ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 4: MORE ABOUT VARIETIES AND REGULAR FUNCTIONS.


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1 ALGERAIC GEOMETRY COURSE NOTES, LECTURE 4: MORE AOUT VARIETIES AND REGULAR FUNCTIONS. ANDREW SALCH. More about some claims from the last lecture. Perhaps you have noticed by now that the Zariski topology has very few open sets compared to the familiar, standard topology on R n or C n or a real or complex manifold. In fact, every open set in the Zariski topology on a variety is dense. Since the open sets in the Zariski topology are so big, it is reasonable to ask not only about the regular functions defined on all of a variety X, but also the regular functions defined almost everywhere, in particular, defined on an open subset of a variety X. We can then put an equivalence relation on that set by letting two such functions be equivalent if they agree on the (necessarily open) overlap between their domains. This leads us to the definition of the function field of X: Definition.. Let X be a variety. y the function field of X, written K(X), we mean the ring of equivalence classes of pairs (U, f ), where U is an open subset of X, and f : U k is a regular function; the equivalence relation is given by letting (U, f ) (V, g) if and only if f = g on the overlap U V. The addition and multiplication on these pairs is defined as follows: given elements (U, f ), (V, g) of K(X), we let the sum (U, f ) + (V, g) be (U V, ( f + g) U V ), and we let the product be (U V, ( f g) U V ). Observation.2. Suppose f : X k is a regular function. Then f may or may not be invertible in O(X). If f is not invertible, then this is because there exists some set of zeroes Z X of f, so we cannot divide by f since it would involve dividing by zero at the points in Z. ut (this is the whole point of the Zariski topology!) the set of zeroes of f is closed, so the complement X\Z is open, and (X, f ) is equivalent to (Z, f Z ) in the equivalence relation we have defined, and f Z is invertible. Hence the function field of X really is a field! Suppose that k is algebraically closed. Quoted in the last lecture: Theorem.3. Let X A n k be an affine variety. Then O(X) A(X), that is, the ring of regular functions on X is isomorphic to the coordinate ring of X. Furthermore, we get a onetoone correspondence between points of X and maximal ideals of A(X), and for each maximal ideal m A(X), the local ring A(X) m has Krull dimension equal to the dimension of X. Finally, the ring K(X) of rational functions on X is isomorphic to the field of fractions of A(X), and K(X) is a field extension of k of transcendence degree equal to the dimension of X. Proof. Suppose f k[x,..., x n ]. Then f is a regular function on A n k, and by composition of f with the inclusion X A n k, f is also a regular function on X. Hence we have a map of Date: January 205.
2 2 ANDREW SALCH rings (.0.) k[x,..., x n ] O(X). Clearly I(X) k[x,..., x n ] vanishes on X, so the map.0. factors through the projection k[x,..., x n ] k[x,..., x n ]/I(X) = A(X). Hence A(X) maps by a onetoone ring homomorphism to O(X). For a proof that this ring homomorphism is surjective, consult Hartshorne, Theorem 3.2; take a look at the same theorem for a proof that the local rings of X all have the claimed Krull dimension, and the claim about K(X), as well. To each point x X we can associate the ideal m x A(X) consisting of the functions which vanish at x. First, I claim that m x is maximal. The proof is as follows: if m x I for some maximal ideal I A(X), then I corresponds to some maximal ideal J k[x,..., x n ] containing I(X). The vanishing set V J (k) of J is some subset of A n k contained in {x}, hence is either {x} or is empty. If V J (k) is empty, then J = k[x,..., x n ] by Hilbert s Nullstellensatz, so I = A(X), which contradicts the assumption that I is maximal (maximal ideals are proper ideals, by definition). If V J (k) = {x}, then m x = I({x}) = I(V J (k)) = rad(j) = J, with the last equality rad(j) = J due to J being assumed maximal. So J = m x. So m x is maximal. On the other hand, if m is a maximal ideal of A(X), then it corresponds to a maximal ideal I k[x,..., x n ] containing I(X), so the vanishing set V I (k) must be a minimal nonempty subset of X, i.e., a single point. It is easily seen that the function sending m to that single point is inverse to the function sending x X to m x. For projective varieties, basically the opposite is true, that is, basically nothing about the variety can be read off from its ring of regular functions: Theorem.4. Let X P n k be a projective variety. Then O(X) k, that is, the ring of regular functions on X is isomorphic to the ground field k. Furthermore, the ring K(X) of rational functions on X is isomorphic to the field of fractions of the coordinate ring PA(X). Consult Hartshorne s Theorem 3.4 for a proof. I will say a bit about it in class. A very simple one to begin with: 2. A few small examples. Example 2.. Let k be an algebraically closed field. I claim that the algebraic set V (xy ) (k) A 2 k is an affine variety. To prove this, we just need to show that (xy ) is a prime ideal, i.e., that xy is irreducible. Now this is an elementary problem! Suppose xy = f g for some polynomials f, g k[x, y]. Since deg( f g) = deg( f ) + deg(g) = deg(xy ) = 2, either x, y are both linear polynomials, or one is quadratic and one is constant. Clearly, if one is quadratic and one is constant, then the constant one is a unit, and the factorization f g = xy has no bearing on whether xy is irreducible if one of the polynomials f, g is a unit constant. So suppose that x, y are both linear. Now we can write with α, β, γ, δ, ɛ, θ k. Now we solve! f (x, y) = αx + βy + γ, g(x, y) = δx + ɛy + θ, f g = αδx 2 + (αɛ + βδ)xy + βɛy 2 + (αθ + γδ)x + (βθ + γɛ)y + γθ = xy +,
3 ALGERAIC GEOMETRY COURSE NOTES, LECTURE 4: MORE AOUT VARIETIES AND REGULAR FUNCTIONS.3 so either α or δ are zero, and either β or ɛ are zero. If α = β = 0 then αɛ + βδ, and similarly for the case δ = ɛ = 0. So suppose α = ɛ = 0. Solving the remaining equations, we get that γ = 0 and θ = 0, which means that γθ = is impossible. We get a similar contradiction if β = δ = 0. Consequently the equation f g = xy has no solutions except when f or g is a unit constant. So V (xy ) (k) A 2 k is an affine variety. Example 2.2. Let k be an algebraically closed field. I claim that the quasiaffine variety X A k consisting of all points in A k except zero (i.e., X is the open complement of the variety V (x) (k) A k ) is isomorphic to the affine variety V (xy )(k) A 2 k. An isomorphism is as follows: with inverse f : X V (xy ) (k) ( f (x) = x, ), x g : V (xy ) (k) X g(x, y) = x. Since f, g are given by quotients of polynomials which are everywhere defined, composition with f and g sends regular functions to regular functions, hence f and g are each morphisms of varieties. Clearly f and g are mutually inverse, i.e., f g = id V(xy ) (k) and g f = id X. That s it! Example 2.2 shows something interesting: the variety X is quasiaffine, but not affine, since it is clearly not the vanishing set of any collection of polynomials in k[x]. Yet it is isomorphic (but not equal to) an affine variety, V (xy ) (k) A 2 k. In other words, X looks like (for most purposes) the vanishing set of a set of polynomials, but in A 2 k, not in A k. This is one of the important reasons to think about isomorphism of varieties and not just equalities: sometimes the particular way that a variety is given as embedded in affine (or projective) space obscures some important geometry of that variety that would be more visible if we saw the variety as embedded in a different way in affine or projective space (perhaps an affine or projective space of larger dimension, as we saw in Example 2.2). One of the main reasons we will develop cohomological tools in this course (when we reach chapter 3 of Hartshorne s textbook) is to have ways of determining intrinsic properties of varieties (such as whether or not a variety is isomorphic to an affine variety; we will see that this is equivalent to the vanishing of all its positivedimensional cohomology with quasicoherent coefficients!) that are very difficult to determine if we have to fool around with trying to find the right embedding in affine or projective space. Exercise 2.3. Let k be an algebraically closed field, and let A, k be nonzero. Prove that the algebraic set V (x 2 +Ay 2 2 )(k) A 2 k is an affine variety if and only if the characteristic of k is not 2. (Hint: this comes down a bruteforce algebra problem, that of finding when a certain polynomial factors.) In exercise. in Hartshorne s book, you are supposed to determine which quadratic curves in A 2 k are isomorphic to A k, and which ones are isomorphic to V (xy )(k) A 2 k. One way that you can study this question is as follows: suppose you choose an irreducible quadratic f (x, y) k[x, y]. You want to know if k[x, y]/( f (x, y)) is isomorphic to k[t] or to k[s, t]/(st ). What you need is an invariant of commutative kalgebras which isn t too hard to compute and which distinguishes k[t] from k[s, t]/(st ). One such invariant
4 4 ANDREW SALCH is the group of units: notice that k[t] k, while (k[s, t]/(st )) k Z. Hence, if k[x, y]/( f (x, y)) k[t], then the only units in k[x, y]/( f (x, y)) will be units in k multiplied by the element k[x, y]/( f (x, y)); if instead k[x, y]/( f (x, y)) k[s, t]/(st ), then k[x, y]/( f (x, y)) will have more units than just units in k multiplied by k[x, y]/( f (x, y)). This leads to, for example, the following computation: Example 2.4. Let k be an algebraically closed field of characteristic not equal to 2. If A, k and A, are nonzero, then I claim that the affine variety V (x 2 +Ay 2 2 )(k) A 2 k, an ellipse, is isomorphic to the affine variety V (xy ) (k) A 2 k. First, I note that ( (x + ) ( Ay) (x ) Ay) = x2 + Ay 2 2 mod (x 2 + Ay 2 2 ), so (x + Ay) is a unit in k[x, y]/(x 2 + Ay 2 2 ), with inverse (x y). This is exactly what we need to get a ring homomorphism! Here it is: f : k[s, t]/(st ) k[x, y]/(x 2 + Ay 2 2 ) f (s) = (x + Ay) f (t) = (x y). Notice that, if you restrict the domain and codomain of f, you get a klinear homomorphism from the kvector space of homogeneous degree polynomials in k[s, t]/(st ) to the homogeneous degree polynomials in k[x, y]/(x 2 + Ay 2 2 ), given by the matrix A. This matrix has inverse 2 2 A so now we check that g : k[x, y]/(x 2 + Ay 2 2 ) k[s, t]/(st ) g(x) = (s + t) 2 g(y) = 2 (s t) A
5 ALGERAIC GEOMETRY COURSE NOTES, LECTURE 4: MORE AOUT VARIETIES AND REGULAR FUNCTIONS.5 is a ring homomorphism, as follows: g(x) 2 + Ag(y) 2 = ( ) 2 ( 2 (s + t) + = 2 st = 2 = g(x 2 + Ay 2 ), 2 A ) 2 (s t) as desired. So g is a ring homomorphism. Clearly f, g are mutually inverse. So V (x 2 +Ay 2 2 )(k) is isomorphic to V (st ) (k).
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