ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ.

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1 ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ. ANDREW SALCH 1. Hilbert s Nullstellensatz. The last lecture left off with the claim that, if J k[x 1,..., x n ] is an ideal, then rad(j) = I(V J (k)). This fact, which depends on k being algebraically closed, was first proven by D. Hilbert, and is called the strong form of Hilbert s Nullstellensatz. Theorem 1.1. (Hilbert s Nullstellensatz, strong form.) Let k be an algebraically closed field, and let J k[x 1,..., x n ] be an ideal. Then I(V J (k)) = rad(j). Proof. If f k[x 1,..., x n ] is in the radical f rad(j), then f n J for some positive integer n, so f n (a 1,..., a n ) = 0 for all (a 1,..., a n ) V J (k). Since k is a field, f (a 1,..., a n ) = 0 as well. So f I(V J (k)). So rad(j) I(V J (k)). (This part did not depend on k being algebraically closed.) Showing that rad(j) I(V J (k)) requires more work. Suppose f rad(j). Since rad(j) is the intersection of the prime ideals of k[x 1,..., x n ] containing J, there exists some prime ideal p of k[x 1,..., x n ] such that f p and such that p J. Choose a maximal ideal m of the ring k[x 1,..., x n ]/p[ f 1 ]. The k-algebra k[x 1,..., x n ]/p[ f 1 ] is a finitely generated k-algebra (by x 1,..., x n, f 1 ), so k[x 1,..., x n ]/p[ f 1 ]/m is a finitely generated k-algebra and also a field. Hence k[x 1,..., x n ]/p[ f 1 ]/m is a finite field extension. (The observation that a finitely generated k-algebra which is also a field is a finite field extension of k is called Zariski s lemma, and a careful proof of the Nullstellensatz would include a proof of Zariski s lemma.) Since k is assumed algebraically closed, any finite field extension of k is equal to k itself. Let (a 1,..., a n ) A n k denote the image of the elements x 1,..., x n of k[x 1,..., x n ] under the composite k[x 1,..., x n ] k[x 1,..., x n ]/p[ f 1 ]/m k. Since J p, we have (a 1,..., a n ) V J (k), and since f is inverted in k[x 1,..., x n ]/p[ f 1 ]/m, we have f (a 1,..., a n ) 0. Hence f I(V J (k)). Hence I(V J (k)) rad(j). Example 1.2. It is important that the Nullstellensatz depends on k being algebraically closed! For example, let k = R (which is clearly not algebraically closed), let n = 1, and let J = (x 2 + 1). Then V J (R) =, so I(V J (R)) = R[x]. So for J = (x 2 + 1) we have but the inclusion is not an equality! J = rad(j) I(V J (R), Corollary 1.3. If k is an algebraically closed field and n is a nonnegative integer, then the operations V and I establish a one-to-one correspondence between radical ideals Date: January

2 2 ANDREW SALCH in k[x 1,..., x n ] and algebraic sets in A n k. This one-to-one correspondence reverses the partial-ordering (given by inclusion) on these two collections. Definition 1.4. To each algebraic set X, we associate the commutative k-algebra k[x 1,..., x n ]/I(X), for which we write A(X). We call A(X) the (affine) coordinate ring of X. Note that A(X) is always a reduced k-algebra, since you showed that R/I is reduced if and only if I is radical, and I(X) is always radical! 2. The Zariski topology. Proposition 2.1. V on products of ideals. Let I, J k[x 1,..., x n ] be ideals. Then: V I (k) V J (k) = V IJ (k). Proof. We have an inclusion of ideals IJ I, so we get an inclusion of algebraic sets in the opposite direction V IJ (k) V I (k). Swap I and J to get a similar inclusion V IJ (k) V J (k). Since both V I (k) and V J (k) are contained in V IJ (k), we have V I (k) V J (k) V IJ (k). Now suppose (a 1,... a n ) V IJ (k), i.e., f (a 1,..., a n ) = 0 for all f IJ. If (a 1,..., a n ) is not contained in V I (k), then there is some element i I such that i(a 1,..., a n ) 0. Then, for all j J, i j IJ, so (i j)(a 1,..., a n ) = i(a 1,..., a n ) j(a 1,..., a n ) = 0 but i(a 1,..., a n ) 0. So j(a 1,..., a n ) = 0, for all j J. So (a 1,..., a n ) V J (k). So every element (a 1,... a n ) V IJ (k) is contained in either V I (k) or V J (k). So V IJ (k) V I (k) V J (k). Theorem 2.2. Let k be an algebraically closed field, and let n be a nonnegative integer. There exists a topology on A n k in which the closed sets are exactly the algebraic sets. Proof. First, the empty set A n k is algebraic, since V k[x 1,...,x n ](k) =. Second, the subset A n k of An k is algebraic, since V (0)(k) = A n k. Third, suppose that X, Y A n k are algebraic sets. Let X = V I(k) and let Y = V J (k). Then X Y = V IJ (k), by Proposition 2.1. So X Y is also an algebraic set. So a union of finitely many algebraic sets is algebraic. Lastly, suppose that {X s } s S is a set of algebraic sets in A n k. For each s S, choose an ideal I s in k[x 1,..., x n ] such that X s = V Is (k). Let J be the ideal of k[x 1,..., x n ] generated by the union s S I s of the ideals I s. Then every element f k[x 1,..., x n ] which vanishes on J also vanishes on s S I s, hence vanishes on each ideal I s, hence f V Is (k) for all s S ; and conversely, if f V Is (k) for all s S, then f vanishes on each I s, hence f vanishes on the union s S I s, hence f vanishes on J and hence f V J (k). We conclude that V J (k) = s S V Is (k) = s S X s. So any intersection of algebraic sets is an algebraic set. Definition 2.3. The topology on A n k in which the closed sets are the algebraic sets is called the Zariski topology. If X A n k is an algebraic set, we give X the subspace topology that it inherits from the Zariski topology on A n k, and we call this subspace topology the Zariski topology on X. 3. Affine varieties. Definition 3.1. Let k be algebraically closed. An affine variety is an algebraic set X A n k such that, if X is equal to the union Y Z of two algebraic sets Y, Z A n k, then either Y = X or Z = X.

3 ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ. 3 In other words, algebraic varieties are those algebraic sets that cannot be written nontrivially as a union of two smaller algebraic sets. Geometrically it makes sense to study such things: if you can answer some question about algebraic varieties, in principle you probably ought to be able to answer the same question about algebraic sets, since every algebraic set is a union of finitely many algebraic varieties. (This last fact, that every algebraic set is a union of finitely many algebraic varieties, is true, but it isn t obvious or trivial! We give a proof, below, in Theorem 4.4.) The algebraic varieties are distinguished among the algebraic sets by an appealingly simple algebraic property: Theorem 3.2. Let k be algebraically closed. Suppose X A n k is an algebraic set. Then X is an affine variety if and only if its coordinate ring A(X) is an integral domain. Proof. The algebraic set X failing to be an affine variety is equivalent to X splitting as a union X = Y Z where Y and Z are both algebraic sets which are proper subsets of X. Suppose X fails to be an affine variety. Then I(X) = I(Y) I(Z). Since neither Y nor Z are equal to X, there exists an element y Y such that y Z, and there exists an element z Z such that z Y. So we can choose an element f I(Y) such that f (z) 0, and we can choose an element g I(Z) such that g(y) 0. So both f and g have nonzero image under the map f, g k[x 1,..., x n ] k[x 1,..., x n ]/ (I(Y) I(Z)) = A(X). But now consider the product f g in k[x 1,..., x n ]. If we have an element x Y, then f g(x) = f (x)g(x) = 0 since f vanishes on Y, and if we have an element x Z, then f g(x) = f (x)g(x) = 0 since g vanishes on Z. Hence f g vanishes on Y Z = X. Hence f g is zero in the quotient algebra k[x 1,..., x n ]/I(X) = A(X). So f, g are zero divisors in A(X). For the converse: suppose that X A n k is an affine variety, and suppose that f, g k[x 1,..., x n ] are such that f g I(X). So ( f g) I(X). So: V ( f ) (k) V (g) (k) = V ( f g) (k) V I(X) (k) So X = (X V ( f ) (k)) (X V (g) (k)). As an intersection of two algebraic sets is algebraic, X V ( f ) (k) = V J1 (k) and X V (g) (k) = V J2 (k) for some ideals J 1, J 2 k[x 1,..., x n ]. The assumption that X is an affine variety together with the fact that X = V J1 (k) V J2 (k) tells us that either X = V J1 (k) or X = V J2 (k). Now V ( f ) (k) V J1 (k), so rad(( f )) rad(j 1 ), so ( f ) rad(j 1 ), so some power of f is contained in J 1. So, if X = V J1 (k), then f I(X). By the same argument, in X = V J2 (k), then g I(X). Either X = V J1 (k) or X = V J2 (k) is true, so either f I(X) or g I(X). Hence, if f g I(X), then either f I(X) or g I(X). So I(X) is a prime ideal. So A(X) = k[x 1,..., x n ]/I(X) is an integral domain. We showed in the previous lecture that there is an order-reversing one-to-one correspondence between algebraic sets in A n k and ideals in k[x 1,..., x n ], when k is algebraically closed. Now Theorem 3.2 tells us that the affine varieties correspond, under this one-to-one correspondence, with the prime ideals: Corollary 3.3. Suppose k is an algebraically closed field. The operations V and I give us a one-to-one correspondence between algebraic varieties in A n k and prime ideals in k[x 1,..., x n ]. This correspondence reverses the partial-ordering given by inclusion. = X.

4 4 ANDREW SALCH 4. Chain conditions and dimension. Definition 4.1. If R is a commutative ring, by the Krull dimension of R we mean the largest integer n such that there exists a chain of proper inclusions of prime ideals of R. p 0 p 1 p n Some examples of Krull dimension: Any field k has Krull dimension 0, since the only prime ideal in k is (0). Suppose R is a commutative ring such that there exists a surjective ring homomorphism f : R k, with k a field, such that the kernel of f is a nilpotent ideal of f. Then the only prime ideal of R is ker f, and the Krull dimension of R is still zero. Principal ideal domains R (such as Z) that are not fields have Krull dimension 1, since you have the chain (0) (r) for every non-unit r R, and no longer chains of inclusions of prime ideals (you can prove this as an easy exercise if you want). The ring k[x 1,..., x n ] has Krull dimension n. Observation 4.2. Suppose k is algebraically closed. Since the prime ideals of k[x 1,..., x n ]/I are in inclusion-preserving one-to-one correspondence with the prime ideals of k[x 1,..., x n ] containing I, the Krull dimension of k[x 1,..., x n ]/I is less than or equal to the Krull dimension of k[x 1,..., x n ]. In particular, the Krull dimension of k[x 1,..., x n ]/I is finite. So the Krull dimension of A(X) is finite, for all algebraic sets X A n k. Proposition 4.3. Let k be algebraically closed. Let X A n k be an affine variety. Then the length i of the longest chain of proper inclusions X = X 0 X 1 X i = A n k of affine varieties is equal to the Krull dimension of A(X). Proof. Affine varieties containing X are in order-reversing bijection with prime ideals of k[x 1,..., x n ] contained in I(X), by Corollary 3.3. That s it! We refer to the Krull dimension of A(X) as the dimension of X. Lemma 4.4. Let k be algebraically closed. There are no infinite chains of proper inclusions X 0 X 1 X 2... of algebraic sets in A k n. Proof. Any such chain corresponds to a chain of proper inclusions I 0 I 1 I 2... of ideals of k[x 1,..., x n ]. But k[x 1,..., x n ] is a Noetherian ring, so every ascending chain of inclusions of ideals in k[x 1,..., x n ] eventually stabilizes, so there are no infinite ascending chains of proper inclusions of ideals in k[x 1,..., x n ]. Theorem 4.5. Let k be algebraically closed. Then every algebraic set X A n k is a union of finitely many affine varieties. Proof. Suppose X is an algebraic set but not an affine variety. Then X = X 0 X 1 for some algebraic sets X 0, X 1 with X X 0 and X X 1. Either X 0 and X 1 each are a union of finitely many affine varieties (in which case X is, as well), or at least one of the two algebraic sets X 0, X 1 fails to be a union of finitely many affine varieties. Without loss of generality, we can suppose that X 0 is the one that fails to be a union of finitely many affine varieties. Then X 0 = X 00 X 01 for some algebraic sets X 00, X 01 with neither X 00 nor X 01 equal to all of X 0.

5 ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ. 5 Now the process repeats: either each X 00 and X 01 are each a union of finitely many affine varieties (contradicting our assumption about X 0 ), or at least one of the two algebraic sets X 00, X 01 fails to be a union of finitely many affine varieties. Without loss of generality, we can suppose that X 00 is the one that fails to be a union of finitely many affine varieties. Then X 00 = X 000 X 001 for some algebraic sets X 000, X 001 with neither X 000 nor X 001 equal to all of X 00. We iterate this process to get a chain of equalities X = X 0 X 1 = X 00 X 01 X 1 = X 000 X 001 X 01 X 1 =... If X is not a union of finitely many affine varieties, then this chain of equalities never ends, i.e., at no (finite) stage in this process do we get that each X i0...i n is an affine variety. But this means that, stripping off all but the first space in each union, we have an infinite chain of proper inclusions X X 0 X 00 X with each X i an algebraic set in A n k which is impossible by Lemma 4.4. So X must be a union of finitely many affine varieties.