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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:
First we discuss the naive Zariski topology on affine spaces $k^n$, consider the classical proofs and discover thereby the special role of prime ideals and maximal ideals;
then we turn to the modern definition of the Zariski topology on affine varieties $Spec(R)$ which takes the concept of prime (and maximal) ideals as primary, and again we provide the classical arguments;
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 $k^n$, 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 $k$ in $n$ variables, hence the solution sets to equations of the form
for $f_i \in k[X_1, \cdots, X_n]$ 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 $k^n$, 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. and prop. below), and that the closed points correspond to the maximal ideals among these (prop. ).
This motivates the modern refinement of the concept of the Zariski topology, where one considers any commutative ring $R$ and equips its set of prime ideals with a topology, by direct analogy with the previously naive affine space $k^n$, which is recovered with $R$ a polynomial ring and restricting attention to the maximal ideals (example below).
These sets of prime ideals of a ring $R$ equipped with the Zariski topology are called the (topological spaces underlying) the prime spectrum of a commutative ring, denoted $Spec(R)$.
The Zariski topology is in general not Hausdorff (example below) which makes it sometimes be regarded as an “exotic” type of topology. But it is in fact sober (prop. below) and hence as well-behaved in this respect as general locales are.
We consider here, for $k$ a field, the vector space $k^n$ 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 below we reconsider the naive case of interest in this subsection here from that more refined perspective.
(Zariski topology on affine space)
Let $k$ be a field, let $n \in \mathbb{N}$, and write $k[X_1, \cdots, X_n]$ for the set of polynomials in $n$ variables over $k$.
For $\mathcal{F} \subset k[X_1, \cdots, X_n]$ a subset of polynomials, let the subset $V(\mathcal{F}) \subset k^n$ of the $n$-fold Cartesian product of the underlying set of $k$ (the vanishing set of $\mathcal{F}$) be the subset of points on which all these polynomials jointly vanish:
These subsets are called the Zariski closed subsets.
Write
for the set of complements of the Zariski closed subsets. These are called the Zariski open subsets of $k^n$.
(Zariski topology is well defined)
Assuming excluded middle, then:
For $k$ a field and $n \in \mathbb{N}$, then the Zariski open subsets of $k^n$ (def. ) form a topology. The resulting topological space
is also called the $n$-dimensional affine space over $k$.
We need to show for $\{\mathcal{F}_i \subset k[X_1, \cdots, X_n]\}_{i \in I}$ a set of subsets of polynomials that
$\underset{i \in I}{\cup} \left(k^n \backslash V(\mathcal{F}_i)\right) = k^n \backslash V(\mathcal{F}_\cup)$ for some $\mathcal{F}_\cup \subset k[X_1, \cdots, X_n]$;
if $I$ is finite then $\underset{i \in I}{\cap} \left( k^n \backslash V(\mathcal{F}_{\cap})\right) = k^n \backslash \mathcal{F}_\cap$ for some $\mathcal{F}_{\cap} \subset k[X_1, \cdots, X_n]$.
By de Morgan's law for complements (and using excluded middle) this is equivalent to
$\underset{i \in I}{\cap} V(\mathcal{F}_i) = V(\mathcal{F}_\cup)$ for some $\mathcal{F}_\cup \subset k[X_1, \cdots, X_n]$;
if $I$ is finite then $\underset{i \in I}{\cup} V(\mathcal{F}_i) = V(\mathcal{F}_{\cap})$ for some $\mathcal{F}_{\cap} \subset k[X_1, \cdots, X_n]$.
We claim that we may take
$\mathcal{F}_\cup = \underset{i \in I}{\cup} \mathcal{F}_i$
$\mathcal{F}_{\cap} = \underset{i \in I}{\prod} \mathcal{F}_i \coloneqq \left\{ \underset{i \in I}{\prod} f_i \,\vert\, f_i \in \mathcal{F}_i \right\}$.
(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 $\mathcal{F}_i$.)
Regarding the first point:
Regarding the second point, in one direction we have the immediate implication
For the converse direction we need to show that
hence that
By excluded middle, this is equivalent to its contraposition, which by de Morgan's law is
This now is true by the assumption that $k$ is a field: If all factors $f_i(a_1, \dots a_n) \in k$ are non-zero, then their product $\underset{i \in I}{\prod} f_i(a_1, \cdots, a_n) \in k$ is non-zero.
For $k$ a field and $n \in \mathbb{N}$, consider a subset
of the underlying set of the $n$-fold Cartesian product of $k$ with itself. Then the topological closure $Cl(S)$ of this subset with respect to the Zariski topology $\tau_{\mathbb{A}^n_k}$ (def. ) is the vanishing set of all those polynomials that vanish on $S$:
We compute as follows:
Here the first equality is the definition of topological closure, the second is the definition of closed subsets in the Zariski topology (def. ), the third is the expression of intersections of these in terms of unions of polynomials as in the proof of prop. , and then the last one is immediate.
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.
(vanishing ideal of Zariski closed subset)#
Let $k$ be a field, and let $n \in \mathbb{N}$. Then for $V(\mathcal{F}) \subset k^n$ a Zariski closed subset, according to def. , hence for $\mathcal{F} \subset k[X_1, \cdots, X_n]$ a set of polynomials, write
for the maximal subset of polynomials that still has the same joint vanishing set:
This set is clearly an ideal in the polynomial ring $k[X_1, \cdots, X_n]$, called the vanishing ideal of $V(\mathcal{F})$.
With excluded middle then:
Let $k$ be a field, let $n \in \mathbb{N}$, and let $V(\mathcal{F}) \subset k^n$ be a Zariski closed subset (def. ). Then the following are equivalent:
$V(\mathcal{F})$ is an irreducible closed subset;
The vanishing ideal $I(V(\mathcal{F}))$ (def. ) is a prime ideal.
In one direction, assume that $V(\mathcal{F})$ is irreducible and consider $f,g \in k[X_1, \cdots, X_n]$ with $f \cdot g \in I(V(\mathcal{F}))$. We need to show that then already $f \in I(V(\mathcal{F}))$ or $g \in I(V(\mathcal{F}))$.
Now since $k$ is a field, we have
This implies that
and hence that
But since $V(\{f\})$, $V(\{g\})$ and $V(\mathcal{F})$ are all closed, by construction, their intersections are closed and hence this is a decomposition of $V(\mathcal{F})$ as a union of closed subsets. Therefore now the assumption that $V(\mathcal{F})$ is irreducible implies that
Now for the converse, assume that $I(V(\mathcal{F}))$ is a prime ideal, and that $V(\mathcal{F}) = V(\mathcal{F}_1) \cup V(\mathcal{F}_2)$. We need to show that $V(\mathcal{F}) = V(\mathcal{F}_1)$ or that $V(\mathcal{F}) = V(\mathcal{F}_2)$.
Assume on the contrary, that there existed elements
Then in particular the vanishing ideals would not contain each other
and hence there were polynomials
But since a product of polynomials vanishes at some point once one of the factors vanishes at that point, it would follows that
which were in contradiction to the assumption that $I(V(\mathcal{F}))$ is a prime ideal. Hence we have a proof by contradiction.
Proposition gives an injection
The following says that for algebraically closed fields then this is in fact a bijection:
Let $k = \overline{k}$ be an algebraically closed field and let $n \in \mathbb{N}$. Then the function
from prop. is a bijection.
The proof uses Hilbert's Nullstellensatz.
(generalization to affine varieties)
Prop suggests to consider the set of prime ideals of a polynomial ring $k[X_1, \cdots, X_n]$ for general $k$ as more fundamental, in some sense, than the set $k^n$. Morover, the set of prime ideals makes sense for every commutative ring $R$, not just $R = k[X_1, \cdots, X_n]$, 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. below.
If the field $k$ is not a finite field, then the Zariski topology on the affine space (def. ) 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 $V(\mathcal{F})$ 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. below.
(Zariski topology on set of prime ideals)
Let $R$ be a commutative ring. Write $PrimeIdl(R)$ for its set of prime ideals. For $\mathcal{F} \subset R$ any subset of elements of the ring, consider the subsets of those prime ideals that contain $\mathcal{F}$:
These are called the Zariski closed subsets of $PrimeIdl(R)$. Their complements are called the Zariski open subsets.
(Zariski topology well defined)
Assuming excluded middle, then:
Let $R$ be a commutative ring. Then the collection of Zariski open subsets (def. ) in its set of prime ideals
satisfies the axioms of a topology, the Zariski topology.
This topological space
is called (the space underlying) the prime spectrum of the commutative ring.
For $\mathcal{F} \subset R$ write $I(\mathcal{F}) \subset \mathcal{F}$ for the ideal which is generated by $\mathcal{F}$. Evidently the Zariski closed subsets depend only on this ideal
and therefore it is sufficient to consider the $V(\mathcal{F})$ for the case that $\mathcal{F} \subset R$ is not just a subset, but an ideal.
So let $\{F_i \in Idl(R)\}_{i \in I}$ be a set of ideals in $R$ and let $\{V(\mathcal{F}_i) \subset PrimeIdl(R)\}_{i \in I}$ be the corresponding set of Zariski closed subsets. We need to show that there exists $\mathcal{F}_\cup, \mathcal{F}_\cap \subset R$ such that
$\underset{i \in I}{\cap} V(\mathcal{F}_i) = V(\mathcal{F}_\cup)$;
if $I$ is finite set then $\underset{i \in I}{\cup} V(\mathcal{F}_i) = V(\mathcal{F}_\cap)$.
We claim that
$\mathcal{F}_{\cup} = \underset{i \in I}{\sum} \mathcal{F}_i \coloneqq \left\{ \underset{i \in I}{\sum} f_i \,, \in R\;\vert\; f_i \in \mathcal{F}_i \right\}$
$\mathcal{F}_{\cap} = \underset{i \in I}{\prod} \mathcal{F}_i \coloneqq \left\{ \underset{i \in I}{\prod} f_i \, \in R \;\vert\; f_i \in \mathcal{F}_i \right\}$,
Regarding the first point:
By using the various definitions, we get the following chain of logical equivalences:
Regarding the second point, in one direction we have the immediate implication
For the converse direction we need to show that
hence that
By excluded middle, this is equivalent to its contraposition, which by de Morgan's law is
This holds by the assumption that $p$ is a prime ideal.
We discuss some properties of the Zariski topology on prime spectra of commutative rings.
(topological closure of points)
Let $R$ be a commutative ring and consider $Spec(R) = (PrimeIdl(R), \tau_{Spec(R)})$ its prime spectrum equipped with the Zariski topology (def. ).
Then the topological closure of a point $p \in PrimeIdl(R)$ is $V(p) \subset PrimeIdl(R)$ (def. ).
By definition the topological closure of $\{p\}$ is
Hence unwinding the definitions, we have the following sequence of logical equivalences:
Recall:
Assuming the axiom of choice or at least the ultrafilter principle then:
For $R$ a commutative ring and $I \subset R$ a proper ideal, then $I$ 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.).
(maximal ideals are closed points)
Let $R$ be a commutative ring, consider the topological space $Spec(R) = (PrimeIdl(R),\tau_{Spec(R)})$, i.e. its prime spectrum equipped with the Zariski topology from def. .
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 $\mathfrak{m} \in MaxIdl(R) \subset PrimeIdl(R)$ into all prime ideals is precisely the inclusion of the subset of closed points into all points of $Spec(R)$.
and hence we need to show that
precisely if $\mathfrak{m}$ is maximal.
In one direction, assume that $\mathfrak{m}$ is maximal. By definition $V(\mathfrak{m})$ contains all the prime ideals $p$ such that $\mathfrak{m} \subset p$. That $\mathfrak{m}$ is maximal means that it is not contained in a larger proper ideal, in particular not in any larger prime ideal, and hence $V(\mathfrak{m}) = \{\mathfrak{m}\}$.
In the other direction, assume that $\mathfrak{m}$ is a prime ideal such that $V(\mathfrak{m}) = \{\mathfrak{m}\}$. By definition this means equivalently that the only prime ideal $p$ with $\mathfrak{m} \subset p$ is $\mathfrak{m}$ itself. We need to show that more generally $\mathfrak{m} \subset I$ for $I$ any proper ideal implies that $\mathfrak{m} = I$.
But the axiom of choice/ultrafilter principle imply the prime ideal theorem (lemma ), which says that there is a prime ideal $p$ with $I \subset p$, hence a sequence of inclusions $\mathfrak{m} \subset I \subset p$. This implies $\mathfrak{m} \subset p$, hence $\mathfrak{m} = p$, hence $I = \mathfrak{m}$.
(irreducible closed subsets correspond to prime ideals)
With excluded middle then:
Let $R$ be a commutative ring, and let $\mathcal{F} \subset R$ be an ideal in $R$, hence $V(\mathcal{F}) \subset Spec(R)$ be a Zariski closed subset in the prime spectrum of $R$. Then the following are equivalent:
$V(\mathcal{F})$ is an irreducible closed subset;
$\mathcal{F}$ is a prime ideal.
In one direction, assume that $V(\mathcal{F})$ is irreducible, and that $f,g \in R$ with $f \cdot g \in \mathcal{F}$. We need to show that then already $f \in \mathcal{F}$ or $g \in \mathcal{F}$.
To this end, first observe that
This is because
where the implication in the middle uses that $p$ is a prime ideal.
It follows that
This is a decomposition of $V(\mathcal{F})$ as a union of closed subsets, hence the assumption that $V(\mathcal{F})$ is irreducible implies that
Now for the converse. Assume that $\mathcal{F}$ is a prime ideal and that $V(\mathcal{F}) = V(\mathcal{F}_1) \cup V(\mathcal{F}_2)$. Observe (as in the proof of prop. ) that this means equivalently that $\mathcal{F} = \mathcal{F}_1 \cdot \mathcal{F}_2$. We need to show that then $V(\mathcal{F}) = V(\mathcal{F}_1)$ or that $V(\mathcal{F} = V(\mathcal{F}_2))$.
Suppose on the contrary that neither $\mathcal{F}_1$ nor $\mathcal{F}_2$ coincided with $\mathcal{F}$. This means that there were elements $f \in \mathcal{F}_1 \backslash \mathcal{F}$ and $g \in \mathcal{F}_2 \backslash \mathcal{F}$ such that still $f \cdot g \in \mathcal{F}$, in contradiction to the assumption. Hence we have a proof by contradiction.
As a corollary:
(Zariski topology on prime spectra is sober)
With excluded middle and axiom of choice (or at least the ultrafilter principle) then:
Let $R$ be a commutative ring. Then $Spec(R)$ (its prime spectr equipped with the Zariski topology of def. ) is a sober topological space.
We need to show that the function
which sends a point to its topological closure, is a bijection.
By lemma this function is given by sending a prime ideal $p \in PrimeIdl(R)$ to the Zariski closed subset $V(p)$. That this is a bijection is the statement of prop. .
(affine space as prime spectrum)
Reconsider the case where $R = k[X_1,\cdots, X_n]$ is a polynomial ring, for $k$ a field, as in the discussion of the naive affine space $k^n$ above.
Observe that, by , the closed points in the prime spectrum $Spec(k[X_1, \cdots, X_n])$ correspond to the maximal ideals in the polynomial ring. These are of the form
and hence are in bijection with the points of the naive affine space
There is however also prime ideals in $k[X_1, \cdots, X_n]$ which are not maximal. In particular there is the 0-ideal $(0)$.
(Spec(Z))
Let $R = \mathbb{Z}$ be the commutative ring of integers. Consider the corresponding Zariski prime spectrum (prop. ) $Spec(\mathbb{Z})$.
The prime ideals of the ring of integers are
the ideals $(p)$ generated by prime numbers $p$ (this special case is what motivates the terminology “prime ideal”);
the ideal $(0) = \{0\}$.
All the prime ideals $p \geq 2$ are maximal ideals. Hence by prop. these are closed points of $Spec(\mathbb{Z})$.
Only the prime ideal $(0)$ is not maximal, hence the point $(0)$ is not closed. Its closure is the entire space
To see this, notice that in fact $Spec(\mathbb{Z})$ is the only closed subset containing the point $(0)$. This is because
and $V(0) = Spec(\mathbb{Z})$, because
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.
(Galois connection induced from a relation)
Consider two sets $X,Y \in Set$ and a relation
Define two functions between their power sets $P(X), P(Y)$, as follows. (In the following we write $E(x, y)$ to abbreviate the formula $(x, y) \in E$.)
Define
by
Define
by
The construction in def. has the following properties:
$V_E$ and $I_E$ are contravariant order-preserving in that
if $S \subset S'$, then $V_E(S') \subset V_E(S)$;
if $T \subset T'$, then $I_E(T') \subset I_E(T)$
The adjunction law holds: $\left( T \subset V_E(S) \right) \,\Rightarrow\, \left( S \subset I_E(T) \right)$
which we denote by writing
both $V_E$ as well as $I_E$ take unions to intersections.
Regarding the first point: the larger $S$ is, the more conditions that are placed on $y$ in order to belong to $V_E(S)$, and so the smaller $V_E(S)$ will be.
Regarding the second point: This is because both these conditions are equivalent to the condition $S \times T \subset E$.
Regarding the third point: Observe that in a poset such as $P(Y)$, we have that $A = B$ iff for all $C$, $C \leq A$ iff $C \leq B$ (this is the Yoneda lemma applied to posets). It follows that
and we conclude $V_E(\bigcup_{i: I} S_i) = \bigcap_{i: I} V_E(S_i)$ by the Yoneda lemma.
(closure operators from Galois connection)
Given a Galois connection as in def. , consider the composites
and
These satisfy:
For all $S \in P(X)$ then $S \subset I_E \circ V_E(S)$.
For all $S \in P(X)$ then $V_E \circ I_E \circ V_E (S) = V_E(S)
$I_E \circ V_E$ is idempotent and covariant.
and
For all $T \in P(Y)$ then $T \subset V_E \circ I_E(T)$.
For all $T \in P(Y)$ then $I_E \circ V_E \circ I_E (T) = I_E(T)$.
$V_E \circ I_E$ is idempotent and covariant.
This is summarized by saying that $I_E \circ V_E$ and $V_E \circ I_E$ are closure operators (idempotent monads).
The first statement is immediate from the adjunction law (prop. ).
Regarding the second statement: This holds because applied to sets $S$ of the form $I_E(T)$, we see $I_E(T) \subset I_E \circ V_E \circ I_E(T)$. But applying the contravariant map $I_E$ to the inclusion $T \subset V_E \circ I_E(T)$, we also have $I_E \circ V_E \circ I_E(T) \subset I_E(T)$.
This directly implies that the function $I_E \circ V_E$. is idempotent, hence the third statement.
The argument for $V_E \circ I_E$ is directly analogous.
(closed elements)
Given a Galois connection as in def. , then
$S \in P(X)$ is called closed if $I_E \circ V_E(S) = S$;
the closure of $S \in P(X)$ is $Cl(S) \coloneqq I_E \circ V_E(S)$
and similarly
$T \in P(Y)$ is called closed if $V_E \circ I_E(T) = T$;
the closure of $T \in P(Y)$ is $Cl(T) \coloneqq V_E \circ I_E(T)$.
It follows from the properties of closure operators, hence form prop. :
(fixed points of a Galois connection)
Given a Galois connection as in def. , then
the closed elements of $P(X)$ are precisely those in the image $im(I_E)$ of $I_E$;
the closed elements of $P(Y)$ are precisely those in the image $im(V_E)$ of $V_E$.
We says these are the fixed points of the Galois connection. Therefore the restriction of the Galois connection
to these fixed points yields an equivalence
now called a Galois correspondence.
Given a Galois connection as in def. , then the sets of closed elements according to def. are closed under forming intersections.
If $\{T_i \in P(Y)\}_{i: I}$ is a collection of elements closed under the operator $K = V_E \circ I_E$, then by the first item in prop. it is automatic that $\bigcap_{i: I} T_i \subset K(\bigcap_{i: I} T_i)$, so it suffices to prove the reverse inclusion. But since $\bigcap_{i: I} T_i \subset T_i$ for all $i$ and $K$ is covariant and $T_i$ is closed, we have $K(\bigcap_{i: I} T_i) \subset K(T_i) \subset T_i$ for all $i$, and $K(\bigcap_{i: I} T_i) \subset \bigcap_{i: I} T_i$ follows.
We now redo the discussion of the Zariski topology on the affine space $k^n$ from above as a special case of the general considerations of Galois connections.
(Zariski closed subsets in affine space via Galois connection)
Let $k$ be a field and let $n \in \mathbb{N}$, and write $k[X_1, \cdots, X_n]$ for the polynomial ring over $k$ in $n$ variables. Define a relation
by
By def. and prop. we obtain the corresponding Galois connection of the form
(where now $k[X_1, \cdots, X_n]$ and $k^n$ denote their underlying sets).
sends a set $\mathcal{F}$ of polynomials to its corresponding variety,
These are just the Zariski closed subsets from def. .
In the other direction,
sends a set of points $T \subseteq k^n$ to its corresponding vanishing ideal
We may now use the abstract theory of Galois connections to verify that Zariski closed subsets form a topology:
(Zariski topology is well defined)
Using excluded middle, then:
The set of Zariski closed subsets of $k^n$ from example constitutes a topology in that it is closed under
arbitrary intersections;
finite untions.
Regarding the first point: From prop. we know that $V_E$ takes unions to intersections, hence that
Regarding the second point, we exploit the commutative ring structure of $k[x_1, \ldots, x_n]$. 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 $k^n$, which is $V(1)$ (the variety associated with the constant polynomial $1$),
Hence it only remains to see closure under binary unions.
To this end, recall from prop. that we may replace $\mathcal{F}$ with the corresponding ideal
without changing the variety:
With this it is sufficient to show that
where $I \cdot I'$ is the ideal consisting of finite sums of elements of the form $f g$ with $f \in I$ and $g \in I'$.
We conclude by proving this statement:
Applying the contravariant operator $V_E$ to the inclusions $I \cdot I' \subseteq I$ and $I \cdot I' subseteq I'$ (which are clear since $I, I'$ are ideals), we derive $V_E(I) \subseteq V_E(I \cdot I')$ and $V_E(I') \subseteq V(I \cdot I')$, so the inclusion $V_E(I) \cup V_E(I') \subseteq V(I \cdot I')$ is automatic.
In the other direction, to prove $V(I \cdot I') \subseteq V_E(I) \cup V(I')$, suppose $x \in V(I \cdot I')$ and that $x$ doesn’t belong to $V(I)$. Then $f(x) \neq 0$ for some $f \in I$. For every $g \in I'$, we have $f(x)g(x) = (f \cdot g)(x) = 0$ since $f \cdot g \in I \cdot I'$ and $x \in V_E(I \cdot I')$. Now divide by $f(x)$ to get $g(x) = 0$ for every $g \in I'$, so that $x \in V_E(I')$.
Let $k$ be a field, let $n \in \mathbb{N}$ and write $k[X_1, \cdots, X_n]$ for the polynomial ring over $k$ in $n$ variables, and $MaxIdl(k[X_1, \cdots, X_n])$ for the set of maximal ideals in this ring.
Define then a relation
by
For a subset $T \subseteq MaxIdl(k[x_1, \ldots, x_n])$ we calculate
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.)
This is a slight generalization of example since each point $a = (a_1, \ldots, a_n)$ induces a maximal ideal
i.e. the kernel of the function
which evaluates polynomials $f$ at the point $a$, where we have $f(a) = 0$ iff $f \in \mathfrak{m}_a$.
Of course it need not be the case that all maximal ideals $\mathfrak{m}$ are given by points in this way; for example, the ideal $(x^2 + 1)$ is maximal in $\mathbb{R}[x]$ but is not given by evaluation at a point because $x^2 + 1$ does not vanish at any real point. However, if the ground field $k$ is algebraically closed, then every maximal ideal of $k[x_1, \ldots, x_n]$ is given by evaluation at a point $a = (a_1, \ldots, a_n)$. This result is not completely obvious; it is sometimes called the “weak Nullstellensatz”.
The set $S \subseteq k^n$ that are closed under the operator $V_E \circ I_E: P(k^n) \to P(k^n)$ in example form a topology.
The proof is virtually the same as in the proof of prop. : they are closed under arbitrary intersections by our earlier generalities, and they are closed under finite unions by the similar reasoning: $V_E(S) = V_E(I)$ where $I = I_E \circ V_E(S)$ is an ideal, so there is no loss of generality in considering $V_E(I)$ for ideals $I$, and $V_E(I) \cup V_E(I') = V_E(I \cdot I')$. If $\mathfrak{m} \in V_E(I \cdot I')$ (meaning $I \cdot I' \subseteq M$) but $\mthfrak{m}$ doesn’t belong to $V_E(I)$, i.e., $f \notin \mathfrak{m}$ for some $f \in I$, then for every $g \in I'$ we have $f \cdot g \in \mathfrak{m}$. Taking the quotient map $\pi: R \to R/\mathfrak{m}$ to the field $R/\mathfrak{m}$, we have $\pi(f \cdot g) = \pi(f)\cdot \pi(g) = 0$, and since $\pi(f) \neq 0$ we have $\pi(g) = 0$ for every $g \in I'$, hence $\mathfrak{m} \in V_E(I')$.
Thus the fixed elements of $V_E \circ I_E$ on one side of the Galois correspondence are the closed sets of a topology. The fixed elements of $I_E \circ V_E$ on the other side are a matter of interest; in the case where $k$ is algebraically closed, they are the radical ideals of $k[X_1, \ldots, X_n]$ according to the “strong Nullstellensatz.
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.
Lecture notes include
See also
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