NOTES ON FIBER DIMENSION

NOTES ON FIBER DIMENSION SAM EVENS Let φ : X Y be a morphism of affine algebraic sets, defined over an algebraically closed field k. For y Y, the set φ 1 (y) is called the fiber over y. In these notes, I explain some basic results about the dimension of the fiber over y. These notes are largely taken from Chapters 3 and 4 of Humphreys, Linear Algebraic Groups, chapter 6 of Bump, Algebraic Geometry, and Tauvel and Yu, Lie algebras and algebraic groups. The book by Bump has an incomplete proof of the main fact we are proving (which repeats an incomplete proof from Mumford s notes The Red Book on varieties and Schemes ). Tauvel and Yu use a step I was not able to verify. The important thing is that you understand the statements and are able to use the Theorems 0.22 and 0.24. Let A be a ring. If p 0 p 1 p k is a chain of distinct prime ideals of A, we say the chain has length k and ends at p. Definition 0.1. Let p be a prime ideal of A. We say ht(p) = k if there is a chain of distinct prime ideals p 0 p k = p in A of length k, and there is no chain of prime ideals in A of length k +1 ending at p. If B is a finitely generated integral k-algebra, we set dim(b) = dim(f), where F is the fraction field of B. Theorem 0.2. (Serre, Local Algebra, Proposition 15, p. 45) Let A be a finitely generated integral k-algebra and let p A be a prime ideal. Then ht(p) = dim(a) dim(a/p). Theorem 0.3. (see Matsumura, Commutative Ring Theory, Theorem 13.5, p. 100) (i) Let A be a Noetherian ring and let f A be a nonzero nonunit. Then if p is minimal among prime ideals of A containing f, ht(p) 1. (ii) If A is also an integral domain and p is as in (i), then ht(p) = 1. To prove (ii) using (i), note that (0) p is chain of length 1, so ht(p) 1. The above results in commutative algebra are not easy consequences of definitions. You can find approaches to algebraic geometry which get around using them (notably the treatment in Springer s book on Algebraic Groups). My feeling is that it is better to understand these results. For the purposes of this course, you may simply believe them, but if you are going to do mathematics involving algebraic geometry, it is useful to understand these results in commutative algebra. Theorem 0.4. ( Going Down theorem)(matsumura, as above, Theorem 9.4, p. 68, or Dummit and Foote, Theorem 21, p. 668) Let B be an integrally domain and suppose that 1

2 S. EVENS B is integral over a subring A. Suppose that A is integrally closed. Let p 2 p 1 be prime ideals of A and suppose that q 1 is a prime ideal of B such that q 1 A = p 1. Then there exists a prime ideal q 2 of B such that q 2 q 1 and q 2 A = p 2. Theorem 0.5. (see Eisenbud, Commutative Algebras with a View Toward Algebraic Geometry, Corollary 13.13). Let A be a finitely generated integral k-algebra with fraction field K. Let B be the integral closure of A. Then B is a finitely generated A-module. In particular, B is a finitely generated integral k-algebra. The first assertion of the Theorem is that there exist b 1,...,b k B such that B = Ab 1 + + Ab k. It follows easily that B is generated as an algebra by {b 1,...,b k } and any set of generators of A. Lemma 0.6. (Serre, Local Algebra, Lemma 1, p. 8) Let p 1,...,p n be prime ideals of a ring A. If I is an ideal of A and I n i=1p i, then I p i for some i. Proposition 0.7. (Tauvel and Yu, Theorem 3.28) (i) Let A be an integrally closed domain. Then A[t] is an integrally closed domain. (ii) If A is an integrally closed domain, then A[t 1,...,t n ] is an integrally closed domain. Proof : Let K be the fraction field of A and let p(t),q(t) be monic polynomials such that p(t)q(t) A[t]. Then p(t),q(t) A[t]. Indeed, let M be the algebraic closure of K. Factor p(t) into linear factors in M[t] with roots α 1,...,α k and factor q(t) into linear factors with roots β 1,...,β r. If γ is one of the α i or β j, then p q(γ) = 0. Since p q is monic in A[t], it follows that γ is integral over A in M. The coefficients of p(t) and q(t) are polynomials in the α i and β j and hence are integral over A, and in K. Thus, these coefficients are in A, so p(t),q(t) A[t]. Let K and L be the fraction fields of A and A[t] respectively. Let α L be integral over A[t]. Then α is also integral over K[t] A[t], so since K[t] is a UFD and hence integrally closed, α K[t]. Let Q(x) A[t][x] be a monic polynomial with Q(α) = 0 and write Q(x) = x m + p m 1 x m 1 + +p 0, with p i A[t]. Choose s so s is larger than the degree of α = α(t) and also so for all i = 0,...,m 1, s is larger than the degree of p i = p i (t). Let β = α t s K[t], so by the choice of s, β is monic in K[t]. Let R(x) = Q(x + t s ). Then R(β) = Q(β + t s ) = Q(α) = 0. Expand R(x) = x m +q m 1 x m 1 + +q 0 as a polynomial in x with coefficients in A[t]. The term q 0 = t sm +p m 1 t s(m 1) + +p 0 is monic in t by the choice of s. Since R(β) = 0, q 0 = β(β m 1 +q m 1 β m 2 + +q 1 ) = β γ,γ = (β m 1 +q m 1 β m 2 + +q 1 ) K[t]. Since q 0 and β are monic in K[t], it follows that γ is monic in K[t]. Since q 0 A[t], it follows from the assertion of the first paragraph of this proof that β A[t]. Hence, α = β +t s A[t], which establishes the first assertion. The second assertion follows by an easy induction.

Let X be an affine variety and let f k[x] be a nonzero nonunit. Then the zero set V(f) is called a hypersurface. If X is an affine variety and Y X is a closed subvariety, then codim X (Y) = dim(x) dim(y) is the codimension of Y in X. It is clear that if Z Y is a closed subvariety, then codim X (Z) = codim X (Y)+codim Y (Z). Theorem 0.8. Let X be an affine variety and let Z X be a closed subvariety. Then codim X (Z) = 1 if and only if Z is an irreducible component of a hypersurface V(f) X. Proof: Suppose codim X (Z) = 1. Since Z X, I(Z) is a nonzero prime ideal of k[x], so we can choose nonzero f I(Z). Clearly, f is not a unit. Thus, V(f) is a proper subset of X and Z V(f). Let V(f) = Y 1 Y k be the irreducible components of the hypersurface V(f). Then Z Y i for some i. But dim(y i ) < dim(x), so the hypothesis implies that dim(y i ) dim(z). It follows that Y i = Z. Now suppose that Z is an irreducible component of a hypersurface V(f). Thus, f I(Z), and I(Z) is prime. If f J I(Z), where J is prime, then Z V(J) V(f), so since V(J) is irreducible, Z = V(J). Hence, J = I(Z). Thus, I(Z) is minimal among all prime ideals of k[x] containing f. Hence by Theorem 0.3, ht(i(z)) = 1, so dim(z) = dim(k[z]) = dim(k[x]/i(z)) = dim(k[x]) 1 = dim(x) 1 by Thereom 0.2. This last theorem can be proved without using 0.2 and 0.3 if X = k n since then we can reduce to the case when V(f) is irreducible. Corollary 0.9. LetX be anaffine varietyandlety beaclosedsubvariety. Ifcodim X (Y) = r, thenfor eachi = 1,...,r, thereexistsaclosedsubvarietyy i ofx such that codim X (Y i ) = i and Y = Y r Y r 1 Y 1. Proof: This is clear in case r = 1, and we argue by induction on r. Since Y X, there exists nonzero f I(Y). Since f is a nonunit, by Theorem 0.8, all irreducible components of V(f) have codimension 1 in X. Since Y is irreducible and Y V(f), it follows that Y Z 0 for some irreducible component Z 0 of V(f). Take Y 1 = Z 0. Then codim Y1 (Y) = r 1, so by induction, there exists a chain of closed subvarieties Y r = Y Y r 1 Y 2 in Y 1 such that codim Y1 (Y i ) = i 1. Then Y r Y 1 is the desired chain in X. Corollary 0.10. LetX be an affine varietyandlet f 1,...,f r k[x]. IfZ isan irreducible component of V(f 1,...,f r ), then codim X (Z) r. Proof: Proceed by induction on r, noting that the case r = 1 follows by Theorem 0.8. Let Z be an irreducible component of V(f 1,...,f r ). Since Z V(f 1 ), Z Y for some irreducible component Y of V(f 1 ). Let f 2,...,f r be the images of f 2,...,f r in k[y]. Then Z V(f 2,...,f r ) V(f 1,...,f r ). Since Z is a maximal closed irreducible subset 3

4 S. EVENS of V(f 1,...,f r ), it follows that Z is an irreducible component of V(f 2,...,f r ) contained in Y. Apply the inductive assumption with Y in place of X to obtain codim Y (Z) r 1. Since codim X (Y) 1 by Theorem 0.8, we get codim X (Z) r. Corollary 0.11. Let X be an affine variety and let Y X be a closed subvariety. Suppose codim X (Y) = r. Then Y is an irreducible component of V(f 1,...,f r ) for some f 1,...,f r k[x]. Proof: We prove the following stronger statement: (*) if we are given a chain of closed subvarieties Y r Y 1 with codim X (Y i ) = i, then there exists f 1,...,f r k[x] such that for all 1 q r, every irreducible component of V(f 1,...,f q ) has codimension q in X and Y q is an irreducible component of V(f 1,...,f q ). If we assume (*) and take Y r = Y and use 0.9 to find a chain Y r Y 1 as above, we complete the proof. Prove (*) by induction on q. If q = 1, then by Theorem 0.8, Y 1 is an irreducible component of V(f) for some nonzero nonunit f k[x], and every irreducible component of V(f) has codimension one in X. Now assume we have found f 1,...,f q 1 as above. Let Z 1 = Y q 1, Z 2,...,Z m be the irreducible components of V(f 1,...,f q 1 ), so codim X (Z j ) = q 1 for all j. Hence, dim(y q ) < dim(z j ), so no Z j Y q, so I(Y q ) I(Z j ) for all j. Thus, I(Y q ) I(Z 1 ) I(Z m ) by Lemma 0.6. Hence, there exists f q I(Y q ) such that f q I(Z j ) for all j. Thus, f q is a nonzero nonunit in k[z j ] for all j, where f q is the image of f q in k[z j ]. Hence, by Theorem 0.8 applied to f q k[z j ], every irreducible component W of V(f q ) = V(f q ) Z j has codim Zj (W) = 1. Thus, codim X (W) = q. Let V be an irreducible component of V(f 1,...,f q ). Then V V(f 1,...,f q 1 ), and it follows that V is in some irreducible component Z j. Thus, V V(f q ) Z j. Hence V W, where W is an irreducible component of V(f q ) Z j. But codim X (V) q by Corollary 0.10, so dim(v) dim(w), and hence V = W. Finally, note that Y q V(f 1,...,f q ), so since Y q is irreducible, Y q is a subset of one of the above irreducible components V. Since dim(y q ) = dim(v), Y q = V. Theorem 0.12. Let φ : X Y be a dominant morphism of affine varieties. Let r = dim(x) dim(y). Let W Y be a closed subvariety and let Z be an irreducible component of φ 1 (W). If φ(z) = W, then dim(z) dim(w)+r. In particular, if y φ(x), then each irreducible component of the fiber φ 1 (y) has dimension at least r. Proof: Let s = codim Y (W). By Corollary 0.11, W is an irreducible component of V(f 1,...,f s ) for some f 1,...,f s k[y]. Let g i = φ (f i ) for i = 1,...,s. It follows from definitions that Z V(g 1,...,g s ). Since Z is irreducible, Z Z 0, where Z 0 is an irreducible component of V(g 1,...,g s ). Again from definitions, φ(v(g 1,...,g s )) V(f 1,...,f s ), so φ(z) φ(z 0 ) V(f 1,...,f s ). Note that φ(z 0 ) is irreducible by Exercise

8, problemset1. SinceW = φ(z)byassumption, W φ(z 0 ), sow = φ(z 0 )bydefinition of irreducible component. Thus, φ(z 0 ) W, so Z 0 φ 1 (W). But Z is an irreducible component of φ 1 (W), so Z = Z 0. Hence Z is an irreducible component of V(g 1,...,g s ) so by Corollary 0.10, codim X (Z) s. Thus, dim(z) dim(x) s = r+dim(y) s = r+dim(w). Remark 0.13. An affine variety X is called normal if k[x] is integrally closed in k(x). The affine variety k n is normal since a unique factorization domain is integrally closed. Remark 0.14. Let X be a normal affine variety. Then X k n is a normal variety. Indeed, k[x k n ] = k[x][t 1,...,t n ] is integrally closed by Proposition 0.7. Proposition 0.15. Let φ : X Y be a finite dominant morphism of affine varieties. Then (i) φ is surjective. (ii) If y Y, then the fiber φ 1 (y) is finite. (iii) Let Z X be a closed subvariety. Then φ(z) is closed, dim(z) = dim(φ(z)), and φ : Z φ(z) is finite. (iv) Suppose that Y is normal and let W Y be a closed subvariety. For any irreducible component Z of φ 1 (W), φ(z) = W and dim(z) = dim(w). Proof: We leave (i) as an exercise. For (iii), let ψ = φ Z : Z V := φ(z) be the restriction of φ to Z. We claim that ψ is finite. Given the claim, since ψ is clearly dominant, ψ is surjective, so φ(z) = ψ(z) = V, so φ(z) is closed. Thus, k[z] is integral over k[v] so k(z) is algebraic over k(v), and it follows from properties of transcendence bases that dim(z) = dim(v). Thus, we have reduced (iii) to the claim. Fortheclaim, leti : Z X andletj : V Y betheinclusions. Thensinceφ i = j ψ, ψ j = i φ. Thus, if f k[y], (*)φ (f) + I(Z) = ψ (f + I(V)) in k[z] = k[x]/i(z) (this uses i : k[x] k[z] takes the quotient by I(Z) and corresponding result for j ). Since φ is finite, k[x] is a finitely generatedφ k[y]-module. Thus, thereexista 1,...,a k k[x]suchthatforeachf k[x], there exist f 1,...,f q k[y] such that (**) f = k i=1 φ (f i )a i. This identity is still true in k[x]/i(z), so using (*) above, f = k i=1 ψ (f i + I(V)) a i + I(Z) in k[z]. Thus, k[z] is a finitely generated ψ k[v] module, so ψ is finite. This completes the proof of (iii). For (iv), let W Y be a closed subvariety and let Z φ 1 (W) be an irreducible component. We claim that dim(z) dim(w). Thus, by (iii), dim(z) = dim(φ(z)) 5

6 S. EVENS dim(w). Since φ(z) is closed and φ(z) W, it follows that φ(z) = W, and hence dim(z) = dim(w). This reduces (iv) to the claim. To verify the claim, recall that if V X is a closed subvariety, then it follows from definitions that (*) φ 1 I(V) = I(φ(V)). In our situation, φ is injective since φ is dominant, so I(V) φ k[y] = φ (I(φ(V))). Let B = k[x] and let A = φ (k[y]). Let I = I(Z) and let J = φ (I(W)). Since φ(z) W, I(W) I(φ(Z)), so by (*), I(Z) A = φ (I(φ(Z))) J. We apply the Going Down Theorem to the extension B A. Let α X be a point of φ 1 (W) contained in Z but in no other irreducible component of φ 1 (W) and let m α B and m φ(α) k[y] be the corresponding maximal ideals. By (*), m α A = φ (m φ(α) ). Since φ(α) W, I(W) m φ(α), so J φ (m φ(α) ) in A. By the Going Down Theorem, there exists a prime ideal p 1 of B such that p 1 m α and p 1 A = J. Then α V(p 1 ), so since V(p 1 ) is irreducible, by the choice of α, V(p 1 ) Z. Thus, I(Z) p 1, so I A p 1 A = J. Hence, I A = J. and the induced morphism A/J B/I is injective. Hence, φ : k[w] k[z] is injective, and it follows from properties of transcendence degree that dim(w) dim(z), verifying the claim. For (ii), let Z be an irreducible component of φ 1 (y). By (iii), dim(z) = dim(φ(z)) = dim(y) = 0. Therefore, Z is a point, which implies (ii). To proceed, we need to establish some results about normal varieties (see Tauvel and Yu, Chapter 17). Let φ : X Y be a dominant morphism of affine varieties. Then φ : k[y] k[x] is injective, so there is an induced field homomorphism φ : k(y) k(x). Remark 0.16. A dominant morphism φ : X Y is called birational if φ : k(y) k(x) is an isomorphism of fields. For example, the morphism φ : k V(y 2 x 3 ) given by φ(a) = (a 2,a 3 ) is birational, but is not an isomorphism. Proposition 0.17. Let φ : X Y be a birational morphism of affine varieties. Then there exists nonzero s k[y] such that φ : X φ (s) Y s is an isomorphism of affine varieties. In particular, a dominant morphism φ : X Y of affine varieties is birational if and only if there is a nonempty open set U Y such that φ : φ 1 (U) U is an isomorphism. Proof : Let k[x] = k[α 1,...,α m ]. Since k[x] k(x), by hypothesis, there exist b 1,...,b m k(y) such that φ (b i ) = α i. Then each b i = γ i s i for some γ i,s i 0 k[y]. Write b i = β i with s = s s 1 s m, and β i = sγ i s i k[y]. Then φ : k[y] s k[x] φ (s) is easily seen to be onto. Since φ : k(y) k(x) is injective, φ : k[y] s k[x] φ (s) is

injective. Hence φ : X φ (s) Y s is an isomorphism. This implies one direction of the last assertion, taking U = Y s. The other direction is easy. Lemma 0.18. Let X be a normal affine variety and let a k[x] be nonzero. Then X a is normal. Proof : This follows from the following assertion: let A be an integral domain and let B be the integral closure of A in the fraction field of A. Let S A be a multiplicatively closed set and assume 0 S. Then S 1 B is the integral closure of S 1 A in the fraction field of A. This can be proved just as in the proof of (1) implies (2) in Proposition 39, p. 687, of Dummit and Foote. Proposition 0.19. Let X be an affine variety. There exists nonzero a k[x] such that X a is normal. EXERCISE: Prove this. The following steps are useful. (i) Let B be the integral closure of k[x]. Show that B is a finitely generated integrally closed k-algebra (use a commutative algebra result from these notes). (ii) Let Y be the affine variety such that k[y] = B. Show that there exists a birational morphism φ : Y X such that φ : k[x] B is the inclusion. (iii) Find a k[x] such that φ : Y φ (a) X a is an isomorphism. Conclude that X a is normal. Remark 0.20. Let φ : X Y be a dominant morphism of affine varieties and let r = dim(x) dim(y). By the proof of Proposition 3.26 in the book An Introduction to Invariants and Moduli by Mukai, the author shows that there exists a k[y] and η : X φ (a) k r such that the morphism ψ = (φ,η) : X φ (a) Y a k r is finite and dominant, where r = dim(x) dim(y). By Proposition 0.15, it follows that dim(x φ (a)) dim(y a ), so r 0. By Proposition 0.19, there is nonzero f k[y] such that Y f is normal. Hence, Y af = (Y f ) a is normal by Lemma 0.18, so Y af k r is normal by 0.7. Let b = af. Since localization preserves finiteness, it follows that ψ : X φ (b) Y b k r is a finite dominant morphism to a normal variety. Remark 0.21. (EXERCISE) Prove the following. Let Y be a closed subset of a variety X, and let Y = Y 1 Y k be the decomposition of Y into irreducible components. If V Y is an nonempty open set, then the irreducible components of Y V are the Y i V such that Y i V is nonempty. Further, if Y i V is nonempty, then dim(y i ) = dim(y i V). Theorem 0.22. Let φ : X Y be a dominant morphism of affine varieties. Let r = dim(x) dim(y). There exists a nonempty open set U of Y such that if W is a closed subvariety of Y such that W U is nonempty, then for any irreducible component Z of φ 1 (W U), dim(z) = dim(w)+r. 7

8 S. EVENS Proof : Take U = Y b as in Remark 0.20 and let V = φ 1 (U), so U and V are affine varieties. As in Remark 0.20, there exists a morphism η : V k r such that ψ = (φ,η) : V U k r isafinitedominantmorphismtoanormalvariety. Byhypothesis, (W U) k r is a closed subvariety of U k r. Let Z be an irreducible component of φ 1 (W) such that Z V is nonempty. By Remark 0.21, Z V is an irreducible component of φ 1 (W) V = ψ 1 ((W U) k r ). Hence, by Proposition 0.15(iv), dim(z V) = dim((w U) k r ) = dim(w U)+r. It follows that dim(z) = dim(w)+r by Remark 0.21. Corollary 0.23. Let φ : X Y be a dominant morphism of affine varieties. There exists a nonempty open set U φ(x) such that for all y U and any irreducible component Z of φ 1 (y), dim(z) = r := dim(x) dim(y). Theorem 0.24. Let φ : X Y be a morphism of affine varieties. For x X, let φ 1 φ(x) = Z 1 Z j be the irreducible components of φ 1 φ(x). Let e(x) be the maximum of the dimensions of the Z i. Let S n (φ) := {x X : e(x) n}. Then S n (φ) is a closed subset of X. Proof : By replacing Y with φ(x), we may assume that φ is dominant. Let r = dim(x) dim(y) 0 by Remark 0.20. Argue by induction on p := dim(y). If p = 0, then φ 1 (φ(x)) = φ 1 (Y) = X is clearly closed. Assume the assertion is true for j < p. If n r, then S n (φ) = X by Theorem 0.12, so we may suppose n > r. By Corollary 0.23, there exists a nonempty open subset U of Y such that S n (φ) X φ 1 (U). Let W 1,...,W s be the irreducible components of Y U. Then for all i, dim(w i ) < dim(y) since each W i is proper and closed. Let {Z ij } j Ji be the irreducible components of φ 1 (W i ), and let φ ij : Z ij W i be the restriction of φ to Z ij. By the inductive assumption, S n (φ ij ) is closed in Z ij. The reader can check easily that S n (φ) = i,j S n (φ ij ), and the result follows.