ABSTRACT NONSINGULAR CURVES


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1 ABSTRACT NONSINGULAR CURVES Affine Varieties Notation. Let k be a field, such as the rational numbers Q or the complex numbers C. We call affine nspace the collection A n k of points P = a 1, a,..., a n with coordinates a i k. For example, A 1 k = k. Similarly, we call projective nspace the collection P n k of equivalence classes a 1 : : a n : a 0, where we say two nonzero points a1,..., a n, a 0, b1,..., b n, b 0 A n+1 k are equivalent if a i = λ b i for some nonzero λ k. For example, a 1 : a 0 = b1 : b 0 P 1 k formally if and only if a 1 = b 1. We can a 0 b 0 always embed A n k P n k via the map a 1,..., a n a1 : : a n : 1. We will be interested in subsets of these spaces. Denote k[x 1, x,..., x n ] as the collection of polynomials in n variables with coefficients in k. Given a collection of polynomials f 1, f,..., f m }, we denote the ideal } f 1, f,..., f m = g 1 f 1 + g f + + g m f m g 1, g,..., g m k[x 1, x,..., x n ]. Recall that = f 1, f,..., f m is a prime ideal if and only if the quotient ring O = k[x 1, x,..., x n ]/ is an integral domain. Whenever this is the case, we call an affine variety a set in the form } X = P A n k f 1P = f P = = f m P = 0. I claim that any f O is actually a function f : X P 1 k. To see why, say that f = h 1 + = h + for some polynomials h 1, h. Since h 1 h, we have h 1 h = g f, so that h 1 P h P = g P f P = 0. Hence fp = h 1 P : 1 = h P : 1 is a welldefined element. We are motivated by the following philosophy, usually attributed to Alexander Grothendieck: Instead of studying a variety X, study its ring of functions O. Zariski Topology. We wish to place a topology on an affine variety X. We say that a subset V X is closed if there exists a subset T k[x 1, x,..., x n ] such that } V = ZT = P X fp = 0 for all f T. T may be taken to be an ideal containing, but this is not crucial. We say that a subset U X is open if U = X V is the compliment of some closed set V. I claim that: Both and X are open sets. If U } is a arbitrary collection of open sets that U is also open. If U } is a finite collection of open sets, then U is also open. 1
2 Let me explain why. It is easy to see that = X Z 0} and X = X Z 1}. Say that each U = X V for some closed set V = ZT. De Morgan s Laws imply that U = X V = X Z T, U = X V = X Z T. This topology is called the Zariski topology, named after Oscar Zariski. Relation with Complex Analysis. Here are how the definitions we ve studied before match up with the ones gives thus far: Complex Analysis Commutative Algebra C, the ambient space A 1 k, affine space D, region in C n X, affine variety in P n k Ω, open set U = X ZT, Zariski open set f : D C, analytic function f : X P 1 k, rational function S, sheaf of analytic functions O, ring of rational functions Example. We seek a working definition for the types of maps allowed between varieties. To this end, we consider a simple example. Let k be a field such that 1 k. Let f 1 x, y = x + y 1 be a polynomial from the ring k[x, y], and consider the following affine variety: } X = x, y A k x + y 1 = 0. This is simply the unit circle. There is a morphism ψ : P 1 k X defined by a a1 : a 0 1 a 0 a 1 a 0,. a 1 + a 0 a 1 + a 0 Note that this map is welldefined and only involves ratios of polynomials. We will show that this map is birational by exhibiting another morphism ϕ : X P 1 k such that ψ ϕ = id X and ϕ ψ = id Y a maps also involving ratios of polynomials. To this end, consider the following subsets of X: } U 1 = x, y X x, y 1, 0 = X Z x + 1}, U = x, y X } x, y +1, 0 = X Z x 1}. These are open sets such that X = U 1 U, so they define an open cover of the unit circle. Define now the following maps: ϕ 1 : U 1 P 1 } k x, y 1 + x : y which send ϕ : U P 1 k x, y y : 1 x
3 Using the formal relation x + y = x y = a 1 a 0 = y 1 x a x, y = 1 a 0, a 1 + a 0 a 1 a 0 a 1 + a 0 we immediately see that ϕ U1 1 U = ϕ U1 U. In particular, the pairs U 1, ϕ 1 U, ϕ are equivalent, so they define a morphism 1 + x : y if x, y U ϕ : X P 1 1, and k which sends x, y y : 1 x if x, y U. It is easy to verify that ψ ϕ = id X and ϕ ψ = id Y. Note how we glued two maps together to find a morphism. Abstract Nonsingular Varieties Points as Primes. Recall that Grothendieck s philosophy is to replace X with O whenever possible. But what exactly is the relationship between them? Assume now that k is algebraically closed. Fix a prime ideal = f 1, f,..., f m in k[x 1, x,..., x n ]; then O = k[x 1, x,..., x n ]/ is an integral domain. Continue to denote X as the collection of P A n k such that fp = 0 for all f. Proposition 1. There is a onetoone correspondence between points P = a 1, a,..., a n on X and maximal ideals m P = x 1 a 1, x a,..., x n a n in mspec O. Proof. Fix a point P = a 1, a,..., a n on X. We show that m = x1 a 1, x a,..., x n a n is a maximal ideal. Recall that an ideal m O is maximal if and only if O/m is a field. Since O/m k[a 1,..., a n ] = k, we see that m is indeed maximal. Conversely, fix a maximal ideal m in mspec O. We show that m P = x 1 a 1, x a,..., x n a n for some point P = a 1, a,..., a n on X. Since O/m is a finite extension of k and k is algebraically closed, we have a map O O/m k. Let a i k be the image of x i O, and consider the ldeal m P = x 1 a 1, x a,..., x n a n. Since each polynomial x i a i is in the kernel of this map, we must have m P m O. But m P is a maximal ideal, so m = m P. Abstract Varieties. To recap, there is an injective map X mspec O Spec O, for some integral domain O, which is defined by P m P. Even though is map not surjective, for any integral domain O we say that Spec O is a abstract variety. Note that X mspec O is really an affine variety, although we will abuse notation and call Spec O an affine variety as well. Example. Let O = Z. The only prime ideals in O are m = p Z and 0} for any rational prime p Z, because Fp for m = p Z, and O/m Z for m = 0}. In particular, mspec O, 3, 5,..., p,... } consists of the rational primes and Spec O = 0} mspec O. Hence the affine variety mspec O has points which are the rational primes, 3
4 yet the abstract variety Spec O is strictly larger. These are varieties but we cannot express the points as zeroes of some polynomial! Zariski s Notion of Nonginsgularity. Let s continue to denote X as the collection of P A n k such that fp = 0 for all f. We say that X is a nonsingular affine variety if we can define a tangent at every point P X. We can make this rigorous using derivatives as follows: Since X is the collection of zeroes for the set = f 1, f,..., f m, we define the Jacobian of X at a point P to be the m n matrix f 1 f 1 f 1 J P P P x 1 x x n f 1, f,..., f m P = f m x 1 P f m x P f m x n P Since is a prime ideal, the number of polynomials cannot be greater than the number of variables: m n. The rank of this matrix should be as large as possible, so we say that X is a nonsingular affine variety of rank k J f 1, f,..., f m P = m for every P X. Points as DVRs. Keeping Grothendieck s philosophy in mind, we wish to translate this idea into one involving the ring O. We give some motivation: Say that X is the collection of P A n k such that fp = 0 for all f. If P = a 1, a,..., a n is a point on X, then any f in O = k[x 1, x,..., x n ]/ has a Taylor series expansion fx 1, x,..., x n = fp + + n i=1 n i=1 f x i x 1 a 1 n j=1 1. f x i x j x i a i x j a j + Modulo the maximal ideal m P = x 1 a 1, x a,..., x n a n, we have fq + P fp mod m P and fq + P fp + fp Q mod m P for any Q X. This means f mod m P is the same as computing fp. f mod m P is the same as computing fp. For now, assume that n = and m = 1, i.e., O = k[x, y]/ f 1. The following proposition explains then importance of m P /m P in associating P X with a discrete valuation ring. It is a restatement of Proposition 9. on pages in AtiyahMacdonald. Theorem. For each P X, the following are equivalent: i. X is nonsingular at P, i.e., f 1 P 0, 0. ii. dim k mp /m P = 1. iii. m P O P is a principal ideal. iv. O P, the localization of O at m P, is a discrete valuation ring. v. O P is integrally closed. Proof. i ii. We make a general observation, following Zariski. Say that we have O = k[x 1, x,..., x n ]/, as before. For each point P X, we define the Jacobian to be the 4
5 m n matrix f 1 f 1 f 1 J P P P x 1 x x n f 1, f,..., f m P = f m f m f m. P P P x 1 x x n Hence multiplication by this matrix induces a short exact sequence 0} T P X A n k A m k where T P X is the tangent space of X at P. We have a perfect i.e., bilinear and nondegenerate pairing T P X m P /m P k defined by Q, f fp Q. For this reason, we call m P /m P the cotangent space of X at P. Hence dim k mp /m P = dimk T P X. But dim k T P X = n m if and only if J f 1, f,..., f m P has rank m. The case of interest follows with n = and m = 1. ii iii. We will abuse notation and think of m P as an ideal of O P. As m P is a maximal ideal, Nakayama s Lemma states that we can find t m P where t / m P. Consider the injective map O/m P m P /m P defined by x t x. Clearly this is surjective if and only if m P = t O P is principal. Recall now that dim k O/mP = 1. iii = iv. Say that m P = t O P as a principal ideal. In order to show that O P is a discrete valuation ring, it suffices to show that any nonzero x O P is in the form x = t m y for some m Z and y O P. Consider the radical of the ideal generated by x: x = y OP y n x O P for some nonnegative integer n }. As O P has a unique nonzero prime ideal, we must have x = m. But then there is largest nonnegative integer m such that t m 1 / x O P yet t m x O P. Hence y = x/t m O P but y / m P. iv = v. Say that O P is a discrete valuation ring. Say that x K is a root of a polynomial equation x n + a 1 x n a n = 0 for some a i O P. Assume by way of contradiction that x / O P. Then v P x < 0, so that v P 1/x > 0, hence y = 1/x is an element of O P. Upon dividing by x n 1 we have the relation x = a 1 +a y+ +a n y n 1 O P. This contradiction shows that O P is indeed integrally closed. v = iii. Say that O P is integrally closed. We must construct an element t O P such that m P = t O P. Fix a nonzero x m P. By considering the radical x and noting that m P is a finitely generated O P module, we see that there exists some m Z such that m m P x O P yet m m 1 P x O P. Choose y m m 1 P such that y / x O P, and let t = x/y be an element in K. Consider the module 1/t m P O P ; we will show equality. As y / O P, we have 1/t / O P, so that 1/t is not integral over O P. Then 1/t m P cannot be a finitely generated O P module, we have 1/t m P m P. As there is an element of 1/t m P which is not in m P, we must have equality: 1/t m P = O P. Hence m P = t O P as desired. Examples. Take f 1 x, y = y x 3 + x, and P = 0, 0. Then f 1 P = 1, 0 is nonzero. The maximal ideal is m P = x, y and m P = x, x y, y, so that x = x x + 1 y m P. Hence m P /m P is a 1dimensional kvector space spanned by y alone. Now take f 1x, y = y x 3, and P = 0, 0. Then f 1 P = 0, 0 is zero. The quotient m P /m P is a dimensional kvector space spanned by both x and y. 5
6 Dedekind Domains. Here is an application to the rings one studies in Algebraic Number Theory. Corollary 3. The following are equivalent: i. X is a nonsingular curve, i.e., f 1 P 0, 0 for all points P X. ii. dim k m/m = 1 for all maximal ideals m of O. iii. O is a Dedekind domain. Proof. i ii. Recall that the map X mspec O defined by P m P is a bijection. Now use the previous theorem. ii iii. A Dedekind domain is a Noetherian integral domain of dimension 1 that is integrally closed. Using the previous theorem, it suffices show that the localization O m is integrally closed for each maximal ideal m if and only if O is integrally closed. But this is clear. If not, consult Theorem 5.13 on page 63 of Atiyah Macdonald. Abstract Nonsingular Curves. Let O be a Noetherian integral domain with Krull dimension dim O = 1. We say that X = Spec O is an abstract curve. Note that X is nonsingular precisely when O is a Dedekind domain. For instance, when O = Z, we think of X = Spec Z as a nonsingular curve! Abstract Nonsingular Varieties. This can be generalized considerably. Let O be a Noetherian integral domain. As O is an integral domain, we have O K; as O is Noetherian, the Krull dimension, dim O, is finite. We say that the abstract variety X = Spec O is nonsingular if O is integrally closed. It is not hard to show that dim k m/m = dim O for any maximal ideal m of O; we define this as the dimension of X. 6
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