Chapter 1 - Roots of Commutative Algebra

Question 1

Discuss number theory motivations for commutative algebra

Answer 1

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

Question 2

Discuss algebraic geometry motivations for commutative algebra

Answer 2

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

Question 3

Discuss invariant theory motivations for commutative algebra

Answer 3

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.

Question 4

Define: ring, module

Answer 4

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

Question 5

Examples of non-commutative rings?

Answer 5

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

Question 6

Discuss constructions which turn out to be special cases of modules

Answer 6

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

Question 7

How does a group G act on polynomials?

Answer 7

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)

Question 8

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

Answer 8

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

Question 9

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

Answer 9

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

Question 10

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

Answer 10

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

B 22

Question 11

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

Answer 11

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

Question 12

Example of ideal that is not f.g?

Answer 12

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

Question 13

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

Answer 13

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

B 32

Question 14

Discuss 3 equivalent definitions of Noetherian rings and prove equivalence

Answer 14

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

B33-34

Question 15

Example of non-Noetherian ring and failure of conditions

Answer 15

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

Question 16

Define: Artinian ring and give nonexample

Answer 16

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) > …

Question 17

Give sequence of examples of Noetherian/non-Noetherian rings

Answer 17

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

Question 18

Is a subring of a Noetherian ring Noetherian?

Answer 18

No

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

Question 19

Discuss how quotient rings behave with Noetherian property

Answer 19

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

Question 20

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

Corollaries?

Answer 20

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

Question 21

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

Answer 21

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

Question 22

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?

Answer 22

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

Question 23

Define: Noetherian module

Answer 23

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

Question 24

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

Corollaries?

Answer 24

=> 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

Question 25

# Define: graded ring, homogeneous element, idea, component, irrelevant ideal Examples?

Answer 25

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.

Question 26

Define: reduced ring, radical ideal

Answer 26

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.

Question 27

What is Nullstellensatz? 4 Corollaries? Define: affine k-algebra, affine ring

Answer 27

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.

Question 28

Define: graded module, Hilbert function, Hilbert polynomial

Answer 28

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.

Question 29

Three ways to draw picture of a ring?

Answer 29

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

Question 30

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

Answer 30

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

Question 31

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

Answer 31

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

Question 32

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

Answer 32

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

Question 33

Define: Spec(R) and check that topological space

Answer 33

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

Question 34

Examples of Spec(R)

Answer 34

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)...

Question 35

Exercise 2

Question 36

Exercise 4

Question 37

Exercise 5

Question 38

Exercise 11

Question 39

Exercise 14

Question 40

Exercise 18

Question 41

Exercise 19

Question 42

Exercise 24

Question 43

Exercise 25

Question 44

Why is Spec(R) called spectrum?

Answer 44

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

Question 45

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

Answer 45

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

Question 46

Discuss relationship between Spec Z and Spec Z[i]

Answer 46

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

Question 47

Discuss Spec C[x,y]

Answer 47

max ideals <=> point in C^2 prime ideals <=> (f) irreducible polynomial generic point <=> (0) pg 87

Question 48

Prove Spec R is quasicompact = compact

Answer 48

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

Question 49

Is Spec R connected?

Answer 49

Sometimes yes, sometimes no. 1. If R integral domain, then Spec R is connected Pf. Spec R = closure of (0)

Question 50

Define: irreducible | Examples?

Answer 50

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

Question 51

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

Answer 51

Spec over Z is refinement of Spec over Q.

Question 52

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

Answer 52

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.

Question 53

Definitions of Noetherian topological space? Relationship with Noetherian rings?

Answer 53

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

Question 54

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

Answer 54

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