Download last updated 2012-02-25 with Set 8

Set 1.
Please study the following Problems 1-10 by January 13 (Friday).
In Problems 1-10, let X be a compact Hausdorff space and let A be the ring of
all C-valued continuous functions on X. We consider how X is reflected in A and
how A is reflected in X like the relation of a flower and its image in the water.
For a subset S of X, let I(S) be the ideal {f ∈ A | f (p) = 0 for all p ∈ S} of A.
For a subset E of A, let V (E) be the closed subset {p ∈ X | f (p) = 0 for all f ∈ E}
of X.
1. Prove that if S is a subset of X, then V (I(S)) = S̄ where S̄ is the closure of
S.
(You can use the following property of a compact Hausdorff space X. Let C be
a closed subset of X and let p ∈ X, p ∈
/ C. Then there is f ∈ A such that f (p) = 1
and f has value 0 at any point of S.)
2. Prove that for a subset S of X, S is closed if and only if there is an ideal I of
A such that S = V (I).
3. Prove that if I is an ideal of A such that V (I) = ∅, then I = A.
This is not easy to prove. Here is a suggestion for the proof. Note that
n
∩i=1 V (fi ) = V (f1 f¯1 + · · · + fn f¯n ) for f1 , · · · , fn ∈ A, and use the assumption
∩f ∈I V (f ) = V (I) = ∅ and the compactness of X. Here V (f ) for f ∈ A means
V ({f }) (i.e. V of the subset {f } of A), and f¯ denotes the function on X
p 7→ the complex conjugate of f (p).
4. Prove that the map X → max(A) ; p 7→ {f ∈ A | f (p) = 0} is bijective. Here
max(A) denotes the set of all maximal ideals of A.
Suggestion for the proof. To get the surjectivity of this map, apply Problem 3
to a maximal ideal I of A. For the injectivity, use Problem 1.
5. Let HomC (A, C) be the set of all ring homomorphisms over C. Prove that the
map
X → HomC (A, C)
is bijective.
Here the meaning of ”ring homomorphism over C” is the following. Let k be
a commutative ring. A commutative ring over k is a commutative ring endowed
with a fixed ring homomorphism form k. For commutative rings R1 , R2 over k, a
ring homomoprphism ϕ : R1 → R2 is called a ring homomorphism over k if the
ϕ
composition k → R1 → R2 of ϕ and the fixed k → R1 coincides with the fixed
k → R2 .
Suggestion for the proof for Problem 5. You can use the fact that the map
X → max(A) ; p 7→ {f ∈ A | f (p) = 0} is bijective.
1
6. Let Y be another compact Hausdorff space and let B be the ring of all Cvalued continuous functions on Y . Let ϕ : A → B be a ring homomorphism over C.
Prove that there is a unique continuous map Φ : Y → X such that ϕ(f ) = f ◦ Φ for
Φ
f
any f ∈ A. Here f ◦ Φ denotes the composition Y → X → C.
(Remark: Using the terminology of category theory, this can be said as follows.
The contravariant functor
{Compact Hausdorff spaces} → {Commutative rings over C} ;
X 7→ {continuous C-valued functions on X}
is fully faithful. )
7. In Problem 6, prove that ϕ : A → B is injective if and only if Φ : Y → X is
surjective.
8. In Problem 6, prove that ϕ : A → B is surjective if and only if Φ : Y → X is
injective. (You can use the property of a compact Hausdorff space X that a C-valued
continuous function on a closed subset C of X extends to a C-valued continuous
function on X.)
9. Prove that X is connected if and only if there is no f ∈ A such that f 2 = f ,
f 6= 0, f 6= 1.
10. Assume X is a finite set. Prove that S 7→ I(S) is a bijection from the set of
all subsets of X to the set of all ideals of A, and that J 7→ V (J) (here J is an ideal
of A) is the inverse map of this bijection.
Suggestion for the proof. You can use the fact that for a product A1 × A2 of
commutative rings (the addition and the multiplication are component-wise), any
ideal of A1 × A2 has the form I1 × I2 where I1 is an ideal of A1 and I2 is an ideal
of A2 . As a commutative ring, A in Problem 10 is isomorphic to the product of n
copies of C where n is the order of the finite set X.
2
Set 2.
Please study the following Problems 11–20 by January 20 (Friday).
√
√
√
This
time,
we
consider
friends
of
Z
;
rings
Z[i],
Z[
−2],
Z[(1+
−11)/2],
Z[
2],
√
Z[ 3].
For Problems 11-13, you can use the following things about the ring Z[i] (which
I mentioned in the course) freely . Z[i] is a principal ideal domain and hence is a
unique factorization domain. If p is a prime number and p ≡ 1 mod 4, p = αᾱ for
some α ∈ Z[i], the ideals (α) and (ᾱ) of Z[i] are maximal ideals of Z[i] (that is,
α and ᾱ are prime elements of Z[i]), and (α) 6= (ᾱ). If p is a prime number and
p ≡ 3 mod 4, the ideal (p) of Z[i] is a maximal ideal (that is, p is a prime element
in Z[i]). 2 has a factorization 2 = −i(1 + i)2 , and (1 + i) is a maximal ideal (that
is, 1 + i is a prime element of Z[i]). Any maximal ideal of Z[i] coincides with one of
those maximal ideals. Z[i]× = {±1, ±i}.
11. By regarding the set {x2 + y 2 | x, y ∈ Z} as the image of the map Z[i] →
Z ; α 7→ αᾱ, which preserves the multiplication, and using the knowledges on the
prime factorization in Z[i] explained above, prove the following fact.
Let n be an integer and assume n > 0. Then n has the form x2 + y 2 (x, y ∈ Z)
if and only if n has the following property (i).
(i) For any prime number p such that p ≡ 3 mod 4, ordp (n) is an even integer.
Here ordp (n) is the largest integer e such that pe divides n.
(Example. 21 can not be expressed as x2 + y 2 , for ord7 (21) = 1 is odd.)
12. Using the uniqueness of the prime factorization in Z[i], explain the fact that
if tan(α) = 1, tan(β) = 3/2, and tan(γ) = 2, then α, β, γ are linearly independent
over Q.
13. Using the prime factorization of 65 in Z[i], explain why there can be four
right triangles with the following property: The length of each side is an integer
and the length of its longest side is 65. (In fact, we have right triangles with sides
(16, 63, 65), (33, 56, 65), (25, 60, 65), (39, 52, 65).)
√
14. By considering prime factorizations in the ring Z[ −2], prove
{(x, y) ∈ Z × Z | y 2 = x3 − 2} = {(3, ±5)}.
√
You can use the fact that√Z[ −2] is a principal ideal domain
and hence
is a
√
√
×
unique factorization domain, −2 is a prime element in Z[ −2], and Z[ −2] =
{±1}.
I am afraid that it is not so simple to solve this problem, and so I divide it√into
two steps.
Prove first the following
√
√ (1) and (2). Let π be a prime divisor of y + −2
in Z[√ −2]. Assume that in Z[ −2], π appears a times in the prime factorization
of
√
y + −2 (so we assume a > 0), b times in the prime factorization of y − −2, and
3
√
√
c times in the prime factorization of x in Z[ −2]. Prove (1) If b > 0, (π) = ( −2)
and a = b. Prove the following (2) by using a + b = 3c. (2) c is a multiple of 3.
√
2
3
15. (Continuation
of
14.)
Prove
that
if
x,
y
∈
Z
and
y
=
x
−2,
then
y
+
−2 =
√
3
2
3
α for some α ∈ Z[ −2]. Obtain all integral solutions of y = x − 2 from this fact.
Remark. Fermat stated the result of 14. (Precisely speaking, he did not consider
negative integers so (3, 5) was the only solution for him.)
√
16. By considering prime factorizations in the ring Z[(1 + −11)/2], prove
{(x, y) ∈ Z × Z | y 2 = x3 − 11} = {(3, ±4), (15, ±58)}.
√
You can use the fact that A := Z[(1 + −11)/2]
is a principal ideal domain and
√
hence is a unique factorization domain, 2 and −11 are prime
√ elements in A, and
A× = {±1}. Like in the case of Problem 14, show that y + −11 = α3 for some
α ∈ A.
√
17. Prove that there is a bijection √
between the two sets Z[ 2]× and {(x, y) ∈
Z × Z | x2 − 2y 2 = ±1} given by x + y 2 ↔ (x, y) (x, y ∈ Z).
√
√
√
18. For x + y 2 ∈ Z[ 2]× (x, y ∈ Z), prove that x + y 2 > 1 if and only if
x > 0 and y > 0. Using the fact (x, y) = (1, 1) is the smallest
integral solution of
√
2
2
x −√
y = ±1 such that x > 0 and y > 0, prove that 1 + 2 is the smallest element
of Z[ 2]× which is > 1.
√
√
19. By using Problems 17 and 18, prove that Z[ 2]× = {±(1 + 2)n | n ∈ Z}.
√
√
20. Prove that Z[ 3]× = {±(2 + 3)n | n ∈ Z}.
4
Set 3. Please study the following Problems 21–30 by January 27 (Friday).
21. Let A be a ring, let M be an A-module, and let N be an A-submodule of
M . Prove that if N and M/N are finitely generated, then M is finitely generated.
Prove also that if M is finitely generated, then M/N is finitely generated.
22. Let A be a ring, let M be an A-module, and let N be an A-submodule of
M . By using Problem 21, prove that M is Noetherian if and only if N and M/N
are Noetherian.
23. Let A be a ring, let M1 , . . . , Mn be A-modules, and let M = M1 ⊕· · ·⊕Mn . By
using Problem 22, prove that if M1 , . . . , Mn are Noetherian, then M is Noetherian.
24. Let A be the ring of all C-valued continuous functions on the closed interval
[0, 1] = {x ∈ R | 0 ≤ x ≤ 1}. For each integer n ≥ 1, let fn ∈ A be the function
whose value at x ∈ [0, 1] is the n-th root x1/n ≥ 0 of x. Prove that the ideal of A
generated by all fn (n = 1, 2, 3, . . . ) is not finitely generated, and hence that A is
not a Noetherian ring.
25. Let A be the ring of all holomorphic functions on the complex plane C. For
each integer n ≥ 1, let fn ∈ A be the function defined by
fn (z) = (e
2πiz
n
Y
− 1) · (z − r)−1 .
r=1
Prove that the ideal generated by all fn (n = 1, 2, 3, . . . ) is not finitely generated,
and hence that A is not a Noetherian ring.
26. Let M be a Noetherian module over a ring A. Let f : M → M be an
A-homomorphism. Assume f is surjective. Prove that f is injective (and hence is
an isomorphism from M onto M ).
(Remark. Consider the case A = Z and M = Z. Then Z → Z ; x 7→ 2x
is injective but not surjective. So you can not exchange the surjectivity and the
injectivity in this problem.)
Suggestion for the proof. Consider the sequence
Ker(f ) ⊂ Ker(f 2 ) ⊂ Ker(f 3 ) ⊂ · · ·
of A-submodules of M .
27. Let A be a ring and let M be an A-module. An A-module M is called an
Artinian module over A if for any sequence N1 ⊃ N2 ⊃ N3 ⊃ · · · of A-submodules
of M , there is n such that Nn = Nn+1 = Nn+2 = · · · . Let M be an Artinian module
over a ring A. Let f : M → M be an A-homomorphism. Assume f is injective.
Prove that f is surjective (and hence is an isomorphism from M onto M ).
Suggestion for the proof. Consider the sequence
Image(f ) ⊃ Image(f 2 ) ⊃ Image(f 3 ) ⊃ · · ·
5
of A-submodules of M .
Remark. An example of an Artinian module is the Z-module Z[1/p]/Z for a
prime number p, where Z[1/p] is the ring generated over Z by 1/p (that is, Z[1/p] =
{m/pn | m, n ∈ Z, n ≥ 0}). The map x 7→ px on Z[1/p]/Z is surjective but not
injective.
28. Let K be a field. Prove that there are infinitely many irreducible monic
polynomials in K[T ], imitating the proof by Euclid of the fact that there are infinitely
many prime numbers. (Here monic means that the coefficient of the highest degree
term of the polynomial is 1. If you are not aware of the proof of Euclid, please find
it in some reference.) In the case K is an infinite field, give a simpler proof.
29. Let Fq be a finite field consisting of q elements. (Remark: Any finite field is
a commutative field.) Imitating Rieman’s zeta function
∞
X
1
,
s
n
n=1
ζ(s) =
define
ζFq [T ] (s) =
X
f
1
](Fq [T ]/(f ))s
where f ranges over all monic polynomials in Fq [T ] and ](Fq [T ]/(f )) denotes the
order (the number of elements) of the finite set Fq [T ]/(f ). Prove that
ζFq [T ] (s) =
1
.
1 − q 1−s
Remark. It is not interesting to consider the analogue of Riemann hypothesis
for this zeta function ζFq [T ] (s), for this zeta function has no zero. In later homework
problems, you will encounter zeta functions of rings which are friends of Fq [T ], for
which analogues of Riemann hypothesis are interesting.
Before I present Problem 30, I write something about ζ(s). For the fact there are
infinitely many prime numbers, Euler gave the following analytic proof. By using
the unique factorizations of numbers into prime numbers, we have
Y
ζ(s) =
(1 − p−s )−1
p
where p ranges over
numbers (here s > 1). We can prove that when s > 1
P all prime
−s
tends to 1, then ∞
n
tends
to ∞. If there were only finite number
of prime
n=1
Q
Q
−s −1
numbers, when s > 1 tends to 1, p (1 − p ) should converge to p (1 − p−1 )−1
and would not tend to ∞. Thus we have a contradiction and hence there are
infinitely many prime numbers.
30. By using Problem 29 and by using the method of Euler, prove that there are
infinitely many irreducible monic polynomials in Fq [T ].
6
Set 4. Please study the following Problems 31–40 by February 3 (Friday).
31. Let k be a field whose characteristic is not 2, let f (T ) ∈ k[T ], and assume
2
that f (T ) is of degree ≥ 1 and is not
pdivisible by g(T ) for any g(T ) ∈ k[T ] of
degree ≥ 1.p Prove that the ring k[T, f (T )] is the integral closure of k[T ] in the
field k(T )( f (T )).
32. For a commutative ring A which is finitely generated over Z, prove that
ζA[T1 ,...,Tn ] (s) = ζA (s − n). Here ζA (s) denotes the Hasse zeta function of A.
(This is a homework in algebra, and so please do not worry here about the
convergence of the infinite product. Precisely speaking, this problem has sense if we
already know that the (usually infinite) product ζA (s) converges absolutely when
Re(s) is sufficiently large.) Suggestion of the proof. Use induction on n and the
result on ζFq [T ] (s) in Problem 29.
33. By using Problem 32, prove that for any commutative ring A which is
finitely generated over Z, ζA (s) (as a product) converges absolutely when Re(s) is
sufficiently large.
Since this is a homework in algebra, it is fine that analytic arguments are not
perfectly precise.
Preparation for Problems 34–38. I put an introduction to the quadratic reciprocity law proved in 1796 by Gauss (1777–1855). In his time, it was the greatest
theorem in number theory. For a prime number p 6= 2 and for a ∈ Z which is not
divisible by p, define
a
∈ {±1}
p
as follows. If the image of a in Fp = Z/pZ is b2 for some b ∈ Fp , then ap = 1.
b
a
=
(here a, b ∈ Z, ab is not divisible by
Otherwise, ap = −1. We have ab
p
p
p
p). Gauss proved the following (1), (2), (3). ((1) is called the quadratic reciprocity
law).
(1) For any prime number q 6= 2 which is different form p, we have
p−1 q−1
q
p
=
· (−1) 2 · 2 .
p
q
(2)
−1
p
(3)
2
p
= (−1)
= (−1)
p−1
2
p2 −1
8
.
.
It seems that quadratic reciprocity law is a love song in the world of prime numbers. In our world, in the case a girl q shines in the heart of a boy p, unfortunately
it is not clear whether the boy
p shines in the heart of the girl q. However in the
world of prime numbers, pq , that is, whether the girl q shines in the heart of the
boy p, and pq , that is, whether the boy p shines in the heart of the girl q, are closely
7
related as in the quadratic reciprocity law (1). Such strong relation between prime
numbers is surprising.
p−1
34. Let p be a prime number and assume p 6= 2, 3. Using p3 = p3 · (−1) 2 (a
special case √
of the quadratic reciprocity law), prove the following. If p ≡ 1, 11 mod
of two different prime ideals. If p ≡
12, (p) in Z 3] decomposes into the product
√
5, 7 mod 12, p is a prime element in Z[ 3].
√
35. By using the fact Z[ 3] is a PID, prove that for a prime number p 6= 2, 3,
p = ±(x2 − 3y 2 ) for some x, y ∈ Z if p ≡ 1, 11 mod 12, and that p can not be
expressed in that way if p ≡ 5, 7 mod 12.
by using some
36. Let p be a prime number and assume p 6= 2. Computing −2
p
formulae written
at the place where I explained quadratic reciprocity law, and using
√
the fact Z[ −2] is a PID, prove that p = x2 +2y 2 for some x, y ∈ Z if p ≡ 1, 3 mod 8,
and that p can not be expressed in that way if p ≡ 5, 7 mod 8.
37. Let p be a prime number and assume p 6= 2. Computing p2 by using a
formula written
at the place where I explained quadratic reciprocity law, and using
√
the fact Z[ 2] is a PID, prove that p = x2 − 2y 2 for some x, y ∈ Z if p ≡ 1, 7 mod 8,
and that p can not be expressed in that way if p ≡ 3, 5 mod 8. (You may think
2
that p = ±(x2 − 2y
is correct, but improve it to p = x2 − 2y 2 by
√ ) appears. That √
multiplying x + y 2 by the unit 1 + 2 if necessary.)
38. Let p be a prime number and assume p 6= 2, 5. The quadratic reciprocity
law (together with the formula for ( −1
) tells that −5
is 1 if p ≡ 1, 3, 7, 9 mod 20
p
p
and is −1 if p ≡ 11, 13, 17, 19 mod
20.
One
thing
which
did
not appear in Problems
√
√
34–37 is that the integer ring Z[ −5] of Q(√ −5) is not a PID. Prove that if p is a
prime number and p ≡ 3 mod 20, (p) in Z[ −5] is decomposed into the product of
two different prime ideals but p can not be expressed as x2 + 5y 2 (x, y ∈ Z).
Preparation for Problems 39.
A Dirichlet character is a homomorphism χ : (Z/N Z)× → C× where N is an
integer ≥ 1. For a Dirichlet character χ : (Z/N Z)× → C× , the Dirichlet L-function
L(s, χ) is defined as
∞
X
L(s, χ) =
χ(n)n−s
n=1
in which χ(n) means χ(n mod N ) if n is coprime (i.e. relatively prime) to N (then
n mod N ∈ (Z/N Z)× ), and χ(n) is defined to be 0 if n is not coprime to N . This
infinite series converges when the real part Re(s) of the complex number s is > 1.
In the case N =P1 and χ is the trivial homomorphism, L(s, χ) is Riemann’s zeta
−s
function ζ(s) = ∞
and so Dirichlet L-function is a generalization of Riemann
n=1 n
zeta function. Like Riemann zeta function, Dirichlet L-function has the presentation
as a product over prime numbers
Y
L(s, χ) =
(1 − χ(p)p−s )−1 .
p prime number
8
L(s, χ) has an analytic continuation to the whole C as a meromorphic function, and
is holomorphic at any s 6= 1. (In fact, if χ is not the trivial homomorphism, then
L(s, χ) is holomorphic also at s = 1.)
Example. In the case N = 1 and χ : (Z/4Z)× = {1, 3 mod 4} → C× is given by
χ(1) = 1 and χ(3) = −1, we have
L(s, χ) = 1 −
1
1
1
1
+ s − s + s − ....
s
3
5
7
9
39. By using the formula (2) in Preparation for Problems 34–38, prove that
ζZ[i] (s) = ζ(s)L(s, χ) where ζ(s) is Riemann zeta function and χ is as in the above
Example.
Preparation 1 for Problem 40.
The quadratic reciprocity law has the following analogue for the polynomial ring
Fq [T ] over a finite field Fq of q elements whose characteristic is not 2. This is one
example of the mysterious analogies between numbers and polynomials.
Let f, g ∈ Fq [T ] be irreducible monic polynomials and assume f 6= g. Then
q−1
f
g
=
· (−1) 2 ·deg(f )deg(g) .
f
g
(deg means the degree). Here fg is defined to be 1 if the image of g in the residue
field Fq [T ]/(f ) is b2for some
b element b of Fq [T ]/(f ), and is defined to be −1 otherab
a
wise. (We have f = f f for a, b ∈ Fq [T ] with ab is not divisible by f .)
Preparation 2 for Problems 40. Now we consider an analogue of Dirichlet Lfunction for Fq [T ]. Take a monic polynomial g ∈ Fp [T ] and let χ be a homomorphism
(Fq [T ]/(g))× → C× . We define
X
L(s, χ) =
χ(f ) · ](Fq [T ]/(f ))−s
f monic
where f ranges over all monic polynomials in Fq [T ] (note that the constant polynomial 1 is regarded as a monic polynomial) and ](Fq [T ]/(f )) is the order (the
number of elements) of the finite ring Fq [T ]/(f ). Here χ(f ) means χ(f mod g)
if f is coprime
to g, and χ(f ) means 0 otherwise. We have the presentation
Q
L(s, χ) = h (1 − χ(h)](Fq [T ]/(h))−s )−1 as product, where h ranges over all monic
irreducible polynomials in Fq [T ].
40. Prove that ζF5 [T,√T 3 +1] (s) = ζF5 [T ] (s)L(s, χ), where χ is the homomorphism
(F5 [T ]/(T 3 + 1))× → C× defined by
χ(f mod T 3 + 1) =
f f
,
T + 1 T2 − T + 1
where T f+1 ∈ {±1} etc. are the quadratic residue symbol defined in Preparation
1 for Problem 40.
9
Set 5. Please study the following Problems 41–50 by February 10 (Friday).
We use the following (*) and (**) explained in Set 4.
(*) For a finite field Fq of characteristic 6= 2, we have an analogue of quadratic
reciprocity law for Fq [T ] explained in Preparation 1 for Problem 40.
(You will prove that formula in Problem 50. )
(**) For a homomorphism χ : (Fq [T ]/(g))× → C× , where Fq is a finite field and g
is a monic polynomial in Fq [T ], we have the L-function L(s, χ) (analogue of Dirichlet
L-function) defined as in Preparation 2 for Proposition 40.
41. In the above (**), consider the case g = T and χ is not the trivial homomorphism. Prove L(s, χ) = 1.
P
Suggestion for the proof. Prove that if d ≥ 1, a1 ,...,ad ∈Fq χ(T d + a1 T d−1 + · · · +
ad ) = 0.
The fact there are infinitely many prime numbers p such that p ≡ 3 mod 4 is
proved as follows by using Dirichlet L-function. Consider the Dirichlet L-function
L(s, χ) where χ is the Dirichlet character (Z/4Z)× → C× defined by χ(1 mod 4) = 1
and χ(3 mod 4) = −1. If there were only finitely many prime numbers p such that
p ≡ 3 mod 4, in the product presentation of L(s, χ), almost all factors (1−χ(p)p−s )−1
(called the Euler factor at p) should be (1 − p−s )−1 , that is, ζ(s) and L(s, χ) would
have the same Euler factors at almost all p. Since ζ(s) diverges to ∞ when s > 1
tends to 1, L(s, χ) should diverge to ∞ when s > 1 tends to 1. But it can be seen
that when s > 1 and s → 1, L(s, χ) = 1 − 1/3s + 1/5s − 1/7s + 1/9s − 1/11s + · · ·
converges to 1 − 1/3 + 1/5 − 1/7 + 1/9 − 1/11 + · · · = π/4 < ∞. Contradiction.
Hence there are infinitely many prime numbers p such that p ≡ 3 mod 4. (The fact
there are infinitely many prime numbers such that p ≡ 1 mod 4 can be also proved
by using this L(s, χ).)
42. By using Problem 41, prove that there are infinitely many irreducible monic
polynomials f ∈ F3 [T ] whose constant term is 2 ∈ F3 .
43. Consider a finite field Fq of order q such that q ≡ 1 mod 4. By using the
formula (*), and by considering the√relation between maximal ideals of Fq [T ] and
maximal ideals of the extension Fq [ T ] of Fq [T ] (similarto Problems 39 and 40)
prove that ζFq [√T ] (s) = ζFq [T ] (s)L(s, χ) where χ(f ) = Tf for f ∈ Fq [T ] which is
relatively prime to T . By using this, prove that L(s, χ) = 1 (this is already proved
in 41 but here another proof is obtained for this χ).
44. Prove that in (**), if g is of degree n, L(s, χ) is a polynomial of q −s of degree
< n.
45. Let Fq be a finite field of characteristic 6= 2 of order q, let g be a monic
polynomial in Fq [T ] of degree n ≥ 1 which is not divisible by h2 for any element
10
√
h ∈ Fq [T ] of degree ≥ 1, and let A = Fq [T, g]. By using Problem 44 and (*), prove
that the Hasse zeta function of A satisfies
a polynomial of q −s of degree < n
ζA (s) =
.
1 − q 1−s
√
46. Let A = F5 [T, T 3 + 1]. By using Problem 40, prove that
ζA (s) =
1 + 51−2s
.
1 − 51−s
Prove Riemann hypothesis for this zeta function.
Comment for the proof. I am afraid that you may waste your precious time just
to compute χ(f ) for χ in Problem 40. So I write here the result of my computation.
For a monic polynomial f in F5 [T ] of degree < 3 which is coprime to T 3 + 1, χ(f )
is as follows.
χ(1) = 1,
χ(T ) = 1,
2
χ(T ) = 1,
χ(T 2 + T + 1) = 1,
χ(T 2 +2T ) = −1,
χ(T 2 + 3T ) = 1,
χ(T + 2) = −1,
2
χ(T + 1) = −1,
χ(T + 3) = 1,
2
χ(T + 2) = −1,
χ(T 2 + T + 2) = 1,
χ(T 2 +2T +2) = −1,
χ(T 2 + 3T + 1) = 1,
χ(T 2 + 4T ) = −1,
χ(T + 4) = −1,
χ(T 2 + 3) = 1,
χ(T 2 + T + 3) = 1,
χ(T 2 + T + 4) = 1,
χ(T 2 +2T +3) = −1,
χ(T 2 +2T +4) = 1,
χ(T 2 + 3T + 3) = −1,
χ(T 2 + 4T + 2) = 1,
χ(T 2 + 3T + 4) = 1,
χ(T 2 + 4T + 4) = 1.
47.p [As in my mail to you, this problem should be forgotten.]
F3 [T, T (T − 1)(T − 2)]. Prove that
ζA (s) =
Let A =
1 + 31−2s
.
1 − 31−s
Prove Riemann hypothesis for this zeta function.
Comment for the proof. I am a little afraid that you waste your precious time
just for the computation of χ(f ) for Problem 47, where χ : (F3 [T ]/(g))× → C× with
g = T (T − 1)(T − 2) is defined by
χ(f mod T (T − 1)(T − 2)) = (−1)deg(f )
f f f .
T T −1 T −2
But this time there are only four monic polynomials in F3 [T ] of degree < 3 which
are coprime to T (T − 1)(T − 2), so the computation should be not so terrible. They
are
1, T 2 + 1, T 2 + T + 2, T 2 + 2T + 2.
11
48. Let A = F5 [T,
p
T (T − 1)(T − 2)]. Prove that
ζA (s) =
1 + 2 · 5−s + 51−2s
.
1 − 51−s
Prove Riemann hypothesis for this zeta function.
Comment for the proof. I am again afraid that you waste your precious time just
for the computation of χ(f ) where χ : (F5 [T ]/(g)× → C× with g = T (T − 1)(T − 2)
is defined by
χ(f mod T (T − 1)(T + 1)) =
f f f .
T T −1 T +1
For a monic polynomial f in F5 [T ] of degree < 3 which is coprime to T (T −1)(T +1),
χ(f ) is as follows.
χ(1) = 1,
χ(T + 2) = 1,
χ(T 2 + 1) = 1,
χ(T + 3) = 1,
χ(T 2 + 2) = −1,
χ(T 2 + T + 1) = −1,
χ(T 2 + 3) = −1,
χ(T 2 + T + 2) = 1,
χ(T 2 + T + 4) = 1,
χ(T 2 + 2T + 3) = 1,
χ(T 2 + 2T + 4) = 1, ,
χ(T 2 + 3T + 3) = 1,
χ(T 2 + 3T + 4) = 1,
χ(T 2 + 4T + 1) = −1,
χ(T 2 + 4T + 2) = 1,
χ(T 2 + 4T + 4) = 1.
Preparation for Problems 49 and 50. In these problems, you prove the analogue
for Fq [T ] of the quadratic reciprocity law.
49. Let k be a commutative field. For monic polynomials f, g ∈ k[T ] which are
coprime, let [ fg ] ∈ k × be the image of
g mod f ∈ (k[T ]/(f ))×
under the norm map (k[T ]/(f ))× → k × . Prove
g
f
[ ] = [ ] · (−1)deg(f )deg(g) .
f
g
Here for a commutative ring A over k which is finite dimensional as a k-vector
space, the norm map N : A → k is defined as follows. For a ∈ A, N (a) is the
determinant of the k-linear map A → A ; x 7→ ax. In the case A is a field, N : A → k
is the norm map which appears in
Qthe text book of field theory. If A is a field and is
a Galois extension of k, N (a) = σ∈G σ(a) where G is the Galois group Gal(A/k).
To prove 49, I suggest to reduce the problem to the case k is an algebraically closed
12
field, and write f = (T − a1 )Q
· · · (T − am ) and g = (T − b1 ) · · · (T − bn ), and suggest
to prove that [ fg ] is equal to 1≤i≤m,1≤j≤n (ai − bj ).
50. Let Fq be a finite field of characteristic 6= 2 of order q. By using Problem
49, prove the analogue of the quadratic reciprocity law
q−1
g
f
=
· (−1) 2 ·deg(f )deg(g)
f
g
for monic irreducible polynomials f, g ∈ Fq [T ] such that f 6= g .
Suggestions for the proof. It may be helpful to use the following facts. (1) For a
finite field k of characteristic not 2, k × /(k × )2 is of order 2. Here (k × )2 = {x2 | x ∈
k × }. (2) For a finite field k and for a finite field K which is an extension of k,
∼
=
the norm map K × → k × is surjective and induces an isomorphism K × /(K × )2 →
k × /(k × )2 .
13
Set 6. Please study the following Problems 51–60 by February 17 (Friday).
As I told in the course, Grothendieck introduced the notation Spec(A) for the
set of all prime ideals of a commutative ring A, for it remained him of the spectrum
of a linear operator. Before he came to algebraic geometry, Grothendieck studied
linear operators on infinite dimensional vector spaces.
51. Let Mn (C) be the ring of all (n, n)-matrices over C and let ϕ ∈ Mn (C). Let
A be the commutative subring of Mn (C) over C generated over C by ϕ. That is,
A = {a0 + a1 ϕ + a2 ϕ2 + · · · + an ϕn | n ≥ 0, ai ∈ C}.
Prove that there is a bijection between Spec(A) and the set of all eigen values of ϕ.
(The set of all eigen values of ϕ is called the spectrum of ϕ.)
In the following Problems 52, 53, 54, we compare what happens for
(a) (X, A) where X is a compact Hausdorff space and A is the ring of C-valued
continuous functions on X
and what happens for
(b) (X, A) where A is any commutative ring and X = Spec(A).
We have the following facts (a1)–(a3) for (X, A) in (a).
(a1) For f ∈ A, f is invertible if and only if f (x) 6= 0 for any x ∈ X.
(a2) (This appeared in Problem 1.) For a subset S of X, let I(S) be the ideal
{f ∈ A | f (x) = 0 for any x ∈ S} of A. For an ideal J of A, let V (J) be the
closed subset {x ∈ X | f (x) = 0 for any f ∈ J} of X. Then for any subset S of X,
V (I(S)) coincides with the closure of S.
(a3) let Y and Z be closed subsets of X. Assume Y ∩Z = ∅. Let f be a C-valued
continuous function on Y and let g be a C-valued continuous function on Z. Then
there is a C-valued continuous function h on X such that the restriction of h to Y
is f and the restriction of h to Z is g.
In the following Problems 52, 53, 54, we consider the analogue of (a1), (a2),
(a3) for (X, A) in (b), respectively. For the Problems 52 and 54, you may use the
fact that if I is an ideal of a commutative ring A such that I 6= A, then there is a
maximal ideal m of A such that I ⊂ m.
52. Let (X, A) be as in the above (b). Let f ∈ A. Prove that the following
conditions (i)–(iii) are equivalent.
(i) f is invertible.
(ii) f (p) 6= 0 for any p ∈ Spec(A).
(iii) f (p) 6= 0 for any p ∈ max(A).
14
Here as in the course, f (p) for p ∈ Spec(A) denotes the image of f in the residue
field of p. (The residue field of p = the field of fractions Q(A/p) of the integral
domain A/p.)
53. Let (X, A) be as in the above (b). For a subset S of X, let I(S) be the
ideal {f ∈ A | f (x) = 0 for any x ∈ S} of A. For an ideal J of A, let V (J) be the
subset {x ∈ X | f (x) = 0 for any f ∈ J} of X. Recall that as in the course, Zariski
topology on X is defined by taking all subsets of X of the form V (J) (for ideals J
of A) as closed sets. Prove that for any subset S of X, V (I(S)) coincides with the
closure of S for Zariski topology.
54. Let (X, A) be as in the above (b). Let I and J be ideals of A. Assume
V (I) ∩ V (J) = ∅. Prove that I + J = A. Prove that for any f ∈ A/I and g ∈ A/J,
there is h ∈ A whose image in A/I is f and whose image in A/J is g.
55. Let (X,
√ in the above (b). Prove that for any ideal J of A, we have
√ A) be as
I(V (J)) = J. Here J = {f ∈ A f n ∈ J for some n ≥ 1}. (You can use the
fact, which was proved in the course, that the intersection of all prime ideals of a
commutative ring coincides with the set of all nilpotent elements.)
56. Let (X, A) be as in the above (b). Prove that the maps V and I give a
bijection (one
√ is the inverse map of the other) between the set of all ideals J of A
such that J = J and the set of all closed subsets of X (for Zariski topology).
Let (X, A) be as in the above (b). A closed subset S of X (for Zariski topology)
is said to be irreducible if the following conditions (i) and (ii) are satisfied: (i) S 6= ∅.
(ii) If S = Y ∪ Z with Y , Z closed subsets of X, we have either S = Y or S = Z.
57. Let (X, A) be as in the above (b). Prove that the bijection in Problem
56 induces a bijection between the set of all prime ieals of X and the set of all
irreducible closed subsets of X.
58. In the class, I explained that for an element f of a commutative ring A, the
following two conditions are equivalent.
(i) f is nilpotent. That is, f n = 0 for some n ≥ 1.
(ii) f (p) = 0 for any p ∈ Spec(A).
Give an example in which these conditions are not equivalent to the following
condition.
(iii) f (p) = 0 for any p ∈ max(A).
Let A be the coordinate ring of the algebraic set X = {(x, y) ∈ C2 | y 2 = x3 + 1}
∼
=
in C2 . We have an isomorphism C[T1 , T2 ]/(T22 − T13 − 1) → A by sending T1 (resp.
T2 ) to the function x (resp. y) on X which has value x (resp. y) at (x, y) ∈ X. In
the course, I told (without proof) the following. A is not PID, but the local ring
of A at any prime ideal is PID. In the following Problems 59 and 60, let p be the
maximal ideal f ∈ A | f (0, 1) = 0} of A. Note that we have p = (x, y − 1) and that
(y − 1)(y + 1) = x3 .
15
59. Note that any element f of A is written in the form f0 (y) + f1 (y)x + f2 (y)x2 ,
where fi (y) (i = 0, 1, 2) are polynomials in y. For i = 0, 1, 2, let mi be the (y − 1)adic order of fi (y). (This means that in the case fi (y) 6= 0, fi (y) is divisible by
(y − 1)mi but not divisible by (y − 1)mi +1 . In the case fi (y) = 0, mi is defined to be
∞.) Let m = min{3mi + i | i = 0, 1, 2}. Prove that if f 6= 0, in the local ring Ap of
A at p, f is xm times a unit.
60. Let the notation be as in 59. Prove that any non-zero ideal of Ap is written
in the form (xm ) for some m ≥ 0, and hence Ap is a PID.
16
Set 7. Please study the following Problems by February 24 (Friday).
As I told you in the course, the 5-adic completion Z5 = limn Z/5n Z of Z contains
←−
a square root of −1. We consider this fact in Problems 61–63.
Recall that we have the Taylor expansion
a
(1 + x) =
∞ X
a
n=0
n
xn
(1)
for x ∈ C such that |x| < 1, where
a(a − 1) . . . (a − (n − 1))
a(a − 1)
a
a
a
a
,
=
.
= 1,
= a,
=
n
0
1
2
2
n!
In the case a = 1/m (m ≥ 1), this gives an m-th root of 1 + x. For example,
1
1
(1 + x)1/2 = 1 + x − x2 + . . . .
2
8
(2)
Now let p bea prime number. It can be shown that if a ∈ Z(p) = { mr | r, m ∈
a
Z, p 6 |m}, then
∈ Z(p) for any n ≥ 0. For m ≥ 1 which is prime to p and for
n
x ∈ pZ, we can obtain an m-th root of 1 + x in Zp by the above same formula (1)
for a = 1/m ∈ Z(p) .
61. Obtain a square root 68 mod Z/53 Z of −1 = 22 (1 − 45 ) in Z/53 Z by applying
the above formula (2) to x = − 54 .
(In the computation, if 1/4 appears, a good method is to expand it as 1/4 =
−1/(1 − 5) = −1 − 5 − 52 − . . . .)
62. Prove that for any n ≥ 1, the canonical ring homomorphism Z/5n Z →
Z[i]/(2 − i)n is an isomorphism. By taking limn , deduce that Z5 contains a square
←−
root of −1.
63. (Here assume that you already know that Z5 has a square root of −1.)
Prove that there are two ring homomorphisms Z[i] → Z5 . (You can use the fact
Z5 is an integral domain.) Show that the inverse image of 5Z5 ⊂ Z5 under one
homomorphism is (2 − i) ⊂ Z[i], and the inverse image of 5Z5 under the other
homomorphism is (2 + i) ⊂ Z[i].
In Problems 64-70, you prove quadratic reciprocity law and related things.
64. Let K be a commutative field whose characteristic is not 2 and which contains
a primitive 8-th root ζ8 of 1 (a primitive n-th root of 1 is an element α such that
αn = 1 and αi 6= 1 for 1 ≤ i < n).
Prove that ζ8 + ζ8−1 is a square root of 2.
17
65. Let p be a prime number which is not 2. In Problem 64, take the algebraic
closure of Fp as K. By using the fact Fp = {x ∈ K | xp = x} and by using Problem
64, prove the formula
p2 −1
2
) = (−1) 8 ,
p
that is, a square root of 2 exists in Fp if and only if p ≡ 1, 7 mod 8 and does not
exist if p ≡ 3, 5 mod 8.
66. Let K be a commutative field and let q be a prime number which is different
from 2. Assume that the characteristic of K is not 2, q, and assume that K contains
a primitive q-th root ζq of 1. Let q ∗ = q if q ≡ 1 mod 4, and let q ∗ = −q if
q ≡ 3 mod 4. Prove that
X a
ζqa is a square root of q ∗ .
q
×
a∈Fq
This is not easy to prove. Problems 68 and 69 may be useful.
67. Let p and q be prime number which are not 2 and assume p 6= q. In Problem
66, take the algebraic closure of Fp as K. By using the fact Fp = {x ∈ K | xp = x}
and by using Problem 66, prove the formula
q∗
p
)=
.
p
q
From this, deduce the quadratic reciprocity law
p−1 q−1
q
p
)=
· (−1) 2 · 2 .
p
q
p−1
(Please use the fact −1
= (−1) 2 freely.)
p
The following Problems 68 and 69 are put here as preparations for the proof of
Problem 66.
68. Let N ≥ 1 be an integer, let K be a commutative field, and assume that K
contains a primitive N -th root ζN of 1. For a function f : Z/N Z → K, define the
Fourier transform F(f ) of f as the function Z/N Z → K defined by
X
(F(f ))(x) =
f (y)ζNxy .
y∈Z/N Z
(This is an analogue of the Fourier transform of a function on R.) Let g = F(F(f )).
Prove g(x) = N f (−x).
69. In Problem 68, in the case N is a prime number q and f : Fq = Z/qZ → K
is defined by f (x) = xq if x 6= 0 and f (0) = 0, prove that F(f ) = G · f where
P
G = a∈F×q aq ζqa .
70. By using Problem 64 and Problem 66 taking the algebraic closure of Q as K,
prove that if L is a quadratic field (an extension of Q of degree 2), then L ⊂ Q(ζN )
for some N ≥ 1. Here ζN denotes a primitive N -th root of 1.
18
Set 8. Please study the following Problems 71–80 by March 2 (Friday).
As in the class, consider a commutative field k, a finite extension K of k(T ),
the integral closure A of k[T ] in K, the integral closure A0 of k[1/T ] in K, and the
integral closure A00 of k[T, 1/T ] in K.
For example:
√
If k = √
C and K is the extension
C(T )( T n + 1) with n ≥ 2 an
p
√ even integer, then
0
00
n
n
A = C[T, T + 1], A = C[1/T, (1/T ) + 1], A = C[T, 1/T, T n + 1] (note that
p
√
(1/T )n + 1 = T −n/2 T n + 1).
If k = C and K is the extension C(T )((T n + 1)1/n ) with n ≥ 1 an integer, then
A = C[T, (T n + 1)1/n ], A0 = C[1/T, ((1/T )n + 1)1/n ], A00 = C[T, 1/T, (T n + 1)1/n ]
(note that ((1/T )n + 1)1/n = T −1 (T n + 1)1/n ).
71. Assume k = k̄ (that is, k is algebraically closed). Prove A ∩ A0 = k.
(This is a difficult problem. Here is a suggestion for the proof. Let α ∈ A ∩ A0 .
Let h(T ) be the monic minimal polynomial of α over k(T ). Since α is integral over
A, all coefficients of h(T ) belong to k[T ], . . . .)
The dimension of the k-vector space A00 /(A + A0 ) coincides with the invariant
of K called the genus of K. If k = C, the genus of K is the number of holes in
X = max(A) ∪ max(A0 ) endowed with the topology coming from the topology of C
(which is homeomorphic to the surface of a doughnut with several holes).
√
72. Let k = C and K is the extension C(T )( T n + 1) of C(T ) with n ≥ 2 an
even integer. Computing A00 /(A + A0 ), show that the genus of K is (n − 2)/2.
In Problems 73–76, you consider Dirichlet’s unit theorem and a similar subject
in the geometric situation.
For a number field K, Dirichlet’s unit theorem says that
(OK )× = Zr−1 ⊕ (a finite cyclic group)
where r is the number of all infinite places of K. (An infinite place of K is either an
embedding of K (as a field) into R or the complex conjugacy class of an embedding
of K (as a field) into C with dense image.
√ √
For example, if K = Q( 2, 3), then r = 4 (there are four embeddings K → R,
and there is no embedding K → C with dense image).
√ √
√
√
√ √
73. Show that in the case K = Q( 2, 3), 1− 2, 2− 3, and 3− 2 generate
a subgroup Γ of (OK )× such that Γ ∼
= Z3 = Zr−1 .
(Remark. Kato is not sure whether (OK )× is equal to {±1} × Γ.)
74. Let k, K, A, A0 , A00 be at the beginning of this Set 8, and assume k = k̄.
Let S = max(A0 ) − max(A00 ) = X − max(A) (X is the union of max(A) ∪ max(A0 ),
P − Q means the complement of a subset Q of a set P ), and let r be the number of
elements of S. Prove that A× /k × is a free Z-module of rank ≤ r − 1.
19
This is hard to prove. Here is a suggestion of the proof. We have a homomorphism A× → ⊕x∈S Z whose kernel coincides with k × , which sends f ∈ A× to
(ordx (f ))x∈S ∈ ⊕x∈S Z. Here for f ∈ K × and for x ∈ X, ordx (f ) is the integer
n (which can be negative) such that f = ux · πxn where ux is a unit of the local
ring at x and πx is a prime element of the local ring at x. (If x ∈ max(A) (resp.
x ∈ max(A0 )), the local ring at x means the local ring of A (resp. A0 ) at x.) You
can use the following known fact
X
ordx (f ) = 0 for f ∈ K × .
x∈X
√
75. In Problem 75, in the case k = C and K = C(T )( T n + 1) with n ≥ 2 an
even integer, prove that the order of S is 2 and A× /C× ∼
= Z.
76. In Problem 75, in the case k = C and K = C(T )((T 4 + 1)1/4 ), prove that
the order of S is 4 and A× /C× ∼
= Z3 .
Remark. The Z-rank of A× /k × is ≤ r − 1 as above, but it need not coincide
with r − 1 in general. However, if k is a finite field, we have
A× = Zr−1 ⊕ (a finite cyclic group)
just as in Dirichlet’s unit theorem.
Note that for a sequence an (n = 1, 2, 3, . . . ) of rational numbers, for a prime
number p, and for c ∈ Q, an converges to c in the p-adic number field Qp if and only
if the p-adic order ordp (an − c) tends to ∞.
77. Prove that 1 − (2/3)n! (resp. 1 − (3/4)n! , resp. 1 − 6n! ) (n = 1, 2, 3, . . . )
converges to 0 in Qp for any prime number p 6= 2, 3, and converges to 1 in Q2 and
in R (resp. in Q3 and in R, resp. in Q2 and in Q3 ).
78. Let n ≥ 1 and consider a finite field Fq of q-elements such that q ≡ 1 mod 2n.
For monic irreducible polynomials f and g over Fq such that f 6= g, prove that
f mod g is an n-th power (= xn for some element x) in Fq [T ]/(g) if and only if
g mod f is an n-th power in Fq [T ]/(f ).
Suggestion for a proof. Use Problem 49 and use the known fact that for a finite
field k and for a finite extension k 0 of k, the norm map (k 0 )× → k × is surjective and
∼
=
induces an isomorphism (k 0 )× /{xn | x ∈ (k 0 )× } → k × /{xn | x ∈ k × }.
Next week, I will talk about the dimension of a commutative ring and will talk
about the following fact. For the polynomial ring k]T1 , . . . , Tn ] in n variables over a
commutative field k, the ideals pi = (T1 , . . . , Ti ) (i = 0, 1, . . . , n) are prime ideals of
k[T1 , . . . , Tn ] (p0 means the ideal (0)), and for each i = 0, . . . , n−1, there is no prime
ideal q of A such that pi ( q ( pi+1 . The last fact is deduced from the following
Problem 79 applied to A = k[T1 , . . . , Tn ]/pi ∼
= k[Ti+1 , . . . , Tn ] and f = Ti+1 .
20
79. Let A be a Noetherian integral domain. Let f be a prime element of A (this
means that f 6= 0 and the ideal (f ) of A is a prime ideal). Prove that there is no
prime ideal p of A such that 0 ( p ( (f ).
A suggestion for the proof: Assume such p exists. Take a non-zero element g of
p. By using the fact f ∈
/ p and by some argument, get g = g1 f for some g1 ∈ A,
g1 = g2 f for some g2 ∈ A, g2 = g3 f for some g3 ∈ A, . . . , and use the Noetherian
property of A looking at the ideals (g) ⊂ (g1 ) ⊂ (g2 ), . . . of A.
80. Let A be a unique factorization domain (UFD). Let p be a prime ideal of
A. Assume that there is no prime ideal q of A such that (0) ( q ( p. Prove that
p = (f ) for some prime element f of A.
A suggestion for the proof. Take a non-zero element of p and consider the prime
factorization of it.
21