Copyright (©) 1997 by Joel Roberts.
Let X be a variety. Thus, we can be considering X An or X 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 Xa = V(Sa), where a runs through some index set and the Sa are subsets of K[x1, ...,xn], then:
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 X2 = V(T) where:
THE NOETHERIAN PROPERTY. By definition, a topological space is Noetherian if every descending sequence of closed subsets eventually stabilizes. Thus, if:
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, ...,xn].
TO PROVE THE LEMMA, we show that if S is a subset of K[x1, ...,xn], 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:
TO PROVE THE PROPOSITION, we consider a descending sequence:
DISTINGUISHED OPEN SETS (AFFINE CASE). If f is an element of the polynomial ring K[x1,...,x n], we define:
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).
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:
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)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 (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 (U) by its obvious name, specifically the ring of regular functions on U. As a particular case we have (X), the ring of globally regular functions on X.
On the other hand, for X An, we define the coordinate ring of X to be the quotient A(X) = K[x 1, ...,xn]/I(X), where I(X) is the defining ideal of X.
We have a map K[x1 ,...,xn] - > (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:
As observed in the text (with appropriate sketches), we can understand this isomorphism geometrically by observing that if X An, then we have a bijective correspondence between D(f) and the subvariety Y An, whose defining ideal is:
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 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 U«V. If we think about it briefly, we see that germs can be added and multiplied in a natural way, so that X,p is indeed a commutative ring.
A basic property of X,p is that the set of nonunit (or equivalently, noninvertible) elements in 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 X,p is one way to characterize a ocal ring:
PROPOSITION. Let be a commutative ring with 1. Then the following statements are equivalent:
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) m, which forces m to coincide with the set of nonunits.
A SPECIFIC CASE. Let X = An. For any p in An, elements of An, p can be identified as quotients of polynomials:
FOR A GENERAL AFFINE VARIETY X An, we also can identify elements of 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:
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:
IT IS A PLEASANT SURPRISE that the identification of 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 (X). Explicitly, we map our ring of quotients to 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.
Let X Pn be a projective variety. There are two ways to define the concept of a regular function on an open subset U X. These two approaches are easily seen to be equivalent.
METHOD 1. Consider the standard affine covering {U0,...,Un } of Pn. If f is defined on U X, we say that f is regular on U if f is regular on UUi for i = 0,..., n.
METHOD 2. Again, let f be defined on U 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,..., Xn) and H (X0,..., Xn) with degree(G) = degree(H) such that f(q) = G(q)/H(q) for every point q in V.
Explicitly, if q = [a0,..., an], we are requiring that
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,...,xn 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 P2 by the following quotient:
Setting x = X/Z and y = Y/Z, we transform this into the following quotient of homogeneous polynomials:
SOME RINGS ASSOCIATED TO A PROJECTIVE VARIETY. We can define the local ring of germs of regular functions 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 Pn and define I(X) to be the ideal in K[X0,..., Xn] generated by all homogeneous polynomials that vanish identically along X. We define the homogeneous coordinate ring to be
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]:
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.
We can work either with affine varieties or with projective varieties. Thus, if we consider an affine variety X An, we say that X is irreducible if it is nonempty and is not the union of two properly smaller nonempty affine varieties. Thus:
PROOF OF 4). We will consider the case of An, since the case of Pn is similar.Suppose that An = XY, 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
LEMMA. Let K be an algebraically closed field, and let f(x1,..., xn) be a nonzero element of K[x1,..., xn]. Then V(f) is nonempty, and V(f) 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 xn actually occurs nontrivially in f. Therefore, we can write: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 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 An be an irreducible affine variety. If X YZ, where Y and Z are affine varieties, then either X Y or X Z.
PROOF. If X YZ, then X = (XY) ( XZ). By irreducibility,
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:
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 = YZ, 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 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:
Consider a particular subset Yi from the first decomposition. Since Yi Z1 ... Zp, we can use Proposition 2 to show that Yi Zj for some j. Similarly, we show that Zj Yk for some k. Thus, Yi Zj Yk . Because there are no non trivial inclusions among the Y's, it follows that i = k, and Yi = Zj . Hence, all of the Yi occur among Z1, ... ,Zp . Now, we can show by similar methods that all of the Zj occur among Z1, ... , Zp . Therefore, the same irreducible closed subsets occur in the two decompositions.
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.]
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