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Math 418 Algebraic Geometry Notes Math 418 Algebraic Geometry Notes 1 Affine Schemes Let R be a commutative ring with 1. Definition 1.1. The prime spectrum of R, denoted Spec(R), is the set of prime ideals of the ring R. Spec(R) = fP ⊆ R : P is primeg: Example 1.2. Let R = Z. Then Spec(Z) = f(p): p 2 Z is primeg [ f(0)g. Notice that the ideals (p) are maximal as well as prime. The ideal (0) is not maximal, and is contained in every other element (p) of Spec(Z). Example 1.3. Let R = F a field. Then Spec(F ) = f(0)g. Example 1.4. Let R = F [x], where F is a field. This ring is a principal ideal domain, and so its prime ideals are of the form (f(x)) where f(x) 2 F [x] is irreducible. That is, Spec(F [x]) = f(f(x)) : f(x) 2 F [x] is irreducibleg [ f(0)g: The structure of Spec(F [x]) will depend on F . Example 1.5. Let R = Q[x]. Then we saw that Spec(Q[x]) = f(f(x)) : f(x) 2 Q[x] is irreducibleg[ f(0)g: For any q 2 Q, the linear polynomial xp− q is irreducible, and so (x − q) 2 Spec(Q[x]). The 2 2 polynomial x +bx+c is irreducible whenever b2 − 4c 62 Q. In this case (x +bx+c) 2 Spec(Q[x]). n There are polynomials of arbitrary degree in Spec(Q[x]). For example, (x − 2) 2 Spec(Q[x]) for all n 2 N. Example 1.6. Let R = C[x] (or F [x] where F is algebraically closed). In this case, the irreducible polynomials are exactly the linear polynomials. So Spec(C[x]) = f(x − z): z 2 Cg [ f(0)g. Note that the maximal ideals of C[x] are in one-to-one correspondence with the elements of C. Example 1.7. Let R = C[x; y]. This ring is not a principal ideal domain, however it is Noetherian, and so its ideals are finitely generated. The maximal ideals of R are of the form (x − z; y − w) 2 where z; w 2 C, and are in bijection with the points of C . Other prime ideals are generated by sets of irreducible polynomials, for example (x − y) and (x − y2) are prime. Of course, (0) is always a prime ideal in an integral domain. We would like to view X = Spec(R) as a geometric space. The points in our space will be the elements of Spec(R). We will distinguish the points corresponding to maximal ideals, and call them the closed points of X. In order to view X as a geometric object, we will need what is called a topology on X. In other words, we will need to have some concept of an open neighborhood of a point in X. 1 Definition 1.8. For any subset I ⊆ R, the vanishing set of I is the set of prime ideals of R containing I, that is V (I) := fP 2 Spec(R): I ⊆ P g: We consider these sets to be closed subsets of X. For any element f 2 R, the complement c Xf := V (ffg) ⊆ Spec(R) is called the basic open set of f. A subset U ⊂ X is said to be open if it is a union of basic open sets. The open and closed sets described above are called Zariski open and Zariski closed. The structure they give on X is called the Zariski topology on X. The map V takes subsets of R to subsets of X. It is not, however, injective. Lemma 1.9. If S is a subset of R, and I(S) is the ideal generated by S in R, then V (S) = V (I(S)). Proof. This is homework problem AM 15i. Based on this result, if we want to understand open and closed subsets of X, we can restrict our attention to ideals of R. The following lists some handy properties of V . Lemma 1.10. Let I;J be ideals of R. Then 1. V (I \ J) = V (I) [ V (J). 2. V (I [ J) = V (I) \ V (J). 3. If I ⊆ J, then V (J) ⊆ V (I). Proof. Suppose P is prime and contains I and J. If P contains neither I nor J, then there exist i 2 I and j 2 J, neither of which are in P . This contradicts that ij 2 I \ J ⊆ P . Therefore V (I \ J) ⊆ V (I) [ V (J). If P contains I or J then it certainly contains I \ J, so V (I) [ V (J) ⊆ V (I \ J). Suppose now P contains I [ J. Then P contains both I and J, so P lies in V (I) \ V (J) On the other hand, if P lies in V (I) \ V (J), it contains both I and J, so it lies in V (I [ J). The third property is left as an exercise. Note that although the second property above holds for general subsets of R, the first does not. For example, if we take R to be the integers, I to be the natural numbers and 0, and J to be the negative numbers and 0, then all prime ideals contain I \ J = f0g. But no proper ideals contain I or J. In fact, even if we restrict to ideals of R, the map V will not be injective. For example, if 2 2 we consider R = C[x], then V ((x)) = V ((x )). This is because any prime ideal containing x contains the product x · x, and hence contains x itself. In fact, we could show by induction that V ((x)) = V ((xn)) for any positive integer n. In order to remedy this problem, we will need the following notion. Definition 1.11. Let I be an ideal of R. The radical of I is m r(I) := fr 2 R : r 2 I for some k 2 Ng: As a special case, the nilradical of R is r((0)), in other words, it is the set of nilpotent elements of R. 2 Note that r(r(I)) = r(I) for every ideal I of R, so we may call an ideal whose radical is itself a radical ideal. Lemma 1.12. For any ideal I of R, V (r(I)) = V (I). Proof. This proof is homework problem AM 15i. There is another characterization of radical ideals which will be useful to us. Proposition 1.13. If I is an ideal in R, then r(I) is the intersection of all prime ideals containing I. In particular, the nilradical of R is the intersection of all prime ideals of R. Proof. First, if x 2 r(I) then xm 2 I for some positive integer m. Hence x lies in every prime ideal containing I. This shows one containment. Now we must show that the intersection of all prime ideals containing I lies in r(I). Suppose first the I = (0), and r(I) is the nilradical of R. We wish to show that the intersection of all prime ideals in R contains only nilpotent elements. Suppose then that f is not nilpotent. Consider the set of ideals Σ which contain no power of f. Since f is not nilpotent, the 0 ideal is in Σ. Since Σ is not empty, by Zorn's Lemma it contains a maximal element under inclusion, call it M. If x; y 62 M then the ideal (x) + M contains M, so f m 2 (x) + M for some positive integer m. Similarly, f n 2 (y) + M for some positive integer n. Therefore f n+m 2 (xy) + M, and xy 62 M. This shows M is prime, so f is not in the nilradical of R. Now for a general ideal I, we can apply the previous argument to the ring R=I to get the result. Corollary 1.14. If I and J are ideals of R such that V (I) = V (J), then r(I) = r(J). In particular, the map V is an injective map from the set of radical ideals of R to subsets of X. Definition 1.15. The space X = Spec(R) together with the Zariski topology is called an affine scheme. We may now wish to consider maps between different affine schemes. There is a close relationship between such morphisms and ring homomorphisms. Definition 1.16. Let X = Spec(R) and Y = Spec(S). A morphism of affine schemes f : X ! Y is a map of sets given by a ring homomorphism f ∗ : S ! R as follows: For any prime ideal P 2 X, f(P ) := (f ∗)−1(P ). The map f ∗ is called the pull-back map of f. Note that this map is well defined, since if xy 2 (f ∗)−1(P ), then f ∗(xy) 2 P , hence at least one of (f ∗)−1(x), (f ∗)−1(y) lies in (f ∗)−1(P ): It is important to note, however, that if Q 2 Y is prime, f ∗(Q) need not be prime. ∗ Example 1.17. Suppose we start with the quotient map f : C[x; y] ! C[x; y]=(xy). We can write any ideal of C[x; y]=(xy) as I + (xy) for an ideal I of C[x; y] containing (xy). Therefore, the map f is simply the inclusion map of prime ideals of C[x; y] containing (xy) back into C[x; y]. Note that the maximal ideals containing (xy) are those of the form (x; y − a) and (x − b; y) for some a; b 2 C.
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