Math 326 - Algebra 2
Lectures by Kazuya Kato
Notes by Zev Chonoles
The University of Chicago, Winter 2013
Lecture 1 (2013-01-07) 1 Lecture 16 (2013-02-13) 43
Lecture 2 (2013-01-09) 4 Lecture 17 (2013-02-15) 46
Lecture 3 (2013-01-11) 7 Lecture 18 (2013-02-18) 47
Lecture 4 (2013-01-14) 9
Lecture 19 (2013-02-20) 50
Lecture 5 (2013-01-16) 11
Lecture 20 (2013-02-22) 52
Lecture 6 (2013-01-18) 14
Lecture 21 (2013-02-25) 53
Lecture 7 (2013-01-23) 16
Lecture 22 (2013-02-27) 54
Lecture 8 (2013-01-25) 19
Lecture 23 (2013-03-01) 56
Lecture 9 (2013-01-28) 23
Lecture 24 (2013-03-04) 58
Lecture 10 (2013-01-30) 26
Lecture 25 (2013-03-06) 59
Lecture 11 (2013-02-01) 29
Lecture 26 (2013-03-08) 61
Lecture 12 (2013-02-04) 32
Lecture 13 (2013-02-06) 35 Lecture 27 (2013-03-11) 63
Lecture 14 (2013-02-08) 37 Lecture 28 (2013-03-13) 64
Lecture 15 (2013-02-11) 40 Lecture 29 (2013-03-15) 66
,Introduction
Math 326 is one of the nine courses offered for first-year mathematics graduate students at the
University of Chicago. It is the second of three courses in the year-long algebra sequence.
These notes are being live-TeXed, though I edit for typos and add diagrams requiring the Tik Z
package separately. I am using the editor TeXstudio.
I am responsible for all faults in this document, mathematical or otherwise; any merits of the
material here should be credited to the lecturer, not to me.
Please email any corrections or suggestions to .
,Lecture 1 (2013-01-07)
The main subject of this course is commutative ring theory, and its relation to algebraic number
theory and algebraic geometry. We will also see some more advanced topics such as class field
theory and the Weil conjectures, though we will not go into them in depth.
There will be no exams. The assignments will appear on each Friday, on the Chalk site if it can be
set up; otherwise, on each Thursday evening, the assignments will be sent by email. They will then
be due the following Friday.
If you look on the street, you never meet a commutative ring; that’s rather strange. They are rather
shy I think. We need to ask them to come to this room. Rings, rings, please come! Rings, rings
please come! *shuffles along the floor, playing the part of the ring* Finite fields, come! Rings of
functions, come! *hops* I think they are here now.
The Mysterious Analogy Between Prime Numbers and Points
Let A be a commutative ring. We define
max(A) = {maximal ideals of A}.
There is a bijection
max(Z) {prime numbers}
∈
∈
(p) = pZ p
There is also a bijection
max(C[T ]) C
∈
∈
(T − α) α
More generally, there is a bijection
max(C[T1 , . . . , Tn ]) Cn
∈
∈
(T1 − α1 , . . . , Tn − αn ) (α1 , . . . , αn )
Note that (T1 − α1 , . . . , Tn − αn ) = {f ∈ C[T1 , . . . , Tn ] | f (α1 , . . . , αn ) = 0}.
We can walk on C, but I think it is hard walking on the prime numbers. I hope someday I can find
a pair of shoes that can help. One difficulty is that the points in C are all the same size, but the
primes are like stones of different sizes; 3 is a little bigger than 2, 5 is a little bigger than 3. . .
Let X be a compact Hausdorff space, and let A = C(X) = {continuous maps X → R}. Then there
is a bijection
max(A) X
∈
∈
{f ∈ A | f (p) = 0} p
Last edited Math 326 - Algebra 2 Page 1
2013-08-09 Lecture 1
, If X is not compact, this is false. We can see this by considering X = R for example. In A =
{continuous maps R → R}, we can consider the ideal
I = {f ∈ A | there is some c such that f (x) = 0 if x > c}.
There is a proposition from commutative ring theory:
Proposition. If A is a commutative ring and I ( A is a proper ideal of A, then there is some
m ∈ max(A) such that I ⊂ m.
This proposition requires the Axiom of Choice, so we will skip the proof for the moment. But the
proposition implies that there is some maximal ideal of A containing I, and no ideal of the form
{f ∈ A | f (p) = 0} can contain I, so there must be other maximal ideals.
Now consider X = {(x, y) ∈ C2 | y 2 = x3 + 1}, and let A = {polynomial functions on X}. For
example, the function x sends (a, b) ∈ X to a ∈ C, and the function y sends (a, b) ∈ X to b ∈ C.
Note that
A = {C-valued functions on X written as a polynomial over C in the functions x, y}.
√
Then A ∼ = C[T, T 3 + 1], where T is the function x. This ring is a quadratic extension √ of C[T ],
which you should think of as being similar to an extension of Z, such as for example Z[ 26].
√
There is an isomorphism A ∼= C[T1 , T2 ]/(T22 − T13 − 1), where T1 7→ T and T2 7→ T 3 + 1. There is
also a bijection
max(A) X
∈
∈
{f ∈ A | f (p) = 0} p
which can be deduced from the correspondence between max(C[T1 , T2 ]) and C2 ,
max(C[T1 , T2 ]) C2
⊂
⊂
max(A) X
(Recall that there is a bijection
max(A/I) ←→ {m ∈ max(A) | I ⊆ m},
where M ∈ max(A/I) corresponds to {x ∈ A | x mod I ∈ M }.)
√
In the ring Z[ −26], note that we do not have unique factorization:
√ √
33 = 27 = (1 + −26)(1 − −26).
√ √
Writing p = (3, 1 + −26) and p0 = (3, 1 − −26), we can recover unique factorization for ideals:
√ √ 3
(3) = pp0 , (1 + −26) = p3 , (1 − −26) = p0
3
(pp0 )3 = (27) = p3 p0 .
(recall that for ideals I and J, their product is IJ = { ni=1 ai bi | ai ∈ I, bi ∈ J}.)
P
Last edited Math 326 - Algebra 2 Page 2
2013-08-09 Lecture 1
Lectures by Kazuya Kato
Notes by Zev Chonoles
The University of Chicago, Winter 2013
Lecture 1 (2013-01-07) 1 Lecture 16 (2013-02-13) 43
Lecture 2 (2013-01-09) 4 Lecture 17 (2013-02-15) 46
Lecture 3 (2013-01-11) 7 Lecture 18 (2013-02-18) 47
Lecture 4 (2013-01-14) 9
Lecture 19 (2013-02-20) 50
Lecture 5 (2013-01-16) 11
Lecture 20 (2013-02-22) 52
Lecture 6 (2013-01-18) 14
Lecture 21 (2013-02-25) 53
Lecture 7 (2013-01-23) 16
Lecture 22 (2013-02-27) 54
Lecture 8 (2013-01-25) 19
Lecture 23 (2013-03-01) 56
Lecture 9 (2013-01-28) 23
Lecture 24 (2013-03-04) 58
Lecture 10 (2013-01-30) 26
Lecture 25 (2013-03-06) 59
Lecture 11 (2013-02-01) 29
Lecture 26 (2013-03-08) 61
Lecture 12 (2013-02-04) 32
Lecture 13 (2013-02-06) 35 Lecture 27 (2013-03-11) 63
Lecture 14 (2013-02-08) 37 Lecture 28 (2013-03-13) 64
Lecture 15 (2013-02-11) 40 Lecture 29 (2013-03-15) 66
,Introduction
Math 326 is one of the nine courses offered for first-year mathematics graduate students at the
University of Chicago. It is the second of three courses in the year-long algebra sequence.
These notes are being live-TeXed, though I edit for typos and add diagrams requiring the Tik Z
package separately. I am using the editor TeXstudio.
I am responsible for all faults in this document, mathematical or otherwise; any merits of the
material here should be credited to the lecturer, not to me.
Please email any corrections or suggestions to .
,Lecture 1 (2013-01-07)
The main subject of this course is commutative ring theory, and its relation to algebraic number
theory and algebraic geometry. We will also see some more advanced topics such as class field
theory and the Weil conjectures, though we will not go into them in depth.
There will be no exams. The assignments will appear on each Friday, on the Chalk site if it can be
set up; otherwise, on each Thursday evening, the assignments will be sent by email. They will then
be due the following Friday.
If you look on the street, you never meet a commutative ring; that’s rather strange. They are rather
shy I think. We need to ask them to come to this room. Rings, rings, please come! Rings, rings
please come! *shuffles along the floor, playing the part of the ring* Finite fields, come! Rings of
functions, come! *hops* I think they are here now.
The Mysterious Analogy Between Prime Numbers and Points
Let A be a commutative ring. We define
max(A) = {maximal ideals of A}.
There is a bijection
max(Z) {prime numbers}
∈
∈
(p) = pZ p
There is also a bijection
max(C[T ]) C
∈
∈
(T − α) α
More generally, there is a bijection
max(C[T1 , . . . , Tn ]) Cn
∈
∈
(T1 − α1 , . . . , Tn − αn ) (α1 , . . . , αn )
Note that (T1 − α1 , . . . , Tn − αn ) = {f ∈ C[T1 , . . . , Tn ] | f (α1 , . . . , αn ) = 0}.
We can walk on C, but I think it is hard walking on the prime numbers. I hope someday I can find
a pair of shoes that can help. One difficulty is that the points in C are all the same size, but the
primes are like stones of different sizes; 3 is a little bigger than 2, 5 is a little bigger than 3. . .
Let X be a compact Hausdorff space, and let A = C(X) = {continuous maps X → R}. Then there
is a bijection
max(A) X
∈
∈
{f ∈ A | f (p) = 0} p
Last edited Math 326 - Algebra 2 Page 1
2013-08-09 Lecture 1
, If X is not compact, this is false. We can see this by considering X = R for example. In A =
{continuous maps R → R}, we can consider the ideal
I = {f ∈ A | there is some c such that f (x) = 0 if x > c}.
There is a proposition from commutative ring theory:
Proposition. If A is a commutative ring and I ( A is a proper ideal of A, then there is some
m ∈ max(A) such that I ⊂ m.
This proposition requires the Axiom of Choice, so we will skip the proof for the moment. But the
proposition implies that there is some maximal ideal of A containing I, and no ideal of the form
{f ∈ A | f (p) = 0} can contain I, so there must be other maximal ideals.
Now consider X = {(x, y) ∈ C2 | y 2 = x3 + 1}, and let A = {polynomial functions on X}. For
example, the function x sends (a, b) ∈ X to a ∈ C, and the function y sends (a, b) ∈ X to b ∈ C.
Note that
A = {C-valued functions on X written as a polynomial over C in the functions x, y}.
√
Then A ∼ = C[T, T 3 + 1], where T is the function x. This ring is a quadratic extension √ of C[T ],
which you should think of as being similar to an extension of Z, such as for example Z[ 26].
√
There is an isomorphism A ∼= C[T1 , T2 ]/(T22 − T13 − 1), where T1 7→ T and T2 7→ T 3 + 1. There is
also a bijection
max(A) X
∈
∈
{f ∈ A | f (p) = 0} p
which can be deduced from the correspondence between max(C[T1 , T2 ]) and C2 ,
max(C[T1 , T2 ]) C2
⊂
⊂
max(A) X
(Recall that there is a bijection
max(A/I) ←→ {m ∈ max(A) | I ⊆ m},
where M ∈ max(A/I) corresponds to {x ∈ A | x mod I ∈ M }.)
√
In the ring Z[ −26], note that we do not have unique factorization:
√ √
33 = 27 = (1 + −26)(1 − −26).
√ √
Writing p = (3, 1 + −26) and p0 = (3, 1 − −26), we can recover unique factorization for ideals:
√ √ 3
(3) = pp0 , (1 + −26) = p3 , (1 − −26) = p0
3
(pp0 )3 = (27) = p3 p0 .
(recall that for ideals I and J, their product is IJ = { ni=1 ai bi | ai ∈ I, bi ∈ J}.)
P
Last edited Math 326 - Algebra 2 Page 2
2013-08-09 Lecture 1
