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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... |
s the product gh of g; h 2 AŒX, then it divides g or h in F ŒX. Suppose the first, and write f q D g with q 2 F ŒX. Then c.q/ D c.f /c.q/ D c.f q/ D c.g/ 2 A, and so q 2 AŒX. Therefore f divides g in AŒX. We have shown that every element of AŒX is a product of irreducible elements and that every irreducible element of ... |
egral over S 1A. Then n b s C a1 s1 b s n1 C C an sn D 0: for some ai 2 A and si 2 S . On multiplying this equation by sns1 sn, we find that s1 snb 2 A0, and therefore that b s 2 S 1A0. D s1snb ss1sn COROLLARY 1.48. Let A B be rings, and let S be a multiplicative subset of A. If A is integrally closed in B, then S 1A i... |
! Homk-alg.kŒX1; : : :; R/ Homk-alg.kŒXmC1; : : :; R/ induced by the inclusions is a bijection. But this map can be identified with the obvious bijection RmCn ! Rm Rn: In terms of the constructive definition of tensor products, the isomorphism is f ˝ g 7! fgW kŒX1; : : : ; Xm ˝k kŒXmC1; : : : ; XmCn ! kŒX1; : : : ; XmC... |
t the finite sets and the whole space, and so the topology is not Hausdorff (in fact, there are no disjoint nonempty open subsets at all). We shall see in 2.68 below that the proper closed subsets of k2 are finite 2 are much coarser unions of points and curves. Note that the Zariski topologies on C and C (have fewer op... |
lized as the intersection of Y D X 2 and Y D X 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 radi... |
dical 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 / in Example 2.29. i i i. Regular functions; the coordinate ring of an algebraic set Let V be an algebraic subset of A n, and let I.... |
dent over k, and such that A is finite over kŒx1; : : : ; xd . It is not necessary to assume that A is reduced in Theorem 2.45, nor that k is algebraically closed, although the proof we give requires 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.... |
; xn/ be the field of fractions of kŒx1; : : : ; xn. As f is not the zero polynomial, some Xi , say, Xn, occurs in it. Then Xn occurs in every nonzero multiple of f , and so no nonzero polynomial in X1; : : : ; Xn1 belongs to .f /. This means that x1; : : : ; xn1 are algebraically independent. On the other hand, xn is ... |
h g with gh0Cg 0h hh0 on U 0, 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 i... |
e image of f in A=m D k, i.e., f .m/ D f (mod m/. This allows us to identify elements of A with functions spm.A/ ! k. We attach a ringed space .V; For f 2 A, let O V / to A by letting V be the set of maximal ideals in A. D.f / D fm j f .m/ ¤ 0g D fm j f … mg: Since D.fg/ D D.f / \ D.g/, there is a topology on V for whi... |
is a closed subset of spm.A/, and every closed subset is of this form. Now let a be a radical ideal in A, so that A=a is again reduced. Corresponding to the homomorphism A ! A=a, we get a regular map Spm.A=a/ ! Spm.A/: The image is V .a/, and spm.A=a/ ! V .a/ is a homeomorphism. Thus every closed subset of spm.A/ has ... |
that, after renumbering, zd C1 will be separably algebraic over k.z1; : : : ; zd /, and this implies that fz1; : : : ; zd g is a transcendence basis for ˝ over k (1.63). l. Noether Normalization Theorem DEFINITION 3.39. The dimension of an affine algebraic variety is the dimension of the underlying topological space (... |
1994 (I 6, Exercise 8; see also Hartshorne 1977, I, Exercise 2.17). 3 — see Apparently, it is not known whether two polynomials always suffice to define a curve in A 3 can’t be defined by two polynomials (ibid. Kunz 1985, p136.5 The union of two skew lines in P 3 can be defined by two polynomials. p. 140), but it is u... |
double point is called a node. There are many names for special types of singularities — see any book, especially an old book, on algebraic curves. Examples The following examples are adapted from Walker, Robert J., Algebraic Curves. Princeton Mathematical Series, vol. 13. Princeton University Press, Princeton, N. J., ... |
: When we identify Tb.A regarded as a regular map) with the differential of F (F regarded as a regular function). 1/; f. Tangent spaces to affine algebraic varieties 89 LEMMA 4.24. Let 'W A km into W D V .b/ kn, then .d'/a maps Ta.V / into Tb.W /, b D '.a/. n be as at the start of this subsection. If ' maps V D V .a/ m... |
imension of the geometric tangent cone at P is the same as the dimension of V (because the Krull dimension of a noetherian local ring is equal to P / is a polynomial ring in dim V variables if and that of its graded ring). Moreover, gr. P / is a only if polynomial ring in d variables, in which case CP .V / D TP .V /. P... |
d connected solvable normal subgroup (except feg). Such a group G may have a finite nontrivial centre Z.G/, and we call two semisimple groups G and G0 locally isomorphic if G=Z.G/ G0=Z.G0/. For example, SLn is semisimple, with centre n, the set of diagonal matrices diag.; : : : ; /, n D 1, and SLn =n D PSLn. A Lie alge... |
e subset of Z on which '1 and '2 agree is closed. PROOF. There are regular functions xi on V such that P 7! .x1.P /; : : : ; xn.P // identifies n (take the xi to be any set of generators for kŒV as a k-algebra). V with a closed subset of A Now xi ı '1 and xi ı '2 are regular functions on Z, and the set where '1 and '2 ... |
of two sets (in the category of sets) is the usual cartesian product of the sets, and the product of two topological spaces (in the category of topological spaces) is the product of the underlying sets endowed with the product topology. We shall show that products exist in the category of algebraic varieties. Suppose, ... |
ct topology, then every pair of points .x; y/ … has an open neighbourhood U U 0 such that .U U 0/ \ D ¿. In other words, if x and y are distinct points in V , then there are open neighbourhoods U and U 0 of x and y respectively such that U \ U 0 D ¿. Thus V is Hausdorff. Conversely, if V is Hausdorff, the reverse argum... |
Let V and W be closed subvarieties of A ducible component Z of V \ W , n; for any (nonempty) irre- that is, dim.Z/ dim.V / C dim.W / nI codim.Z/ codim.V / C codim.W /: PROOF. In the course of the proof of Theorem 5.29, we saw that V \ W is isomorphic to \ .V W /, and this is defined by the n equations Xi D Yi in V W . ... |
re complicated. To get a section to ', it is necessary to remove a line in C from 0 to infinity, which is not closed for the Zariski topology. |0VA1ϕ n. ´Etale maps 119 It is not possible to fit the graph of the complex curve Y 2 D X into 3-space, but the picture at right is an early depiction of it (from Neumann, Carl... |
, then we get a diagram as in the rank theorem. From it we get maps UP ! A m An U'.P / ! U'.P /: The first is ´etale, and the second is the projection of A mn U'.P / onto U'.P /. COROLLARY 5.62. Let V and W be nonsingular varieties. If 'W V ! W is smooth at P , then it is smooth on an open neighbourhood of V . PROOF. ... |
areful to ensure that O.U / is a set; for example, restrict U and A category of k-algebras of the form kŒX0; X1; : : :=a for a fixed family of symbols .Xi / indexed by N. 1 to the r. Rational and unirational varieties 127 A little history In his first proof of the Riemann hypothesis for curves over finite fields, Weil ... |
pology of P n. As in the opening paragraph of this chapter, we let Ui D D.Xi /. To each polynomial f .X1; : : : ; Xn/, we attach the homogeneous polynomial of the same degree f .X0; : : : ; Xn/ D X deg.f / 0 X1 X0 f ; : : : ; Xn X0 ; and to each homogeneous polynomial F .X0; : : : ; Xn/, we attach the polynomial F.X1; ... |
this field (the subscript 0 is short for “subfield of elements of n/ D k.X0; : : : ; Xn/0. Note that for F D G degree 0”), so that k.P H in k.X0; : : : ; Xn/0; .a0 W : : : W an/ 7! G.a0; : : : ; an/ H.a0; : : : ; an/ W D.H / ! k, is a well-defined function, which is obviously regular (look at its restriction to Ui /. W... |
jective space P n;m can be written .: : : W bi0:::in m vW P n ! P n;m, .a0 W : : : W an/ 7! .: : : W bi0:::in W : : :/; bi0:::in D ai0 0 : : : ain n : In other words, the Veronese mapping sends an n C 1-tuple .a0W : : : W an/ to the set of monomials in the ai of degree m. For example, when n D 1 and m D 2, the Veronese... |
surjective regular map A n. Consider the map n ! A .x0W : : : W xn/ 7! .x2 0 W W x2 n/W P n ! P n: It is mW 1 with m > 1 except over the points .0W W 1W W 0/. If H is a general hyperplane avoiding these points, then P n. For example, when we take n still maps onto P n X H A we obtain the surjective map H W x0 C C xn D... |
sen for E, the condition becomes the vanishing of the minors of order n d C 1 of a linear map E ! Vd C1 E), and Gd .E/ D P.Vd E/ \ W: 150 6. PROJECTIVE VARIETIES with Ei a subspace of E of dimension di . The map Gd.E/ F 7!.E i / ! Q i Gdi .E/ Q i P.Vdi E/ realizes Gd.E/ as a closed subset8 Q i Gdi .E/, and so it is a p... |
. The problem becomes that of showing V is not a finite union of proper subspaces E_ i by a hyperplane Hi containing it. Then Hi is defined by a nonzero linear form Li . We have to show that Q Lj is not identically zero on V . But this follows from the statement that a polynomial in n variables, with coefficients not ... |
regular map f W V ! A 7.8. 1 P 1, to which we can apply 7.10. A regular map 'W V ! W from a complete connected variety to an affine variety has image equal to a point. In particular, every complete connected affine variety is a point. n, and write ' D .'1; : : : ; 'n/, where 'i is the composite Embed W as a closed sub... |
mials P1 and Q1 satisfying (33). A root ˛ of P must be also be a root of P1 or of Q. If the former, cancel X ˛ from the left hand side of (33), and consider a root ˇ of P1=.X ˛/. As deg P1 < deg P , this argument eventually leads to a root of P that is not a root of P1, and so must be a root of Q. d. Elimination theory... |
taking inverses, we find that gh D hg. Since the negative map, a 7! aW A ! A, takes the identity element to itself, the preceding corollary shows that it is a homomorphism. f. Chow’s Lemma The next theorem is a useful tool in extending results from projective varieties to complete varieties. It shows that a complete va... |
Q to 0 and p and q are the projections V W ! V; W . CHAPTER 8 Normal Varieties; (Quasi-)finite maps; Zariski’s Main Theorem We begin by studying normal varieties. These varieties have some of the good properties of nonsingular varieties, and it is easy to show that every variety is birationally equivalent to a normal v... |
Z is locally principal at P if there exists an open affine neighbourhood U of P and an f 2 kŒU such that I.Z \ U / D .f /; the regular function f is then called a local equation for Z at P . If P … Z, then Z is locally principal at P because then we can choose U so that Z \ U D ;, and I.Z \ U / D .1/. 1See, for example... |
morphisms is finite. PROOF. (a) Let Z be a closed subvariety of a variety V , and let U be an open affine subvariety of V . Then Z \ U is a closed subvariety of U . It is therefore affine, and the map Z \ U ! U corresponds to a map A ! A=a of rings, which is obviously finite. This proves (a). As to be finite is a loca... |
C a1.P / T m1 C C am.P / D 0 (*) has r D deg ' distinct roots, and so m D r. Consider the discriminant disc F of F . Because (*) has distinct roots, disc.F /.P / ¤ 0, and so disc.F / is nonzero on an open neighbourhood U of P . The factorization kŒV ! kŒV ŒT =.F / T 7!f! kŒW gives a factorization W ! Spm.kŒV ŒT =.F // ... |
e normal (thereby excluding example (b)), and we require the map to be quasi-finite (thereby excluding example (c)), then we are left with (a). PROPOSITION 8.57. Let 'W W ! V be a birational regular map of irreducible varieties. If V is normal and the map ' is quasi-finite, then ' is an isomorphism from W onto an open ... |
t resolution of singularities holds for V . Note that with “nonsingular” replaced by “normalization”, the normalization of V (see 8.5) provides such a map (resolution of abnormalities). Nagata’s embedding theorem 7.50 shows that it suffices to prove resolution of singularities for complete varieties, and Chow’s lemma 7... |
mits an infinite set. combined using only “and” and “or” (or, better, statements of the form f D 0 combined using “and”, “or”, and “not”). The next proposition shows that a constructible set C that is dense in an irreducible variety V must contain a nonempty open subset of V . Contrast Q, which is dense in R (real topo... |
ious). (c) The composite of two flat maps is flat (obvious). PROPOSITION 9.14. Let A ! A0 be a homomorphism of rings. If A ! B is flat, then so also is A0 ! B ˝A A0. PROOF. For any A0-module M , .B ˝A A0/ ˝A0 M ' B ˝A .A0 ˝A0 M / ' B ˝A M: In other words, tensoring an A0-module M with B ˝A A0 is the same as tensoring M... |
pen neighbourhood of P and W with an open subset intersecting Z, we may suppose that both V and W are affine. Let V V1 Vm D fP g be a maximal chain of distinct irreducible closed subsets of V (so m D dim.V /). Now '.Z/ D fP g, and so (see 9.24) there exists a chain of irreducible closed subsets W W1 Wm D Z such that '.... |
osed subset of ˘ P projective variety. The dimension of m is m.mC1/.mC5/ C 3. 6 ; it is therefore a EXAMPLE 9.45. For m D 1; m is the set of pairs consisting of a plane in P 3 and a line on the plane. The theorem says that the dimension of 1 is 5. Since there are 13 planes in P 3, and each has 12 lines on it, this seem... |
strictly lower dimension. Deduce that each orbit is a nonsingular subvariety of V , and that there exists at least one closed orbit. 9-2. Let G D GL2 D V , and let G act on V by conjugation. According to the theory of Jordan canonical forms, the orbits are of three types: (a) Characteristic polynomial X 2 C aX C b; di... |
tion Pai ci D 0; Pbi ci ¤ 0: The .n C 1/-tuples .c0; : : : ; cn/ satisfying these conditions form a nonempty open subset of the hyperplane H W P ai Xi D 0 in A nC1. On applying this remark to the pairs .P0; Pi /, we find that the .n C 1/-tuples c D .c0; : : : ; cn/ such that P0 lies on the hyperplane Lc but not P1; : :... |
“copy” of X we work in, the rest of ˜f has to follow what f does. Proof. First we show it is open. Let y be such that ˜f1(y) = ˜f2(y). Then there is an evenly covered open neighbourhood U ⊆ X of f (y). Let ˜U be such that ˜f1(y) ∈ ˜U , p( ˜U ) = U and p| ˜U : ˜U → U is a homeomorphism. Let V = ˜f −1 2 ( ˜U ). We will ... |
roup of a space is non-trivial. We can be more ambitious, and try to actually find π1(X, x0). In the best possible scenario, we would have π1( ˜X, ˜x0) trivial. Then we have a bijection between π1(X, x0) and p−1(x0). In other words, we want our covering space ˜X to be simply connected. Definition (Universal cover). A cov... |
be “the same”, this is a one-to-one correspondence — each subgroup corresponds to a covering space. We can have the following table of correspondences: Covering spaces Fundamental group (Based) covering spaces ←→ Subgroups of π1 Number of sheets ←→ Index Universal covers ←→ Trivial subgroup We now want to look at some ... |
emma. Lemma. If G is a group and φ : S → G is a set map, then there exists a unique homomorphism f : F (S) → G such that the following diagram commutes: F (S) S f φ G where the arrow not labeled is the natural inclusion map that sends sα (as a symbol from the alphabet) to sα (as a word). Proof. Clearly if f exists, the... |
ct with amalgamation π1(X, x0) = π1(A, x0) ∗ π1(A∩B,x0) π1(B, x0). The Seifert-van Kampen theorem says that, under mild hypotheses, this guess is correct. Theorem (Seifert-van Kampen theorem). Let A, B be open subspaces of X such that X = A ∪ B, and A, B, A ∩ B are path-connected. Then for any x0 ∈ A ∩ B, we have π1(X,... |
e. Proof. Let S = {a1, · · · , am} and R = {r1, · · · , rn}. We start with a single point, and get our X (1) by adding a loop about the point for each ai ∈ S. We then get our 2-cells e2 j for j = 1, · · · , n, and attaching them to X (1) by fi : S1 → X (1) given by a based loop representing ri ∈ F (S). Since all groups... |
dent. Now suppose Then we can write n i=0 tiai = 0, n i=0 ti = 0. t0 = − n i=1 ti. Then the first equation reads 0 = − n i=1 ti a0 + t1a1 + · · · + tnan = n i=1 ti(ai − a0). So linear independence implies all ti = 0. 56 6 Simplicial complexes II Algebraic Topology The relevance is that these can be used to define simplic... |
−1)). Proposition. |K| = |K | and K really is a simplicial complex. Proof. Too boring to be included in lectures. 62 6 Simplicial complexes II Algebraic Topology We now have a slight problem. Even though |K | and |K| are equal, the identity map from |K | to |K| is not a simplicial map. To solve this problem, we can cho... |
−1 0 1 −1 0 1 0 1 −1 68 7 Simplicial homology II Algebraic Topology We have now everything we need to know about the homology groups, and we just need to do some linear algebra to figure out the image and kernel, and thus the homology groups. We have H0(K) = ker(d0 : C0(K) → C−1(K)) im(d1 : C1(K) → C0(K)) ∼= C0(K)... |
= C1(K) dL n−1 Cn−1(L) dL n Cn(L) = Cn−1(K) Cn(K) = 0 dK n−1 · · · · · · 74 7 Simplicial homology II Algebraic Topology For k < n − 1, we have Ck(K) = Ck(L) and Ck+1(K) = Ck+1(L). Also, the boundary maps are equal. So Hk(K) = Hk(L) = 0. We now need to compute Hn−1(K) = ker dK n−1 = ker dL n−1 = im dL n . We get the la... |
n−1 An−2 in−2 Bn−2 jn−2 Cn−2 0 0 0 We want to check that dn−1(z) = 0. We will use the commutativity of the diagram. In particular, we know in−2 ◦ dn−1(z) = dn−1 ◦ in−1(z) = dn−1 ◦ dn(y) = 0. By exactness at An−2, we know in−2 is injective. So we must have dn−1(z) = 0. (iii) (a) First, in the proof, suppose we picked a ... |
> 0 such that each ball of radius 2ε in |L| lies in some star StL(w). Now apply the Lebesgue number lemma again to {f −1(Bε(y))}y∈|L|, an open cover of |K|, and get δ > 0 such that for all x ∈ |K|, we have f (Bδ(x)) ⊆ Bε(y) ⊆ B2ε(y) ⊆ StL(w) for some y ∈ |L| and StL(w). Now since g and f differ by at most ε, we know g(B... |
However, the consequence is that we will lose some information. Fortunately, the way in which we lose information is very well-understood and rather simple. Lemma. If Hn(K) ∼= Zk ⊕ F for F a finite group, then Hn(K; Q) ∼= Qk. Proof. Exercise. Hence, when passing to rational homology groups, we lose all information abou... |
exact sequence of complexes 0 A· i· B· q· C· 0 . Then there are maps such that there is a long exact sequence ∂ : Hn(C·) → Hn−1(A·) · · · Hn(A) Hn−1(A) i∗ i∗ Hn(B) ∂∗ Hn−1(B) q∗ q∗ Hn(C) . Hn−1(C) · · · The method of proving this is sometimes known as “diagram chasing”, where we just “chase” around commutative diagrams... |
([0, 1] × σ) ◦ i1 − H ◦ ([0, 1] × σ) ◦ i0 − n j=0 (−1)jH# ◦ (([0, 1] × σ) ◦ δj)#(Pn−1 (−1)jH# ◦ (([0, 1] × σ) ◦ δj)#(Pn−1) j=0 = g ◦ σ − f ◦ σ − Fn−1(dσ) = g#(σ) − f#(σ) − Fn−1(dσ). So we just have to show that Pn exists. We already had a picture of what it looks like, so we just need to find a formula that represents i... |
e points a, b, c, whose topology is {∅, {a}, {a, b, c}}. However, these spaces are horrible. We want to restrict our attention to nice spaces. Spaces that do feel like actual, genuine spaces. The best kinds of space we can imagine would be manifolds, but that is a bit too strong a condition. For example, the union of t... |
Algebraic Topology 6 (Co)homology with coefficients Recall that when we defined (co)homology, we constructed these free Z-modules from our spaces. However, we did not actually use the fact that it was Z, we might as well replace it with any abelian group A. Definition ((Co)homology with coefficients). Let A be an abelian gr... |
eorems of similar flavour — K¨unneth’s theorem and the universal coefficients theorem. They are both fairly algebraic results that relate different homology and cohomology groups. They will be very useful when we prove things about (co)homologies in general. In both cases, we will not prove the “full” theorem, as they requ... |
undle). Let π : E → X be a vector bundle, and F ⊆ E a subspace such that for each x ∈ X there is a local trivialization (U, ϕ) E|U ϕ π U U × Rd , π1 such that ϕ takes F |U to U × Rk, where Rk ⊆ Rd. Then we say F is a vector sub-bundle. Definition (Quotient bundle). Let F be a sub-bundle of E. Then E/F , given by the fibe... |
erminant on each fiber, then E is orientable for any R. Note that we don’t have to check the determinant at each point on Uα ∩ Uβ. By continuity, we only have to check it for one point. Proof. Choose a generator u ∈ H d(Rd, Rd \ {0}; R). Then for x ∈ Uα, we define εx by pulling back u along Ex E|Uα ϕα Uα × Rd π2 Rd . (†α... |
of the above computation. We have K = γR 1,n+1 → RPn = Gr1(Rn+1). The previous trick doesn’t work, and indeed this isn’t Z-orientable. However, it is F2-oriented as every vector bundle is, and by exactly the same argument, we know S(L) ∼= Sn. So by the same argument as above, we will find that H ∗(RPn, F2) ∼= F2[e(L)]/(... |
B | L) · · · ∂ The next step is to take the direct limit over K ∈ K(A) and L ∈ K(B). We need to make sure that these do indeed give the right compactly supported cohomology. The terms H n(A | K) ⊕ H n(B | L) are exactly right, and the one for H n(A ∩ B | K ∩ L) also works because every compact set in A ∩ B is a compact... |
ity III Algebraic Topology So we get a duality map DM = lim −→ (µK · ) : lim −→ H (M | K; R) → Hd−(M ; R). Now we can prove Poincar´e duality. Proof. We say M is “good” if the Poincar´e duality theorem holds for M . We now do the most important step in the proof: Claim 0. Rd is good. The only non-trivial degrees to che... |
following result: Lemma. Let X be a space and V a vector bundle over X. If V = U ⊕ W , then orientations for any two of U, W, V give an orientation for the third. Proof. Say dim V = d, dim U = n, dim W = m. Then at each point x ∈ X, by K¨unneth’s theorem, we have an isomorphism H d(Vx, V # x ; R) ∼= H n(Ux, U # x ; R)... |
this is equivalent to requiring that for each fixed point x, the map Dx∆ ⊕ DxF : TxM ⊕ TxM → TxM ⊕ TxM is an isomorphism. We can write the matrix of this map, which is I Dxf . I I Doing a bunch of row and column operations, this is equivalent to requiring that I Dxf 0 I − Dxf is invertible. Thus the condition is equiva... |
be invertible, i.e. a unit, as they cannot all lie in the Jacobson radical. We may wlog assume α1 ◦ β1 is a ∼= M 1. unit. If this is the case, then both α1 and β1 have to be invertible. So M1 Consider the map id −θ = φ, where θ : M α−1 1 M1 M 1 M m i=2 Mi M. Then φ(M 1) = M1. So φ|M 1 looks like α1. Also m φ i=2 Mi = m... |
(A)] → A [A, A] , by linear algebra. Now if P is finitely generated projective. It is a direct summand of some An. Thus we can write An = P ⊕ Q, for P, Q projective. Moreover, projection onto P corresponds to an idempotent e in Mn(A) = EndA(An), and that and we have P = e(An). EndA(P ) = eMn(A)e. Any other choice of ide... |
ut is rather messy. It requires a careful induction. 33 2 Noetherian algebras III Algebras Example. Let g be a Lie algebra, and consider A = U(g). This has generating set x1, · · · , xn, which is a vector space basis for g. Again using the filtration for finitely-generated algebras, we get that if ai ∈ Ai and aj ∈ Aj, th... |
0. Also, we know that d(M ) is an integer for cases A = An or U(g), since it is the degree of a polynomial. However, for general M = A, we can get non-integral values. In fact, the values we can get are 0, 1, 2, and then any real number ≥ 2. We can also have ∞ if the lim sup doesn’t exist. Example. If A = kG, then we ... |
By transitivity of essential extensions, F/V1 is an essential extension of M , but E is maximal. So E ∼= F/V1. In other words, F = E ⊕ V1. 44 2 Noetherian algebras III Algebras We can now make the following definition: Definition (Injective hull). A maximal essential extension of M is the injective hull (or injective env... |
g, given a k-algebra A and an A-A-bimodule M , Hochschild cohomology is an infinite sequence of k-vector spaces HH n(A, M ) indexed by n ∈ N associated to the data. While there is in theory an infinite number of such vector spaces, we are mostly going to focus on the cases of n = 0, 1, 2, and we will see that these group... |
-simple. Thus, as a bimodule, A ⊗ A is completely reducible. So the quotient of AAA is a direct summand (of bimodules). So we know that whenever char k |G|, then kG is k-separable. The obvious question is — is this notion actually a generalization of separable field extensions? This is indeed the case. Fact. Let L/K be ... |
maps V ⊗ V → V . For convenience, we will write F0(a, b) = ab. The only non-trivial requirement f has to satisfy is associativity: f (f (a, b), c) = f (a, f (b, c)). What condition does this force on our Fi? By looking at coefficients of t, this implies that for all λ = 0, 1, 2, · · · , we have Fm(Fn(a, b), c) − Fm(a, F... |
: v ∈ V . The product is given by the wedge, and the Schouten bracket is given by [λ1 ∧ · · · ∧ λm] = (−1)(m−1)(n−1) (−1)i+j[λi, λj] λ1 ∧ · · · λm ith missing ∧ λ 1 ∧ · · · ∧ λ n jth missing . i,j For any Gerstenhaber algebra H = H i, there is a canonical homomorphism of Gerstenhaber algebras H 1 → H. Theorem (Hochschi... |
algebra. Definition (Drinfeld double). Let G be a finite group. We define D(G) = (kG)∗ ⊗k kG as a vector space, and the algebra structure is given by the crossed product (kG)∗ G, where G acts on (kG)∗ by Then the product is given by f g(x) = f (gxg−1). (f1 ⊗ g1)(f2 ⊗ g2) = f1f g−1 1 2 ⊗ g1g2. The coalgebra structure is t... |
> 0)(∀y ∈ A) |y − a| < δ ⇒ |f (y) − f (a)| < ε. Intuitively, f is continuous at a if we can obtain f (a) as accurately as we wish by using more accurate values of a (the definition says that if we want to approximate f (a) by f (y) to within accuracy ε, we just have to get our y to within δ of a for some δ). For example... |
ontinuous induction. Finally, we can prove a theorem that we have not yet proven. Definition (Cover of a set). Let A ⊆ R. A cover of A by open intervals is a set {Iγ : γ ∈ Γ} where each Iγ is an open interval and A ⊆ γ∈Γ Iγ. A finite subcover is a finite subset {Iγ1 , · · · , Iγn } of the cover that is still a cover. Not ... |
heorems, including but not limited to Taylor’s theorem, which gives us Taylor’s series. Theorem (Rolle’s theorem). Let f be continuous on a closed interval [a, b] (with a < b) and differentiable on (a, b). Suppose that f (a) = f (b). Then there exists x ∈ (a, b) such that f (x) = 0. It is intuitively obvious: if you mov... |
evidenced by the above examples, the ratio test can be used to find the radius of convergence. We also have an alternative test based on the nth root. Lemma. The radius of convergence of a power series anzn is R = 1 lim sup n|an| . Often n|an| converges, so we only have to find the limit. Proof. Suppose |z| < 1/ lim sup... |
i = sup x∈[xi−1,xi] f (x), mi = inf x∈[xi−1−xi] f (x). The upper sum is the total area of the red rectangles, while the lower sum is the total area of the black rectangles: y · · · · · · a x1 x2 x3 xi xi+1 · · · b x Definition (Refining dissections). If D1 and D2 are dissections of [a, b], we say that D2 refines D1 if eve... |
mum and infimum. We find what we pretend is the difference between the upper and lower sum: xi − xi−1)(f (vi)g(vi) − f (ui)g(ui) (xi − xi−1)f (vi)(g(vi) − g(ui)) + (f (vi) − f (ui))g(ui) n i=1 n i=1 n = ≤ C(Li − i) + (Mi − mi)C i=1 = C(UDg − LDg + UDf − LDf ). Since ui and vi are arbitrary, it follows that UD(f g) − LD(f ... |
n+1)(t) dt. Proof. Induction on n. When n = 0, the theorem says f (b) − f (a) = b a f (t) dt. which is true by the fundamental theorem of calculus. Now observe that b a (b − t)n n! f (n+1)(t) dt = = b f (n+1)(t) a f (n+1)(t) dt −(b − t)n+1 (n + 1)! b + a (b − a)n+1 (n + 1)! (b − t)n+1 (n + 1)! f (n+1)(a) + b a (b − t)n... |
emann integral*). Let f : [a, b] → R be a bounded function, and let Df be the set of points of discontinuities of f . Then f is Riemann integrable if and only if Df has measure zero. Using this result, a lot of our theorems follow easily of these. Apart from the easy ones like the sum and product of integrable function... |
nly if it converges to x with respect to ∥ · ∥′. Proof. (i) This is direct from definition of equivalence. (ii) Say we have a, b such that a∥y∥ ≤ ∥y∥′ ≤ b∥y∥ for all y. So a∥xk − x∥ ≤ ∥xk − x∥′ ≤ b∥xk − x∥. So ∥xk − x∥ → 0 if and only if ∥xk − x∥′ → 0. So done. What if the norms are not equivalent? It is not surprising... |
is a limit point of E if and only if (Br(y) \ {y}) ∩ E ̸= ∅ for every r. Proof. (⇒) If y is a limit point of E, then there exists a sequence (xk) ∈ E with xk ̸= y for all k and xk → y. Then for every r, for sufficiently large k, xk ∈ Br(y). Since xk ̸= {y} and xk ∈ E, the result follows. (⇐) For each k, let r = 1 k . ... |
by d(x, y) = ∥x−y∥. Then we can define xk → x to mean d(xk, x) → 0. Hence, given any function d : V × V → R, we can define a notion of “convergence” as above. However, we want this to be well-behaved. In particular, we would want the limits of sequences to be unique, and any constant sequence xk = x should converge to ... |
ence has a convergent subsequence, then the original sequence converges to the same limit. Proof. (i) If xk → x, then as m, n → ∞. d(xm, xn) ≤ d(xm, x) + d(xn, x) → 0 (ii) Suppose xkj → x. Since (xk) is Cauchy, given ε > 0, we can choose an 2 for all n, m ≥ N . We can also choose j0 such N such that d(xn, xm) < ε that ... |
, (x2, y2)) Recall that at the very beginning, we proved that a continuous map from a closed, bounded interval is automatically uniformly continuous. This is true whenever the domain is compact. Theorem. Let (X, d) be a compact metric space, and (X ′, d′) is any metric space. If f : X → X ′ be continuous, then f is uni... |
see that both f = 0 and f (t) = 1 4 t2 are both solutions. In fact, for any α ∈ [0, b], the function fα(tt − α)2 α ≤ t ≤ b is also a solution. So we have an infinite number of solutions. We are now going to use the contraction mapping theorem to prove this. In general, this is a very useful idea. It is in fact possibl... |
uld relax this condition and consider “one-sided” derivatives instead, but we will not look into these in this course. We can interpret the definition of differentiability as saying we can find a “good” linear approximation (technically, it is affine, not linear) to the function f near a. While the definition of the de... |
fact that linear maps are differentiable and hence continuous. We can use this fact to define the norm of linear maps. Since L is finitedimensional (it is isomorphic to the space of real m × n matrices, as vector spaces, and hence have dimension mn), it really doesn’t matter which norm we pick as they are all Lipschit... |
n (Diffeomorphism). Let U, U ′ ⊆ Rn are open, then a map g : U → U ′ is a diffeomorphism if it is C 1 with a C 1 inverse. Note that different people have different definitions for the word “diffeomorphism”. Some require it to be merely differentiable, while others require it to be infinitely differentiable. We will sti... |
s → Dif (a + θtei + stej), there is some η ∈ (0, 1) such that gij(t) = t2DjDif (a + θtei + ηtej). We can do the same for gji, and find some ˜θ, ˜η such that gji(t) = t2DiDjf (a + ˜θtei + ˜ηtej). Since gij = gji, we get t2DjDif (a + θtei + ηtej) = t2DiDjf (a + ˜θtei + ˜ηtej). Divide by t2, and take the limit as t → 0. B... |
definition of a weak = U ηε(x − y)Dαu(y) dy = ηε ∗ Dαu. It is an exercise to verify that we can indeed move the derivative past the integral. Thus, if we fix V U . Then by the previous parts, we see that Dαuε → Dαu in Lp(V ) as ε → 0 for |α| ≤ k. So uε − up W k.p(V ) = |α|≤k Dαuε − Dαup Lp(V ) → 0 as ε → 0. Theorem (Glo... |
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