NOTES on FIBER DIMENSION Let Φ : X → Y Be a Morphism of Affine

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 p0 ⊂ p1 ⊂···⊂ pk 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 p0 ⊂···⊂ pk = 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 p2 ⊂ p1 be prime ideals of A and suppose that q1 is a prime ideal of B such that q1 ∩ A = p1. Then there exists a prime ideal q2 of B such that q2 ⊂ q1 and q2 ∩ A = p2. 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 b1,...,bk ∈ B such that B = Ab1 + ··· + Abk. It follows easily that B is generated as an algebra by {b1,...,bk} and any set of generators of A.

Lemma 0.6. (Serre, “Local Algebra”, Lemma 1, p. 8) Let p1,..., pn be prime ideals of n a ring A. If I is an ideal of A and I ⊂∪i=1pi, then I ⊂ pi 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[t1,...,tn] 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 α1,...,αk and factor q(t) into linear factors with roots β1,...,βr. If γ is one of the αi or βj, 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 αi and βj 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) = xm + m−1 pm−1x + ··· + p0, with pi ∈ 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 pi = pi(t). Let β = α − ts ∈ K[t], so by the choice of s, −β is monic in K[t]. Let R(x) = Q(x + ts). Then R(β) = Q(β + ts) = Q(α) = 0. Expand R(x) = m m−1 x + qm−1x + ··· + q0 as a polynomial in x with coefficients in A[t]. The term q0 = sm s(m−1) t + pm−1t + ··· + p0 is monic in t by the choice of s. Since R(β)=0, m−1 m−2 m−1 m−2 q0 = −β(β + qm−1β + ··· + q1)= −β · γ,γ =(β + qm−1β + ··· + q1) ∈ K[t].

Since q0 and −β are monic in K[t], it follows that γ is monic in K[t]. Since q0 ∈ A[t], it follows from the assertion of the ﬁrst paragraph of this proof that −β ∈ A[t]. Hence, α = β + ts ∈ A[t], which establishes the ﬁrst assertion. The second assertion follows by an easy induction. Q.E.D. 3

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 codimX (Y ) = dim(X)− dim(Y ) is the codimension of Y in X. It is clear that if Z ⊂ Y is a closed subvariety, then codimX (Z) = codimX (Y )+codimY (Z). Theorem 0.8. Let X be an aﬃne variety and let Z ⊂ X be a closed subvariety. Then codimX (Z)=1 if and only if Z is an irreducible component of a hypersurface V (f) ⊂ X.

Proof: Suppose codimX (Z) = 1. Since Z =6 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)= Y1 ∪···∪ Yk be the irreducible components of the hypersurface V (f). Then Z ⊂ Yi for some i. But dim(Yi) < dim(X), so the hypothesis implies that dim(Yi) ≤ dim(Z). It follows that Yi = 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. Q.E.D. This last theorem can be proved without using 0.2 and 0.3 if X = kn since then we can reduce to the case when V (f) is irreducible.

Corollary 0.9. Let X be an aﬃne variety and let Y be a closed subvariety. If codimX (Y )= r, then for each i =1,...,r, there exists a closed subvariety Yi of X such that codimX (Yi)= i and Y = Yr ⊂ Yr−1 ⊂···⊂ Y1.

Proof: This is clear in case r = 1, and we argue by induction on r. Since Y =6 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 ⊂ Z0 for some irreducible component Z0 of V (f). Take Y1 = Z0.

Then codimY1 (Y ) = r − 1, so by induction, there exists a chain of closed subvarieties

Yr = Y ⊂ Yr−1 ⊂···⊂ Y2 in Y1 such that codimY1 (Yi) = i − 1. Then Yr ⊂···⊂ Y1 is the desired chain in X. Q.E.D.

Corollary 0.10. Let X be an aﬃne variety and let f1,...,fr ∈ k[X]. If Z is an irreducible component of V (f1,...,fr), then codimX (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 (f1,...,fr). Since Z ⊂ V (f1), Z ⊂ Y for some irreducible component Y of V (f1). Let f2,..., fr be the images of f2,...,fr in k[Y ]. Then Z ⊂ V (f2,..., fr) ⊂ V (f1,...,fr). Since Z is a maximal closed irreducible subset 4 S. EVENS of V (f1,...,fr), it follows that Z is an irreducible component of V (f2,..., fr) contained in Y . Apply the inductive assumption with Y in place of X to obtain codimY (Z) ≤ r −1. Since codimX (Y ) ≤ 1 by Theorem 0.8, we get codimX (Z) ≤ r. Q.E.D. Corollary 0.11. Let X be an aﬃne variety and let Y ⊂ X be a closed subvariety. Suppose codimX (Y ) = r. Then Y is an irreducible component of V (f1,...,fr) for some f1,...,fr ∈ k[X].

Proof: We prove the following stronger statement:

(*) if we are given a chain of closed subvarieties Yr ⊂···⊂ Y1 with codimX (Yi)= i, then there exists f1,...,fr ∈ k[X] such that for all 1 ≤ q ≤ r, every irreducible component of V (f1,...,fq) has codimension q in X and Yq is an irreducible component of V (f1,...,fq). If we assume (*) and take Yr = Y and use 0.9 to ﬁnd a chain Yr ⊂···⊂ Y1 as above, we complete the proof.

Prove (*) by induction on q. If q = 1, then by Theorem 0.8, Y1 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 f1,...,fq−1 as above. Let Z1 = Yq−1, Z2,...,Zm be the irreducible components of V (f1,...,fq−1), so codimX (Zj)= q − 1 for all j. Hence, dim(Yq) < dim(Zj), so no Zj ⊂ Yq, so I(Yq) 6⊂ I(Zj) for all j. Thus, I(Yq) 6⊂ I(Z1) ∪···∪ I(Zm) by Lemma 0.6. Hence, there exists fq ∈ I(Yq) such that fq 6∈ I(Zj) for all j. Thus, fq is a nonzero nonunit in k[Zj] for all j, where fq is the image of fq in k[Zj]. Hence, by Theorem 0.8 applied to fq ∈ k[Zj], every irreducible component W of V (fq)= V (fq) ∩ Zj has codimZj (W ) = 1. Thus, codimX (W )= q. Let V be an irreducible component of V (f1,...,fq). Then V ⊂ V (f1,...,fq−1), and it follows that V is in some irreducible component Zj. Thus, V ⊂ V (fq) ∩ Zj. Hence V ⊂ W , where W is an irreducible component of V (fq) ∩ Zj. But codimX (V ) ≤ q by Corollary 0.10, so dim(V ) ≥ dim(W ), and hence V = W . Finally, note that Yq ⊂ V (f1,...,fq), so since Yq is irreducible, Yq is a subset of one of the above irreducible components V . Since dim(Yq) = dim(V ), Yq = V . Q.E.D. 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 = codimY (W ). By Corollary 0.11, W is an irreducible component of ∗ V (f1,...,fs) for some f1,...,fs ∈ k[Y ]. Let gi = φ (fi) for i = 1,...,s. It follows from deﬁnitions that Z ⊂ V (g1,...,gs). Since Z is irreducible, Z ⊂ Z0, where Z0 is an irreducible component of V (g1,...,gs). Again from deﬁnitions, φ(V (g1,...,gs)) ⊂ V (f1,...,fs), so φ(Z) ⊂ φ(Z0) ⊂ V (f1,...,fs). Note that φ(Z0) is irreducible by Exercise 5

8, problem set 1. Since W = φ(Z) by assumption, W ⊂ φ(Z0), so W = φ(Z0) by deﬁnition −1 of irreducible component. Thus, φ(Z0) ⊂ W , so Z0 ⊂ φ (W ). −1 But Z is an irreducible component of φ (W ), so Z = Z0. Hence Z is an irreducible component of V (g1,...,gs) so by Corollary 0.10, codimX (Z) ≤ s. Thus, dim(Z) ≥ dim(X) − s = r + dim(Y ) − s = r + dim(W ). Q.E.D.

Remark 0.13. An aﬃne variety X is called normal if k[X] is integrally closed in k(X). The aﬃne variety kn is normal since a unique factorization domain is integrally closed.

Remark 0.14. Let X be a normal affine variety. Then X × kn is a normal variety. n Indeed, k[X × k ]= k[X][t1,...,tn] 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. For the claim, let i : Z → X and let j : V → Y be the inclusions. Then since φ◦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 ∗ generated φ k[Y ]-module. Thus, there exist a1,...,ak ∈ k[X] such that for each f ∈ k[X], there exist f1,...,fq ∈ k[Y ] such that k ∗ (**) f = Pi=1 φ (fi)ai. This identity is still true in k[X]/I(Z), so using (*) above, k ∗ ∗ f = Pi=1 ψ (fi + I(V )) · ai + 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)) ≥ 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∗−1I(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 −1 −1 φ (W ) contained in Z but in no other irreducible component of φ (W ) and let mα ⊂ B ∗ and mφ(α) ⊂ k[Y ] be the corresponding maximal ideals. By (*), mα ∩ A = φ (mφ(α)). ∗ Since φ(α) ∈ W , I(W ) ⊂ mφ(α), so J ⊂ φ (mφ(α)) in A. By the Going Down Theorem, there exists a prime ideal p1 of B such that p1 ⊂ mα and p1 ∩ A = J. Then α ∈ V (p1), so since V (p1) is irreducible, by the choice of α, V (p1) ⊂ Z. Thus, I(Z) ⊂ p1, so I ∩ A ⊂ p1 ∩ 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). Q.E.D. 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 (y2 − x3) given by φ(a)=(a2,a3) 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 : Xφ∗(s) → Ys 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[α1,...,αm]. Since k[X] ⊂ k(X), by hypothesis, there exist ∗ γi b ,...,bm ∈ k(Y ) such that φ (bi) = αi. Then each bi = for some γi,si =6 0 ∈ k[Y ]. 1 si β sγ ∗ i i ∗ Write bi = s with s = s1 ··· sm, and βi = s ∈ k[Y ]. Then φ : k[Y ]s → k[X]φ (s) is ∗ i ∗ easily seen to be onto. Since φ : k(Y ) → k(X) is injective, φ : k[Y ]s → k[X]φ∗(s) is 7 injective. Hence φ : Xφ∗(s) → Ys is an isomorphism. This implies one direction of the last assertion, taking U = Ys. The other direction is easy. Q.E.D.

Lemma 0.18. Let X be a normal aﬃne variety and let a ∈ k[X] be nonzero. Then Xa 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 6∈ S. Then S−1B is the integral closure of S−1A 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. Q.E.D. Proposition 0.19. Let X be an affine variety. There exists nonzero a ∈ k[X] such that Xa 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 : Yφ∗(a) → Xa is an isomorphism. Conclude that Xa 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 : r r Xφ∗(a) → k such that the morphism ψ =(φ,η): Xφ∗(a) → Ya × k is finite and dominant, where r = dim(X) − dim(Y ). By Proposition 0.15, it follows that dim(Xφ∗(a)) ≥ dim(Ya), so r ≥ 0. By Proposition 0.19, there is nonzero f ∈ k[Y ] such that Yf is normal. Hence, r Yaf = (Yf )a is normal by Lemma 0.18, so Yaf × k is normal by 0.7. Let b = af. Since r localization preserves finiteness, it follows that ψ : Xφ∗(b) → Yb × k 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 = Y1 ∪···∪ Yk 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 Yi ∩ V such that Yi ∩V is nonempty. Further, if Yi ∩V is nonempty, then dim(Yi) = dim(Yi ∩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. 8 S. EVENS

−1 Proof : Take U = Yb as in Remark 0.20 and let V = φ (U), so U and V are affine varieties. As in Remark 0.20, there exists a morphism η : V → kr such that ψ =(φ,η): V → U ×kr is a finite dominant morphism to a normal variety. By hypothesis, (W ∩ U) × kr is a closed subvariety of U × kr. 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) × kr). Hence, by Proposition 0.15(iv), dim(Z ∩ V ) = dim((W ∩ U) × kr) = dim(W ∩ U)+ r. It follows that dim(Z) = dim(W )+ r by Remark 0.21. Q.E.D. 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 −1 −1 φ φ(x)= Z1∪···∪Zj be the irreducible components of φ φ(x). Let e(x) be the maximum of the dimensions of the Zi. Let Sn(φ) := {x ∈ X : e(x) ≥ n}. Then Sn(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

If n ≤ r, then Sn(φ) = X by Theorem 0.12, so we may suppose n>r. By Corollary −1 0.23, there exists a nonempty open subset U of Y such that Sn(φ) ⊂ X − φ (U). Let W1,...,Ws be the irreducible components of Y − U. Then for all i, dim(Wi) < dim(Y ) since each Wi is proper and closed. Let {Zij}j∈Ji be the irreducible components of −1 φ (Wi), and let φij : Zij → Wi be the restriction of φ to Zij. By the inductive assumption, Sn(φij) is closed in Zij. The reader can check easily that Sn(φ)= ∪i,jSn(φij), and the result follows. Q.E.D.