MATH 763: ALGEBRAIC GEOMETRY ANDREI CALDARARU Notes by Connor Simpson Disclaimer: there may be typos. Use these notes at your own risk. 1. September 6, 2018: Motivation Site: www.math.wisc.edu/~andreic/teaching/763/index.html Essentially all information is there. Logistics: There will be weekly HW posted on website. Backgroud: Primary decomposition, tensor products, and valuation rings will all be important in this class. Book: Hartshorne. We will follow fairly closely for the first two thirds of the course, and then will switch to Riemann-Roch, divisors, etc. 1.1. A little history. AG was practiced a lot in Italy in the early 1900's, but their work was built on very weak foundations. Serre helped solidify them with Faisceaux Algebriques Coherents and Grothendieck later developed this work into the theory of schemes. The problem for the Italians was essentially that the global sections functor is left but not right exact. Serre solved this problem by extending the short (not necessarily right exact) sequence gotten by applying global sections to a long ex- act sequence with cohomology groups. Grothendieck generalized this into abelian categories and created homological algebra in the process. Grothendieck and Serre also wanted to absorb number theory into AG. This led to EGA and SGA. Unfortunately, EGA and SGA were way too technical for anyone to learn from. Hartshorne wrote his book to fix this problem. To make his book comprehensible, he reduced the level of generality (and hence the level of technicality) and picked the most important results instead of everything. 1.2. Two motivating theorems that give the right flavor but that we don't need much background to understand or appreciate. Theorem 1.1. Let X be a compact smooth manifold and let A = C1(X) be the ring of real-valued smooth functions on X. There exists a bijection between ∼= fponts of Xg ! fmp ⊆ A : mp is a maximal idealg p 7−! ff 2 A : f(p) = 0g Proof. We first show that the map is well-defined; that is, that the image of a point is a maximal ideal. It is clear that the set of functions corresponding to a point is an ideal. To see that mp is maximal, let f 2 A n mp. Then f(p) 6= 0. Let I = mp + hfi. Pick g 2 A. We then have g(p) g − f 2 m =) g 2 I =) I = A: f(p) p 1 2 ANDREI CALDARARU We now need to show that all maximal ideals of A take this form. For I an ideal, define Z(I) := fp 2 X : f(p) = 0; 8f 2 Ig. Let m be a maximal ideal of A. We will first show that Z(m) is nonempty. Assume towards a contradiction that Z(M) is empty. Then for all p 2 X, there exists fp 2 m such that fp(p) 6= 0. Now, let Ufp be the open set of points at which fp is nonzero. By compactness, finitely many such sets fU1;:::;Ukg, corresponding to functions ff1; : : : ; fkg ⊆ m, P 2 cover X. Let g = i fi 2 m. This function is nowhere-zero (we are working with 1 real-valued functions!). This implies that g · g = 1 2 m, a contradiction. Finally, different points give different ideals because, given two distinct points, we can find a bump function that vanishes at one but not the other. Remark 1.2. In fact, the topology on X generated by defining closed sets to be the zero sets of elements of A also recovers the topology on X. This is rather strange, and is a feature of the fact that we are not taking a ring of algebraic functions, but the bigger ring of smooth functions. Remark 1.3. A is also a little bit nicer than the general case in that it has no nilpotents. However, in the general case, we do have rings with nilpotents, which do not correspond to natural geometric constructions. Schemes will let us examine these using the same machinery that we use on everything else. Since we will not be doing schemes in this class, we will be studying mostly nilpotent-free rings. Exercise 1.4. If I is an ideal in A and p 2 Z(I), show that I ⊆ mp. Exercise 1.5. Show that A is a direct sum when X has multiple connected compo- nents. A loose version of the second theorem is as follows Theorem 1.6. We can parameterize all the rational points on the unit circle by the rational points on a line. A more precise version is Theorem 1.7. Let Q be a smooth projective quadric over a field k with a rational point. Then Q is isomorphic to P1. We don't have time to prove this today, but the proof is captured in the picture below. MATH 763: ALGEBRAIC GEOMETRY 3 R Q (−1; 0) L 2. September 11, 2018: Algebraic sets and the Zariski topology We'll now prove Theorem 1.6. Proof of Theorem 1.6. x2 + y2 = 1 is the equation of the unit circle and let Q be a rational point on the circle. Let L be the line x = 1. If Q is a rational point of S1, then the line determined by (−1; 0) and Q has rational slope so the intersection point of L and of the line determined by (−1; 0) and Q is a rational point R. Conversely, suppose that R = (1; r2) is a rational point of L. Then Q = (q1; q2) r2 satisfies 2 (q1 + 1) = q2 so q1 is a root of the quadratic equation r x2 + 2 (x + 1) = 1: 2 This equation already has a rational root x = −1, so q1 is rational and it follows that q2 is as well. (Note that this actually holds over any field: we could have taken L to be the y-axis instead, which case we would not need a 1=2 in the relation between q1 and q2.) The lesson here is that we can often convert geometric arguments into rigorous algebraic ones as long as our spaces are determined by polynomal equations. The other lesson is that if we work in algebraic geometry instead of just geometry, we can often extend our arguments to work for more general spaces (i.e. over other fields). By the end of the course, we will be able to fully understand all the words in the more general Theorem 1.7 and we should be able to prove it. Remark 2.1. Theorem 1.7 is false if the curve in question is cubic instead of being a quadric. In general, cubics are much more complicated than quadrics. Now that we've seen these motivating examples, we can move to proper content. Definition 2.2. Fix an algebraically closed field k = k¯ and let n n n A = Ak = f(x1; : : : ; xn) 2 k g 4 ANDREI CALDARARU be n-dimensional affine space. n Definition 2.3. The ring of regular functions on A is A = k[x1; : : : ; xn]. Each element of f 2 A defines a function f : An ! k Remark 2.4. From a modern perspective, we never want to think about a space without thinking about what functions on it we want to consider. For example, when we think about topological manifolds, we want to think about continuous functions. When we think about differentiable manifolds, we want to think about smooth functions, and so on. Remark 2.5. Why do we consider An instead of kn? Because we want to allow affine transformations and writing kn brings along the mental baggage of thinking of kn as a vector space, where we allow only linear transformations. Definition 2.6. If f 2 A, let Z(f) := fp 2 An : f(p) = 0g. More generally, if T ⊆ A then n \ Z(T ) = fp 2 A : f(P ) = 0; 8f 2 T g = Z(f) f2T is the vanishing set or zero set of T . If Y ⊆ An is of the form Y = Z(T ) for some T ⊆ A, then we call Y an algebraic set. Theorem 2.7. The algebraic sets form the closed sets of a topology called the Zariski topology. Proof. Both An and ; are closed because An = Z(f0g) and ; = Z(f1g). Arbitrary intersections are fine because ! \ [ Z(Tα) = Z Tα ; α α and finite unions are fine because Z(T1) [ Z(T2) = Z(ffg : f 2 T1; g 2 T2g) That the LHS is contained in the RHS is clear. For the reverse containment, let P 62 Z(T1 [ Z(T2). Then there exists f 2 T1 and g 2 T2 such that f(P ); g(P ) 6= 0. But then f(P )g(P ) 6= 0, so P is not in the RHS. Observe that Z(T ) = Z(hT i) where hT i ⊆ A is the ideal generated by the elements of T . Furthermore, since A is noetherian, hT i = hf1; : : : ; fki for some finite k; therefore, we need only concern ourselves with the vanishing sets of finite sets of polynomials. Example 2.8 (The Zariski topology on A1). The closed sets are the whole space and finite subsets of it. Remark 2.9. In higher dimensions there are infinite algebraic sets; for example, the coordinate x-axis of A2 is V (x) and is an infinite set. However, when we define curves later, we will see that the closed sets of curves are just finite sets and the whole space, just like for A1. Beware: not all these curves will be the same for us, even though they will all be homeomorphic in the Zariski topology! The lesson to take here is that the Zariski topology contains very little information about the underlying space.