ABSTRACT
Title of Dissertation: CLASS GROUPS OF CHARAC-
TERISTIC-p FUNCTION FIELD
ANALOGUES OF Q(n1/p)
Steven Reich
Doctor of Philosophy, 2021
Dissertation Directed by: Professor Lawrence Washington
Department of Mathematics
In the theory of cyclotomic function fields, the Carlitz module ?M associated
to a polynomial M in a global function field of characteristic p provides a strong
analogy to the roots of unity ?p in a number field. In this work, we consider a
natural extension of this theory to give a compatible analogue of the p-th root of an
integer n.
The most fundamental case, and the one which most closely mimics the number
field situation, is when the Carlitz module is defined by a linear polynomial (which
can be assumed to be T ) in k = Fq(T ). The Carlitz module ?T generates a degree-
(q ? 1) extension k(?T ) which shares many properties with the field Q(?p), where
?p is the module of p-th roots of unity.
?
To form the analogue of Q( p n), we define a degree-q extension F/k associated
to a polynomial P (T ) ? k, for which the normal closure is formed by adjoining ?T .
In the introduction, we describe in detail the parallels between this construction and
that in the number field setting. We then compute the class number hF for a large
number of such fields. The remainder of the work is concerned with proving results
about the class groups and class numbers of this family of fields. These are:
? a formula relating the class number of F to that of its normal closure, along
with a theorem about the structure of the class group of the normal closure
? a formula relating the class number of a compositum of such F to the class
numbers of the constituent fields
? conditions on P (T ) for when the characteristic, p, of F divides its class number,
along with bounds on the rank of the p-part of the class group.
CLASS GROUPS OF CHARACTERISTIC-p
FUNCTION FIELD ANALOGUES OF Q(n1/p)
by
Steven Reich
Dissertation submitted to the Faculty of the Graduate School of the
University of Maryland, College Park in partial fulfillment
of the requirements for the degree of
Doctor of Philosophy
2021
Advisory Committee:
Professor Lawrence Washington, Chair/Advisor
Professor Patrick Brosnan
Professor Niranjan Ramachandran
Professor James Schafer
Professor William Gasarch
? Copyright by
Steven Reich
2021
Dedication
This work is dedicated to my family: my parents Robert and Elizabeth, and my
brother Daniel. This would not have been possible without your love and support.
?Behold, I am doing a new thing;
now it springs forth, do you not perceive it?
I will make a way in the wilderness
and rivers in the desert.? ? Isaiah 43:19
ii
Acknowledgments
I would first like to thank my advisor, Professor Lawrence Washington, for his
encouragement and insight throughout my graduate career. His zeal for mathemat-
ics, and more importantly his care for the people around him, continue to be an
inspiration to me.
I would also like to recognize all members of the Mathematics Department:
the faculty, staff, and students, who have all contributed to making this experience
a worthwhile one. I am especially thankful to my friends and peers who helped to
make these years enjoyable.
I am grateful as well to the Human Language Technology Center of Excellence
at Johns Hopkins University, which provided funding for three semesters of my
graduate studies. In particular, I thank Dr. Nicholas Andrews, who has consistently
advocated in my support.
There are many other people responsible for shaping my identity as a mathe-
matician and as a person, including teachers who nurtured my curiosity, and friends
who have enriched my life. I regret that I cannot thank each of them individually,
and surely I am indebted to some in ways that I am not even aware. However, I
would be remiss to not recognize my dear friend Dr. Patrick Devlin, who has been
inspiring my interest in mathematics since we were kids.
Most of all, I want thank my wonderful family, and praise God.
iii
Table of Contents
Dedication ii
Acknowledgements iii
Table of Contents iv
Chapter 1: Introduction and background 1
1.1 Carlitz modules and??cyclotomic? function fields . . . . . . . . . . . . 1
1.2 The analogue of Q( p n) . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Results to be presented . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Chapter 2: Explicit computation of class numbers 7
Chapter 3: Relation to the Galois closure 14
3.1 Proof of the class number relation . . . . . . . . . . . . . . . . . . . . 15
3.2 Proof of class group structure theorem . . . . . . . . . . . . . . . . . 19
3.2.1 The non-p part of Cl0L . . . . . . . . . . . . . . . . . . . . . . 19
3.2.2 The p part of Cl0L . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.2.3 Galois cohomology of Cl0L . . . . . . . . . . . . . . . . . . . . 24
Chapter 4: Class numbers of composite fields 29
4.1 The case q = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
4.2 The case q 6= 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
Chapter 5: p-divisibility of the class number 35
5.1 The case q = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
5.2 The case q 6= 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
Chapter 6: Future Work 50
Bibliography 52
iv
Chapter 1: Introduction and background
1.1 Carlitz modules and ?cyclotomic? function fields
We begin by describing the basic theory of cyclotomic function fields. We
generally follow the constructions given by Goss [7] (cf. also [27]).
Let k = Fq(T ), with q a power of a prime p, and ksep a separable closure. The
Frobenius map F : ksep ? ksep defined by F (x) = xq is additive, as are its powers F i
under composition (including the identity map F 0). Multiplication by an element
of Ok = Fq[T ] also represents an additive endomorphism of ksep. We may thus
define an injective homomorphism C : Ok ? End(ksep), where C(T ) = F + TF 0,
C(?) = ?F 0 for ? ? Fq, and C(M) for arbitraryM ? Ok is determined by extending
linearly from these relations. As a nontrivial example,
C(T 2+1) = C(T )2+C(1) = (F+TF 0)?(F+TF 0)+F 0 = F 2+(T q+T )F+(T 2+1)F 0.
Using this construction, we can associate to a polynomialM the Carlitz module
?M = {? ? ksep |C(M)(?) = 0} (first introduced by Carlitz in [4]). This is isomor-
phic as an O deg(M)k-module to Ok/(M), which has cardinality q , and generates an
abelian extension k(?M)/k with Galois group isomorphic to (Ok/M)?.
1
We will focus on the case M = T , for which the analogy with the number field
setting is most clear. The Carlitz module ?T is generated as an Fq-module by a
fixed nonzero root ? of C(T )(x) = xq + Tx. The extension k(?)/k has Galois group
isomorphic to Ok[T ]/(T ) ?= F?q , and the integral closure of Ok in k(?) is Ok[?].
Additionally, this extension is unramified outside of T and 1/T (which we consider
as the infinite prime of k), and these primes are totally ramified. In particular,
(T ) = (?)q?1 and (1/T ) = (1/?)q?1.
This should be reminiscent of the p-th roots of unity generated by a primitive
root ?p, with the multiplicative structure ?p replaced by the additive structure ?T .
The extension Q(?p)/Q has Galois group (Z/pZ)? and its ring of integers is Z[?p].
The only finite prime that ramifies is p, and its ramification is total.
The parallel between the behavior of the respective infinite primes is slightly
more nuanced. The maximal real subfield Q(? )+ = Q(? + ??1p p p ) is the fixed field
of the image of Z? = {?1} in (Z/pZ)?. The infinite prime of Q splits completely
in Q(? +p) /Q, and the primes above it ramify in Q(?p)/Q(? +p) . In the function field
setting, the maximal ?real? subfield k(?)+ is the fixed field of O? ?k = F?q , which in this
case is the full Galois group (i.e., k(?)+ = k), and 1/T is totally ramified in k(?)/k.
(We note that in the more general case that M is a power of a monic irreducible
polynomial, k(? )+M is a non-trivial extension of k in which 1/T splits completely.)
Hayes [11] gives an even deeper connection between Carlitz modules and roots
of unity: an analogue of the Kronecker-Weber theorem. He shows that any finite
abelian extension of k in which 1/T is not wildly ramified is contained in a constant
field extension of k(?M) for some M ? Ok.
2
?
1.2 The analogue of Q( p n)
We now describe the construction of a function field which parallels that of
?
Q( p n), continuing the analogy (and notation) of the previous section. Much of
the following can be done for a general Carlitz module ?M , but for concreteness we
continue to focus on ?T , since this is the situation we consider in later sections.
Let P (T ) ? Ok be such that P (T ) 6= Q(T )q + TQ(T ) for any Q(T ) ? Ok,
and fix a root ? ? ksep of C(T )(x) ? P (T ) = xq + Tx ? P (T ). We will show in
Proposition 1.1 below that this is irreducible. By the linearity of C(T ), the full set
of roots in ksep is {? + ?? |? ? Fq}. Thus the extension k(?)/k is of degree q, and
its normal closure is formed by adjoining ?. The Galois group Gal(k(?, ?)/k) has
normal subgroup G = Gal(k(?, ?)/k(?)) ?= F+q , which permutes the roots ? + ??,
and the subgroup ? = Gal(k(?, ?)/k(?)) ?= Gal(k(?)/k) ?= F?q , which acts on G
via a cyclotomic character. This closely mirrors the number field situation, in which
? ?
Q(? pp, n) is the normal closure of Q( p n), and the Galois subgroup structure is
essentially the same.
By a change of variables, the polynomial defining k(?, ?)/k(?) can be put in
q?1
the form xq?x? P (?? )q , whereby this is an Artin-Schreier extension (but note that?
the extension k(?)/k is not1, as Artin-Schreier extensions are necessarily Galois).
Hasse [10] considered general Artin-Schreier extensions as an analogue of adjoining
p-th roots of n, but we argue that our construction gives a more precise analogy
because of the relationship it enjoys with the cyclotomic theory.
1Except when q = 2, where the Carlitz module generates a trivial extension of k, in analogy
with the square roots of unity which of course lie in Q.
3
Proposition 1.1. Suppose P (T ) ? Ok is not of the form Q(T )q + TQ(T ) for any
Q(T ) ? Ok. Then xq + Tx? P (T ) is irreducible over k.
Proof. Clearly this is true for q = 2, so we assume q > 2. The condition shows that
xq + Tx? P (T ) does not have a root in k. Now, if ? ? ksep is a fixed root, the set
of all roots in ksep is {? + ?? |? ? Fq}, where ? is a nonzero root of xq + Tx (i.e.,
a generator of the Carlitz module ?T ). If one of the roots ? is in K = k(?), we
would have F = k(?) is Galois over k (since K/k is abelian). But this would mean
F contains another root ? + ??, and thus that F = K. Since [K : k] = q ? 1, this
would mean xq + Tx ? P (T ) has an irreducible factor of degree q ? 1, and thus a
linear factor, contradicting the above.
Now, let L = k(?, ?), the splitting field of xq + Tx ? P (T ). Since Gal(L/k)
has Gal(K/k) ?= F?q as a quotient, we can obtain an element ? ? Gal(L/k) of
order q ? 1 by lifting a generator of Gal(K/k) and (if necessary) raising it to a
suitable power. We observe that ? cannot fix all the roots of xq + Tx ? P (T ) (in
fact, it cannot fix more than one, otherwise it would fix all of L, since both ? and
? can be obtained as Fq-linear combinations of any two distinct roots). Without
loss of generality, suppose ? is not fixed by ?, i.e. that ?(?) = ? + ?? for some
? ? F?q . Then ?i(?) = ?+?(1 +?+ ? ? ?+?i?1)(?). If ?i fixes ?, then we must have
(1??i)(?) = (1??)(1 +?+ ? ? ?+?i?1)(?) = (1??)(0), and thus ?i fixes ? as well.
But then ?i fixes L, so i must be a multiple of q ? 1. This says that the orbit of ?
under ??? has order q ? 1. Thus the degree of the minimal polynomial for ? must
be at least q ? 1, and divides xq + Tx ? P (T ) which cannot have a linear factor.
4
Therefore xq + Tx? P (T ) is the minimal polynomial, and so it is irreducible.
This proposition contrasts with what one might expect from the number field
situation, where requiring that xq ?n does not have a root in Q does not guarantee
irreducibility. This (as well as the simpler fact that [k(?T ) : k] = q ? 1 rather than
?(q)) illustrates an interesting subtlety: even when q is not prime, our situation
more closely resembles that of the p-th roots of unity (and n) than that of q-th
roots. In fact, the correct analogue of the latter is the Carlitz module ?T r of a
power of T (or some other irreducible polynomial), but this goes beyond the scope
necessary for our purposes.
We summarize the correspondence between the number field and function field
settings in the following table.
Table 1.1: Dictionary of relevant concepts for number fields and function fields.
Number fields Function fields
? = {? ip p}, roots of xp ? 1 ?T = {?? |? ? Fq}, roots of xq + Tx
Gal(Q(?p)/Q) ?= (Z/pZ)? Gal(k(?)/k) ?= F?q
Z[?p], the ring of integers in Q(?p) Ok[?], the integral closure of Ok in k(?)
p totally ramified in Q(?p)/Q T totally ramified in k(?)/k
Infinite prime splits in Q(? )+p /Q, 1/T splits in k(?)+/k,
totally ramified in Q(?p)/Q(? )+p totally ramified in k(?)/k(?)+
Kronecker-Weber theorem Hayes? explicit class field theory
?
p n, a root of xp ? n ?, a root of xq + Tx? P (T )
where n is not a p-th power where P (T ) 6= Q(T )q + TQ(T )
?
Q(?p, p n)?, the normal k(?, ?), the normal
closure of Q( p n), has Galois closure of k(?), has Galois
group ?= Z/pZ o (Z/pZ)? group ?= F+q o F?q
?
Gal(Q(?p, p n)/Q(?p)) ?=?Z/pZ Gal(k(?, ?)/k(?))
?= F+q
permutes the roots ? i pp n permutes the roots ? + ??
5
1.3 Results to be presented
In this work, we are primarily concerned with the degree-0 divisor class group
of fields in the family we have described above. We take the divisor group DF of a
function field F to mean the free abelian group indexed by its (finite and infinite)
primes, and the degree-0 divisor group D0F to be the subgroup of such elements with
total degree 0. Then the class group Cl0F is the quotient of D0F by the principal
divisors PF . The class number is its cardinality hF .
As in previous sections, let F be the field generated over Fq(T ) by a root ? of
xq +Tx?P (T ), and let L = F (?) denote its normal closure. We briefly summarize
the results that will be presented in subsequent chapters.
? Chapter 2: we compute the class number of many F with q = p a small prime
and P (T ) of low degree.
? Chapter 3: we show that h = hq?1L F , and when q = p is prime, that Cl
0
L is a
(p? 1)-st power of a group.
? Chapt?er 4: we consider composita K = F1F2 . . . Fn of such fields, showing that
hK = hF .i
? Chapter 5: we give complete conditions on P (T ) that determine when hF
is divisible by the characteristic p. For the case q = p = 2, we show that
the 2-class group is always cyclic and give additional conditions for 4- and
8-divisibility of the class number.
6
Chapter 2: Explicit computation of class numbers
The theoretical results presented in subsequent chapters were motivated by
direct computations of the class numbers of members of the family of fields. In
this chapter, we describe the approach by which these calculations were made and
provide tables of the class numbers of over 100 distinct fields.
We employ standard theory about zeta functions of curves and their function
fields (see e.g. [9, 22, 24, 26]) to compute the class number of a field F . First, the
zeta function1 of F/Fq(T ) of genus g can be expressed as
G(t)
Z(F, t) = ,
(1? t)(1? qt)
where G(t) a polynomial with the following properties:
? de(g(G()t)) = 2g
1
? G = qgt2gG(t)
qt
? If z is a root of G(t), then |z| = ?1
q
? G(1) = hF .
1The zeta function is actually ?F (s) = Z(F, q?s), but we refer to Z(F, t) as a zeta function for
brevity.
7
In order to find G(t), we express the zeta function in terms of rational points
on a curve that determines F :
(? ) ( )? ?ti ? ti ? 1? tdx
Z(F, t) = exp #C?(Fqi) = exp #C(Fqi) ,
i i 1? tex
i=1 i=1 x
where C? is a non-singular model of a curve C with function field F . Each term of
the rightmost product is a ratio of Euler factors corresponding to a singular point
x, and dx|ex. Thus we arrive at the expression
(? )? ti (1? t)(1? qt)
G(t) = exp #C(Fqi) ,
i f(t)
i=1
where f(t) is a polynomial (possibly equal to 1) all of whose roots are roots of unity.
We outline the computational algorithm as applied to xq + Tx? P (T ) below.
Algorithm 1: Computation of class number
Input: q = pr, p prime
P (T ) ? Fq[T ]
1 Set f(x, T ) = xq + Tx? P (T )
2 Set K = (q ? 1)deg(P (T )) ? 2g
3 for k = 1, . . . , K do
4 nk = 1 (accounting for the point at infinity)
5 for ? ? Fqk do
6 nk ?( knk + deg()gcd(xq ? x, f(x, ?)))?K nk
7 G(t) = exp tk (1? t)(1? qt) mod tK+1
k
k=1
8 if ?z such that G(z) = 0 and |z| ?/ {1, ?1 } then
q
9 Increase K
10 go to 3
11 while ?z such that G(z) = 0, |z| = 1 do
12 G(t)? G(t)/(t? z)
13 return G(t) and h = G(1)
8
The results of these computations are given in the following tables. We omit
cases of low degree (deg(P (T )) < 3 for q = 2, and < 2 otherwise) since these are
genus 0. We also note that by a change of variables, fields of this kind can always be
generated by a P (T ) with no terms of degree divisible by q, except for q = 2 when
there may be a quadratic term, and hence we only list P (T ) with this property.
Table 2.1: Class numbers of extensions F/F2(T ) determined by various P (T ).
P (T ) GF (t) hF
T 3 1 1
T 3 + T 1 + t+ 2t2 4
T 3 + T 2 1 1
T 3 + T 2 + T 1? t+ 2t2 2
T 5 1 + 2t2 3
T 5 + T 1 + t+ 2t3 + 4t4 8
T 5 + T 2 1 + 2t2 3
T 5 + T 2 + T 1? t? 2t3 + 4t4 2
T 5 + T 3 1 + 2t+ 2t2 5
T 5 + T 3 + T 1? t+ 2t2 ? 2t3 + 4t4 4
T 5 + T 3 + T 2 1? 2t+ 2t2 1
T 5 + T 3 + T 2 + T 1 + t+ 2t2 + 2t3 + 4t4 10
T 7 1 + 4t4 5
T 7 + T 1 + t+ 2t2 + 4t3 + 4t4 + 4t5 + 8t6 24
T 7 + T 2 1 + 4t4 5
T 7 + T 2 + T 1? t+ 2t2 ? 4t3 + 4t4 ? 4t5 + 8t6 6
T 7 + T 3 1 + 2t+ 4t2 + 4t3 + 4t4 15
T 7 + T 3 + T 1? t? 4t5 + 8t6 4
T 7 + T 3 + T 2 1? 2t+ 4t2 ? 4t3 + 4t4 3
T 7 + T 3 + T 2 + T 1 + t+ 4t5 + 8t6 14
T 7 + T 5 1 + 2t+ 2t2 + 4t3 + 4t4 13
T 7 + T 5 + T 1? t+ 2t2 ? 2t3 + 4t4 ? 4t5 + 8t6 8
T 7 + T 5 + T 2 1? 2t+ 2t2 ? 4t3 + 4t4 1
T 7 + T 5 + T 2 + T 1 + t+ 2t2 + 2t3 + 4t4 + 4t5 + 8t6 22
T 7 + T 5 + T 3 1 + 2t2 + 4t4 7
T 7 + T 5 + T 3 + T 1 + t? 2t3 + 4t5 + 8t6 12
T 7 + T 5 + T 3 + T 2 1 + 2t2 + 4t4 7
T 7 + T 5 + T 3 + T 2 + T 1? t+ 2t3 ? 4t5 + 8t6 6
9
Table 2.2: Class numbers of extensions F/F2(T ) determined by various P (T ) (cont.)
P (T ) GF (t) hF
T 9 1? 2t3 + 8t6 7
T 9 + T 1 + t+ 2t2 + 2t3 + . . . 2 48
T 9 + T 2 1 + 2t3 + 8t6 11
T 9 + T 2 + T 1? t+ 2t2 ? 2t3 + 6t4 ? . . . 18
T 9 + T 3 1 + 2t+ 4t2 + 6t3 + 8t4 + 8t5 + 8t6 37
T 9 + T 3 + T 1? t+ 2t302t4 + 4t5 ? 8t7 + 16t8 12
T 9 + T 3 + T 2 1? 2t+ 4t2 ? 6t3 + 8t4 ? 8t5 + 8t6 5
T 9 + T 3 + T 2 + T 1 + t? 2t3 ? 2t4 ? 4t5 + 8t7 + 16t8 18
T 9 + T 5 1 + 2t+ 2t2 + 2t3 + 4t4 + 8t5 + 8t6 27
T 9 + T 5 + T 1? t+ 2t2 ? 4t3 + 2t4 ? . . . 8
T 9 + T 5 + T 2 1? 2t+ 2t2 ? 2t3 + 4t4 ? 8t5 + 8t6 3
T 9 + T 5 + T 2 + T 1 + t+ 2t2 + 4t3 + 2t4 + . . . 50
T 9 + T 5 + T 3 1 + 2t2 + 2t3 + 4t4 + 8t6 17
T 9 + T 5 + T 3 + T 1 + t+ 2t4 + 8t7 + 16t8 28
T 9 + T 5 + T 3 + T 2 1 + 2t2 ? 2t3 + 4t4 + 8t6 13
T 9 + T 5 + T 3 + T 2 + T 1? t+ 2t4 ? 8t7 + 16t8 10
T 9 + T 7 1 + 2t+ 4t2 + 6t3 + 8t4 + 8t5 + 8t6 37
T 9 + T 7 + T 1? t? 2t3 + 6t4 ? 4t5 ? 8t7 + 16t8 8
T 9 + T 7 + T 2 1? 2t+ 4t2 ? 6t3 + 8t4 ? 8t5 + 8t6 5
T 9 + T 7 + T 2 + T 1 + t+ 2t3 + 6t4 + 4t5 + 8t7 + 16t8 38
T 9 + T 7 + T 3 1 + 2t3 + 8t6 11
T 9 + T 7 + T 3 + T 1 + t+ 2t2 + 2t3 + 2t4 + . . . 44
T 9 + T 7 + T 3 + T 2 1? 2t3 + 8t6 7
T 9 + T 7 + T 3 + T 2 + T 1? t+ 2t2 ? 2t3 + 2t4 ? . . . 14
T 9 + T 7 + T 5 1 + 2t2 ? 2t3 + 4t4 + 8t6 13
T 9 + T 7 + T 5 + T 1 + t? 2t4 + 8t7 + 16t8 24
T 9 + T 7 + T 5 + T 2 1 + 2t2 + 2t3 + 4t4 + 8t6 17
T 9 + T 7 + T 5 + T 2 + T 1? t? 2t4 ? 8t7 + 16t8 6
T 9 + T 7 + T 5 + T 3 1 + 2t+ 2t2 + 2t3 + 4t4 + 8t5 + 8t6 27
T 9 + T 7 + T 5 + T 3 + T 1? t+ 2t2 + 2t4 + 8t6 ? 8t7 + 16t8 20
T 9 + T 7 + T 5 + T 3 + T 2 1? 2t+ 2t2 ? 2t3 + 4t4 ? 8t5 + 8t6 3
T 9 + T 7 + T 5 + T 3 + T 2 + T 1 + t+ 2t2 + 2t4 + 8t6 + 8t7 + 16t8 38
2We truncate the polynomial for space in some cases, but note that the remaining terms can
be recovered using the functional equation for G(t).
10
Table 2.3: Class numbers of extensions F/F3(T ) determined by various P (T ).
P (T ) GF (t) hF
T 2 1 1
T 2 + T 1 + 2t+ 3t2 6
T 2 + 2T 1? t+ 3t2 3
T 4 1 + 9t4 10
T 4 + T 1 + 2t+ 6t2 + 9t3 + 18t4 + 18t5 + 27t6 81
T 4 + 2T 1? t? 9t5 + 27t6 18
T 4 + T 2 1 + 3t2 + 9t4 13
T 4 + T 2 + T 1? t+ 6t3 ? 9t5 + 27t6 24
T 4 + T 2 + 2T 1 + 2t+ 3t2 + 6t3 + 9t4 + 18t5 + 27t6 66
T 4 + 2T 2 1 + 3t+ 6t2 + 9t3 + 9t4 28
T 4 + 2T 2 + T 1? t+ 3t3 ? 9t4 + 27t6 21
T 4 + 2T 2 + 2T 1? t+ 3t2 ? 6t3 + 9t4 ? 9t5 + 27t6 24
T 5 1 + 27t6 28
T 5 + T 1? t? 27t7 + 81t8 54
T 5 + 2T 1 + 2t+ 6t2 + 9t3 + 18t4 + . . . 252
T 5 + T 2 1 + 3t+ 6t2 + 9t3 + 18t4 + 27t5 + 27t6 91
T 5 + T 2 + T 1? t+ 3t2 + 3t3 + 9t5 + 27t6 ? 27t7 + 81t8 96
T 5 + T 2 + 2T 1? t+ 3t3 ? 9t4 + 9t5 ? 27t7 + 81t8 57
T 5 + 2T 2 1 + 3t2 + 9t4 + 27t6 40
T 5 + 2T 2 + T 1 + 2t+ 3t2 + 6t3 + 18t4 + . . . 210
T 5 + 2T 2 + 2T 1? t? 3t3 + 9t4 ? 9t5 ? 27t7 + 81t8 51
T 5 + T 4 1 + 3t+ 9t2 + 15t3 + 27t4 + 27t5 + 27t6 109
T 5 + T 4 + T 1? t? 3t3 + 12t4 ? 9t5 ? 27t7 + 81t8 54
T 5 + T 4 + 2T 1? t+ 6t3 ? 6t4 + 18t5 ? 27t7 + 81t8 72
T 5 + T 4 + T 2 1? 3t3 + 27t6 25
T 5 + T 4 + T 2 + T 1 + 2t+ 3t2 + 6t3 + 12t4 + . . . 204
T 5 + T 4 + T 2 + 2T 1? t+ 3t2 ? 3t3 + 3t4 ? . . . 75
T 5 + T 4 + 2T 2 1 + 6t3 + 27t6 34
T 5 + T 4 + 2T 2 + T 1? t+ 3t2 ? 3t3 + 12t4 ? . . . 84
T 5 + T 4 + 2T 2 + 2T 1 + 2t+ 3t2 + 6t3 + 12t4 + . . . 204
T 5 + 2T 4 1 + 3t3 + 27t6 31
T 5 + 2T 4 + T 1 + 2t+ 6t2 + 12t3 + 24t4 + . . . 270
T 5 + 2T 4 + 2T 1? t+ 3t3 ? 3t4 + 9t5 ? 27t7 + 81t8 63
T 5 + 2T 4 + T 2 1 + 3t2 + 3t3 + 9t4 + 27t6 43
T 5 + 2T 4 + T 2 + T 1? t+ 6t4 ? 27t7 + 81t8 60
T 5 + 2T 4 + T 2 + 2T 1 + 2t+ 3t2 ? 3t4 + 27t6 + 54t7 + 81t8 165
T 5 + 2T 4 + 2T 2 1 + 3t+ 6t2 + 12t3 + 18t4 + 27t5 + 27t6 94
T 5 + 2T 4 + 2T 2 + T 1? t? 3t3 + 6t4 ? 9t5 ? 27t7 + 81t8 48
T 5 + 2T 4 + 2T 2 + 2T 1? t+ 3t2 ? 3t3 + 15t4 ? . . . 87
11
Table 2.4: Class numbers of extensions F/F5(T ) determined by various P (T ).
P (T ) GF (t) hF
T 2 1 + 5t2 6
T 2 + T 1 + 4t+ 10t2 + 20t3 + 25t4 60
T 2 + 2T 1? t? 5t3 + 25t4 20
T 2 + 3T 1? t+ 5t2 ? 5t3 + 25t4 25
T 2 + 4T 1? t? 5t3 + 25t4 20
T 3 1 + 125t6 126
T 3 + T 1? t+ 5t2 + 10t3 + 50t5 + 125t6 ? 125t7 + 625t8 690
T 3 + 2T 1? t? 5t3 + 25t4 ? 25t5 ? 125t7 + 625t8 495
T 3 + 3T 1? t+ 5t3 ? 25t4 + 25t5 ? 125t7 + 625t8 505
T 3 + 4T 1 + 4t+ 15t2 + 40t3 + 100t4 + . . . 1860
T 3 + T 2 1 + 5t+ 15t2 + 35t3 + 75t4 + 125t5 + 125t6 381
T 3 + T 2 + T 1? t+ 5t2 ? 5t3 + 40t4 ? . . . 640
T 3 + T 2 + 2T 1? t+ 5t2 ? 5t3 + 15t4 ? . . . 615
T 3 + T 2 + 3T 1? t? 5t3 + 15t4 ? 25t5 ? 125t7 + 625t8 485
T 3 + T 2 + 4T 1? t? 5t3 + 40t4 ? 25t5 ? 125t7 + 625t8 510
T 3 + 2T 2 1 + 5t2 ? 5t3 + 25t4 + 125t6 151
T 3 + 2T 2 + T 1 + 4t+ 15t2 + 30t3 + 80t4 + . . . 1780
T 3 + 2T 2 + 2T 1? t? 5t3 + 5t4 ? 25t5 ? 125t7 + 625t8 475
T 3 + 2T 2 + 3T 1? t+ 10t3 ? 20t4 + 50t5 ? 125t7 + 625t8 540
T 3 + 2T 2 + 4T 1? t? 20t4 ? 125t7 + 625t8 480
T 3 + 3T 2 1 + 5t3 + 125t6 131
T 3 + 3T 2 + T 1? t+ 5t2 ? 15t3 + 20t4 ? . . . 560
T 3 + 3T 2 + 2T 1 + 4t+ 10t2 + 15t3 + 20t4 + . . . 1500
T 3 + 3T 2 + 3T 1? t+ 5t2 ? 5t3 + 45t4 ? . . . 645
T 3 + 3T 2 + 4T 1? t+ 20t4 ? 125t7 + 625t8 520
T 3 + 4T 2 1 + 15t3 + 125t6 141
T 3 + 4T 2 + T 1? t+ 10t2 ? 15t3 + 60t4 ? . . . 730
T 3 + 4T 2 + 2T 1? t+ 5t3 ? 15t4 + 25t5 ? 125t7 + 625t8 520
T 3 + 4T 2 + 3T 1 + 4t+ 10t2 + 25t3 + 60t4 + . . . 1600
T 3 + 4T 2 + 4T 1? t? 5t3 + 10t4 ? 25t5 ? 125t7 + 625t8 480
T 4 1 + 75t5 + 3125t10 3201
12
Table 2.5: Class numbers of extensions F/F7(T ) determined by various P (T ).
P (T ) GF (t) hF
T 2 1 + 49t4 50
T 2 + T 1 + 6t+ 21t2 + 56t3 + 147t4 + 294t5 + 343t6 868
T 2 + 2T 1? t+ 7t2 + 7t3 + 49t4 ? 49t5 + 343t6 357
T 2 + 3T 1? t+ 7t2 + 49t4 ? 49t5 + 343t6 350
T 2 + 4T 1? t? 14t3 ? 49t5 + 343t6 280
T 2 + 5T 1? t+ 14t3 ? 49t5 + 343t6 308
T 2 + 6T 1? t+ 7t2 ? 14t3 + 49t4 ? 49t5 + 343t6 336
T 3 1 + 16807t10 16808
We remark on some trends observed in the above tables. For all examples here,
the field F/Fq(T ) generated by P (T ) has class number divisible by p = char(F ) if
and only if P (T ) has a linear term and deg(P (T )) ? 2 (or ? 3, for q = 2). For
q = 2, we notice some additional behavior: 4|hF if and only if 2|hF and P (T ) has
no quadratic term, and 8|hF if and only if 4|hF and P (T ) has no cubic term. These
observations are the basis for Theorems 5.1 and 5.8, which show that these trends
hold in general.
Similar calculations to the above can be performed to find the class number
of two types of extensions: the normal closure of one of these fields, or of the com-
positum of multiple such fields. Studying the results of such computation led to
our initial conjecture of Theorems 3.1 and 3.2 (for the normal closure), and Theo-
rems 4.1 and 4.3 (for composita). Since these theorems allow us to exactly determine
the class numbers (indeed, the G(t)) of such extensions in terms of subfields of the
kind we have already presented, we forego listing examples here.
13
Chapter 3: Relation to the Galois closure
Let k = Fq(T ) where q = pr and p is prime. Let ? be a non-zero root of xq+Tx,
and ? be a root of xq +Tx?P (T ), where P (T ) ? Fq[T ] but P (T ) 6= Q(T )p+TQ(T )
for any Q(T ) ? k. We have the following lattice of fields:1
L = k(?, ?)
? G
F = k(?) ? K = k(?)
?= ?
k = Fq(T )
The main objective of this chapter is to prove the following theorems about
Cl0L, the group of degree-0 divisor classes group of L:
Theorem 3.1. The class numbers of L and F are related via hL = hq?1F .
Theorem 3.2. When q = p is prime, Cl0L is isomorphic to a (p? 1)-st power of a
finite abelian group.
In many cases (for instance, when the non-p part of the class number is
squarefree and the p part of the class group is cyclic), these combine to say that
1The edge labels indicate the Galois groups of the corresponding extensions, which will be
described in Section 3.1
14
Cl0 = (Cl0 )p?1L F . However, this is not known in general.
Theorem 3.1 is inspired by one proved by Honda about pure cubic number
fields [14]:
? ?
Theorem 3.3. Let F = Q( 3 n), and L = Q( 3 n, ?3) be its normal closure. Then
hL = h
2
F or h
1
L = h
2
F .3
Theorem 3.2 is an analogue of a recent result of Schoof (which itself was also
inspired by Honda?s). In [25], he proves the following:
Theorem 3.4. Let p > 2 be a regular prime and n ? Z not a p-th power. Suppose
that all prime divisors l 6= p of n are primitive roots mod p. Then the ideal class
?
group ClL of L = Q(? pp, n) and the kernel of the norm map NL/Q(?p) fit into the
exact sequences
0? V ? ker(N p?1L/Q(?p))? A ? 0 and
0? ker(NL/Q(?p))? ClL ? ClQ(?p) ? 0,
(wher)e A is a finite abelian group and V is an Fp-vector space of dimension at most
p?3 2. In particular, if #ClQ(?p) = 1, then ClL/V is a (p? 1)-st power of a finite2
abelian group.
3.1 Proof of the class number relation
Let k, F , K, and L be as described above, and define ? = Gal(L/k), G =
Gal(L/K), and ? = Gal(L/F ) ?= Gal(K/k). These groups have the presentations
? G = ??1, . . . , ?r | ? pi = 1, ?i?j = ?j? ?i? = F+q ,
15
? ? = ?? |?q?1 = 1? ?= F?q ,
? ? = ?? ? ?, ? ? G |????1 = ??(?)? ?= Go? ?.
The isomorphism G ? F+q is given by ?, where ?(?) is the element of Fq
such that ?(?) = ? + ?(?)?, and ? : ?? F?q is the cyclotomic character defined by
?(?) = ?(?)?. Refer to the field diagram above for a depiction of these relationships.
Now, ? can be conveniently realized as the matrix group
{( ) ? }
a x ??? a, x ? Fq, a 6= 0 ,0 1
( ) ( )
with ? ?(?) 0 for ? and ? 1 ?(?)? ? ? ? for ? ? G. Thus the elements
0 1 0 1
with a = 1 are identified with the elements of G, and those with x = 0 with the
elements of ?.
We are interested in four characters of ? which we will show fit an arithmetic
relation. These are:
? ?L, for the regular representation (permutation representation on ?)
? ?K , for the permutation representation on ?/G
? ?F , for the permutation representation on ?/?
? ?k, for the trivial representation (permutation representation on ?/?).
Proposition 3.5.
?L ? ?k = ?K ? ?k + (q ? 1)(?F ? ?k).
16
Proof. We know of course that ?k takes the value 1 on every element of ?, and that
?L takes |?(| = q(q)? 1)(on th)e (identit)y and(0 els)ew(here.? )1
Now, a x 1 x a 0 a 0 1 a x= = . This says that each
0 1 0 1 0 1 0 1 0 1
coset of ?/? can be represented by a unique element of G, and each coset of ?/G
by a unique ele(ment o)f(?. ) ( )( )
We have a x b 0 ab 0 1 (ab)
?1x
= , so an element of ? fixes
0 1 0 1 0 1 0 1 ( )
a coset of a x?/G if and only if a = 1. This says that ?K = |?/G| = q ? 1 for0 1
a = 1, and 0 for a 6= 1. ( )( ) ( )( )
On the other hand, a x 1 y 1 ax+ y a 0= , so an(elemen)t of0 1 0 1 0 1 0 1
? fixes a coset of ?/? if and only if x = y(1 ? a). This means that a x?F =0 1
|?/?| = q for a = 1, x = 0, 0 for a = 1, x 6= 0, and 1 for a 6= 1. ( )
Using the values ascertained above, the relation holds for each element a x
0 1
of ? as follows:
? For a = 1, x = 0: q(q ? 1)? 1 = (q ? 1)? 1 + (q ? 1)(q ? 1).
? For a = 1, x 6= 0: 0? 1 = (q ? 1)? 1 + (q ? 1)(0? 1).
? For a 6= 1: 0? 1 = 0? 1 + (q ? 1)(1? 1).
This arithmetic relation between characters gives rise to a corresponding mul-
tiplicative relation between L-functions, and thus zeta functions [26]:
17
Corollary 3.5.1. Let ?? denote the zeta function for the field ?. Then
( )q?1
?L ?K ? ?F= .
?k ?k ?k
Schmidt [24] gives the residue formula
q1?gh
lim(s? 1)?(s) =
s?1 (q ? 1) log q
and the functional equation
?(1? s) = q(g?1)(2s?1)?(s)
for the zeta function of a function field in positive characteristic, where g and h
denote the genus and class number, respectively2. These combine to give a formula
for the residue of ?(1? s) at 1:
lim(s? 1)?(1? s) = lim(s? h1)q(g?1)(2s?1)?(s) = .
s?1 s?1 (q ? 1) log q
As K and k have trivial class group, applying this equation to the relation in
Corollary 3.5.1 gives hL = hq?1F , completing the proof of Theorem 3.1.
2See Roquette [22] for an English reference summarizing these results, but note a typo in the
numerator of the residue formula (reversing the sign of the exponent).
18
3.2 Proof of class group structure theorem
We now restrict to the case that q = p is prime, and set k = Fp(T ), K = k(?),
F = k(?), and L = k(?, ?). As before, ? = Gal(L/k), G = Gal(L/K), and
? = Gal(L/F ) ?= Gal(K/k). We remark that now G = ?? | ? p = 1? ?= F+p is cy?clic.
Naturally, there is a Galois action of ? on Cl0L. The norm element NG = G ?
of G gives a map Cl0L ? Cl0L that factors through Cl0K , which is trivial. Thus Cl0L
is a module over the group ring Z[?]/(NG), or alternately a module over Z[G]/(NG)
with a twisted action of ? (by which we mean that ? acts on Cl0L in a way that
is consistent with the action of ? on Z[G]/(NG)). Now, Z[G]/(N ) ?G = Z[?p] as a
?-module (because NG is the p-th cyclotomic polynomial evaluated at ?), so we
may freely apply standard facts about ?p to ? . (We opt to keep the notation in
terms of ? rather than ?p, to maintain coherence with the function field setting.)
We proceed by separately considering the p part and the non-p part of Cl0L.
3.2.1 The non-p part of Cl0L
In this section, let M denote the non-p part of the degree-0 divisor class group
of L. The following proposition describes the ?-module structure of M .
Proposition 3.6. The map
? : M? ?Z Z[G]/(NG)?M
? ?
given by m ? [? i]?7 ? ii i i mi is an isomorphism of ?-modules.
19
?
Proof. Suppose first that ? ii mi = 0 (and note tha?t since NG acts trivially, this
sum can be assumed to be over 1 ? i ? p? 1). Then ? i?(?)i mi = 0 for all ?, and
thus for 1 ? j ? p? 1,
? ?
0 = ??j?(?)(1? ? j?(?)) ? i?(?)mi
????? i
= (1? ? j?(?))? (i?j)?(?)mi.
i ???
?
(1 ? ? j?(?))? (i?j)?(?)? acts as p ? 1 + (1 ? NG) = p for i = j, and as 0 for i 6= j,
so this says pmj = 0, and thus mj = 0, for all j (since M has order prime to p).
Therefore ? is injective. ?
Now suppose m ? M . Then for any i, N (? im) = ? i?(?)? ? ?(m) ? M?, and
accordingly,
?p?1 ? ?
??i?(? )(1? ? i?(??)) ? i?(?)?(m)
?i=1 ? ?p?1 ?
= (1? ? i?(? ))? i(?(?)??(??))?(m) ? im(?) for all ?? ? ?.
? i=1
?
(1? ? i?(??))? i(?(?)??(??))i acts as p for ?? = ? and as 0 for ?? =6 ?, so this says that
each p??(m), and in particular pm, is in im(?). Therefore ? is surjective (again
using that M has order prime to p).
Ignoring the module structure gives M ?= (M?)p?1 as abelian groups, which
settles the non-p part of Theorem 3.2.
20
3.2.2 The p part of Cl0L
From this section forward,M will denote the p part of the degree-0 divisor class
group of L. Having only p-power torsion allows us to strengthen the Z[G]/(NG)-
module structure previously described to a Zp[G]/(NG)-module structure, still with
the twisted ?-action as before. A = Zp[G]/(NG) is a discrete valuation ring, and
its maximal ideal is generated by ? ? 1 (just as ?p ? 1 generates the maximal ideal
of Zp[?p]). Furthermore, (? ? 1)p?1 = (p) as ideals of A, which will be key to this
section?s results.
As before, ? acts on ? , and thus on ? ? 1, by the cyclotomic character ? :
?? F?p . We have a filtration of ideals
A ? (? ? 1)A ? (? ? 1)2A ? . . .
with successive quotients F , F (?), F (?2p p p ), . . . , where X(?i) denotes that the
default action of ? on the module X is twisted by the character ?i.
We now prove two results which characterize the structure of A-modules with
particularly ?nice? ?-action:
Lemma 3.7. [25, Prop. 3.1] Let M ? be a finite A-module with twisted ?-action.
Then ? acts trivially on M ?/(? ? 1)M ? if and only if there exist n1, n2, . . . , nt ? 1
such that ?t
M ? ?= A/(? ? 1)niA.
i=1
Proof. Suppose first that the isomorphism holds. Then M ?/(? ? 1)M ? is a direct
21
sum of A/(? ? 1)A ?= Fp terms with trivial ? action.
Conversely, suppose that ? acts trivially on M ?/(? ? 1)M ?. Then the map
M ?? ? (M ?/(? ? 1)M ?)? = M ?/(? ? 1)M ? is surjective (its cokernel is the first ?-
cohomology group of (? ? 1)M ?, which is trivial because ? and M ? have comprime
orders). This says that there are ?-invariant elements v1, . . . , vt which generate
M ? over A, i.e. that there is a surjective map from At ? M ? taking 1 in the i-th
coordinate to vi. By the finiteness of M ?, this descends to a surjective map
?t
? : A/(? ? 1)niA?M ?.
i=1
If ? is not injective?, there is a nonzero element x in the kernel, and we may assume
x = (x1, . . . , xt) ? i(? ?1)ni?1A/(? ?1)niA. Now, ? acts on x by ?m for some m,
but by ?ni?1 on each of these summands, so xi is zero unless ni? 1 = m+ ki(p? 1)
for some ki. Reordering if necessary, let the first s coordinates of x be exactly those
which are nonzero, and choose n1 to be minimal among n1, . . . , ns. For i = 1, . . . , s,
define ?i such that x m kii = (??1) p ?i, and mi ? Z such that mi ? ?i mod (??1)N ,
where N is the maximum of the ni. Notice that ?i, and thus mi, is a unit in A.
We are now able to construct a new map
?t
?? : A/(? ? 1)n1?1A? A/(? ? 1)niA?M ?,
i=2
?
which takes the basis vector e1 = (1, 0, . . . , 0) to
s m pki?k1i=1 i vi, and ei to vi for
22
i 6= 1. This is well-defined because
?s
??((? ? 1)mpk1e ) = (? ? 1)mpki1 mivi
?i=1s ?s
= (? ? 1)mpki?ivi = xivi = ?(x) = 0.
i=1 i=1
Furthermore, ?? is surjective because, by the surjectivity of ?, m1v ?1 ? im(? ), and
m ?1 is invertible. If ? is not injective, we repeat this procedure until we reach a
map that is, at which point we will have found a direct sum of finite quotients of A
which is isomorphic to M ?.
Proposition 3.8. Let M ? be a finite A-module with twisted ?-action, such that ?
acts trivially on M ?/(? ? 1)M ? and by ??1 on M ?[? ? 1]. Then there exists a finite
abelian p-group H such that
M ?= H ?Zp A.
Proof. Suppose M ? ?= A/(? ? 1)nA for some positive integer n. Then M ?[? ? 1] =
(? ? 1)n?1A/(? ? 1)nA ?= F (?n?1p ). By assumption, this requires n = (p? 1)m for
some positive integer m, whereby A/(? ? 1)nA ?= A/pmA ?= Z/pmZ?Zp A.
By the previous lemma, a generalM ? satisfying the condition onM ?/(??1)M ?
is a direct sum of such Z/pmZ?Zp A, proving the proposition.
It remains to show that this proposition can be applied to (a twist of) M . We
will achieve this by exploring the Galois cohomology of Cl0L and related objects.
23
3.2.3 Galois cohomology of Cl0L
In this section, H? i(X) is the i-th Tate G-cohomology group of X. We fix the
notations
? PL, the principal L-divisors
? D0L, the L-divisors of degree 0
? I0L, the ideles of total valuation 0
? C0L = I0 ?L/L , the idele classes of total valuation 0
? UL, the product of the local unit groups UL of L.
We have the following commutative diagram of ?-modules in which the rows
and columns are exact:
1 1 0
? ? ?
1 ? F?p ? L? ? PL ? 0
? ? ?
1 ? U ? I0L L ? D0L ? 0
? ? ?
1 ? U /F? ? C0 ? Cl0L p L L ? 0
? ? ?
1 1 0
24
We will make use of various long exact sequences induced in cohomology by
the above diagram in order to study H? i(Cl0L) for i = ?1, 0. We remark that a
G-cohomology group of an ?-module is an Fp[?]-module (because it is killed by
p and is G-invariant), and any map induced in cohomology by any of the maps
in the diagram is ?-equivariant. Furthermore, since G is a cyclic group, we have
?-isomorphisms H? i(X) ? H? i?2(X)(??1) for every i ? Z and ?-module X, given
by cupping with a generator of H??2(Z) ?= H (Z) ?1 = G ?= Z/pZ(?).
Lemma 3.9. C0 ?L and Fp have trivial Tate G-cohomology.
Proof. The degree map on the idele class group gives rise to the sequence
0? C0L ? CL ? Z? 0.
This gives rise to a long exact sequence which includes
H0(C )? H0L (Z)? H?1(C0L)? H?1(CL),
where the first two terms are standard (i.e. non-Tate) cohomology. We have that
H0(CL) = C
G
L = CK [1, p. 2]. The leftmost map, then, is the degree map on CK
as a subgroup of CL. An idele class of CK which has valuation 1 at an inert prime
and valuation 0 elsewhere maintains this property when extended to CL. Since
G is cyclic, there are (infinitely many) primes inert in L/K by the Chebotarev
density theorem, and thus the map H0(CL) ? H0(Z) is surjective. We also have
H?1(CL) = 0 [1, p. 19], and so H?1(C0L) = 0.
25
Passing the first exact sequence to Tate cohomology and recognizing that
H?0(C ) ?= Z/pZ [1, p. 19], H?0(Z) ?= Z/pZ, and H??1(Z) ?= H?1L (Z) = 0, we have that
H?0(C0L) = 0 as well.
As for F?p , it is a finite module of order prime to the order of G, and so has
trivial Tate G-cohomology.
Applying this lemma to the long exact sequences induced by the leftmost
column and bottom row of the diagram immediately gives:
Corollary 3.9.1. H??1(Cl0L) ?= H?0(UL) and H?0(Cl0 ? 1L) = H? (UL).
Lemma 3.10. ? acts trivially on H?1(UL) and H?2(UL).
Proof. We write ` for a prime of k, l for a prime of K above `, and L for a prime
of L above l. We have decomposition groups ?L = D(L/`), GL = D(L/l), and
?l = D(l/`). For each i, H? i(UL) can be expressed as a product of local cohomology
groups:
?
H? i(U iL) = H? ( UL)
L?of L?
= H? i( UL)
?l of K?L|l ?
= H? i( UL)
` ra?m in L?l|` L|l
= H? i(GL, UL),
` ram in L l|`
?
with the last equality by applying Shapiro?s Lemma to GL|l UL = IndG UL. Fur-L
thermore, as discussed in Section 1.1, any prime that ramifies in L ramifies totally
26
in K, which means that ?l = ? and any action it has on H?1(UL) is on a single
summand H? i(GL, UL). Thus for i = 1, 2, it is sufficient to show that ? acts trivially
on H? i(GL, UL).
We look first at i = 1. Since G and ? have coprime orders, the inflation-
restriction sequence
0? H?1(?l, Ul)? H?1(?L, U )? H?1L (G ?lL, UL) ? 0
is exact. Now, from local class field theory, H?1(GL, UL) is cyclic of order equal
to the ramification index eL/l [1, p. 9], and likewise H?1(? ?L, UL) = Z/eL/`Z and
H?1(?l, Ul) ?= Z/el/`Z. This forces H?1(GL, U ?l 1L) = H? (GL, UL), and so the action of
? = ?l on H?1(UL) is trivial.
Next we take i = 2. Again by the coprime orders of G and ?, and also using
that H?1(GL, LL) = 0 by Hilbert?s Theorem 90, the sequence
0? H?2(?l, Kl)? H?2(?L, LL)? H?2(GL, L )?lL ? 0
is exact. But local class field theory gives us that these cohomology groups are dual
to the decomposition groups that define them [1, p. 9], and so by order considera-
tions, we must have H?2(GL, L ?l 2L) = H? (GL, LL). Since the inclusion-induced map
H?2(G ,U )? H?2L L (GL, LL) is injective (its kernel is H?1(GL,Z), which is trivial), we
conclude that H?2(UL) is ?-invariant as well.
We are now ready to connect the cohomological theory back to M , the p-part
27
of Cl0L.
Corollary 3.10.1. ? acts trivially on M [? ? 1] and via ? on M/(? ? 1)M .
Proof. By Corollary 3.9.1 and the fact that NG kills M , we have
M [? ? 1] ?= H?0(Cl0 ) ?L = H?1(UL) and
M/(? ? 1)M ?= H??1(Cl0 ? 0 ? 2L) = H? (UL) = H? (UL)(?),
which have the claimed actions by Lemma 3.10.
Finally, we can prove the p part of our result:
Proposition 3.11. M is a (p? 1)-st power of some finite abelian p-group.
Proof. We consider the twist M ? = M(??1). The previous result says that ? acts
trivially on M ?/(? ? 1)M ? and by ??1 on M ?[? ? 1]. Thus Proposition 3.8 may be
applied to M ?. As abelian groups, M ?= M ?, so we are done.
Combining this with Proposition 3.6, we have that the p part and non-p part of
Cl0L are each the (p?1)-st power of an abelian group, and so the proof of Theorem 3.2
is complete.
28
Chapter 4: Class numbers of composite fields
We now consider composita of multiple members of this family of fields. Let
{P (T )}ni i=1 be a set of linearly independent polynomials in k = Fq(T ) whose span
does not contain a polynomial of the form P (T ) = Q(T )q + TQ(T ). Then the
extensions Fi, each given by a fixed root of xq + Tx? Pi(T ), are independent, i.e.,
their compositum K is a degree qn
n
extension of k. The N = q ?1? monic linearq 1
combinations of the Pi determine N distinct subfields of degree q over k. We will
show in this chapter that the class number of K is exactly the product of the class
numbers of these subfields.
The corresponding situation has been studied in the number field setting, espe-
cially for quadratic and cubic extensions. Herglotz [12] proves that the class number
? ?
of Q( m1, . . . mn) is almost the product of the class numbers of its quadratic sub-
fields, with a possible deficiency in the 2-part of the class group related to the unit
group of the composite field. Parry [18] finds a similar result for bicubic fields: the
? ?
class number of Q( 3 m , 31 m2) is the product of the class numbers of its cubic sub-
fields, possibly divided by some power of 3. For general p, Ohta [17] (following an
earlier result of Nakagoshi [16]) proves a similar statement about the non-p part of
the class groups in an abelian number field extension of type (p, p, . . . , p).
29
4.1 The case q = 2
Let {Pi(T )}ni=1 be a set of polynomials in k = F2(T ) whose span does not
contain any polynomial of the form Q(T )2 + TQ(T ). Then the extensions Fi each
given by a root of x2 + Tx ? Pi(T ) are independent, i.e., their compositum K is a
degree 2n extension of k. The Galois group of K/k is G ?= (Z/2Z)n.
There are N = 2n? 1 quadratic subfields1 of K, which correspond to the non-
trivial linear combinations of the Pi (and the index-2 subgroups Gi of G of which
they are the fixed fields). The situation is illustrated in the following diagram:
K = k
G1 GN
G
F1 F2 . . . FN
k = F2(T )
?
Theorem 4.1. The class numbers of these fields are related by hK =
N
i=1 hF .i
We write ?F to denote the character of the permutation representation of G
acting on G/Gal(K/F ).
?
Lemma 4.2. ?K ? ? Nk = i=1(?F ? ?k)i
Proof. Clearly ?k(G) = 1, ?K(1) = |G| = 2n, and ?K(G\1) = 0.
Now, each ?F is the character of the permutation representation of G oni
G/Gi. Since G is abelian, ?F (Gi) = 2 and ?F (G\Gi) = 0. Furthermore, any giveni i
1We denote the complementary subfields Fn+1, . . . , FN , but their ordering is unimportant.
30
non-zero element of G is contained in 2n?1?1 of the subgroups Gi and not contained
in 2n?1 of them. Thus for g 6= 1,
?N ? ?
(? n?1 n?1F (g)? ?k(g)) = 1 + ?1 = 2 ? 1? 2 = ?1 = ?K(g)? ?k(g),i
i=1 g?Gi g?/Gi
and for g = 1,
?N
(? nF (1)? ?k(1)) = N(2? 1) = 2 ? 1 = ?K(1)? ?k(1).i
i=1
As in Section 3.1, this character relation gives a multiplicative relation of class
numbers, and since hk = 1, this proves Theorem 4.1.
4.2 The case q 6= 2
Let q =6 2 be a power of a prime p, and let {Pi(T )}ni=1 be a set of polynomials
in k = Fq(T ) whose span does not contain a polynomial of the form Q(T )q+TQ(T ).
Then the extensions Fi, each given by a fixed root of xq+Tx?Pi(T ), are independent.
n
We note there are N = q ?1? subfields (including the Fi) of K determined by theq 1
projective linear combinations of the Pi.
The Galois closure of each Fi (and thus their compositum) is formed by ad-
joining a non-zero root ? of xq + Tx = 0. The Galois closure L of K has Galois
group ? = Go? ?, where G ?= (F+)n, ? ?= F?q q , and ? acts on G via the cyclotomic
31
character ?. The fields and Galois groups are illustrated in the following diagram:
L = K(?)
G1
?
GN o ?
F1(?) K
k(?) F1 F2 . . . FN
?= ?
k = Fp(T )
?
Theorem 4.3. h = NK i=1 hFi
Lemma 4.4. hL = h
q?1
K
Proof. Following the proof of Proposition 3.5, we get the relations
?F1(?) = ?k(?) + (q ? 1)(?F1 ? ?k)
?F1F2(?) = ?F1(?) + (q ? 1)(?F1F2 ? ?F1)
. . .
?L = ?F1...Fn?1(?) + (q ? 1)(?K ? ?F1...Fn?1).
Recalling that hk = hk(?) = 1, converting to class numbers and successively substi-
tuting gives
h q?1F1(?) = hF1
32
( )q?1
hF F
h = h 1 2F1F2(?) F1(?) = h
q?1
h F1F2F1
. . .
( )q?1
hK
hL = h
q?1
F1...Fn?1(?) = hh KF1...Fn?1
?
Lemma 4.5. ?L ? ? Nk(?) = (q ? 1) i=1(?F ? ?k).i
Proof. Clearly ?k(?) = 1, ?L(1) = |?| = qn(q ? 1), and ?L(?\1) = 0.
?k(?) is the permutation representation of ? on ?/G ?= ?. But G acts trivially
on ?, from which we see ?k(?)(G) = q ? 1, and ?k(?)(?\G) = 0.
For a given i, ?F is the permutation representation of ? on ?/(Gio?). Thei
cosets can be represented by {? s}s?Fq for a fixed ? ? G\Gi. Since G is abelian, it is
easy to see that ?F (Gi) = q and ?F (G\Gi) = 0.i i
Now we consider an element of ?\G, which can be represented by ? t??? , with
t ? Fq, 1 6= ? ? ?, and ?? ? Gi. Its action sends the coset represented by ? s to
? t+s?(?). This is equal to ? s exactly when s = t? , and so ?F (?\G) = 1.1 ?(?) i
We can now show the character relation holds for each element of ?. For 1:
?N
(q ? 1) (?F (1)? ?k(1)) = (q ? 1)N(q ? 1)i
i=1
= (q ? 1)(qn ? 1)
= qn(q ? 1)? (q ? 1)
= ?L(1)? ?k(?)(1).
33
For g ? ?\G:
?N
(q ? 1) (?F (g)? ?k(g)) = (q ? 1)N(1? 1)i
i=1
= 0? 0
= ?L(g)? ?k(?)(g).
For g ? G, g 6 n?1= 1, we note that g is contained in q ?1 = N?1? of the subgroupsq 1 q
Gi, and not contained in qn?1 of them. Thus:
? ?? ? ?N
(q ? 1) (?F (g)? ?k(g)) = (q ? 1)? (q ? 1) + ?1?i
i=1 g?Gi g?/Gi
= (q ? 1)(qn?1 ? 1? qn?1)
= 0? (q ? 1)
= ?L(G)? ?k(?)(g).
Therefore the relation holds for all of ?.
As before, we?convert this to a multiplicative class number relation, which
simplifies to h = ( NL i=1 h
q?1
F ) . Combining this with Lemma 4.4 completes thei
proof of Theorem 4.3.
34
Chapter 5: p-divisibility of the class number
In this chapter, we consider the question of when the class number of the field
generated over Fq(T ) by a root of xq + Tx? P (T ) is divisible by the characteristic
p. The analogous question for quadratic number fields has been studied (albeit
originally in the language of genera of quadratic forms) dating back to Gauss? genus
theory [6]. In modern terms, Gauss proves that the 2-torsion of the narrow class
group has rank t? 1, generated by the t ramified primes (see e.g. [5, 13]).
Red?i and Reichardt extend this theory to study the 2n-torsion of the class
group of quadratic number fields. Reichardt [21] characterizes the rank of the 2n-
torsion in terms of the number of ?D-decompositions of the n-th kind?, the number
of ways of splitting the discriminantD into the product of two discriminants meeting
certain criteria. They also derive residue conditions for a D-decomposition to be of
the second kind [20], and of the third kind (for certain cases) [19]. One of the cases
about which the most is known is D = ?4p, which gives an imaginary quadratic
field in which either one or two primes ramify (and thus cyclic 2-class group). In
this case, the 2- and 4- divisibility conditions reduce to:
35
2|h(?4p) ? p = x2 + 4y2
4|h(?4p) ? p = x2 + 8y2
and Barrucand and Cohn [3] show that
8|h(?4p) ? p = x2 + 32y2
(see [15] for an overview). We will see that this case closely resembles ours when
q = 2, where T and 1/T play the role of of p and 2, respectively, in terms of their
ramification / splitting behavior. We give conditions on the polynomial P (T ) that
determine when the class number of the corresponding field is divisible by 2, 4, and
8, and show that the 2-class group is always cyclic.
The more general question of when p divides the class number of a pure degree-
p field has not been studied as thoroughly, but Honda [14] determines the conditions
for p = 3. Given our somewhat ?nicer? setting, we are able to give a complete
description of when the class number of the extension of Fq(T ) corresponding to
P (T ) is divisible by the characteristic.
5.1 The case q = 2
Let k = F2(T ) and P (T ) ? Ok = F2[T ] such that P (T ) =6 Q(T )2 + TQ(T ) for
any Q(T ) ? Ok. By a change of variables, it suffices to consider P (T ) to comprise
36
only terms of odd degree, except possibly a quadratic term.
Theorem 5.1. Let F = k(?), where ? is a root of x2 + Tx ? P (T ), with P (T ) in
the form described above.1 We denote by Cl0F its degree-0 divisor class group, and
hF = |Cl0F |. Then:
1. The 2-Sylow subgroup of Cl0F is cyclic.
2. 2|hF if and only if ordT (P (T )? T ) ? 2 and deg(P (T )) ? 3.
3. 4|hF if and only if ordT (P (T )? T ) ? 3.
4. 8|hF if and only if ordT (P (T )? T ) ? 5.
As we will show momentarily, the infinite prime 1/T of k ramifies in F in all but
a few cases. When it does, Rosen [23] shows that there is an isomorphism between
Cl0F and the ideal class group of OF , the integral closure of Ok in F (alternatively,
the ring of functions with poles only at infinity), given by:
? Cl
0 ?
F = C?l(OF )
np[p]?7 pnp
? ? p?=6 p?
np[p]? ( n npp)[p?]?[ p ,
p=6 p?
where p? is the prime above 1/T . He also shows that this permits a characterization
of the Hilbert class field H/F as the maximal abelian unramified extension of F in
1Despite being in characteristic 2, we will use subtraction in our notation when it is helpful for
giving intuition or remaining consistent with the more general case.
37
which p? splits completely, which enjoys many of the same properties as the Hilbert
class field of a number field.
The following lemmas identify the behavior of T and 1/T in F/k in terms of
P (T ). We reiterate that both deg(P (T )) and ordT (P (T )) are odd or equal to 2.
Lemma 5.2. T ramifies in F if ordT (P (T )) = 1, is inert if ordT (P (T )) = 2, and
splits if ordT (P (T )) > 2.
Proof. Suppose ordT (P (T )) = 1. Then x2 +Tx?P (T ) is an Eisenstein polynomial
at T , and thus T ramifies in the extension it generates (which is F ).
Now suppose ordT (P (T )) = 2. Then the change of variables Tx ?[ x and
division by T 2 changes the polynomial defining F to x2 + x ? P (T )2 , which hasT
discriminant 1. Thus denoting a root by ?, the integral closure of Ok in F is given
by OF = F2[T, ?]. Since ?2 + ? ? 1 mod T ,
O /TO ?F F = F4,
so T is inert in F .
Finally, suppose ordT (P (T )) > 2. The change of variables Tx ? [ x and
division by T 2 changes the polynomial defining F to x2 +x? P (T )2 . This polynomialT
splits into (distinct) factors mod T , so T splits in F .
Lemma 5.3. 1/T ramifies in F if deg(P (T )) > 2, is inert if deg(P (T )) = 2, and
splits if deg(P (T )) = 1 (i.e., P (T ) = T ).
d+1
Proof. Suppose d = deg(P (T )) > 2. The change of variables T 2 x ? [ x and
38
( ) d?1
division by T d+1 changes the polynomial defining F to x2 + 1 2 x ? P (T )
T T d+1
. This
is Eisenstein at 1/T , and thus 1/T ramifies.
Both cases when d ? 2 follow from the corresponding facts for T in the previous
lemma, noting that the change of variables T 2x?[ x gives x2 + 1 x? P (T )4 .T T
Lemma 5.4. If deg(P (T )) > 2, then O?L = {1}.
Proof. [This also follows from the analogue of the S-unit theorem, stated e.g. in [23];
we provide this elementary proof for the sake of self-containment.]
If ordT (P (T )) = 1, T ramifies in F , so OL = F2[T, ?] is the full integral closure
of Ok in L, as it has discriminant ?T 2. If ordT (P (T )) ? 2, then OL = F2[T, ? ], asT
in Lemma 5.2.
For the first case, suppose f(T ) + ?g(T ) ? O?L with f, g 6= 0. Then
1 = N(f(T ) + ?g(T )) = f(T )2 + P (T )g(T )2 + Tf(T )g(T ).
Let d,m, n denote the degrees of P, f, g respectively. To achieve the necessary can-
cellation, we must have one of the following:
1. 2m = 2n+ d ? m+ n+ 1
2. 2m = m+ n+ 1 ? 2n+ d
3. 2n+ d = m+ n+ 1 ? 2m.
(1) is impossible since d is assumed to be odd. Case (2) would imply m = n + 1,
and thus 2n+ 2 ? 2n+ d, contradicting the condition on d. Finally, (3) would give
39
n + 1 ? m, and thus 2n + d > 2n + 2 ? m + n + 1, a contradiction. Therefore one
of f, g must be 0, and clearly this requires f = 1, g = 0.
In the case that ordT (P (T )) ? 2, a similar degree argument shows that if
? P (T )
1 = N(f(T ) + g(T )) = f(T )2 + g(T )2 + f(T )g(T ),
T T 2
we must have f = 1, g = 0.
Proof of Theorem 5.1(1,2).
In the cases when 1/T does not ramify in F (P (T ) = T , T 2, or T 2 + T ),
F is genus 0, so the theorem is true. We henceforth assume 1/T ramifies in F
(i.e., deg(P (T )) > 2). This permits us to use the isomorphism Cl0 ?F = Cl(OF ) and
consider the behavior of finite ideals.
Suppose I ? I 2O such that I ? PO , and let ? be the non-trivial element ofF F
Gal(F/k). Then ?(I)/I = N(I)/I2 = (a), with N(a) = 1. By Hilbert?s Theorem
90, a = b/?(b) for some b ? F , and so the class of I contains the fixed ideal J = bI.
Thus J is a product of (principal) ideals of Ok and primes of OF above ramified
primes. This shows that the 2-torsion of Cl(OF ) is generated by the ramified primes.
Since T is the only prime which may ramify, the 2-Sylow subgroup is cyclic, and
trivial if T is unramified.
Now, suppose T does ramify. If the prime above T is the principal ideal (b),
we would thus have (b)2 = TOF , which by Lemma 5.4 requires b2 = T . But this
contradicts the separability of F , and thus the prime above T is non-principal.
Therefore 2|hF in this case, which occurs when ordT (P (T )? T ) ? 2.
40
The Hilbert class field theory of Rosen [23] tells us that a rational (degree-
1) prime p of OF splits completely in a subextension H ? of the Hilbert class field
H if and only if the class [p] is contained in the subgroup of Cl(OF ) isomorphic to
Gal(H/H ?). This will be critical for the following proposition, which is the analogue
for our situation of one of Reichardt [21, Thm. 1], and follows approximately the
same proof.
Proposition 5.5. Let F be a field as above whose class group has 2-rank 1. Then
the 2n-rank is 1 if and only if there exists an unramified cyclic extension Hn?1/F of
degree 2n?1 in which the primes of F above T and 1/T split completely.
Proof. Suppose such an extension Hn?1/F exists, and let H ? Hn?1 be the Hilbert
class field of F , so that Cl(O ?F ) = Gal(H/F ). Let Gn?1 denote the subgroup of
Cl(OF ) corresponding to Gal(H/Hn?1). Since the prime p above T splits completely
in Hn?1, its class lies in Gn?1. Recalling that this class is the unique 2-torsion
element of Cl(OF ), we must have that Gn?1 has 2-rank 1. Since Cl(OF )/Gn? ?1 =
Z/2n?1Z, Cl(OF ) must therefore have 2n rank 1.
For the converse, suppose H contains a subfield Hn which is cyclic of degree
2n over F , and denote by Gn the subgroup of Cl(OF ) corresponding to Gal(H/Hn).
Since p2 = (T ) is principal, its class is trivial in the quotient Cl(OF )/Gn. Thus
[p] is contained in a subgroup Gn?1 of Cl(OF ) such that G ?n?1/Gn = Z/2Z. Since
[p] ? Gn?1, p splits completely in the fixed field Hn?1 of the corresponding subgroup
of Gal(H/F ), and Gal(Hn?1/F ) ?= Cl(O ? n?1F )/Gn?1 = Z/2 Z.
Proof of Theorem 5.1(3).
41
Suppose ordT (P (T )?T ) ? 2 and deg(P (T )) ? 3. Let F1 be the field generated
by a root ? of x2 + Tx ? (P (T ) ? T ), and F2 by a root of x2 + Tx ? T . By
Lemmas 5.2 and 5.3, T is unramified and 1/T is ramified in F1, while T is ramified
and 1/T splits in F2. Thus the compositum K = F1F2 is an unramified degree-2
extension of F in which the (unique) prime above 1/T splits.
Since the 2-rank of the class group is 1, K is the unique quadratic subextension
of the Hilbert class field of F . In the case that ordT (P (T ) ? T ) = 2, T is inert in
K/F , so by Proposition 5.5, 4 - hF . On the other hand, if ordT (P (T ) ? T ) ? 3,
then T splits in K/F , so 4|hF .
We now construct explicitly the degree-4 subextension of the Hilbert class field
of F (when it exists), which we will use to prove the last statement of Theorem 5.1.
Let ? be a root of x2 +Tx? (P (T )?T ), where ordT (P (T )?T ) ? 3. Then T
splits in F1 = k(?), and at one of the primes above T , we can express ? as a powerT
series in T (in particular, a power series in P (T )?T2 ):T
(?)2 ? P (T )? T
+ =
T T T 2
?? ( )? 2k? P (T ) T
= ,
T T 2
k=0
and the conjugate ?+T = 1 + ? . At the other prime above T , these power series are
T T
reversed.
Let s be an integer greater than deg(P (T ))?1 . The power series for ? above
2 T
contains finitely many terms of degree less than s. We denote by f(T ) the polynomial
42
comprising these terms. Finally, we define K? = F1(?), where ? is a root of
2 ? ?/T ? f(T )x + x ,
T s
and similarly the conjugate extension K ? ??+T = F1(? ), where ? is a root of
x2 + x? (? + T )/T ? f(T ) .
T s
We remark that the particular fields these define depend on the choice of
s. We will soon choose a specific s, but the following two lemmas hold for any
s > deg(P (T ))?1 . We also note that this construction requires ordT (P (T ) ? T ) ? 3,2
and we continue to assume that P (T ) has no terms of even degree.
Lemma 5.6. The prime of F1 above 1/T splits in both K? and K?+T .
Proof. Let d = deg(P (T )). Then
( )( )
?/T (? + T )/T P (T )? T
T (d?1)/2 T (d?
= ,
1)/2 T d+1
so the prime of F1 above 1/T is ( ?/T 1(d?1)/2 , ). By construction, s > max{d?1 , deg(f)},T T 2
so ?/T?f(T )s (and its conjugate) have positive valuation at this prime. Therefore modT
this prime, the polynomial x2 + x ? ?/T?f(T )s splits (as does its conjugate), so theT
prime splits in both K? and K?+T .
Lemma 5.7. Let s be as before, and suppose further that it is a power of 2. One
prime of F1 above T is unramified in K? and ramified in K?+T , and the other prime
43
above T does the reverse. In particular, if ordT (P (T ) ? T ) = 3, the unramified
primes are inert, and if ordT (P (T )? T ) > 3, they split.
? ( )2k
Proof. Consider the prime above T where ? = ? P (T )?Tk=0 2 . The right-handT T
side contains 2kT , for all k ? 0, as a summand if ordT (P (T )?T ) = 3, and contains no
terms of degree a power of two otherwise. Furthermore, by construction ?/T ?f(T )
has only terms of degree ? s. Thus we see that
?/T ? f(T )
T s
has valuation 0 at the prime in question if ordT (P (T )?T ) = 3, and positive valuation
otherwise, corresponding to the prime being inert and splitting, respectively, in K?.
On the other hand, the polynomial defining K?+T is
( )
2 ? (? + T )/T ? f(T ) 2 ? ?/T ? f(T ) 1x + x = x + x + .
T s T s T s
Because s is a power of 2, a change of variables allows us to replace 1s by 1 in theT T
last term, which thus has valuation -1 at the prime in question. Therefore the prime
is ramified2 in K?+T .
At the other prime above T , the power series associated to ? and ?+T are
T T
reversed, and so the behavior in K? and K?+T is reversed.
We are now prepared to show that when s is the least power of 2 greater than
deg(P (T ))?1 , the composite field L = K?K?+T is an unramified degree-4 extension2
2Since a change of variables can make the polynomial Eisenstein; see also [10].
44
of F in which the prime above 1/T splits completely, and complete the proof of
Theorem 5.1.
Proof of Theorem 5.1(4).
Let L = K?K?+T , with K?, K?+T defined by s, the least power of 2 greater
than deg(P (T ))?1 . The linearity of x2 + x means that L contains a solution to
2
2 (? + T )/T ? f(T ) ? ?/T ? f(T ) 1x + x = = .
T s T s T s
By a change of variables, this is equivalent to containing a root of x2 + Tx ? T ,
so L contains F2. Since it is defined as an extension of F1, it thus contains K =
F1F2. More specifically, K is the third and final intermediate subfield of L/F1. The
situation is illustrated in the diagram below.
L
K? K?+T K
F1 = k(?) F2 = k(?) F = k(?)
k = F2(T )
We recall the splitting/ramification behavior of T and 1/T in the lower right
diamond from Lemmas 5.2 and 5.3 and the proof of Theorem 5.1(3): the (unique)
prime of F above each of T and 1/T splits in K/F ; the primes of F1 above T ramify
and the prime above 1/T splits in K/F1.
45
From the preceding two lemmas, we also have that each prime of F1 above
T is unramified in one of K?/F1 and K?+T/F1, and the prime above 1/T splits in
both. Thus the primes above T are unramified in L/K, and the primes above 1/T
split in L/K. Therefore L is an unramified degree-4 extension of F in which the
prime above 1/T splits completely3.
Furthermore, the previous lemma shows that the primes above T split com-
pletely in L/F if and only if ordT (P (T ) ? T ) > 3. By Proposition 5.5, this means
that 8|hF if and only if ordT (P (T )? T ) > 3, which completes the proof.
5.2 The case q 6= 2
Let q be a power of a prime p. Let k = Fq(T ) and P (T ) ? Ok = Fq[T ] such
that P (T ) 6= Q(T )q+TQ(T ) for any Q(T ) ? Ok. By a change of variables, it suffices
to consider P (T ) to be monic and comprise only terms of degree not divisible by q.
Theorem 5.8. Let F = k(?), where ? is a root of xq+Tx?P (T ), with P (T ) in the
form described above. Then p|hF if and only if deg(P (T )) ? 2 and ordT (P (T )) = 1.
It will be convenient at times to work with the Galois closure F (?) of F , where
?q?1 = ?T . By Theorem 3.1, hF (?) = hq?1F , so the question of p-divisibility is the
same for each, and since the degrees [F : k] = [F (?) : k(?)] = q and [k(?) : k] =
[F (?) : F ] = q ? 1 are relatively prime, it is easy to relate ramification in F/k to
ramification in F (?)/k(?) (or /k). Obviously (T ) = (?)q?1 and (1/T ) = ( 1 )q?1, i.e.
?
both ramify completely in k(?)/k.
3Since the 2-rank of the class group of F is 1, it is the unique such extension, and cyclic.
46
Lemma 5.9. T is totally ramified in F if ordT (P (T )) = 1, otherwise it is unramified
(and ? splits completely in F (?)/k(?)).
Proof. If ordT (P (T )) = 1, xq + Tx ? P (T ) is Eisenstein at T , and thus ramifies
completely.
Suppose ordT (P (T )) ? 2. Then F (?)/k(?) is given by xq??q?1x?P (??q?1),
q?1
or by a change of variables, the Artin-Schreier polynomial xq ? x ? P (?? ) . This
?q
polynomial splits into (distinct) linear factors mod ?, and thus ? splits completely
in F (?) (and T is unramified in F/k).
Lemma 5.10. 1/T is totally ramified in F if deg(P (T )) ? 2, otherwise it is un-
ramified (and 1/? splits completely in F (?)/k(?)).
Proof. The only case when deg(P (T )) < 2 is P (T ) = ?T , ? ? F?q . In this case, a
change of variables gives xq?x? 1? as the polynomial which generates the extension?
F (?)/k(?). As in the previous lemma, this polynomial splits mod 1/?, and so 1/?
splits completely in F (?) (and 1/T is unramified in F/k).
Now suppose that ordT (P (T )) ? 2, so that by the previous lemma, ? splits
completely in F (?)/k(?), and consider the subfield of F (?) fixed by the inertia
group of a prime above 1/?. This is an unramified extension of k(?) in which ?
splits completely. But k(?) has class number 1, so the extension is trivial, and thus
1/? must be totally ramified in F (?)/k(?) (and so must 1/T in F/k).
Finally, suppose ordT (P (T )) = 1 and deg(P (T )) ? 2. Then the compositum
of F with the extension given by xq + Tx? T (which, from above, is unramified at
1/T ) contains a root of xq+Tx?(P (T )??T ), where ?T is the linear term of P (T ).
47
By the above, this generates an extension in which 1/T has ramification index q,
and therefore 1/T must ramify completely in F/k.
Proof of Theorem 5.8.
F has genus 0 if deg(P (T )) < 2, and so the theorem holds. We henceforth
assume deg(P (T )) ? 2, which means that 1/T is totally ramified in F (?)/k, and
thus that Cl(O ? 0 ? 0F (?)) = ClF (?) and Cl(OF ) = ClF .
Suppose ordT (P (T )) > 1, meaning ? is unramified in F (?)/k(?), and that
there exists a degree-p unramified extension K/F (?) in which the prime above 1/?
splits completely. Then the subfield of K fixed by the decomposition group of (a
prime above) 1/? is a nontrivial extension of k(?) which is unramified everywhere
and split completely at 1/?. But k(?) is genus 0, so such an extension must be
trivial. Thus p does not divide hF (?).
Now suppose that ordT (P (T )) = 1. Then ordT (P (T ) ? ?T ) ? 2 for some
? ? F?q . Let F1 be the field generated by xq + Tx ? (P (T ) ? ?T ) and F2 by
xq + Tx ? T . Then (similar to the proof of Theorem 5.1(3)), the compositum
F1F2(?) is a degree-q unramified extension of F (?) in which the (unique) prime
above 1/T splits completely. Therefore q|hF (?).
Appealing to Theorem 3.1, this completes the proof of Theorem 5.8.
Now, in the case that q = p, we can say a bit more about the structure of Cl0F ,
using a method similar to the proof of Theorem 5.1(1). We continue to assume that
1/T ramifies completely in F (?)/k.
Suppose I ? IO such that Ip ? PO , and let ? be a generator for G =F (?) F (?)
48
Gal(F (?)/k(?)). Then for some g ? Z[G], we have (? ? 1)p?1(I) = N(I)/g(Ip) ?
P , i.e. (? ? 1)p?1O kills Cl(OF (?))[p]. For 0 ? k ? p ? 2, we have the followingF (?)
exact sequence, in which we denote A = Cl(OF (?))[p]:
0? A[? ? 1] ? (? ? 1)kA? (? ? 1)k ????A ?1 (? ? 1)kA.
If the class of I is in A[? ? 1], then ?(I)/I = (a) with N(a) = ? ? O? ?k(?) = Fp (all
units of Ok(?) are constant by the S-unit theorem, since there is only one prime at
infinity). By Hilbert?s Theorem 90, a/? = b/?(b) for some b ? F (?), and so the class
of I contains the fixed ideal J = bI. This means J is a product of (principal) ideals
of Ok(?) and primes of OF (?) above those which ramify in F (?)/k(?). Consequently,
A[??1] is cyclic, and dimFq((??1)kA/(??1)k+1A) = dimFq(A[??1]?(??1)kA) ? 1
for each k. Since (? ? 1)p?1A is trivial, this means A has rank at most p ? 1. (In
fact, by Theorem 3.2, A must have rank exactly 0 or p? 1.)
Because the prime above T is totally ramified in F (?)/F , the map Cl(OF (?))?
Cl(OF ) induced by the norm is onto (see [23, Prop 2.2]). Therefore the rank of the
p-Sylow subgroup of Cl(O ) ?F = Cl0F is at most p? 1.
49
Chapter 6: Future Work
Some interesting questions remain, both about this family of fields and gener-
alizations of it. We now explore some of these possible directions for future research
and the accompanying challenges they present.
For one, do the class number relations of Theorems 3.1 and 4.3 extend to
structural relations between class groups? For instance, if F/Fp(T ) is a member
of this family of fields and L is its normal closure, can we say in general that
Cl0 = (Cl0 )p?1L F ? By the discussion following the proof of Theorem 5.8, this relation
would imply that the p-part of Cl0F is always cyclic. Thus it may be possible to find
a counterexample by determining the class group structure of various F with p2|hF .
An obvious generalization is to higher-rank Carlitz modules. In this work we
only considered ?T and a root ? of C(T )(x) ? P (T ) = xq + Tx ? P (T ) as the
?
analogues of roots of ?p and p n, respectively. This can certainly be expanded
to instead consider ?M and a root of C(M)(x) ? P (T ), for a general polynomial
M ? k. Many of the techniques presented here can likely be applied to some extent
(especially for M monic and irreducible, or a power of such, the cases which have
been most studied in the cyclotomic function field literature). However, a significant
obstacle is that while k(?T ) has trivial class group (a property which is key to most
50
of our results), this is not true of k(?M) in general. Nonetheless, we expect our
results to largely generalize, albeit to less simple statements.
Finally, the theory of cyclotomic function fields leads naturally to Iwasawa
theoretic questions about towers of extensions. The extension ??i=1k(?M i)/k(?M),
for M monic and irreducible, can be seen as an analogue of the Zp-extension
Q(?p?)/Q(?p). The growth of the p-part of the class group in this tower, and
congruence relations on the real and relative parts of the class group, have been
studied [2,8]. But as in the number field case, ??i=1k(?M i) can be used to construct
a ?cyclotomic Zp-extension?1 of any function field over Fq(T ). It would be interesting
to study such towers over members of the family of fields we consider here, and we
hope that the results of this work will be useful to that end.
1We remark that unlike in the number field case, the Galois group of such an extension is not
in fact isomorphic to Zp, but rather to a countably infinite product of such groups.
51
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