Zariski Topology and Regular Functions

Math 8203 (Algebraic Geometry)
Fall 1997

The Zariski Topology and Regular Functions

The Zariski topology
Regular functions: the affine case
Regular functions: the projective case
Irreducible varieties

The Zariski topology.

Let X be a variety. Thus, we can be considering X $\subset$ Aⁿ or X $\subset$ Pⁿ. In either case, we claim that the subsets of X which are algebraic (as subsets of Aⁿ or Pⁿ respectively) are the closed sets of a topology. In order to verify this fact, it is clearly sufficient to consider the case where X is Aⁿ or Pⁿ itself.

PROOF IN THE AFFINE CASE. We have V(0) = Aⁿ while V(1) is empty. If X_a = V(S_a), where a runs through some index set and the S_a are subsets of K[x₁, ...,x_n], then:

$\cap V(S_a) = V(\cup S_a).$

Thus, an arbitrary intersection of algebraic sets is algebraic. To show that a finite union of algebraic sets is algebraic, the key step is to understand the union of two algebraic sets. So, if X₁ = V(S₁) and X₂ = V(S₂), then we can check that X₁ $\cup$ X₂ = V(T) where:

T = {fg | f in S₁ and g in S₂ }. The key step is to observe that if p lies in V(T) but p is not in X₁ , then f(p) is nonzero for some f₀ in S₁. Since f₀(p)g(p) = 0 for every g in S₂, it follows that p is a point of X₂.

THE NOETHERIAN PROPERTY. By definition, a topological space is Noetherian if every descending sequence of closed subsets eventually stabilizes. Thus, if:

X₁ $\supseteq$ X₂ $\supseteq$ . . . $\supseteq$ X_k $\supseteq$ X_k+1 $\supseteq$ . . . , then there exists M such that X_M = X_M+1 = . . .

PROPOSITION. The Zariski topology is Noetherian.

Before proving this in the affine case, we state and prove a lemma.

LEMMA. Every algebraic set in Aⁿ is of the form X = V(I), where I is an ideal in K[x₁, ...,x_n].

TO PROVE THE LEMMA, we show that if S is a subset of K[x₁, ...,x_n], then V(S) = V(I), where I is the ideal generated by S. This follows from the observation that elements of I are sums of the form:

g₁f₁ + . . . + g_mf_m , with m varying, f_j in S and g_j in K[x₁, ...,x_n].

TO PROVE THE PROPOSITION, we consider a descending sequence:

X₁ $\supseteq$ X₂ $\supseteq$ . . . $\supseteq$ X_k $\supseteq$ X_k+1 $\supseteq$ . . . , with X_k = V(I_k). We may not have started with an ascendending sequence of ideals, but can arrange that by noting that X_k = V(J_k), where J_k = I₁ + ... + I_k . Note that X_k = X₁ $\cap$ ... $\cap$ X_k .) We now have an ascending sequence of ideals: J₁ $\subseteq$ J₂ $\subseteq$ . . . $\subseteq$ J_k $\subseteq$ J_k+1 $\subseteq$ . . . It is known that K[x₁,...,x_n] is Noetherian, so that every ascending sequence of ideals eventually stabilizes. Therefore our descending sequence of closed subsets also does.

DISTINGUISHED OPEN SETS (AFFINE CASE). If f is an element of the polynomial ring K[x₁,...,x _n], we define:

D(f) = {p | p lies in Aⁿ and f(p) is nonzero} This is clearly an open subset, since D(f) is the complement in Aⁿ of the closed set V(f). These are of interest because of the following result.

PROPOSITION. The family of distinguished open sets is a basis of the Zariski topology of Aⁿ.

PROOF. Every ideal is a sum of principal ideals. Therefore, the closed set V(I) is the intersection of closed sets V(f), where f runs through some set of generators. Therefore the complement of V(I) is the union of the distinguished open sets V(f).

Regular functions: affine case.

Let X be a variety, and let U be an open subset. If f is a K-valued function defined on U, we will say that f is regular at a point p of U if there is an open neighborhood V on which f is expressible as a quotient of polynomials f = g/h, where h(p) is nonzero. We say that f is regular on U if it is regular at every point of U.

FOR INSTANCE:

Let X = A². Then f(x,y) = (x + y)/ (x² + 3xy) is regular on the open set U = D(x² + 3xy ).
Let X = V(xy - zw) in A³. We define two functions:
- f₁(p) = x/z on the distinguished open set D(z);
- f₂(p) = y/w on the distinguished open set D(w).

Now, let U = D(z) D(w). Since xy - zw = 0 at every point of X, we have x/z = y/w at points of D(z) $\cap$ D(w). Therefore, f₁ and f₂ can be "glued together" to give a function f which is regular at every point of U.

SOME RINGS ASSOCIATED TO AN AFFINE VARIETY: If X and U are as above, we define $\Oh$ (U) to be the set of regular functions on U. It is a commutative ring, since we can add and multiply the values of functions. We will usually call $\Oh$ (U) by its obvious name, specifically the ring of regular functions on U. As a particular case we have $\Oh$ (X), the ring of globally regular functions on X.

On the other hand, for X $\subset$ Aⁿ, we define the coordinate ring of X to be the quotient A(X) = K[x ₁, ...,x_n]/I(X), where I(X) is the defining ideal of X.

We have a map K[x₁,...,x_n] - > $\Oh$ (X), by evaluating polynomials at points of X. One checks easily that this is a homomorphism. The kernel is exactly I(X), which follows easily from the definition of I(X). Therefore, we at least have an injective homomorphism:

A(X) - > $\Oh$ (X). More generally, let U = D(f). Then we have a homomorphism: K[x₁, ...,x_n, 1/f] - > $\Oh$ (U), by evaluating expressions of the form g/f^d at points of U. This provides an injective homomorphism A(X) [1/f] - > $\Oh$ (U). To give a naïve but rigorous definition of A(X) [1/f] we can observe that: K[x₁, ...,x_n, 1/f] $\cong$ K[x ₁,...,x_{n
+ 1}]/(1 - x_{n +
1}f(x₁, ...,x_n)) so that: A(X) [1/f] = K[x₁, ...,x_{n +
1}]/(I(X) + (1 - x_{n
+ 1}f)).

As observed in the text (with appropriate sketches), we can understand this isomorphism geometrically by observing that if X $\subset$ Aⁿ, then we have a bijective correspondence between D(f) and the subvariety Y $\subset$ Aⁿ, whose defining ideal is:

(I(X) + (1 - x_{n +
1}f)). The map is given by: (x₁, ...,x_n) - > (x₁, ...,x_n , 1/f(x₁, ...,x_n)).

THE LOCAL RING AT A POINT. Let p be a point of an affine variety X. We define the local ring of X at p to be the set $\Oh$ _X,p of germs of regular functions at p. By definition, a germ is represented by a pair, (U,g), where U is an open neighborhood of p, and g is regular on U. Two pairs (U,g), and (V,h) represent the same germ if g and h agree on some open neighborhood of p contained in UŤV. If we think about it briefly, we see that germs can be added and multiplied in a natural way, so that $\Oh$ _X,p is indeed a commutative ring.

A basic property of $\Oh$ _X,p is that the set of nonunit (or equivalently, noninvertible) elements in $\Oh$ _X,p is an ideal. Specifically, the germ represented by (U,g) is invertible if and only if g(p) is nonzero: the inverse is represented by 1/g (on a possibly smaller neighborhood). Hence the nonunit elements are represented by germs (U,g) such that g(p) = 0. The germs of functions which vanish at p obviously form an ideal. This property of $\Oh$ _X,p is one way to characterize a ocal ring:

PROPOSITION. Let $\Oh$ be a commutative ring with 1. Then the following statements are equivalent:

The set of nonunit elements in $\Oh$ is an ideal.
$\Oh$ has exacly one maximal ideal.

THE PROOF that (1) implies (2) involves the observation that an ideal in a commutative ring with 1 which is distinct from the unit ideal consists entirely of nonunit elements. To prove that (2) implies (1), let m be the unique maximal ideal, and let a be a nonunit element. Then (a) is a nonunit ideal, which is contained in some maximal ideal, by Zorn's lemma (etc. ...). So, (a) $\subset$ m, which forces m to coincide with the set of nonunits.

A SPECIFIC CASE. Let X = Aⁿ. For any p in Aⁿ, elements of $\Oh$ _An_,
p can be identified as quotients of polynomials:

f(x₁, ...,x_n)/ g(x₁, ...,x_n), where g(p) is nonzero. The set of quotients of this type obviously is a subring of the fraction field of K[x₁, ...,x_n]. This fraction field is called the field of rational functions in n variables and denoted as follows: K(x₁, ...,x_n) = frac(K[x₁, ...,x_n])

FOR A GENERAL AFFINE VARIETY X $\subset$ Aⁿ, we also can identify elements of $\Oh$ _X,p as quotients of the form f/g where f and g are elements of the coordinate ring A(X) and g(p) is nonzero, provided that we do the following:

identify elements of the coordinate ring as regular functions on X by means of the inclusion of A(X) in $\Oh$ (X) discussed above;
consider two quotients g₁/ h₁ and g₂/ h₂ as being equal if and only if there exists f in A(X) such that f(p) is nonzero and: f(g₁ h₂ - g₂ h₁) = 0.

While we can consider condition (2) as a technical algebraic condition that adapts the construction of field of fractions to a situation where zerodivisors may exist, it also has the intuitive content of saying that the difference of the two quotients g₁/ h₁ - g₂/ h₂ vanishes identically in some neighborhood of p, specifically D(f).

TO VERIFY THAT this set of quotients is a ring, we work formally with the set of ordered pairs and the equivalence relation between two pairs (g₁, h₁) and (g₂, h₂) given as in condition (2). This involves some calculations, but they are fairly routine except for transitivity. Thus, suppose that g₁/ h₁ = g₂/ h₂ and g₂/ h₂ = g₃/ h₃ as formal quotients. Then f( g₁ h₂ - g₂ h₁) = 0 and f* (g₁ h₂ - g₂ h₁) = 0 in A(X), where f and f* are elements such that f(p) and f*(p) are nonzero. After some calculation, we find that:

ff*h₂ (g₁ h₃ - g₃ h₁) = 0. ¡¡Please check this for yourself!! Since h₂ is a denominator, h₂(p) is nonzero, so that g₁/ h₁ = g₃/ h₃ as formal quotients.

IT IS A PLEASANT SURPRISE that the identification of $\Oh$ _X,p with this ring of quotients follows fairly immediately from the definitions (or at least without major obstacles), even though we will need the Nullstellensatz to verify the "global" identification of A(X) and $\Oh$ (X). Explicitly, we map our ring of quotients to $\Oh$ _X,p by sending the quotient g/h to the germ represented by (U,f), where U = D(h) and f is given by the quotient g/h at points of U. Surjectivity follows from the definitions (more or less). In verifying injectivity, we use the fact that distinguished open subsets form a basis of the topology.

Regular functions: the projective case

Let X $\subset$ Pⁿ be a projective variety. There are two ways to define the concept of a regular function on an open subset U $\subset$ X. These two approaches are easily seen to be equivalent.

METHOD 1. Consider the standard affine covering {U₀,...,U_n } of Pⁿ. If f is defined on U $\subset$ X, we say that f is regular on U if f is regular on U $\cap$ U_i for i = 0,..., n.

METHOD 2. Again, let f be defined on U $\subset$ X. We say that f is regular at a point p in U if there is an open neighborhood V of p in U and homogeneous polynomials G (X₀,..., X_n) and H (X₀,..., X_n) with degree(G) = degree(H) such that f(q) = G(q)/H(q) for every point q in V.

Explicitly, if q = [a₀,..., a_n], we are requiring that

$\f(q) = {G(a_0,...,a_n \over H(a_),...,a_n)}$

ONE CAN CHECK that the value of this expression does not depend on the choice of homogeneous coordinate vector used to represent q.

TO SEE WHY the two methods are equivalent, note that the numerator and denominator in a quotient of (usual) polynomials in x₁,...,x_n can be "homogenized", and that we can adjust degrees by multiplying by a suitable power of X₀. For instance, suppose that f is represented in a neighborhood of some point of U₀ $\subset$ P² by the following quotient:

$<x^3 + 5xy + y^3 \over x^2 - 3y}$ .

Setting x = X/Z and y = Y/Z, we transform this into the following quotient of homogeneous polynomials:

homogenization of the above expression ...

SOME RINGS ASSOCIATED TO A PROJECTIVE VARIETY. We can define the local ring of germs of regular functions $\Oh$ _X,p at a point p in X by methods very similar to what we did in the affine case. So, we won't discuss the details right now.

A concept related to the coordinate ring of an affine variety is the homogeneous coordinate ring of a projective variety. To get started, consider X $\subset$ Pⁿ and define I(X) to be the ideal in K[X₀,..., X_n] generated by all homogeneous polynomials that vanish identically along X. We define the homogeneous coordinate ring to be

S(X) = K[X₀,..., X_n]/I(X) . Since homogeneous polynomials do not define regular functions on Pⁿ, we do not attempt to describe elements of S(X)) as regu lar functions on X.

A SPECIFIC INSTANCE. Let X be the twisted cubic curve in P³. One can show that I(X) is generated by the following three homogeneous polynomials of degree 2 in the polynomial ring K[W, X,Y,Z]:

X² - WY ,
Y² - XZ ,
WZ - XY .

We won't try to verify this right now. Let's just note that it was shown in the text (and also in class) that these polynomials vanish at all points of the twisted cubic (and in fact that their common zeroset is exactly the twisted cubic). Showing that they generate I(X) requires somewhat more work.

ONE MORE DEFINITION: We define a quasiprojective variety to be a nonempty open subset of a projective variety. This includes affine varieties as a special case. The concept of a regular function is the same as in the projective case, since the definition of regular function is essentially local.

Irreducible varieties

We can work either with affine varieties or with projective varieties. Thus, if we consider an affine variety X $\subset$ Aⁿ, we say that X is irreducible if it is nonempty and is not the union of two properly smaller nonempty affine varieties. Thus:

if X = Y $\cup$ Z, where Y and Z are nonempty affine varieties,
then either Y = X or Z = X. The definition in the case of a projective variety X $\subset$ Pⁿ is completely parallel.

Some examples:

A set consisting of one point is obviously irreducible.
A¹ is irreducible Indeed, the proper closed subsets of A¹ are finite sets, while A¹ is infi nite. (We are assuming that our field is algebraically closed, and thus infinite.)
By extension, any curve with a rational parametrization is irreducible .
Aⁿ and Pⁿ are irreducible.

PROOF OF 4). We will consider the case of Aⁿ, since the case of Pⁿ is similar.Suppose that Aⁿ = X $\cup$ Y, where X and Y are affine varieties, both nonempty and distinct from Aⁿ. Let f and g be nonzero elements of I(X) and I(Y) respectively, i.e. f and g are nonzero polynomials which vanish identically along X and Y respectively. It follows that

Aⁿ = V(f) $\cup$ V(g), so that Aⁿ = V(fg). Therefore, we can deduce the conclusion from the following:

LEMMA. Let K be an algebraically closed field, and let f(x₁,..., x_n) be a nonzero element of K[x₁,..., x_n]. Then V(f) is nonempty, and V(f) $\neq$ Aⁿ.

PROOF. We proceed by induction on n, the case n = 1 being immediate. So, in the inductive step, we can assume (by renumbering the variables) that x_n actually occurs nontrivially in f. Therefore, we can write: f(x₁,..., x_n) = g₀( x₁,..., x_n-₁) + g₁( x₁,..., x_n-₁) x_n + ... + g_d( x₁,..., x_n-₁) x_n^d, where g_i is homogeneous of degree = i, while d > 0 and g_d is nonzero. By the inductive hypothesis, there exists (a₁,..., a_n-₁) in A^n-1 such that g_d( a₁,..., a_n-₁) $\neq$ 0. Hence, $\varphi$ (x) := f (a₁,..., a_n-₁,x) is a nonzero polynomial of degree = d > 0. It follows that there exist finitely many values of a_n in K such that f(a₁,..., a_n) = $\varphi$ (a_n) = 0, and infinitely many values of the last coordinate such that the corresponding value of f is nonzero. This proves our claim.

We note that irreducibility is really a property of closed sets in the Zariski topology, and does not really involve any other aspects of varieties. Here are a couple of basic properties of irreducible varieties.

PROPOSITION 1. If X is an affine or projective variety, then X is irreducible if and only if every nonempty open subset U $\subset$ X is dense in X.

PROOF. Every open subset is dense if and only if every two nonempty open subsets have a nonempty intersection. By taking complements, this is equivalent to the statement that the union of two proper closed subsets is not equal to X.

PROPOSITION 2. Let X $\subset$ Aⁿ be an irreducible affine variety. If X $\subseteq$ Y $\cup$ Z, where Y and Z are affine varieties, then either X $\subseteq$ Y or X $\subseteq$ Z.

PROOF. If X $\subseteq$ Y $\cup$ Z, then X = (X $\cap$ Y) $\cup$ ( X $\cap$ Z). By irreducibility,

either X = X $\cap$ Y or X = X $\cap$ Z. This proves our claim.

Now we can state and prove our main result.

THEOREM. Let X be an affine variety (respectively a projective variety). Then X is the union of finitely many irreducible affine varieties (respectively finitely many irreducible projective varieties). Thus:

X = Y₁ $\cup$ ... $\cup$ Y_m, where Y₁ ,..., Y_m are irreducible. If we require that Y_i does not contain Y_j when i $\neq$ j, then Y₁ ,..., Y_m are uniquely determined. The corresponding statements are also true in the projective case.

DEFINITION: The irreducible closed subsets which occur in this uniquely determined decomposition are called the irreducible components of X.

PROOF OF THE THEOREM. In proving the existence of the decomposition for Zariski closed subsets of Aⁿ, we use the fact that the Zariski topology of Aⁿ is Noetherian. By definition, this means that every descending closed subsets stabilizes after fin itely many steps. An equivalent characterization is that every nonempty collection of closed subsets has a minimal member, i.e., there is a member of the collection which is not contained in any other set in the collection.

So, assume that there exists a Zariski-closed subset of Aⁿ which is not the union of finitely many irreducible closed subsets, and let S be the collection of closed subsets which are not finite unions of irreducibles. Then S contains a minimal member, say X₀ . If X₀ were irreducible, this would immediately contradict the fact that no member of S is a union of finitely many irreducible closed subsets. So, we have X₀ = Y $\cup$ Z, where Y and Z are proper closed subsets of X₀ . Since X₀ is a minimal member of S, neither Y nor Z is a member of S; therefore, both Y and Z are unions of finitely many irreducible closed subsets. This implies that S is a union of finitely many irreducible closed subsets, thus contradicting the hypothesis that S is nonempty and thereby proving the first part of the theorem.

To prove uniqueness, suppose that some X $\subset$ Aⁿ has two different decompositions as a union of irreducible closed subsets, both satisfying the condition that there be no inclusion relation among different subsets in the collection. Specifically, suppose that:

X = Y₁ $\cup$ ... $\cup$ Y_m and also: X = Z₁ $\cup$ ... $\cup$ Z_p , where there is no inclusion relation among the various Y's and also no such relation among the Z's

Consider a particular subset Y_i from the first decomposition. Since Y_i $\subseteq$ Z₁ $\cup$ ... $\cup$ Z_p, we can use Proposition 2 to show that Y_i $\subseteq$ Z_j for some j. Similarly, we show that Z_j $\subseteq$ Y_k for some k. Thus, Y_i $\subseteq$ Z_j $\subseteq$ Y_k . Because there are no non trivial inclusions among the Y's, it follows that i = k, and Y_i = Z_j . Hence, all of the Y_i occur among Z₁, ... ,Z_p . Now, we can show by similar methods that all of the Z_j occur among Z₁, ... , Z_p . Therefore, the same irreducible closed subsets occur in the two decompositions.

Another example.

We would now like to claim that if f is an irreducible polynomial in n variables (respectively, if F is an irreducible homogeneous polynomial in n + 1 variables), then the zeroset V(f) is an irreducible closed subset of Aⁿ (respectively that the zeroset V(F) is an irreducible closed subset of Pⁿ ). A little work is required for this, so that it will be saved for Exercises 7 and 8, which are linked below. It will follow from this claim that if f is an arbitrary polynomial in n variables, then the irreducible components of the hypersurface V(f) correspond bijectively to the distinct irreducible factors of f [and similarly in the projective case.]

Related Exercises

Last updated November 18, 1997.

Please send comments and/or corrections to:

Joel Roberts
351 Vincent Hall
625-1076
e-mail: roberts@math.umn.edu
http://www.math.umn.edu/~roberts

Math 8203 (Algebraic Geometry) Fall 1997