p\mathfrak{m} \subset p is 𝔪\mathfrak{m} itself. We need to show that more generally 𝔪I\mathfrak{m} \subset I for II any proper ideal implies that 𝔪=I\mathfrak{m} = I.

But the axiom of choice/ultrafilter principle imply the prime ideal theorem (lemma 2), which says that there is a prime ideal pp with IpI \subset p, hence a sequence of inclusions 𝔪Ip\mathfrak{m} \subset I \subset p. This implies 𝔪p\mathfrak{m} \subset p, hence 𝔪=p\mathfrak{m} = p, hence I=𝔪I = \mathfrak{m}.

Irreducible closed subsets as prime ideals

Proposition

(irreducible closed subsets correspond to prime ideals)

With excluded middle then:

Let RR be a commutative ring, and let R\mathcal{F} \subset R be an ideal in RR, hence V()Spec(R)V(\mathcal{F}) \subset Spec(R) be a Zariski closed subset in the prime spectrum of RR. Then the following are equivalent:

  1. V()V(\mathcal{F}) is an irreducible closed subset;

  2. \mathcal{F} is a prime ideal.

Proof

In one direction, assume that V()V(\mathcal{F}) is irreducible, and that f,gRf,g \in R with fgf \cdot g \in \mathcal{F}. We need to show that then already ff \in \mathcal{F} or gg \in \mathcal{F}.

To this end, first observe that

V()V((f))V((g)). V(\mathcal{F}) \subset V((f)) \cup V((g)) \,.

This is because

pV() p fgp (fp)or(gp) (pV(g))or(pV(f)) pV(f)V(g), \begin{aligned} & p \in V(\mathcal{F}) \\ \Leftrightarrow\; & \mathcal{F} \subset p \\ \Rightarrow \; & f \cdot g \in p \\ \Rightarrow\; & \left( f \in p \right) \,\text{or}\, \left( g \in p \right) \\ \Leftrightarrow\; & \left( p \in V(g) \right) \,\text{or}\, \left( p \in V(f) \right) \\ \Leftrightarrow\; & p \in V(f) \cup V(g) \,, \end{aligned}

where the implication in the middle uses that pp is a prime ideal.

It follows that

V()=(V(f)V())(V(g)V()). V(\mathcal{F}) \;=\; \left( V(f) \cap V(\mathcal{F}) \right) \cup \left( V(g) \cap V(\mathcal{F}) \right) \,.

This is a decomposition of V()V(\mathcal{F}) as a union of closed subsets, hence the assumption that V()V(\mathcal{F}) is irreducible implies that

(V()=V(f)V())or(V()=V(g)V()) (V()V(f))or(V()V(g)) (f)or(g). \begin{aligned} & \left( V(\mathcal{F}) = V(f) \cap V(\mathcal{F}) \right) \,\text{or}\, \left( V(\mathcal{F}) = V(g) \cap V(\mathcal{F}) \right) \\ \Leftrightarrow & \left( V(\mathcal{F}) \subset V(f) \right) \,\text{or}\, \left( V(\mathcal{F}) \subset V(g) \right) \\ \Leftrightarrow\, & \left( f \in \mathcal{F} \right) \,\text{or}\, \left( g \in \mathcal{F} \right) \,. \end{aligned}

Now for the converse. Assume that \mathcal{F} is a prime ideal and that V()=V( 1)V( 2)V(\mathcal{F}) = V(\mathcal{F}_1) \cup V(\mathcal{F}_2). Observe (as in the proof of prop. 5) that this means equivalently that = 1 2\mathcal{F} = \mathcal{F}_1 \cdot \mathcal{F}_2. We need to show that then V()=V( 1)V(\mathcal{F}) = V(\mathcal{F}_1) or that V(=V( 2))V(\mathcal{F} = V(\mathcal{F}_2)).

Suppose on the contrary that neither 1\mathcal{F}_1 nor 2\mathcal{F}_2 coincided with \mathcal{F}. This means that there were elements f 1\f \in \mathcal{F}_1 \backslash \mathcal{F} and g 2\g \in \mathcal{F}_2 \backslash \mathcal{F} such that still fgf \cdot g \in \mathcal{F}, in contradiction to the assumption. Hence we have a proof by contradiction.

As a corollary:

Proposition

(Zariski topology on prime spectra is sober)

With excluded middle and axiom of choice (or at least the ultrafilter principle) then:

Let RR be a commutative ring. Then Spec(R)Spec(R) (its prime spectr equipped with the Zariski topology of def. 3) is a sober topological space.

Proof

We need to show that the function

Cl({}):PrimeIdl(R)IrrClSub(Spec(R)) Cl(\{-\}) \;\colon\; PrimeIdl(R) \longrightarrow IrrClSub(Spec(R))

which sends a point to its topological closure, is a bijection.

By lemma 1 this function is given by sending a prime ideal pPrimeIdl(R)p \in PrimeIdl(R) to the Zariski closed subset V(p)V(p). That this is a bijection is the statement of prop. 7.

Examples

Example

(affine space as prime spectrum)

Reconsider the case where R=k[X 1,,X n]R = k[X_1,\cdots, X_n] is a polynomial ring, for kk a field, as in the discussion of the naive affine space k nk^n above.

Observe that, by 6, the closed points in the prime spectrum Spec(k[X 1,,X n])Spec(k[X_1, \cdots, X_n]) correspond to the maximal ideals in the polynomial ring. These are of the form

(a 1,,a n)((X 1a 1)(X 2a 2)(X na n)) (a_1, \cdots, a_n) \coloneqq \left( (X_1 - a_1) \cdot (X_2 - a_2) \cdots (X_n - a_n) \right)

and hence are in bijection with the points of the naive affine space

k nMaxIdl(k[X 1,,X n]). k^n \simeq MaxIdl(k[X_1, \cdots, X_n]) \,.

There is however also prime ideals in k[X 1,,X n]k[X_1, \cdots, X_n] which are not maximal. In particular there is the 0-ideal (0)(0).

Proposition

(Spec(Z))

Let R=R = \mathbb{Z} be the commutative ring of integers. Consider the corresponding Zariski prime spectrum (prop. 5) Spec()Spec(\mathbb{Z}).

The prime ideals of the ring of integers are

  1. the ideals (p)(p) generated by prime numbers pp (this special case is what motivates the terminology “prime ideal”);

  2. the ideal (0)={0}(0) = \{0\}.

PrimeIdl()={0,2,3,5,7,11,}. PrimeIdl(\mathbb{Z}) = \left\{ 0, \; 2, 3, 5, 7, 11, \cdots \right\} \,.

All the prime ideals p2p \geq 2 are maximal ideals. Hence by prop. 6 these are closed points of Spec()Spec(\mathbb{Z}).

Only the prime ideal (0)(0) is not maximal, hence the point (0)(0) is not closed. Its closure is the entire space

Cl({0})=Spec(). Cl(\{0\}) = Spec(\mathbb{Z}) \,.

To see this, notice that in fact Spec()Spec(\mathbb{Z}) is the only closed subset containing the point (0)(0). This is because

(0)V(I) I(0) I=(0) \begin{aligned} & (0) \in V(I) \\ \Leftrightarrow\; & I \subset (0) \\ \Leftrightarrow\; & I = (0) \end{aligned}

and V(0)=Spec()V(0) = Spec(\mathbb{Z}), because

(pV(0))(0p)true. (p \in V(0)) \Leftrightarrow (0 \subset p) \Leftrightarrow true \,.

In terms of Galois connections

We now discuss how all of the above constructions and statements, and a bit more, follows immediately as a special case of the general theory of what is called Galois connections or adjoint functors between posets.

Background on Galois connections

Definition

(Galois connection induced from a relation)

Consider two sets X,YSetX,Y \in Set and a relation

EX×Y. E \hookrightarrow X \times Y \,.

Define two functions between their power sets P(X),P(Y)P(X), P(Y), as follows. (In the following we write E(x,y)E(x, y) to abbreviate the formula (x,y)E(x, y) \in E.)

  1. Define

    V E:P(X)P(Y) V_E \;\colon\; P(X) \longrightarrow P(Y)

    by

    V E(S){yY|xX((xS)E(x,y))} V_E(S) \coloneqq \left\{ y \in Y \vert \underset{x \in X}{\forall} \left( \left(x \in S\right) \Rightarrow E(x, y) \right) \right\}
  2. Define

    I E:P(Y)P(X) I_E \;\colon\; P(Y) \longrightarrow P(X)

    by

    I E(T){xX|yY((yT)E(x,y))} I_E(T) \coloneqq \left\{x \in X \vert \underset{y \in Y}{\forall} \left( \left(y \in T \right) \Rightarrow E(x, y) \right)\right\}
Proposition

The construction in def. 4 has the following properties:

  1. V EV_E and I EI_E are contravariant order-preserving in that

    1. if SSS \subset S', then V E(S)V E(S)V_E(S') \subset V_E(S);

    2. if TTT \subset T', then I E(T)I E(T)I_E(T') \subset I_E(T)

  2. The adjunction law holds: (TV E(S))(SI E(T)) \left( T \subset V_E(S) \right) \,\Rightarrow\, \left( S \subset I_E(T) \right)

    which we denote by writing

    P(X)V EI EP(Y) op P(X) \underoverset{\underset{V_E}{\longrightarrow}}{\overset{I_E}{\longleftarrow}}{\bot} P(Y)^{op}
  3. both V EV_E as well as I EI_E take unions to intersections.

Proof

Regarding the first point: the larger SS is, the more conditions that are placed on yy in order to belong to V E(S)V_E(S), and so the smaller V E(S)V_E(S) will be.

Regarding the second point: This is because both these conditions are equivalent to the condition S×TES \times T \subset E.

Regarding the third point: Observe that in a poset such as P(Y)P(Y), we have that A=BA = B iff for all CC, CAC \leq A iff CBC \leq B (this is the Yoneda lemma applied to posets). It follows that

TV E( iIS i) iff i:IS iI E(T) iff i:IS iI E(T) iff i:ITV E(S i) iff T i:IV E(S i) \array{ T \subset V_E(\bigcup_{i \in I} S_i) & iff & \bigcup_{i: I} S_i \subset I_E(T) \\ & iff & \forall_{i: I} S_i \subset I_E(T) \\ & iff & \forall_{i: I} T \subset V_E(S_i) \\ & iff & T \subset \bigcap_{i: I} V_E(S_i) }

and we conclude V E( i:IS i)= i:IV E(S i)V_E(\bigcup_{i: I} S_i) = \bigcap_{i: I} V_E(S_i) by the Yoneda lemma.

Proposition

(closure operators from Galois connection)

Given a Galois connection as in def. 4, consider the composites

I EV E:P(X)P(X) I_E \circ V_E \;\colon\; P(X) \longrightarrow P(X)

and

V EI E:P(Y)P(Y). V_E \circ I_E \;\colon\; P(Y) \longrightarrow P(Y) \,.

These satisfy:

  1. For all SP(X)S \in P(X) then SI EV E(S)S \subset I_E \circ V_E(S).

  2. For all SP(X)S \in P(X) then $V_E \circ I_E \circ V_E (S) = V_E(S)

  3. I EV EI_E \circ V_E is idempotent and covariant.

and

  1. For all TP(Y)T \in P(Y) then TV EI E(T)T \subset V_E \circ I_E(T).

  2. For all TP(Y)T \in P(Y) then I EV EI E(T)=I E(T)I_E \circ V_E \circ I_E (T) = I_E(T).

  3. V EI EV_E \circ I_E is idempotent and covariant.

This is summarized by saying that I EV EI_E \circ V_E and V EI EV_E \circ I_E are closure operators (idempotent monads).

Proof

The first statement is immediate from the adjunction law (prop. 10).

Regarding the second statement: This holds because applied to sets SS of the form I E(T)I_E(T), we see I E(T)I EV EI E(T)I_E(T) \subset I_E \circ V_E \circ I_E(T). But applying the contravariant map I EI_E to the inclusion TV EI E(T)T \subset V_E \circ I_E(T), we also have I EV EI E(T)I E(T)I_E \circ V_E \circ I_E(T) \subset I_E(T).

This directly implies that the function I EV EI_E \circ V_E. is idempotent, hence the third statement.

The argument for V EI EV_E \circ I_E is directly analogous.

In view of prop. 11 we say that:

Definition

(closed elements)

Given a Galois connection as in def. 4, then

  1. SP(X)S \in P(X) is called closed if I EV E(S)=SI_E \circ V_E(S) = S;

  2. the closure of SP(X)S \in P(X) is Cl(S)I EV E(S)Cl(S) \coloneqq I_E \circ V_E(S)

and similarly

  1. TP(Y)T \in P(Y) is called closed if V EI E(T)=TV_E \circ I_E(T) = T;

  2. the closure of TP(Y)T \in P(Y) is Cl(T)V EI E(T)Cl(T) \coloneqq V_E \circ I_E(T).

It follows from the properties of closure operators, hence form prop. 11:

Proposition

(fixed points of a Galois connection)

Given a Galois connection as in def. 4, then

  1. the closed elements of P(X)P(X) are precisely those in the image im(I E)im(I_E) of I EI_E;

  2. the closed elements of P(Y)P(Y) are precisely those in the image im(V E)im(V_E) of V EV_E.

We says these are the fixed points of the Galois connection. Therefore the restriction of the Galois connection

P(X)V EI EP(Y) op P(X) \underoverset{\underset{V_E}{\longrightarrow}}{\overset{I_E}{\longleftarrow}}{\bot} P(Y)^{op}

to these fixed points yields an equivalence

im(I E)V EI Eim(V E) op im(I_E) \underoverset{\underset{V_E}{\longrightarrow}}{\overset{I_E}{\longleftarrow}}{\simeq} im(V_E)^{op}

now called a Galois correspondence.

Proposition

Given a Galois connection as in def. 4, then the sets of closed elements according to def. 5 are closed under forming intersections.

Proof

If {T iP(Y)} i:I\{T_i \in P(Y)\}_{i: I} is a collection of elements closed under the operator K=V EI EK = V_E \circ I_E, then by the first item in prop. 11 it is automatic that i:IT iK( i:IT i)\bigcap_{i: I} T_i \subset K(\bigcap_{i: I} T_i), so it suffices to prove the reverse inclusion. But since i:IT iT i\bigcap_{i: I} T_i \subset T_i for all ii and KK is covariant and T iT_i is closed, we have K( i:IT i)K(T i)T iK(\bigcap_{i: I} T_i) \subset K(T_i) \subset T_i for all ii, and K( i:IT i) i:IT iK(\bigcap_{i: I} T_i) \subset \bigcap_{i: I} T_i follows.

Applied to affine space

We now redo the discussion of the Zariski topology on the affine space k nk^n from above as a special case of the general considerations of Galois connections.

Example

(Zariski closed subsets in affine space via Galois connection)

Let kk be a field and let nn \in \mathbb{N}, and write k[X 1,,X n]k[X_1, \cdots, X_n] for the polynomial ring over kk in nn variables. Define a relation

Ek[x 1,,x n]×k n E \hookrightarrow k[x_1, \ldots, x_n] \times k^n

by

E(f,x)(f(x)=0). E(f, x)\coloneqq \left( f(x) = 0\right) \,.

By def. 4 and prop. 10 we obtain the corresponding Galois connection of the form

P(k[X 1,,X n])V EI EP(k n) op P(k[X_1, \cdots, X_n]) \underoverset{\underset{V_E}{\longrightarrow}}{\overset{I_E}{\longleftarrow}}{\bot} P(k^n)^{op}

(where now k[X 1,,X n]k[X_1, \cdots, X_n] and k nk^n denote their underlying sets).

Here by def. 4 the function

V E:P(k[x 1,,x n])P(k n) V_E \;\colon\; P(k[x_1, \ldots, x_n]) \longrightarrow P(k^n)

sends a set \mathcal{F} of polynomials to its corresponding variety,

V E()={xk n| fk[x 1,,x n](f)(f(x)=0)}. V_E(\mathcal{F}) = \{\vec x \in k^n \,\vert\, \forall_{f \in k[x_1, \ldots, x_n]} \; (f \in \mathcal{F}) \Rightarrow (f(x) = 0)\} \,.

These are just the Zariski closed subsets from def. 1.

In the other direction,

I E:P(k n)P(k[x 1,,x n]) I_E \;\colon\; P(k^n) \longrightarrow P(k[x_1, \ldots, x_n])

sends a set of points Tk nT \subseteq k^n to its corresponding vanishing ideal

I E(T)={fk[x 1,,x n]| x:k nxTf(x)=0} I_E(T) = \{f \in k[x_1, \ldots, x_n] \,\vert\, \forall_{x: k^n} \; x \in T \Rightarrow f(x) = 0\}

which we considered earlier in def. 2.

We may now use the abstract theory of Galois connections to verify that Zariski closed subsets form a topology:

Proposition

(Zariski topology is well defined)

Using excluded middle, then:

The set of Zariski closed subsets of k nk^n from example 3 constitutes a topology in that it is closed under

  1. arbitrary intersections;

  2. finite untions.

Proof

Regarding the first point: From prop. 10 we know that V EV_E takes unions to intersections, hence that

iIV E( i)=V E(iI i). \underset{i \in I}{\cap} V_E(\mathcal{F}_i) \;=\; V_E\left( \underset{i \in I}{\cup} \mathcal{F}_i \right) \,.

Regarding the second point, we exploit the commutative ring structure of k[x 1,,x n]k[x_1, \ldots, x_n]. It is sufficient to show that the set of Zariski closed sets is closed under the empty union and under binary unions.

The empty union is the entire space k nk^n, which is V(1)V(1) (the variety associated with the constant polynomial 11),

Hence it only remains to see closure under binary unions.

To this end, recall from prop. 11 that we may replace \mathcal{F} with the corresponding ideal

II EV E() I \coloneqq I_E \circ V_E(\mathcal{F})

without changing the variety:

V E(I)=V E(). V_E(I) = V_E(\mathcal{F}) \,.

With this it is sufficient to show that

V E(I)V E(I)=V(II) V_E(I) \cup V_E(I') = V(I \cdot I')

where III \cdot I' is the ideal consisting of finite sums of elements of the form fgf g with fIf \in I and gIg \in I'.

We conclude by proving this statement:

Applying the contravariant operator V EV_E to the inclusions IIII \cdot I' \subseteq I and IIsubseteqII \cdot I' subseteq I' (which are clear since I,II, I' are ideals), we derive V E(I)V E(II)V_E(I) \subseteq V_E(I \cdot I') and V E(I)V(II)V_E(I') \subseteq V(I \cdot I'), so the inclusion V E(I)V E(I)V(II)V_E(I) \cup V_E(I') \subseteq V(I \cdot I') is automatic.

In the other direction, to prove V(II)V E(I)V(I)V(I \cdot I') \subseteq V_E(I) \cup V(I'), suppose xV(II)x \in V(I \cdot I') and that xx doesn’t belong to V(I)V(I). Then f(x)0f(x) \neq 0 for some fIf \in I. For every gIg \in I', we have f(x)g(x)=(fg)(x)=0f(x)g(x) = (f \cdot g)(x) = 0 since fgIIf \cdot g \in I \cdot I' and xV E(II)x \in V_E(I \cdot I'). Now divide by f(x)f(x) to get g(x)=0g(x) = 0 for every gIg \in I', so that xV E(I)x \in V_E(I').

Example

Let kk be a field, let nn \in \mathbb{N} and write k[X 1,,X n]k[X_1, \cdots, X_n] for the polynomial ring over kk in nn variables, and MaxIdl(k[X 1,,X n])MaxIdl(k[X_1, \cdots, X_n]) for the set of maximal ideals in this ring.

Define then a relation

Ek[x 1,,x n]×MaxIdeal(k[x 1,,x n]) E \hookrightarrow k[x_1, \ldots, x_n] \times MaxIdeal(k[x_1, \ldots, x_n])

by

E(f,M)(fM). E(f, M) \Leftrightarrow (f \in M) \,.

For a subset TMaxIdl(k[x 1,,x n])T \subseteq MaxIdl(k[x_1, \ldots, x_n]) we calculate

I E(T)={fk[x 1,ldots,x n]: 𝔪MaxIdlMSf𝔪}= 𝔪S𝔪 I_E(T) = \{f \in k[x_1, ldots, x_n]: \forall_{\mathfrak{m} \in MaxIdl} M \in S \Rightarrow f \in \mathfrak{m}\} = \bigcap_{\mathfrak{m} \in S} \mathfrak{m}

which is an ideal, since the intersection of any collection of ideals is again an ideal. (However, not all ideals are given as intersections of maximal ideals, a point to which we will return in a moment.)

Remark

This is a slight generalization of example 3 since each point a=(a 1,,a n)a = (a_1, \ldots, a_n) induces a maximal ideal

𝔪 ax 1a 1,,x na n, \mathfrak{m}_a \coloneqq \langle x_1 - a_1, \ldots, x_n - a_n \rangle \,,

i.e. the kernel of the function

k[x 1,,x n] k f f(a) \array{ k[x_1, \ldots, x_n] &\longrightarrow& k \\ f &\mapsto& f(a) }

which evaluates polynomials ff at the point aa, where we have f(a)=0f(a) = 0 iff f𝔪 af \in \mathfrak{m}_a.

Of course it need not be the case that all maximal ideals 𝔪\mathfrak{m} are given by points in this way; for example, the ideal (x 2+1)(x^2 + 1) is maximal in [x]\mathbb{R}[x] but is not given by evaluation at a point because x 2+1x^2 + 1 does not vanish at any real point. However, if the ground field kk is algebraically closed, then every maximal ideal of k[x 1,,x n]k[x_1, \ldots, x_n] is given by evaluation at a point a=(a 1,,a n)a = (a_1, \ldots, a_n). This result is not completely obvious; it is sometimes called the “weak Nullstellensatz”.

Proposition

The set Sk nS \subseteq k^n that are closed under the operator V EI E:P(k n)P(k n)V_E \circ I_E: P(k^n) \to P(k^n) in example 4 form a topology.

Proof

The proof is virtually the same as in the proof of prop. 14: they are closed under arbitrary intersections by our earlier generalities, and they are closed under finite unions by the similar reasoning: V E(S)=V E(I)V_E(S) = V_E(I) where I=I EV E(S)I = I_E \circ V_E(S) is an ideal, so there is no loss of generality in considering V E(I)V_E(I) for ideals II, and V E(I)V E(I)=V E(II)V_E(I) \cup V_E(I') = V_E(I \cdot I'). If 𝔪V E(II)\mathfrak{m} \in V_E(I \cdot I') (meaning IIMI \cdot I' \subseteq M) but mthfrakm\mthfrak{m} doesn’t belong to V E(I)V_E(I), i.e., f𝔪f \notin \mathfrak{m} for some fIf \in I, then for every gIg \in I' we have fg𝔪f \cdot g \in \mathfrak{m}. Taking the quotient map π:RR/𝔪\pi: R \to R/\mathfrak{m} to the field R/𝔪R/\mathfrak{m}, we have π(fg)=π(f)π(g)=0\pi(f \cdot g) = \pi(f)\cdot \pi(g) = 0, and since π(f)0\pi(f) \neq 0 we have π(g)=0\pi(g) = 0 for every gIg \in I', hence 𝔪V E(I)\mathfrak{m} \in V_E(I').

Thus the fixed elements of V EI EV_E \circ I_E on one side of the Galois correspondence are the closed sets of a topology. The fixed elements of I EV EI_E \circ V_E on the other side are a matter of interest; in the case where kk is algebraically closed, they are the radical ideals of k[X 1,,X n]k[X_1, \ldots, X_n] according to the “strong Nullstellensatz.

Applied to affine schemes

We now redo the discussion of the Zariski topology on the prime spectrum of a commutative ring from above as a special case of the general considerations of Galois connections.

References

Lecture notes include

  • Jim Carrell, Zariski topology etc pdf

See also

Revised on November 18, 2017 10:30:41 by Anonymous (98.3.24.49)