Chapter 1 - Roots of Commutative Algebra Flashcards

1
Q

Discuss number theory motivations for commutative algebra

A

Rings of integers

  1. Z
  2. Z[i] integers of quadratic number field
  3. Z[zeta] integers of cyclotomic field

Question: Is this ring a UFD?
Commutative Algebra: Picard Group Pic(R) - invariant measuring how far from UFD

Much of this early development was motivated by the search for a proof of Fermat’s last theorem. Adjoining a root of unity to Z, we can factor x^n + y^n = z^n and arrive at product_i (x - zeta^(2i+1)y) = z^n. If the ring Z[zeta] is a UFD, we are essentially done. Unfortunetly, Z[zeta] is rarely a UFD. Search for generalization of UFD

  1. Facorization of ideals into prime ideals in Dedekind domains
  2. Primary decomposition
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2
Q

Discuss algebraic geometry motivations for commutative algebra

A

Coordinate rings

  • generated by coordinate functions
    1. k[x,y] - ring generated by coordinate functions on k^2
    2. k[x,y] /(y^2 - x^3 +x) coordinate ring of eliptic curve

Question: How are points of curve related to coordinate ring?
Commutative Algebra: points <=> maximal ideals

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3
Q

Discuss invariant theory motivations for commutative algebra

A

Invariant Rings
Consider action of icosahedron on R^3 - group of order 120. Action on R^3 yields an action on R[x,y,z]
Invariants = polynomials fixed by symmetries
Obvious: x^2 + y^2 + z^2 = a
Two other of degree 8 and 10.

Question: Is invariant ring finitely generated?
Commutative Algebra: Hilbert’s Basis Theorem - yes for G reductive

In the 1830s, people became interested in the geometric properties of plane curves invariant under certain classes of transformations. Give a sort of function that associates to a geometric configuration, a number that is invariant under group action. The functions studied were mostly polynomial functions of quantities defining the geometric objects - such as the coefficients of the equations of algebraic plane curves.

Central problem: Given a “nice” action of a group G as automorphisms of a polynomial ring S, find the elements of S that are left invariant by G.

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4
Q

Define: ring, module

A

Set with two operations: + *. Abelian group under +, commutative with identity

Module over a ring is like a vector space over a field - more flexible than ideals
abelian group + scalar multiplication by elements of ring

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5
Q

Examples of non-commutative rings?

A
  1. Matrix rings
  2. Quaternions
  3. Group rings
  4. Rings of differential operators
  5. Non-commutative polynomial rings
  6. Universal Enveloping Algebra of Lie Algebra
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6
Q

Discuss constructions which turn out to be special cases of modules

A
  1. Z-modules = abelian groups
  2. k-module = vector space
  3. k[x]-module = linear transformation
  4. Submodule of R = ideal
  5. R/I as R-module = quotient ring
  6. k[G]-modules = representation theory
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7
Q

How does a group G act on polynomials?

A

Given an action of G on spaces X and Y, get an action of G on the ring of functions f:X –> Y given by (gf)(x) = gf(g^-1 x)

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8
Q

Discuss the invariant ring of S_n acting on k[x1, … , xn]

A

Elementary symmetric polynomials
e_1 = x_1 + x_2 + … + x_n
e_2 = x_1x_2 + x_1x_3 + …

(y-x_1)(y-x_2)…(y-x_n) = y^n - e_1y^n-1 + e_2y^n-2 - …

Thm. Every invariant polynomial in x_1, …, x_n is a polynomial in e_1, … , e_n

Borcherds pg 18 - 20 Eisenbud e1.7

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9
Q

Discuss ring of invariants of A_n acting on k[x1, … , xn]

A

Define A_n as the subgroup fixing Vandermonde determinant. Then clear that invariants of A_n = elementary symmetric polynomials plus VD

Have first order Syzygy

Borcherds 20 - 21

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10
Q

Discuss the invariants of Z/3Z acting on C^2

A

Generated by homogeneous degree 3 polynomials
1st and 2nd order syzygies

B 22

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11
Q

What are the key questions about ring of invariants? Answers?

A
  1. Is ring of invariants R fg as k-Algebra?
  2. Are syzygy modules fg as R-module?
  3. Is chain of modules finite? - finite free resolution

Hilbert: Yes if G is reductive k char 0

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12
Q

Example of ideal that is not f.g?

A

k[x1….] ideal of all polys with constant term 0

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13
Q

Example of f.g. module over Noetherian ring without finite free resolution?

A
R = k[x] / (x^2)
M = R / (x) = k 

B 32

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14
Q

Discuss 3 equivalent definitions of Noetherian rings and prove equivalence

A
  1. All ideals are f.g.
  2. A.c.c
  3. Every nonempty set of ideals has a maximal element

B33-34

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15
Q

Example of non-Noetherian ring and failure of conditions

A

k[x1, x2, …]

  1. (x1, x2, x3, …) not f.g.
  2. (x1) < (x1, x2) < (x1, x2, x3) < … no acc
  3. {(x1), (x1, x2), (x1, x2, x3), … } no maximal element
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16
Q

Define: Artinian ring and give nonexample

A
  1. d.c.c.
  2. Every set of ideals has a minimal element

Much stronger than Noetherian

Z is non-Artinian

(2) > (4) > (8) > (16) > …

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17
Q

Give sequence of examples of Noetherian/non-Noetherian rings

A

R[x] polynomials –> Analytic functions on R –> Analytic on [-1,1] –> Analytic on (-1, 1) –> Analytic at 0 –> Smooth near 0 –> Formal Power Series

In Noetherian rings, zeros are very well behaved. Finite order. Can control them
B36

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18
Q

Is a subring of a Noetherian ring Noetherian?

A

No

  1. Analytic functions < Formal power series
  2. R a non-Noetherian integral domain < Q field of quotients
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19
Q

Discuss how quotient rings behave with Noetherian property

A

If R is Noetherian, so is any quotient R/I. Ideals of R/I <=> Ideals of R containing I

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20
Q

What is Hilbert basis theorem? Noether Proof? Gordan Proof?

Corollaries?

A

Thm. If R is Noetherian, then R[x] is Noetherian.

Pf. (Noether) Suppose I is an ideal of R[x]. Look at ideals I0 < I1 < … where Ik is ideal of leading coefficients of degree k elements. Since R Noetherian, chain stabilizes at some In.

S0, S1, … Sn, where Sk is finite set of polys in I of degree k whose leading coefficients generate Ik.

The union of these finite sets generates I

B40-41

Pf. (Gordan)

  1. Dickson’s Lemma. If S is any set of monomials, it has only a finite number of minimal elements (ordered by divisibility)
  2. Now look at ideal I < k[x1, … , xn]. Order monomials lexicographically. Look at set of leading terms of polys in I. This is a subset of monomials. It has finite number of min elements. Pick a polynomial for each min element. These generate I.

Corollary 1.3. Any homomorphic image of a Noetherian ring is Noetherian. If S is Noetherian and R is a f.g. algebra over S, then R is Noetherian pg 28

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21
Q

Prove R[[x]] is Noetherian if R is Noetherian

A

This is essentially Hilbert basis theorem for formal power series.

Now we can’t look at leading coefficients since no largest elements. Instead, look at smallest term of power series.

B43-44

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22
Q

If G is a group acting on a vector space, when do we know the algebra of invariants is finitely generated?

Proof?

Compact groups? Noncompact?

A
  1. We need G reductive
  2. We need V finite dimensional
  3. We need base field k to have characteristic 0

Say dim V = n. The algebra of all polys is k[x1, … , xn] which is graded by degree.

The key property of ring of invaraints has allowing this is RENOLDS OPERATOR - a homomorphism of I-modules from R –> I. Retraction onto I. Splits s.e.s.
B46-51

pg 31

23
Q

Define: Noetherian module

A

A module M over some ring R is called Noetherian if one of the following hold

  1. Submodules are finitely generated
  2. Every nonempty set of submodules has a maximal element
  3. Every increasing chain M1 < M2 < … terminates acc

Note: R is Noetherian as a ring <=> R is Noetherian. as R-module.
Submodules = Ideals

24
Q

Prove: Given a s.e.s. of R-modules 0 –> A –> B —> C —> 0, B is Noetherian <=> A and C are Noetherian.

Corollaries?

A

=> Easy. Recognize submodules of A and C as submodules of B.

<= Let M be submodule of B. Show M finitely generated. Im M in C is fg and Im A in M is f.g.

Special Case: A,B Noetherian => A (+) B Noetherian.
Even more special: R^n Noetherian if R Noetherian ring

Cor. Any fg module M over a Noetherian ring is Noetherian.

pg 28

Pf. R^n is Noetherian by above. M f.g. implies we have a map R^n –> M –> 0 so M is a quotient of Noetherian => Noetherian.
B52-53

Exercise 3

25
Q

Define: graded ring, homogeneous element, idea, component, irrelevant ideal

Examples?

A

A GRADED RING is a ring R together with a direct some decomposition R = R0 + R1 + R2 + … as abelian groups, such that RiRj = Ri+j for i, j >= 0.

A HOMOGENEOUS ELEMENT of R is an element of one of the groups Ri, and a HOMOGENEOUS ideal of R is an ideal that is generated by homogeneous elements. If f in R, there is a unique expression for f of the form f = f0 + f1 + … with fi in Ri, the fi are called the HOMOGENEOUS COMPONENTS of f.

Although it is the most important ideal of R, the ideal consisting of all elements of degree > 0 is called the IRRELEVANT IDEAL written R_+

Example
1. Polynomials S = k[x1, … , xr] graded by degree
S = S_0 + S_1 + … where S_d is the vector space of all homogeneous polynomials of degree d.

26
Q

Define: reduced ring, radical ideal

A

A ring is REDUCED if its only nilpotent element is 0. rad I = {f in R : f^m in I for some integer m}. An ideal is radical if I = rad I.

R/I is reduced <=> I is radical.

27
Q

What is Nullstellensatz? 4 Corollaries?

Define: affine k-algebra, affine ring

A

I(Z(I)) = rad I. Thus bijection between algebraic sets and radical ideals.

Cor. A system of polynomial equations over an algebraically closed field k has no solution in k^n <=> 1 can be expressed as a linear combo of the polynomials with polynomial coefficients

Cor. If k is an alg closed field and A is a k-algebra then A = A(X) for some algebraic set X <=> A is reduced and f.g. as k-algebra.

Because of this corollary, reduced f.g. k-algebras are often called AFFINE K-ALGEBRAS or AFFINE RINGS.

Cor. Let k be an algebraically closed field and let X < A^n be an algebraic set. Every maximal ideal of A(X) is of the form m_p = (x1 - a1, … , xn-an) / I(X) for some p in X. i.e. the points of X are in 1-to-1 correspondence with the maximal ideals of A(X).

Cor. The category of affine algebraic sets and morphisms over an algebraically closed field k is equivalent to the category of affine k-algebras with the arrows reversed.

28
Q

Define: graded module, Hilbert function, Hilbert polynomial

A

If R = R0 + R1 + … is a graded ring, then a GRADED MODULE OVER R is a module M with a decomposition M = sum_-inf^inf M_i as abelian groups s.t. RiMj = Mi+j

Let M be a finitely generated graded module over k{x1, …, xr] with grading by degree. The numerical function H_M(s) = dim_k M_s is called the HILBERT FUNCTION OF M.

Thm. If M is a finitely generated graded module over k[x1, … , xr], then H_M(s) agrees, for large s, with a polynomial of degree <= r-1.

This polynomial is called the HILBERT POLYNOMIAL OF M.

29
Q

Three ways to draw picture of a ring?

A
  1. Draw a point for each element of the ring
  2. Draw a point for each basis element of the ring
  3. Draw a point for each prime ideal of the ring
30
Q

Examples/applications of drawing a point for each element of ring?

A

Examples: Z, R, C, Z[i] - square lattice in C

Applications: Show Z[i], Z[root(-2)] Euclidean. Z[root(-3)] not Euclidean: show not UFD, show (2, 1+root(-3)) not principal idea.

This method works well for rings that can be embedded in a f.d. vector space – very useful in algebraic number theory

B62-66

31
Q

Examples/applications of drawing a point for each basis element of ring?

A

Assume ring is something like ring of polynomials.

  1. Ideal f.g. ring not f.g.
  2. Ideals of finite codimension
  3. Find dim of k[x,y]/(monomial ideal)
  4. Weierstrass Polynomials => k[[x,y]] is UFD

B67-73

32
Q

Discuss motivation for spectrum of ring in terms of compact Hausdorff space. Problem? Solution?

A

Let X be a compact Hausdorff space and R be the ring of continuous functions on X.
IDEA: X is a good picture of R
The space X can be reconstructed from R.
1. Points of X <=> Maximal ideals of R
x ideal of functions with f(x) = 0.
2. Reconstruct topology on X. Two ways
(a) Basis of open sets U(f) = “points where f != 0” = points m with f not in m.
(b) Take any closed ideal I. Z(I) = set of maximal ideala containing I. CLOSED set (common zeros of all f in I)

We can attempt to copy this for ANY commutative ring R.
X = max ideals of R
Define topology as above.
X is called the MAXIMAL SPECTRUM of R Spec_M(R)

PROBLEM: Want a homomorphism F: R –> S between algebras to induce a continuous map between spaces taking m in Spec_M S to F^-1(m) in Spec_M R. But F^-1(m) might not be maximal. Consider F: Z –> Q. F^-1(0) = 0 not maximal in Z.

SOLUTION: Work with prime ideals instead. F^-1(Prime) = PRIME.

B74-77

33
Q

Define: Spec(R) and check that topological space

A

A topological space associated with any commutative ring.
POINTS = prime ideals of R
TOPOLOGY
1. Basis of open sets: U(f) = {p : f not in p} – points where f is nonzero
2. Closed sets = Z(I) = {p : I <= p} – points where f = 0

Use (2): Z(I) closed under arbitrary intersection
Closed under finite unions

B77-78

34
Q

Examples of Spec(R)

A
  1. R = 0, Spec(R) = empty
  2. R = Field, Spec(R) = {(0)} is a point
  3. R = C[x], Prime: maximal (x-alpha) alpha in C. nonmax (0). Spec(R) = C U generic point - discuss topology B80
  4. Spec(Z) max: (2) (3) (5) …
    Prime/not max (0)
  5. Spec(R[x})
    max: (x-alpha), alpha in R orbits of Gal(k closure)…
35
Q

Exercise 2

36
Q

Exercise 4

37
Q

Exercise 5

38
Q

Exercise 11

39
Q

Exercise 14

40
Q

Exercise 18

41
Q

Exercise 19

42
Q

Exercise 24

43
Q

Exercise 25

44
Q

Why is Spec(R) called spectrum?

A

Consider R = C[A] where A is a matrix in M_n(C). The spectrum of A = Eigenvalues. Given minimum polynomial for A p(x) = (x-a1)^m1(x-a2)^m2 … (x-ak)^mk, the spectrum of A is {a1, … , ak}. Now C[A] is isomorphic to C[x]/(min poly) – prime ideals are (x-ai) so our notion of spectrum from ring theory matches linear algebra

45
Q

Discuss decomposing Spec Z in Z[1/2] and Z_2

A

Z[1/2] kills (2)

Z_2 = {m/n : n odd} kills everything but (2)

Both have generic ideal (0) so not quite a disjoint decomp

pg 84

46
Q

Discuss relationship between Spec Z and Spec Z[i]

A

This is an example of how Spec provides a geometric way of visualizing decomposition of primes in algebraic number fields.

We have a natural inclusion of Z into Z[i]. As usual, this homomorphism of rings induces a continuous map between topological spaces Spec(Z) and Spec(Z[i]) going in opposite direction.

B85

47
Q

Discuss Spec C[x,y]

A

max ideals <=> point in C^2
prime ideals <=> (f) irreducible polynomial
generic point <=> (0)

pg 87

48
Q

Prove Spec R is quasicompact = compact

A

Spec R is covered by open set U(f_i) of basis <=> No prime ideal contains ALL f_i <=> Ideal generated by {f_i} is R <=> 1 = sum_i^n r_if_i <=> Spec R covered by FINITE number of f_i.

B93

49
Q

Is Spec R connected?

A

Sometimes yes, sometimes no.

  1. If R integral domain, then Spec R is connected
    Pf. Spec R = closure of (0)
50
Q

Define: irreducible

Examples?

A

A space X is called IRREDUCIBLE if

  1. X is nonempty
  2. X is NOT a union of 2 proper closed subsets

Notice 2 => Any 2 nonempty open sets intersect. (Otherwise X = (A intersect B)^c = A^c U B^c

Examples

  1. If X is Hausdorff and irreducible, then X is a point. (suppose contains two points a,b. Then nonempty disjoint open sets, contradiction)
  2. Spec Z is irreducible (Spec Z = (0) closure

pg 97

51
Q

Discuss spectrum of the group algebra of Klien 4 group over Q and over Z

A

Spec over Z is refinement of Spec over Q.

52
Q

Prove: In Spec R, any closed irreducible subset is of the form x closure for some x in Spec R.

A

Let Z(I) be an irreducible closed subset, I an ideal. We need to show Z(I) = p closure for some prime p.

  1. Note that Z(I) = Z(rad I) (If I in P, then rad I in P since r^n in P <=> r in P), so we may replace I with rad I and only consider radical ideals.
  2. In general I need not be prime. However, if I radical and Z(I) irreducible, then I prime: Let ab in I. WTS a in I or b in I. Suppose not. (I,a)(I,b) < I. So Z(I,a) U Z(I,b) = Z(I). So one of Z(I,a) or Z(I,b) is the same as Z(I), say Z(I,a). No power of a is in I (since a not in I and I radical). So by familiar lemma, there is a prime disjoint from {1, a, a^2, …} containing I. This contradicts Z(I,a) = Z(I) since p not in Z(I,a) but in Z(I). So I is prime. It is then easy to check that Z(I) = I closure.
53
Q

Definitions of Noetherian topological space?

Relationship with Noetherian rings?

A

TFAE

  1. Every nonempty collection of closed elements has a minimal element
  2. Every nonempty collection of open sets has a maximal element
  3. Every increasing sequence of open sets has a maximal element
  4. Every decreasing sequence of closed sets has a minimal element
  5. Every open set is quasicompaxct (=compact)

R Noetherian => Spec R Noetherian
Spec R Noetherian !=> R Noetherian R= k[x1, …]/ (x1^2, x2^2, …) Spec R is a point… B105-106

54
Q

Prove: In Noetherian top space, any closed set is a union of FINITE number of closed irreducibles

A

Noetherian induction…

So for R Noetherian ring, we know ALL closed subsets of Spec R: Union of finite number of irreducible sets = sets x closure (closure of a point)
B107