Saturday, September 27, 2008
Friday, September 26, 2008
Extending exact sequence
Every chain of short exact sequences of finite length ≥ 2 can be extended at either end by 0→kernel→ and →cokernel→0. If length = 2, this is not possible in general by Snake lemma, which says that the kernel and cokernel link to form a long exact sequence.
Splitting exact sequence
Every exact chain of short exact sequences of length ≥ 2 can be split up into two exact chains of short exact sequences, each ending at kernel(=image) of chain maps.
Thursday, September 25, 2008
Tensor product
Let M, N be R-modules. Take their tensor product as abelian groups. Then it has a natural R-module structure and is isomorphic as an R-module to the R-module obtained by tensor producting M and N as R-modules.
Monday, September 22, 2008
Topological sum
Suppose X is the topological sum of a collection of disjoint topological spaces {Xα}. Then each Xα is open in X, and the open sets in X are precisely the (disjoint) unions of subsets Aα⊆Xα open in Xα. Also, the original topology on Xα is the subspace topology of X on Xα.
Subspace topology
Suppose X, Y are topological spaces and that X⊆Y.
Then X is the subspace topology of Y if and only if the inclusion map j: X →Y is homeomorphism onto its image.
Then X is the subspace topology of Y if and only if the inclusion map j: X →Y is homeomorphism onto its image.
Saturday, September 20, 2008
Union of subcomplexes
Suppose K1, K2 are subcomplexes in a simplicial complex K such that |K1|∩|K2| = |L|, where L is a subcomplex of both K1 and K2. Then K1∪K2 is a subcomplex of K.
pf> Suppose σ1∈K1, σ2∈K2 are simplices. We want to show that they intersect at a face of each. Let I = σ1 ∩ σ2. Then I ⊆ |L|. Let V = {vα}α be a set of vertices of L contained in I. They are contained in σ1 and σ2, too, and they are vertices of σ1 and σ2. In particular, they are finite. Let σ ∈ L be a simplex spanned by these vertices. Then clearly σ ⊆ I. Also, if x∈I then there is a unique simplex f∈L having x as an interior point. Since L is a subcomplex of both K1 and K2, f is none other than the unique face of σ1 having x as an interior point. Similar argument for σ2 shows that x is in a common face of σ1 and σ2. So x is spanned by vertices of this face, which are in V. So I ⊆ σ, and we have I = σ. Since σ is a face of both σ1 and σ2, we proved the theorem.
It is easy to see that the union of an arbitrary collection of subcomplexes every pair of which satisfies above condition is also a subcomplex.
pf> Suppose σ1∈K1, σ2∈K2 are simplices. We want to show that they intersect at a face of each. Let I = σ1 ∩ σ2. Then I ⊆ |L|. Let V = {vα}α be a set of vertices of L contained in I. They are contained in σ1 and σ2, too, and they are vertices of σ1 and σ2. In particular, they are finite. Let σ ∈ L be a simplex spanned by these vertices. Then clearly σ ⊆ I. Also, if x∈I then there is a unique simplex f∈L having x as an interior point. Since L is a subcomplex of both K1 and K2, f is none other than the unique face of σ1 having x as an interior point. Similar argument for σ2 shows that x is in a common face of σ1 and σ2. So x is spanned by vertices of this face, which are in V. So I ⊆ σ, and we have I = σ. Since σ is a face of both σ1 and σ2, we proved the theorem.
It is easy to see that the union of an arbitrary collection of subcomplexes every pair of which satisfies above condition is also a subcomplex.
Intersection of simplices
Suppose σ, σ' are simplices in a simplicial complex and σ' ⊆ σ. Then σ' is a face of σ.
Uniqueness in simplicial complex
Let K be a simplicial complex. If x ∈ |K| then there is a unique simplex in K having x as an interior point.
Simplicial topology
Let K be a simplicial complex, and Δ ∈ K a simplex. Then the topology on Δ as a subspace topology of |K| is the same as the topology on Δ as a subspace topology of the Euclidean space in which K is embedded. So a subset of Δ is |K|-closed if and only if it is Euclidean-closed, and in particular, Δ is a closed subset of |K|.
Wednesday, September 17, 2008
Relative homology
Suppose K0 ⊆ K is a subcomplex.
If ∂ : C*(K) → C*(K) and
∂' : C*(K)/C*(K0) → C*(K)/C*(K0)
are boundary maps, then
Im ∂' = ( Im ∂ + C*(K0) ) / C*(K0),
and ker ∂' = ∂-1(C*(K0) / C*(K0).
In particular, if two *-chains are homologous modulo ∂, then they are also homologous modulo ∂'.
If ∂ : C*(K) → C*(K) and
∂' : C*(K)/C*(K0) → C*(K)/C*(K0)
are boundary maps, then
Im ∂' = ( Im ∂ + C*(K0) ) / C*(K0),
and ker ∂' = ∂-1(C*(K0) / C*(K0).
In particular, if two *-chains are homologous modulo ∂, then they are also homologous modulo ∂'.
Simplicial map
Simplicial map maps each simplex onto a simplex.
It also maps the interior of each simplex onto the interior of a simplex.
It also maps the interior of each simplex onto the interior of a simplex.
Tuesday, September 16, 2008
Regular sequence
Let R be a regular local ring of dimension d and m = (x1, x2, ..., xd) its maximal ideal. Then for each i, R / (x1, x2, ..., xi) is a regular local ring of dimension d - i.
pf> Let S = R / (x1, x2, ..., xd) and k = dim S. Clearly S is local. Since m is generated by d - i elements in S, d - i ≥ k. On the other hand, if m is minimal over (y1, y2, ..., yl) in S, then m is minimal over (x1, x2, ..., xi, y1, y2, ..., xl) in R, so i + l ≥ d, i.e., l ≥ d - i. Since k is the smallest such number l, k ≥ d - i. Hence k = d - i. QED
pf> Let S = R / (x1, x2, ..., xd) and k = dim S. Clearly S is local. Since m is generated by d - i elements in S, d - i ≥ k. On the other hand, if m is minimal over (y1, y2, ..., yl) in S, then m is minimal over (x1, x2, ..., xi, y1, y2, ..., xl) in R, so i + l ≥ d, i.e., l ≥ d - i. Since k is the smallest such number l, k ≥ d - i. Hence k = d - i. QED
Monday, September 15, 2008
Chain condition is preserved upon restriction and quotienting
If M satisfies a chain condition on its submodules, then so do its submodules and quotient modules.
Minimal number of generators of a prime
R a Noetherian ring, P a prime of codimension d.
Then P is generated by at least d generators.
Then P is generated by at least d generators.
Sunday, September 14, 2008
Maximals can see ideal partial orders
If a ⊆ b are ideals in a ring, then a = b if and only if am = bm for every maximal ideal m.
Saturday, September 13, 2008
Fraction field
Suppose an integral domain A is contained in a field K. Then the field of fraction of A in K is the smallest subfield of K containing A. It is given by the set of all a-1b, where a, b are elements of A and a ≠ 0.
The union of two polytopes is a polytope
Suppose a simplicial complex K and two subcomplexes L1 and L2. Let |L1| and |L2| denote their polytopes.
Then |L1 ∩ L2| = |L1| ∩ |L2|.
Then |L1 ∩ L2| = |L1| ∩ |L2|.
The complement of the star at v is the union of all simplices not having v as a vertex
Let {Tβ} β be the simplices in a simplicial complex K not having v as a vertex. Let U be the star at v. Then Int Tβ is in the complement of U and so is their union and closure of the union . By the previous post, it is equal to the union of the closures, i.e., the union L of {Tβ} β. So L is contained in the complement of U. To prove the converse, let {S α} α be the set of simplices having v as a vertex. If x is not in L then x is in the interior of a simplex S α, so x is not contained in the complement of U. i.e. the complement of U is contained in L. QED
The closure and union operations on a set of simplices in a simplicial complex commute
Suppose K is a simplicial complex, and Λ = {S α} α its subset. Let T be the union of {S α} α and U the union of {Int S α} α. Clearly T is contained in the closure of U. But the other inclusion is also true; suppose x is a point in |K| not in T. Then it is in the interior of a simplex not in Λ, hence x is not in the closure of U.
Wednesday, September 10, 2008
Compact subset of a polytope is contained in a finite subcomplex.
Let A be a compact subset of |K|.
Suppose it is not contained in any finite subcomplex of K.
First note that interiors of simplices in K are disjoint. A can't be covered by finite number of interiors of simplices in K, so if we pick an element xs whenever an interior of a simplex s intersects A, then the set B of all such xs is infinite. Furthermore, if B' ⊆ B is any subset, then any simplex in K intersects B' only at finitely many points, so it is closed in |K|, and hence in B. i.e., B is a discrete space. Also B itself is closed in A, so is compact. i.e., we have an infinite compact discrete space B. This can't be true because an infinite compact space must have a limit point. QED
Suppose it is not contained in any finite subcomplex of K.
First note that interiors of simplices in K are disjoint. A can't be covered by finite number of interiors of simplices in K, so if we pick an element xs whenever an interior of a simplex s intersects A, then the set B of all such xs is infinite. Furthermore, if B' ⊆ B is any subset, then any simplex in K intersects B' only at finitely many points, so it is closed in |K|, and hence in B. i.e., B is a discrete space. Also B itself is closed in A, so is compact. i.e., we have an infinite compact discrete space B. This can't be true because an infinite compact space must have a limit point. QED
Subcomplex is a closed subspace
Suppose A ⊆ |L| ⊆ |K|.
1. If Tα∈K, |L| ∩ Tα is a finite union of faces of Tα hence is closed in Tα. So |L| is closed in |K|.
2. If A is closed in the subspace topology of |L| then it is closed in |K| so is closed in |L|.
3. Suppose A is closed in |L|. If Tα∈K, A ∩ Tα = A ∩ (Tα ∩ |L|). Tα ∩ |L| is a finite union of faces fi, i = 1, 2, ..., n, of Tα which are also in L. So A ∩ Tα is a finite union of A ∩ fi , i = 1, 2, ..., n, each of which is closed in fi and hence in Tα. So A is closed in |K|. Hence A is closed in the subspace topology. QED
1. If Tα∈K, |L| ∩ Tα is a finite union of faces of Tα hence is closed in Tα. So |L| is closed in |K|.
2. If A is closed in the subspace topology of |L| then it is closed in |K| so is closed in |L|.
3. Suppose A is closed in |L|. If Tα∈K, A ∩ Tα = A ∩ (Tα ∩ |L|). Tα ∩ |L| is a finite union of faces fi, i = 1, 2, ..., n, of Tα which are also in L. So A ∩ Tα is a finite union of A ∩ fi , i = 1, 2, ..., n, each of which is closed in fi and hence in Tα. So A is closed in |K|. Hence A is closed in the subspace topology. QED
Tuesday, September 9, 2008
Singular chain map
If ∂: Sp(X) → Sp-1(X) is the boundary operator of a singular chain complex, then ∂2 = 0.
Monday, September 8, 2008
Prime minimality over an ideal
If I ⊆ p and I contains some power of p, then
p is a minimal prime over I.
p is a minimal prime over I.
Dimension invariance under localization
Suppose p ⊂ A is disjoint from a multiplicative subset S ⊆ A. Then
dim Ap = dim (S-1A)S-1p.
In particular, dim (S-1A)S-1p is independent of the multiplicative subsets S disjoint from p.
dim Ap = dim (S-1A)S-1p.
In particular, dim (S-1A)S-1p is independent of the multiplicative subsets S disjoint from p.
Atiyah MacDonald Commutative Algebra Error
On page 91, in the last example, m2 is NOT zero. It is in fact generated by one element. The dimension of m/m2, however, is still 2, generated by x2 and x3.
Sunday, September 7, 2008
Polynomial ring
If p is a prime in R, then p[x] is a prime in R[x], and
p[x] ∩ R = p. Hence if p1 ⊂ p2 strictly,
then p1[x] ⊂ p2[x] strictly. Also, p[x] + (x) is a prime in R[x].
If Q is a maximal ideal in R[x], then Q ∩ R is a maximal ideal in R.
p[x] ∩ R = p. Hence if p1 ⊂ p2 strictly,
then p1[x] ⊂ p2[x] strictly. Also, p[x] + (x) is a prime in R[x].
If Q is a maximal ideal in R[x], then Q ∩ R is a maximal ideal in R.
Dimensions
dim R = sup p {dim Rp} = sup m {dim Rm}.
In particular, for any prime p, dim Rp ≤ dim R,
and if dim R is finite, dim Rm = dim R for some m.
In particular, for any prime p, dim Rp ≤ dim R,
and if dim R is finite, dim Rm = dim R for some m.
Prime of codimension n is minimal over an ideal generated by n elements
Here the ring R is assumed to be Noetherian. Suppose P is a prime of codimensin n.
1. Every minimal prime(over (0)) has codimension 0.
2. Every prime contains a minimal prime. If x1 is in P but not in any of the minimal primes, then the primes minimal over (x1) strictly contains a minimal prime, and hence has codimension 1, by Principal Ideal Theorem.
2. If x2 is in P but not in any of the minimal primes over (x1), then the primes minimal over (x1, x2) strictly contains a minimal prime over (x1), and hence has codimension 2 by Principal Ideal Theorem.
3. This process can't continue past k=n, since it would imply P contains a prime of codimension greater than n. The process can't stop before k=n either, because that would imply that P is contained in the unions of primes of codimension less than n, which implies that P is equal to one of them.
4. So this process will stop when k=n, and this implies that P is minimal over (x1, ..., xn). QED
1. Every minimal prime(over (0)) has codimension 0.
2. Every prime contains a minimal prime. If x1 is in P but not in any of the minimal primes, then the primes minimal over (x1) strictly contains a minimal prime, and hence has codimension 1, by Principal Ideal Theorem.
2. If x2 is in P but not in any of the minimal primes over (x1), then the primes minimal over (x1, x2) strictly contains a minimal prime over (x1), and hence has codimension 2 by Principal Ideal Theorem.
3. This process can't continue past k=n, since it would imply P contains a prime of codimension greater than n. The process can't stop before k=n either, because that would imply that P is contained in the unions of primes of codimension less than n, which implies that P is equal to one of them.
4. So this process will stop when k=n, and this implies that P is minimal over (x1, ..., xn). QED
Friday, September 5, 2008
Butterfly & slug
This morning I saw a butterfly sitting on a dead slug. The slug was still juicy but drying up fast under the morning sun. The butterfly was apparently sipping up its juice... Should it be called carnivorous?
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