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1 Affine Varieties 1 Affine Varieties We will begin following Kempf's Algebraic Varieties, and eventually will do things more like in Hartshorne. We will also use various sources for commutative algebra. What is algebraic geometry? Classically, it is the study of the zero sets of polynomials. We will now fix some notation. k will be some fixed algebraically closed field, any ring is commutative with identity, ring homomorphisms preserve identity, and a k-algebra is a ring R which contains k (i.e., we have a ring homomorphism ι : k ! R). P ⊆ R an ideal is prime iff R=P is an integral domain. Algebraic Sets n n We define affine n-space, A = k = f(a1; : : : ; an): ai 2 kg. n Any f = f(x1; : : : ; xn) 2 k[x1; : : : ; xn] defines a function f : A ! k : (a1; : : : ; an) 7! f(a1; : : : ; an). Exercise If f; g 2 k[x1; : : : ; xn] define the same function then f = g as polynomials. Definition 1.1 (Algebraic Sets). Let S ⊆ k[x1; : : : ; xn] be any subset. Then V (S) = fa 2 An : f(a) = 0 for all f 2 Sg. A subset of An is called algebraic if it is of this form. e.g., a point f(a1; : : : ; an)g = V (x1 − a1; : : : ; xn − an). Exercises 1. I = (S) is the ideal generated by S. Then V (S) = V (I). 2. I ⊆ J ) V (J) ⊆ V (I). P 3. V ([αIα) = V ( Iα) = \V (Iα). 4. V (I \ J) = V (I · J) = V (I) [ V (J). Definition 1.2 (Zariski Topology). We can define a topology on An by defining the closed subsets to be the algebraic subsets. U ⊆ An is open iff An n U = V (S) for some S ⊆ k[x1; : : : ; xn]. Exercises 3 and 4 imply that this is a topology. The closed subsets of A1 are the finite subsets and A1 itself. Definition 1.3 (Ideal of a Subset). If W ⊂ An is any subset, then I(W ) = ff 2 k[x1; : : : ; xn]: f(a) = 0 for all a 2 W g Facts/Exercises 1. V ⊆ W ) I(W ) ⊆ I(V ) 2. I(;) = (1) = k[x1; : : : ; xn] 3. I(An) = (0). 1 Definition 1.4 (Affine Coordinate Ring). W ⊂ An is algebraic. Then A(W ) = k[W ] = k[x1; : : : ; xn]=I(W ) We can think of this as the ring of all polynomial functions f : W ! k. Definition 1.5 (Radical Ideal)p . Let R be a ring and I ⊆ R be an ideal, then i the radical of I is the ideal I = ffp2 R : f 2 I for some i 2 Ng We call I a radical ideal if I = I. Exercise p If I is an ideal, then I is a radical ideal. Proposition 1.1. W ⊆ An any subset, then I(W ) is a radical ideal. Proof. We have that I(W ) ⊆ pI(W ). Suppose f 2 pI(W ). Then f i 2 I for some i. That is, for all a 2 W , f i(a) = 0. Thus, f(a)m = 0 = f(a). And so, f(a) 2 I. Exercises 1. S ⊆ k[x1; : : : ; xn], then S ⊆ I(V (S)). 2. W ⊆ An then W ⊆ V (I(W )). 3. W ⊆ An is an algebraic subset, then W = V (I(W )). p p 4. I ⊆ k[x1; : : : ; xn] is any ideal, then V (I) = V ( I) and I ⊆ I(V (I)) Theorem 1.2 (Nullstellensatz)p. Let k be an algebraically closed field, and I ⊆ k[x1; : : : ; xn] is an ideal, then I = I(V (I)). p Corollary 1.3. k[V (I)] = k[x1; : : : ; xn]= I. To prove the Nullstellensatz, we will need the following: Theorem 1.4 (N¨other'sNormalization Theorem). If R is any finitely gener- ated k-algebra (k can be any field), then there exist y1; : : : ; ym 2 R such that y1; : : : ; ym are algebraically independent over k and R is an integral extension of the subring k[y1; : : : ; ym]. Proof is in Eisenbud and other Commutative Algebra texts. Theorem 1.5 (Weak Nullstellensatz). Let k be an algebraically closed field, and I ( k[x1; : : : ; xn] any proper ideal, then V (I) 6= ;. Proof. We may assume without loss of generality that I is actually a maximal ideal. Then R = k[x1; : : : ; xn]=I is a field. R is also a finitely generated k- algebra, and so by Normalization, 9y1; : : : ; ym 2 R such that y1; : : : ; ym are algebraically independent over k and that R is integral over k[y1; : : : ; ym]. −1 Claim: m = 0. Otherwise, y1 2 R is integral over k[y1; : : : ; ym], and so then −p 1−p −1 y1 +y1 f1 +:::+y1 fp−1 +fp = 0 for fi 2 k[y1; : : : ; ym]. Multiplying through 2 p p−1 p by y1 gives 1 = −(y1f1 + ::: + y1 fp−1 + y1 fp) 2 (y1), which contradicts the algebraic independence. Thus, the field R is algebraic over k. As k is algebraically closed, R = k. k ⊆ k[x1; : : : ; xn] ! R = k Let ai = the image in k of xi. Then xi −ai 2 I. Thus, the ideal generated by (x1 −a1; : : : ; xn −an) ⊆ I ( k[x1; : : : ; xn], and so they I = (x1 −a1; : : : ; xn −an), as it is a maximal ideal. V (I) = V (x1 − a1; : : : ; xn − an) = f(a1; : : : ; an)g 6= ; Note: Any maximal ideal of k[x1; : : : ; xn] is of the form (x1 −a1; : : : ; xn −an) with ai 2 k. This is NOT true over R, look at the ideal (x2 + 1) ⊆ R[x]. It is, in fact, maximal. Now, we can prove the Nullstellensatz. p Proof. Let I ⊆ k[x1; : : : ; xnp] be any ideal. We will prove that I(V (I)) = I. It was an exercise that I ⊆ I(V (I)). p Let f 2 I(V (I)). We must show that f 2 I. n+1 Looking at A , we have the variables, x1; : : : ; xn; y. Set J = (I; 1 − yf) ⊆ k[x1; : : : ; xn; y]. n+1 Claim: V (J) = ; ⊂ A . This is as, if p = (a1; : : : ; an; p) 2 V (J), then (a1; : : : ; an) 2 V (I), then (1 − yf)(p) = 1 − bf(a1; : : : ; an). But f(a1; : : : ; an) = 0, so (1 − yf)(p) = 1, and so p2 = V (J). By the Weak Nullstellensatz, J = k[x1; : : : ; xn; y]. Thus 1 = h1g1 + ::: + hmgm + q(1 − yf) where g1; : : : ; gm 2 I and h1; : : : ; hm; q 2 k[x1; : : : ; xn; y]. Set y = f −1, and multiply by some big power of f to get a polynomial equation once more. N ~ ~ ~ N Then f = h1g1 + ::: + hmgm where thephi = f hi(x1; : : : ; xn): And so, we have f N 2 I, and thus, f 2 I, by definition. exercise: V (I(W )) = W in the Zariski Topology. Irreducible Algebraic Sets Recall: V (y2 − xy − x2y + x3) = V (y − x) [ V (y − x2), and V (xz; yz) = V (x; y) [ V (z). Definition 1.6 (Reducible Subsets). A Zariski Closed subset W ⊆ An is called reducible if W = W1 [ W2 where Wi ( W and Wi closed. Otherwise, we say that W is irreducible. Proposition 1.6. Let W ⊆ An be closed. Then W is irreducible iff I(W ) is a prime ideal. Proof. ): Suppose I(W ) is not prime. Then 9f1; f2 2= I(W ) such that f1f2 2 W . Set W1 = W \ V (f1) and W2 = W \ V (f2). As fi 2= I(W ), W 6⊆ V (fi), and so Wi ( W . Now we must show that W = W1 [ W2. Let a 2 W . Assume a2 = W1. Then f1(a) 6= 0, but f1(a)f2(a) = 0, so f2(a) = 0, thus a 2 W2. (: Exercise 3 This gives us the beginning of an algebra-geometry dictionary. Algebra Geometry n k[x1; : : : ; xn] A radical ideals closed subsets prime ideals irreducible closed subsets maximal ideals points In fact, this is an order reversing correspondence. So I ⊆ J () V (I) ⊇ V (J), but this requires I;J to be radical. Definition 1.7 (N¨otherianRing). A ring R is called N¨otherianif every ideal I ⊆ R is finitely generated. Exercise A ring R is N¨otherianiff every ascending chain of ideals I1 ⊆ I2 ⊆ ::: stabilizes, that is, 9N such that IN = IN+1 = :::. Theorem 1.7 (Hilbert's Basis Theorem). If R is N¨otherian, then R[x] is N¨otherian. Corollary 1.8. k[x1; : : : ; xn] is N¨otherian. Definition 1.8 (N¨otherianTopological Space). A topological space X is N¨otherian if every descending chain of closed subsets stabilizes. Corollary 1.9. An is N¨otherian. n W1 ⊇ W2 ⊇ ::: closed in A , then I(W1) ⊆ I(W2) ⊆ ::: ideals in k[x1; : : : ; xn], and so must stabilize. Theorem 1.10. Any closed subset of a N¨otherianTopological Space X is a union of finitely many irreducible closed subsets. Proof. Assume the result is false. 9W a closed subset of X which is not the union of finitely many irreducible closed sets. As X is N¨otherian,we may assume that W is a minimal counterexample. W is not irreducible, and so W = W1 [W2, where Wi ( W and Wi closed. The Wi can't be counterexamples, as W is a minimal one, but then W = W1 [ W2 and each Wi is the union of finitely many irreducible closed sets. Thus, W cannot be a counterexample. Corollary 1.11. Every closed W ⊆ An is union of finitely many irreducible closed subsets. Example: V (xy) = V (x) [ V (y) [ V (x − 1; y). Recall: X is a topological space, then if Y ⊆ X is any subset, it has the subspace topology, that is, U ⊆ Y is open iff 9U 0 ⊆ X open such that U = U 0 \ Y .
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