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checked that these do not depend on the choices of representatives for the equivalence classes, and that we obtain in this way an S 1A-module S 1M D f m s j m 2 M; s 2 S g and a homomorphism m 7! m 1 W M iS! S 1M of A-modules whose kernel is fa 2 M j sa D 0 for some s 2 S g: PROPOSITION 1.17. The elements of S act inve... |
that sm D 0. Therefore a is not contained in m. Since this is true for all maximal ideals m, a D A, and so it contains 1. Now m D 1m D 0. COROLLARY 1.20. An A-module M D 0 if Mm D 0 for all maximal ideals m in A. PROOF. Immediate consequence of the lemma. PROPOSITION 1.21. Let A be a ring. A sequence of A-modules is e... |
; ˛j D ˛i ı ˛i j k (a) M D S (b) mi 2 Mi maps to zero in M if and only if it maps to zero in Mj for some j i. i / if i 2I ˛i.Mi /, and Direct limits of A-algebras are defined similarly. PROPOSITION 1.23. For every multiplicative subset S of A, S 1A'lim! Ah, where h runs over the elements of S (partially ordered by div... |
is irreducible, and the converse holds when A is a unique factorization domain. PROOF. Assume that a is prime. If a D bc, then a divides bc and so a divides b or c. Suppose the first, and write b D aq. Now a D bc D aqc, which implies that qc D 1 because A is an integral domain, and so c is a unit. Therefore a is irred... |
with field of fractions F. If f.X/ 2 AŒX factors into the product of two nonconstant polynomials in F ŒX, then it factors into the product of two nonconstant polynomials in AŒX. PROOF. Let f D gh in F ŒX. For suitable c; d 2 A, the polynomials g1 D cg and h1 D dh have coefficients in A, and so we have a factorization ... |
and let bj0, j0 n, the first coefficient of g not divisible by p. Then all the terms in the sum P i Cj Di0Cj0 ai bj are divisible by p, except ai0bj0, which is not divisible by p. Therefore, p doesn’t divide the.i0 C j0/th-coefficient of fg. We have shown that no irreducible element of A divides all the coefficients o... |
ization in AŒX. It must also be irreducible in F ŒX, because otherwise it would have a nontrivial factorization in AŒX (by 1.27). PROPOSITION 1.31. If A is a unique factorization domain, then so also is AŒX. PROOF. We shall check that A satisfies the conditions of Proposition 1.26. Let f 2 AŒX, and write f D c.f /f1. T... |
lynomial f is the largest total degree of a monomial occurring in f with nonzero coefficient. Since deg.fg/ D deg.f / C deg.g/, kŒX1; : : : ; Xn is an integral domain and kŒX1; : : : ; Xn D k. An element f of kŒX1; : : : ; Xn is irreducible if it is nonconstant and f D gh H) g or h is constant. 26 1. PRELIMINARIES FROM... |
cij xj D 0; i D 1; : : : ; m; with coefficients in a ring A, then det.C / xj D 0; j D 1; : : : ; m; (10) where C is the matrix of coefficients. To prove this, expand out the left hand side of 0 B @ det c11 ::: cm1 : : : : : : c1 j 1 ::: cm j 1 P P i c1i xi ::: i cmi xi c1 j C1 ::: cm j C1 : : : : : : c1m ::: cmm 1 C A... |
0 a21e1 C.˛ a22/e2 a23e3 D 0 D 0: Let C be the matrix of coefficients on the left-hand side. Then Cramer’s formula tells us that det.C / ei D 0 for all i. As M is faithful and the ei generate M, this implies that det.C / D 0. On expanding out the determinant, we obtain an equation ˛n C c1˛n1 C c2˛n2 C C cn D 0; ci 2 A... |
; bnŒ is finite over AŒb1; : : : ; bn, and so it is finite over A. Therefore is integral over A by 1.34. THEOREM 1.38. Let A be a subring of a ring B. The elements of B integral over A form an A-subalgebra of B. 28 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA PROOF. Let ˛ and ˇ be two elements of B integral over A. Then ... |
over A and d 2 A. DEFINITION 1.42. An integral domain A is said to be integrally closed if it is equal to its integral closure in its field of fractions F, i.e., if ˛ 2 F; ˛ integral over A H) ˛ 2 A: An integrally closed integral domain is called an integrally closed domain or normal domain. PROPOSITION 1.43. Unique f... |
conjugate of ˛, i.e., a root of f in some splitting field of f. Then f is also the minimal polynomial of ˛0 over F, and so (see above), there is an F -isomorphism W F Œ˛! F Œ˛0;.˛/ D ˛0: On applying to the above equation we obtain the equation ˛0m C a1˛0m1 C C am D 0; which shows that ˛0 is integral over A. As the coe... |
ic factor g of f in F ŒX. Let ˛ be a root of g in some extension field of F. Then g is the minimal polynomial of ˛. As ˛ is a root of f, it is integral over A, and so g has coefficients in A. PROPOSITION 1.47. Let A B be rings, and let A0 be the integral closure of A in B. For any multiplicative subset S of A, S 1A0 is... |
ideal consisting of the a 2 A such that ac 2 A is not contained in any maximal ideal m, and therefore equals A. Hence 1 c 2 A. Let E=F be a finite extension of fields. Then.˛; ˇ/ 7! TrE=F.˛ˇ/W E E! F (11) is a symmetric bilinear form on E regarded as a vector space over F. LEMMA 1.50. If E=F is separable, then the tra... |
mg be a basis for E over F. According to Proposition 1.40, there exists a d 2 A such that d ˇi 2 B for all i. Clearly fd ˇ1; : : : ; d ˇmg is still a basis for E as a vector space over F, and so we may assume to begin with that each ˇi 2 B. Because the trace pairing is nondegenerate, there is a dual basis fˇ0 g of E ov... |
B be rings with B integral over A. (a) For every prime ideal p of A, there is a prime ideal q of B such that q \ A D p. (b) Let p D q \ A; then p is maximal if and only if q is maximal. 32 1. PRELIMINARIES FROM COMMUTATIVE ALGEBRA PROOF. (a) If S is a multiplicative subset of a ring A, then the prime ideals of S 1A ar... |
q such that q00 \.A=p/ D p0=p. The inverse image q0 of q00 in B has the required properties. ASIDE 1.55. Let A be a noetherian integral domain, and let B be the integral closure of A in a finite extension E of the field of fractions F of A. Is B always a finite A-algebra? When A is integrally closed and E is separable ... |
We write it M ˝A N. Note that HomA-bilinear.M N; T /'HomA-linear.M ˝A N; T /: CONSTRUCTION Let M and N be A-modules, and let A.M N / be the free A-module with basis M N. Thus each element A.M N / can be expressed uniquely as a finite sum X ai.xi ; yi /; ai 2 A; xi 2 M; yi 2 N: Let P be the submodule of A.M N / generat... |
a commutative triangle. 9“An element of the tensor product of two vector spaces is not necessarily a tensor product of two vectors, but sometimes a sum of such. This might be considered a mathematical shenanigan but if you start with the state vectors of two quantum systems it exactly corresponds to the notorious noti... |
equipped with the given map k! B and the identity map B! B, has the universal property characterizing k ˝k B, so k ˝k B'B. In terms of the constructive definition of tensor products, the isomorphism is c ˝ b 7! cbW k ˝k B! B. EXAMPLE 1.57. The ring kŒX1; : : : ; Xm; XmC1; : : : ; XmCn, equipped with the obvious inclus... |
n'˝ŒX1; : : : ; Xn: If A D kŒX1; : : : ; Xn=.g1; : : : ; gm/, then ˝ ˝k A'˝ŒX1; : : : ; Xn=.g1; : : : ; gm/: (c) If A and B are algebras of k-valued functions on sets S and T respectively, then.f ˝ g/.x; y/ D f.x/g.y/ realizes A ˝k B as an algebra of k-valued functions on S T. f. Transcendence bases We review the theor... |
” replaced by “linearly”. 1.62. A transcendence basis for ˝ over k is an algebraically independent set A such that ˝ is algebraic over k.A/: 1.63. Assume that there is a finite subset A ˝ such that ˝ is algebraic over k.A/. Then (a) every maximal algebraically independent subset of ˝ is a transcendence basis; (b) every... |
algebraic subset V.S/ of kn is the set of common zeros of some collection S of polynomials in kŒX1; : : : ; Xn, V.S/ D f.a1; : : : ; an/ 2 kn j f.a1; : : : ; an/ D 0 all f 2 Sg: We refer to V.S/ as the zero set of S. Note that S S 0 H) V.S/ V.S 0/I — more equations means fewer solutions. Recall that the ideal a genera... |
curve. However, this can be misleading — see the examples 4.11 and 4.17 below. 2.3. If S consists of the single equation Z2 D X 2 C Y 2; then V.S/ is a cone. 2.4. A nonzero constant polynomial has no zeros, and so the empty set is algebraic. 2.5. The proper algebraic subsets of k are the finite subsets, because a poly... |
X r C a1X r1 C C ar ; ai 2 A; a0 ¤ 0; a0 is called the leading coefficient of f. Let a be a proper ideal in AŒX, and let a.i / denote the set of elements of A that occur as the leading coefficient of a polynomial in a of degree i (we also include 0). Clearly, a.i/ is an ideal in A, and a.i/ a.i C 1/ because, if cX i C ... |
kŒX is a principal ideal domain, which means that every ideal is generated by a single element. Also, if V is a linear subspace of kn, then linear algebra shows that it is the zero set of n dim.V / polynomials. All one can say in general, is that at least n dim.V / polynomials are needed to define V (see 3.45), but of... |
the Zariski topology on A subset V of A n is called the Zariski topology on V. The Zariski topology has many strange properties, but it is nevertheless of great importance. For the Zariski topology on k, the closed subsets are just the finite sets and the whole space, and so the topology is not Hausdorff (in fact, the... |
k of k-algebras whose kernel contains a, we obtain a point P in V.a/, namely, 2Nullstellensatz = zero-points-theorem. P D.'.X1/; : : : ; '.Xn//: e. The correspondence between algebraic sets and radical ideals 41 Thus, to prove the theorem, we have to show that there exists a k-algebra homomorphism kŒX1; : : : ; Xn=a! ... |
then at least one xi, say x1, is not algebraic over k. Then, kŒx1 is a polynomial ring in one symbol over k, and its field of fractions k.x1/ is a subfield of K. Clearly K is generated as a k.x1/-algebra by x2; : : : ; xr, and so the induction hypothesis implies that x2; : : : ; xr are algebraic over k.x1/. From 1.40,... |
). Alternatively, it follows from the strong Hilbert Nullstellensatz (2.19 below). 42 2. ALGEBRAIC SETS EXAMPLE 2.13. Let P be the point.a1; : : : ; an/, and let mP D.X1 a1; : : : ; Xn an/. Clearly I.P / mP, but mP is a maximal ideal, because “evaluation at.a1; : : : ; an/” defines an isomorphism kŒX1; : : : ; Xn=.X1 a... |
is radical (2.15b), and hence is the smallest radical ideal containing a. If a and b are radical, then a \ b is radical, but a C b need not be: consider, for example, a D.X 2 Y / and b D.X 2 C Y /; they are both prime ideals in kŒX; Y, but X 2 2 a C b, X … a C b. (See 2.22 below.) The strong Nullstellensatz For a poly... |
mC1.1 Y h/ (in the ring kŒX1; : : : ; Xn; Y ). On applying the homomorphism Xi 7! Xi Y 7! h1 W kŒX1; : : : ; Xn; Y! k.X1; : : : ; Xn/ to the above equality, we obtain the identity 1 D m X i D1 fi.X1; : : : ; Xn; h1/ gi.X1; : : : ; Xn/ (*) in k.X1; : : : ; Xn/. Clearly fi.X1; : : : ; Xn; h1/ D polynomial in X1; : : : ; ... |
f.P / D 0 mP a ” P 2 V.a/. If f 2 mP for all P 2 V.a/, then f is zero on V.a/, and so f 2 I V.a/ D rad.a/. We have shown that rad.a/ \ P 2V.a/ mP \ ma m. Remarks 2.19. Because V.0/ D kn, I.kn/ D I V.0/ D rad.0/ D 0I in other words, only the zero polynomial is zero on the whole of kn. In fact, this holds whenever k is ... |
2, it has “multiplicity 2”. P Q! P! and V W be the set of subsets of kn and let 2.23. Let Then I W and (see FT 7.19). It follows that I and V define a one-to-one correspondence between I. and V. ideals, and (by definition) V. 2.17. Q / P / consists exactly of the radical / consists of the algebraic subsets. Thus we re... |
or not a polynomial belongs to the ideal, and hence an algorithm for deciding whether or not a polynomial belongs to the radical of the ideal. There are even algorithms for finding a set of generators for the radical. These algorithms have been implemented in the computer algebra systems CoCoA and Macaulay 2. g. Prope... |
for any pair of distinct points, each has an open neighbourhood not containing the other. A topological space having the property (b) is said to be noetherian. The condition is equivalent to the following: every nonempty set of closed subsets of V has a minimal element. A topological space having property (c) is said ... |
.a/ [ V.b/ (2.10); hence I.W / D a \ b. If W ¤ V.a/, then there exists an f 2 a X I.W /. Let g 2 b. Then fg 2 a \ b D I.W /, and so g 2 I.W / (because I.W / is prime). We conclude that b I.W /, and so V.b/ V.I.W // D W. SUMMARY 2.28. There are one-to-one correspondences, radical ideals in kŒX1; : : : ; Xn $ algebraic s... |
: : [ Wr, etc. PROPOSITION 2.31. Let V be a noetherian topological space. Then V is a finite union of irreducible closed subsets, V D V1 [ : : : [ Vm. If the decomposition is irredundant in the sense that there are no inclusions among the Vi, then the Vi are uniquely determined up to order. PROOF. Suppose that V canno... |
ible. 48 2. ALGEBRAIC SETS COROLLARY 2.32. The radical of an ideal a in kŒX1; : : : ; Xn is a finite intersection of prime ideals, a D p1 \ : : : \ pn. If there are no inclusions among the pi, then the pi are uniquely determined up to order (and they are exactly the minimal prime ideals containing a). PROOF. Write V.a/... |
/ \ : : : \.fr /. 2.36. In a noetherian ring, every proper ideal a has a decomposition into primary ideals: a D T qi (see CA 19). For radical ideals, this becomes a simpler decomposition into prime ideals, as in the corollary. For an ideal.f / with f D Q f mi, the primary decomposition is the decomposition.f / D T.f mi... |
Then b 7! 1.b/ is a bijection from the set of ideals of kŒV to the set of ideals of kŒX1; : : : ; Xn containing a, under which radical, prime, and maximal ideals correspond to radical, prime, and maximal ideals (because each of these conditions can be checked on the quotient ring, and kŒX1; : : : ; Xn= 1.b/'kŒV =b). C... |
h/. Note that D.h/ D.h0/ ” V.h/ V.h0/ ” rad..h// rad..h0// ” hr 2.h0/ some r ” hr D h0g, some g: Some of this should look familiar: if V is a topological space, then the zero set of a family of continuous functions f W V! R is closed, and the set where a continuous function is nonzero is open. Let V be an irreducible a... |
ai 2 kŒX1; : : : ; Xm; a0 ¤ 0; m 2 N: k. Hypersurfaces; finite and quasi-finite maps 51 We assume that m ¤ 0, i.e., that X occurs in f (otherwise, H is a cylinder over a hypersurface n over.t1; : : : ; tn/ 2 kn is the set of points.t1; : : : ; tn; c/ n). The fibre of the map H! A in A such that c is a root of the poly... |
W! V be a regular map of algebraic subsets, and let 'W kŒV! kŒW be the map f 7! f ı '. (a) The map'is dominant if '.W / is dense in V. (b) The map'is quasi-finite if '1.P / is finite for all P 2 V. (c) The map'is finite if kŒW is a finite kŒV -algebra. As we shall see (8.28), finite maps are indeed quasi-finite. As kŒ... |
ideal n of kŒW such that m D n \ kŒV. Because of the correspondence between points and maximal ideals, this implies that'is surjective. l. Noether normalization theorem Let H be a hypersurface in A projection map.x1; : : : ; xnC1/ 7!.x1; : : : ; xn/W A nC1. We show that, after a linear change of coordinates, the n. n ... |
. Then F.X1 C c1T; : : : ; Xn C cnT; T / D F.c1; : : : ; cn; 1/T r C terms of degree < r in T, because the polynomial F.X1 C c1T; : : : ; Xn C cnT; T / is still homogeneous of degree r in X1; : : : ; Xn; T, and so the coefficient of the monomial T r can be obtained by setting each Xi equal to zero in F and T to 1. As F... |
it to be infinite (for the general proof, see CA 8.1). Let A D kŒx1; : : : ; xn. We prove the theorem by induction on n. If the xi are algebraically independent, there is nothing to prove. Otherwise, the next lemma shows that A is finite over a subring B D kŒy1; : : : ; yn1. By induction, B is finite over a subring C ... |
lynomial g.X1; : : : ; Xd ; T / defD f.X1 C c1T; : : : ; Xd C cd T; T / takes the form g.X1; : : : ; Xd ; T / D bT r C b1T C C br with b 2 k (see 2.43). As g.x1 c1xn; : : : ; xd cd xn; xn/ D 0 this shows that xn is integral over kŒx1 c1xn; : : : ; xd cd xn, and so A is finite over kŒx1 c1xn; : : : ; xd cd xn; xd C1; : ... |
: : ; xd cd xn; xd C1; : : : ; xn1g and we can repeat the argument. m. Dimension The dimension of a topological space Let V be a noetherian topological space whose points are closed. DEFINITION 2.48. The dimension of V is the supremum of the lengths of the chains V0 V1 Vd of distinct irreducible closed subsets (the le... |
Then dim.V / D tr degkk.V /: The proof will occupy the rest of this subsection. Let A be an arbitrary commutative ring. Let x 2 A, and let Sfxg denote the multiplicative subset of A consisting of the elements of the form xn.1 ax/; n 2 N; a 2 A: The boundary Afxg of A at x is defined to be the ring of fractions S 1 fxg... |
tr degkF.A/ D n 2 N. We argue by induction on n. We can replace k with its algebraic closure in A without changing tr degkF.A/. Let x 2 A. If x … k, then it is transcendental over k, and so tr degk.x/F.A/ D n 1 by 1.64; since k.x/ Afxg, this implies (by induction) that dim.Afxg/ n 1. If x 2 k, then 0 D 1 x1x 2 Sfxg, a... |
that its elements are parametrized by n-tuples. It is not true in general that the points of an algebraic set of dimension n are parametrized by n-tuples. All we can say is Corollary 2.57. ASIDE 2.59. The inequality in Proposition 2.54 may be strict; for example, A D k.x/ has dimension 0 but its field of fractions k.x... |
r can be expressed as a linear combination of the elements of the other. Therefore, kŒX1; : : : ; Xn D kŒ`1; : : : ; `r ; Xi1; : : : ; Xinr ; and so kŒX1; : : : ; Xn=c D kŒ`1; : : : ; `r ; Xi1; : : : ; Xinr =c'kŒXi1; : : : ; Xinr : EXAMPLE 2.62. If W is a proper algebraic subset of an irreducible algebraic set V, then ... |
Then V.F.X; Y // has dimension 1 by 2.64, and so V.F.X; Y // \ V.G.X; Y // must have dimension zero; it is therefore a finite set. PROPOSITION 2.66. Let W be a closed set of codimension 1 in an algebraic set V ; if kŒV is a unique factorization domain, then I.W / D.f / for some f 2 kŒV. PROOF. Let W1; : : : ; Ws be th... |
a; b 2 k: Exercises 2-1. Find I.W /, where W D.X 2; XY 2/. Check that it is the radical of.X 2; X Y 2/. 2-2. Identify kmn with the set of m n matrices, and let r 2 N. Show that the set of matrices with rank r is an algebraic subset of kmn. 2-3. Let V D f.t; t 2; : : : ; t n/ j t 2 kg. Show that V is an algebraic subse... |
and let M and N be A-modules. Show that if kal ˝k M and kal ˝k N are isomorphic kal ˝k A-modules, then M and N are isomorphic A-modules. 2-8. Show that the subset f.z; ez/ j z 2 Cg is not an algebraic subset of C 2. CHAPTER 3 Affine Algebraic Varieties In this chapter, we define the structure of a ringed space on an a... |
Note that, for disjoint open subsets Ui of V, condition (c) says that O O O V.U /'Q O V.Ui /. i O Examples 3.2. Let V be a topological space, and for each open subset U of V let of all continuous real-valued functions on U. Then V is a sheaf of R-algebras. O V.U / be the set O 3.3. Recall that a function f W U! R on a... |
aves of this one. V.U / be the set V is a sheaf of k-algebras. By definition, all our sheaves of O O b. Ringed spaces O A pair.V; V / consisting of a topological space V and a sheaf of k-algebras on V will be called a k-ringed space (or just a ringed space when the k is understood). For historical reasons, we sometimes... |
on a neighbourhood U 0 of c defines the same power series if and only if f and f 0 agree on some neighbourhood of c contained in U \ U 0 (ibid. I, 4.3). Thus we have a well-defined injective map from the ring of germs of holomorphic functions at c to the ring of convergent power series, which is obviously surjective. ... |
, and so is regular at P. on U 0, and so f C f 0 is regular at P. Similarly, ff 0 agrees with gg 0 h and g 0 hh0 (b,c) The definition is local. We next determine O V.U / when U is a basic open subset of V. LEMMA 3.10. Let g; h 2 kŒV with h ¤ 0. The function P 7! g.P /= h.P /mW D.h/! k is zero if and only if and only if... |
Note that D.aN assume that Vi D D.hi /. i / D D.ai /. Therefore, after replacing gi with gi g0 i and hi with aN i, we can We now have that D.h/ D S D.hi / and that f jD.hi / D gi. Because D.h/ is quasicomhi on D.hi / \ D.hj / D D.hi hj /, pact, we can assume that the covering is finite. As gi hi D gj hj hi hj.gi hj gj... |
agrees with that in Section 2i. COROLLARY 3.12. For every P 2 V, there is a canonical isomorphism where mP is the maximal ideal I.P /. P! kŒV mP, O PROOF. In the definition of the germs of a sheaf at P, it suffices to consider pairs.f; U / with U lying in a some basis for the neighbourhoods of P, for example, the basi... |
˘ For each f 2 A, let D.f / D fm j f … mg; the topology on V is that for which the O sets D.f / form a base. ˘ For f 2 Ah and m 2 D.h/, let f.m/ denote the image of f in Ah=mAh'k; in this V is the V / D Ah for all h 2 A. way Ah becomes identified with a k-algebra of functions D.h/! k, and unique sheaf of k-valued funct... |
INE ALGEBRAIC VARIETIES 3.18. Let U D A 2, but it is not a basic open subset because its complement f.0; 0/g has dimension 0, and therefore can’t be of the form V..f // (see 2.64). Since U D D.X/ [ D.Y /, the ring of regular functions on U is 2 X f.0; 0/g. It is an open subset of A.U; OA2/ D kŒX; Y X \ kŒX; Y Y (inters... |
of k-ringed spaces.U; V jU /!.V; V /. O A morphism of ringed spaces maps germs of functions to germs of functions. More O precisely, a morphism 'W.V; V /!.W; W / induces a k-algebra homomorphism O O OW;'.P /! V;P O for each P 2 V, which sends the germ represented by.U; f / to the germ represented by.'1.U /; f ı '/. In... |
usually denote an affine algebraic variety.V; V / by V. Let.V; V / and.W; W / be affine algebraic varieties. A map 'W V! W is regular (or a morphism of affine algebraic varieties) if it is a morphism of k-ringed spaces. With these definitions, the affine algebraic varieties become a category. We usually shorten “affin... |
m/ V.U / to be the set of functions f W U! k that are 66 3. AFFINE ALGEBRAIC VARIETIES PROPOSITION 3.22. The pair.V; Ah for each h 2 A X f0g. O V / is an affine algebraic variety with.D.h/; V /'O PROOF. Represent A as a quotient kŒX1; : : : ; Xn=a D kŒx1; : : : ; xn. Then.V; morphic to the k-ringed space attached to th... |
omorphism g hm 7! ˛.g/ ˛.h/m W Ah! B˛.h/: For every maximal ideal n of B, m D ˛1.n/ is maximal in A because A=m! B=n D k is an injective map of k-algebras which implies that A=m D k. Thus ˛ defines a map 'W spm B! spm A; '.n/ D ˛1.n/ D m: For m D ˛1.n/ D '.n/, we have a commutative diagram: A A=m ˛'B B=n: Recall that t... |
defines a homomorphism of the associated affine k-algebras kŒW! kŒV. Since these maps are inverse, we have shown: V /!.W; O O PROPOSITION 3.24. For all affine algebras A and B, Homk-alg.A; B/ '! Mor.Spm.B/; Spm.A//I for all affine varieties V and W, Mor.V; W / '! Homk-alg.kŒW ; kŒV /: In terms of categories, Propositi... |
�V. It therefore defines a map spm.kŒV /! spm.kŒW /, and it remains to show that this coincides with'when we identify spm.kŒV / with V and spm.kŒW / with W. Let P 2 V, let Q D '.P /, and let mP and mQ be the ideals of elements of kŒV and kŒW that are zero at P and Q respectively. Then, for f 2 kŒW, ˛.f / 2 mP ” f.'.P /... |
for Q 2 V.a/, ˛.g/.Q/ D g.'.Q// D 0; and so ˛.g/ 2 I V.a/ D a. Conversely, suppose ˛.b/ a, and let P 2 V.a/; for f 2 b, f.'.P // D ˛.f /.P / D 0; and so '.P / 2 V.b/. When these conditions hold,'is the morphism of affine varieties V.a/! V.b/ corresponding to the homomorphism kŒY1; : : : ; Yn=b! kŒX1; : : : ; Xm=a defi... |
, 1! A 2. This is V W Y 2 D X 3; but it is not an isomorphism onto its image because the inverse map is not regular. In view of 3.25, to prove this it suffices to show that t 7!.t 2; t 3/ does not induce an isomorphism on the 1 D kŒT and kŒV D rings of regular functions. We have kŒA kŒX; Y =.Y 2 X 3/ D kŒx; y. The map ... |
Spm.Ah/I PROOF. The map A! Ah defines a morphism spm.Ah/! spm.A/. The image is D.h/, and it is routine (using (1.13)) to verify the rest of the statement. If V D V.a/ A n, then.a1; : : : ; an/ 7!.a1; : : : ; an; h.a1; : : : ; an/1/W D.h/! A nC1; defines an isomorphism of D.h/ onto V.a; 1 hXnC1/. For example, there is ... |
i. Properties of the regular map Spm.˛/ PROPOSITION 3.34. Let ˛W A! B be a homomorphism of affine k-algebras, and let 'W Spm.B/! Spm.A/ be the corresponding morphism of affine varieties. (a) The image of'is dense for the Zariski topology if and only if ˛ is injective. (b) The morphism'is an isomorphism from Spm.B/ ont... |
the choice of a basis for E defines a bijection A.E/! A n, and the inherited structure of an affine algebraic variety on A.E/ is independent of the choice of the basis (because the bijections defined by two different bases differ by an automorphism of A n). We now give an intrinsic definition of the affine variety A.E... |
property: every k-linear map V! A from V into a k-algebra A extends uniquely to a k-algebra homomorphism S.V /! A: V i S.V / k-linear 9Š k-algebra A: (17) As usual, this universal property determines the pair.S.V /; i/ uniquely up to a unique isomorphism. We now define A.E/ to be Spm.S.E_//, where E_ is the dual vecto... |
.V / and k.W /. This allows us to suppose that A and B have a common field of fractions K. Let x1; : : : ; xn generate B as k-algebra. As K is the field of fractions of A, there exists a d 2 A such that dxi 2 A for all i ; then B Ad. The same argument shows that there exists an e 2 B such that Ad Be. Now B Ad Be H) Be ... |
� such that ˝ D k.x1; : : : ; xd C1/. After renumbering, fx1; : : : ; xd g will be a transcendence basis for ˝ over k and xd C1 will be separable over k.x1; : : : ; xd /. PROOF. Let ˝ D k.x1; : : : ; xn/. After renumbering, we may suppose that x1; : : : ; xd are algebraically independent, hence a transcendence basis (1... |
; : : : ; xd /). According to the primitive element theorem (FT 5.1), there exists a y 2 ˝ such that k.x1; : : : ; xd C2/ D k.x1; : : : ; xd ; y/. Now ˝ D k.x1; : : : ; xd ; y; xd C3; : : : ; xn/, contradicting the minimality of n. 1 ; : : : ; X p We have shown that ˝ D k.z1; : : : ; zd C1/ for some zi 2 ˝. The argumen... |
We have to show that dim Zi D dim V 1 for each i. There exists a point P 2 Zi not contained in any other Zj. Because the Zj are closed, there exists an open affine neighbourhood U of P in V not meeting any Zj with j ¤ i. Now V.f jU / D Zi \ U, which is irreducible. Therefore, on replacing V with U, we may assume that ... |
GEBRAIC VARIETIES We can restate Theorem 3.42 as follows: let V be a closed irreducible subvariety of A and let F 2 kŒX1; : : : ; Xn; then n V \ V.F / D 8 < : V ; pure codimension 1 otherwise. if F is identically zero on V if F has no zeros on V COROLLARY 3.43. Let V be an irreducible affine variety, and let Z be a max... |
0 of V.f1; : : : fr1/. By induction, codim.Z0/ r 1. Also Z is an irreducible component of Z0 \ V.fr / because Z Z0 \ V.fr / V.f1; : : : ; fr / and Z is a maximal irreducible closed subset of V.f1; : : : ; fr /. If fr vanishes identically on Z0, then Z D Z0 and codim.Z/ D codim.Z0/ r 1; otherwise, the theorem shows that... |
im Zi D i. We shall show that there exist f1; : : : ; fr 2 kŒV such that, for all s r, Zs is an irreducible component of V.f1; : : : ; fs/ and all irreducible components of V.f1; : : : ; fs/ have codimension s. We prove this by induction on s. For s D 1, take any f1 2 I.Z1/, f1 ¤ 0, and apply Theorem 3.42. Suppose f1; ... |
of codimension 1 is of the form V.f / for some f 2 kŒV (see 2.66). The condition that kŒV be a unique factorization domain is definitely needed. Again consider the cone, in A 4 and let Z and Z0 be the planes V W X1X4 X2X3 D 0 Z D f.; 0; ; 0/g Z0 D f.0; ; 0; /g: Then Z \ Z0 D f.0; 0; 0; 0/g, which has codimension 2 in ... |
/, f D gq for some q 2 A. Because f is irreducible, q is a unit, and so.f / D.g/ D p — the element f is prime. 3.53. Proposition 3.47 says the following: let A be an affine k-algebra, and let p be a prime ideal in A. If p has height r, then there exist elements f1; : : : ; fr 2 A such that p is minimal among the prime ... |
. p. 140), but it is unknown whether all connected curves in P Macaulay (the man, not the program) showed that for every r 1, there is a curve C in A 3 such that I.C / requires at least r generators (see the same exercise in Hartshorne for a curve whose ideal can’t be generated by 2 elements).6 5Kunz, Ernst Introductio... |
variety V A n. Then 3.47 shows that there is a neighbourhood n and functions f1; : : : ; fr on U such that U \ V D V.f1; : : : ; fr / (zero set in U /. Thus U of P in A U \ V is a set-theoretic complete intersection in U. One says that V is a local complete intersection n such that the ideal I.V \ U / can be at P 2 V ... |
the proofs are more complicated. a. Tangent spaces to plane curves Consider the curve V in the plane defined by a nonconstant polynomial F.X; Y /, V W F.X; Y / D 0: We assume that F.X; Y / has no multiple factors, so that.F.X; Y // is a radical ideal and I.V / D.F.X; Y //. We can factor F into a product of irreducible... |
b/ and @F @Y.a; b/ are not both zero, then TP.V / is a line through.a; b/, and we say that P is a nonsingular or smooth point of V. Otherwise, TP.V / has dimension 2, and we say that P is singular or multiple. The curve V is said to be nonsingular or smooth if all its points are nonsingular. Examples For each of the f... |
point.a; b/ is singular if and only if it is ˘ a singular point of V.F /, ˘ a singular point of V.G/, or ˘ a point of V.F / \ V.G/. This follows immediately from the product rule: @.F G/ @X D F @G @X C @F @X G; @.F G/ @Y D F @G @Y C @F @Y G: b. Tangent cones to plane curves 83 The singular locus PROPOSITION 4.7. The n... |
omial F.X; Y / can be written (uniquely) as a finite sum F D F0 C F1 C F2 C C Fm C (21) with each Fm a homogeneous polynomial of degree m. The term F1 will be denoted F` and called the linear form of F, and the first nonzero term on the right of (21) (the homogeneous summand of F of least degree) will be denoted F and ... |
of lines Y D ˙X, and the singularity is a node. 4.11. F.X. The origin is an isolated point of the real locus. It is again a node, but the tangent cone is defined by Y 2 C X 2, which is the pair of lines Y D ˙iX. In this case, the real locus of the tangent cone is just the point (0,0). 4.12. F.X; Y / D X 3 Y 2. Here th... |
locus is the image of R under the analytic map t 7!.t 3 C 2t; t.t 3 C 2//, which is injective, 2 with closed image. See Milnor, J., Singular proper, and immersive, and hence an embedding into R points of complex hypersurfaces. PUP, 1968, or mo98366 (Elencwajg). c. The local ring at a point on a curve PROPOSITION 4.19.... |
the last condition is equivalent to n being a principal ideal. As P has Krull dimension one (2.64), for its maximal ideal to be principal means that it is a regular local ring of dimension 1 (see 1.6). Thus, for a point P on a curve, O P nonsingular ” P regular. O 86 4. LOCAL STUDY PROPOSITION 4.20. Every regular loca... |
subset of A terminology from linear algebra (which should be familiar from advanced calculus). m we review some LINEAR ALGEBRA For a vector space km, let Xi be the i th coordinate function a 7! ai. Thus X1; : : : ; Xm is the dual basis to the standard basis for km. A linear form P ai Xi can be regarded as an element o... |
; : : : ; Xm at a by the equation:.dF /a D.dXi /a: m X i D1 @F @Xi ˇ ˇ ˇ ˇa It is again a linear form on Ta.A m/ defined by the equations: Ta.A m/. In terms of differentials, Ta.V / is the subspace of.dF /a D 0; F 2 a: (23) I claim that, in (22) and (23), it suffices to take the F to lie in a generating subset for a. T... |
m rank J.a/; 88 4. LOCAL STUDY and so a is nonsingular if and only if the rank of Jac.F1; : : : ; Fr /.a/ is equal to m dim.V /. For example, if V is a hypersurface, say I.V / D.F.X1; : : : ; Xm//, then Jac.F /.a/ D @F @X1.a/; : : : ;.a/ ; @F @Xm and a is nonsingular if and only if not all of the partial derivatives @... |
D 0 B B @ @P1 @X1 @Pn @X1.a/; :::.a/; : : : ;.a/ @P1 @Xm ::: : : : ; @Pn @Xm.a/ 1 C C A : For example, suppose a D.0; : : : ; 0/ and b D.0; : : : ; 0/, so that Ta.A kn, and m/ D km and Tb.A n/ D Pi D cij Xj C.higher terms), i D 1; : : : ; n: m X j D1 Then Yi ı.d'/a D P i.e., it is simply t 7!.cij /t. j cij Xj, and the... |
implies d.f ı '/a D.df /b ı.d'/a; b D '.a/: Let t 2 Ta.V /; then.df /b ı.d'/a.t/ D d.f ı '/a.t/; which is zero if f 2 b because then f ı'2 a. Thus.d'/a.t/ 2 Tb.W /. We therefore get a map.d'/aW Ta.V /! Tb.W /. The usual rules from advanced calculus hold. For example,.d /b ı.d'/a D d. ı '/a; b D '.a/: f. Tangent spaces... |
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