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Title: 
Corings in the category of rings 

Physical Description: 
vi, 61 leaves. : illus. ; 28 cm. 

Language: 
English 

Creator: 
Brooks, Burrow Penn, 1943 

Publication Date: 
1970 

Copyright Date: 
1970 
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Subject: 
Categories (Mathematics) ( lcsh ) Rings (Algebra) ( lcsh ) Mathematics thesis Ph. D Dissertations, Academic  Mathematics  UF 

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bibliography ( marcgt ) nonfiction ( marcgt ) 
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Thesis  University of Florida. 

Bibliography: 
Bibliography: leaf 60. 

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Manuscript copy. 

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Vita. 
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UF00097708 

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VID00001 

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University of Florida 

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University of Florida 

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alephbibnum  000570673 oclc  13718547 notis  ACZ7653 

Full Text 
CORINGS IN THE CATEGORY OF RINGS
By
BURROW PENN BROOKS, JR.
A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
1970
To my mother
ACKNOWLEDGEMENTS
The author wishes to express his gratitude and ap
preciation to Dr. W. Edwin Clark for his direction and
patience during the preparation of this dissertation.
iii
TABLE OF CONTENTS
Page
ACKNOWLEDGEMENTS ....***. ..**........*..*.............. iii
ABSTRACT ****************.............................. v
INTRODUCTION .......................................... 1
Chapter
I. CORINGS ***..............*
II. ADJOINT FUNCTORS *********
III. STRUCTURE OF CORINGS **
IV. THE CATEGORY OF CORINGS *
V. THE CATEGORY OF SEMIGROUPS
BIBLIOGRAPHY ........................
BIOGRAPHICAL SKETCH ..*............*
..................
..................
***** * * * **G 0
Abstract of Dissertation Presented to the Graduate Council of
the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy
CORINGS IN THE CATEGORY OF RINGS
By
Burrow P. Brooks, Jr.
August, 1970
Chairman: Dr. W. Edwin Clark
Major Department: Mathematics
A coring in a category G is an object A of G together
with morphisms a,m:A>A*A inducing binary operations on
the morphism set [A,X] such that with these induced opera
tions, [A,X] becomes a ring, denoted by [A,x] ,m for any
a,m
object X in G. In Chapter I, it is shown that many familiar
functors on R, the category of rings, can be represented as
[A,] where (A,a,m) is a coring in R.
a,m
The second chapter presents several examples of functors
on the category of rings which have left adjoints. It also
proves the result, a special case of a theorem of Freyd,
that a functor on R has a left adjoint if and only if it is
representable as [A,] for some coring (A,a,m) in R.
a,m
A coring (C,a,m) in R is said to be standard if C is
the free ring on a set X and a(x) = ul(x) + u2(x) for each
x in X where ulu2 are the injections into the coproduct C*C.
Chapter III deals with the types of functors representable
by standard corings and characterizes those coring functors
which can be expressed as R for some ring R.
The fourth chapter discusses the category of corings in
R and its properties.
The last chapter is concerned with cosemigroups in the
category of semigroups. It is shown that the category of
semigroups has exactly two autoequivalences, the identity
functor and the opposite functor.
INTRODUCTION
A binary operation on an object A of a category is a
morphism m:AxA>A. Dually, if G is a category with finite
coproducts, then a binary cooperation on an object A is a
morphism m:A>A*A, from A to the coproduct of A with itself.
A binary cooperation m induces a binary operation on the
set of morphisms [A,X] for any object X in the category.
The cooperation is said to be associative commutativee,
etc.) if the induced operation on [A,X] has the property for
every X in Q. A coring in G is an object A together with
cooperations a,m on A such that with the operations induced
by a and m, [A,X] is a ring for any object X in G. If we
denote this ring by [A,X] then [A,] becomes a functor
a,m a,m
from G to R, the category of rings.
Many functors on R, for example, the functor which sends
a ring to its n x n matrix ring, can be represented as [A,]a
a,mn
for an appropriate coring (A,a,m) in R. Also, the functors
on which have left adjoints are precisely those which are
representable by coring functors.
This dissertation deals with several aspects of corings
and functors with left adjoints on the category of rings.
The first two chapters show examples of corings and adjoint
functors on R and prove the fact stated abovethat functors
with left adjoints are the representable functors. Chapter
III deals with the structure of corings in R and shows the
relationship between coring functors and functors which can
be expressed as R for some ring R. The fourth chapter
is concerned with the category of corings and its properties.
The last chapter looks at cosemigroups in the category of
semigroups and proves the result, parallel to a result of
Clark for the category of rings, that the category of semi
groups has exactly two automorphisms, the identity functor
and the opposite functor.
CHAPTER I
CORINGS
A binary operation on a set A is a mapping from AxA>A.
If G is a category with finite products, then a binary
operation on an object A in the category would be a morphism
m:AxAA. Dually, if G is a category with finite coprod
ucts, then a binary cooperation on an object A is a morphism
m:A>A*A, from A to the coproduct of A with itself.
If m:AA*A is a cooperation in the category G, and
if X is any object in G, then m induces a binary operation
on the set of morphisms [A,X] in the following manner:
Suppose f,g:A>X. Let ul,u2:AA*A be the injections into
the coproduct, and denote by the unique morphism from
A*A to X such that u = f and u2 = g. Then define
fg = m. The cooperation m is said to be associative
commutativee, have an identity, etc.) if the induced operation
on [A,X] has this property for every object X in G. If
a,m:AA*A are cooperations inducing operations +, on
[A,X] for all X in G, and if ([A,X],+,) is a ring for all
X in G, then (A,a,m) is called a coring in the category G.
Analagously, one could define a cogroup, cosemigroup, co
monoid, etc. We will denote the ring ([A,X],+,.) by
[A,X]am
a,m
If (A,a,m) is a coring, and if a:XY is a morphism in
G, then define [A,]:[A,X] a,[A,Y]a,m by [A,a](f) = af
a,m am 
for f in [A,X].
(1.1) Lemma: Let (A,a,m) be a coring in G. Then
[A,]:GR is a functor, where 2 denotes the category of
rings.
Proof: Suppose O:X>Y is an Gmorphism. We must show
that [A,a] is actually a ring homomorphism. Suppose
f,g E [A,X]
a,m
[A,a](f + g) = [A,a](a) = aa = a =
af + ag = [A,a] (f) + [A,a] (g).
A similar argument holds for multiplication, so [A,a] is
indeed a ring homomorphism.
If 1X:X>X is the identity morphism, then [A,l X (f)
1Xf = f for any f in [A,X] so [A,l ]a,m is the identity
X a,m xam
map on [A,X]a,m. If a:X>Y and 3:Y>Z are Gmorphisms,
a,m
and if f E [A,X] a,m then [A,B] (f) = 8af = EA,B][A,] (f).
Thus, [A,Sa] = [A,3][A,a], and [A,] is a functor.
A functor T:G>R is said to be reuresentable if it is
naturally equivalent to [A,]a,m for some coring (A,a,m) in
SS a,m
CG.
Let S be a semigroup with 0. If R is a ring, then
5
define the ring R0[S] as follows:
R0S] = ( ir sr 6ER and only a finite number of the
0 s /O S S
rs's are nonzero}.
(Er s) + (Zt s) = E(rs + t )s
s S s s
(Er s).(ts s) = Fus, where u = r t
RO s] is called the contracted semigroup ring of S over R.
If S is any semigroup with 0 with the property that every
nonzero element has only a finite number of factors, then
the ring RE[[S]] can be defined just like RO[S], except that
the restriction that finitely many of the r 's be zero is
removed. R [ES]] is called the contracted power series
semigroup ring of S over R. If S is finite, then RO[CS]] and
RO[S] coincide.
If S is the semigroup of n x n matrix units, then R0ES]
2
is R the ring of n x n matrices over R. If S = L0,l,x,x ,
...), the infinite cyclic semigroup with identity and zero,
then R0[ES]] is the ring of formal power series over R,
while R [S] is the ring of polynomials in one indeterminate
over R.
Let R be the category of rings, and let S be a semigroup
with zero such that every nonzero element has only a finite
number of factors. Then there is a functor T from S to R
defined by T (R) = R [Es]], and if f:R>R', then
S 0
6
TS(f) :TS(R)TS(R') by TS(f) (Erss) = 'Ef(r )s. If S is any
semigroup, then there is a functor F such that F (R) = R [S]
and F (f) (Er s) = Ef(r )s. Note that if S is finite, then
S S S
TS = FS'
(1.2) A functor of the type TS can always be represen
ted as CA,] for an appropriate coring (A,a,m). Let
a,m
A = W(S [(0), the free ring on the set of nonzero elements
of S. To define a and m, it is sufficient to define them
only on the generators of A, the nonzero elements of S.
Define:
a(s) = u1(s) + u2(s)
m(s) = Er u (r)u (s)
rt=s 1 2
= 0 if s has no factors.
To show that TS and [A,]a,m are naturally equivalent,
we define the following natural equivalence: If B is a ring,
define BC:[A,B] a>B C[S]] by (f) = sES 7(0 f(s)s. We
must show that TB is a ring homomorphism.
1B(f + g) = fB(a) = Ea(s)s
= E(ul(s) + u2(s))s = E(f(s) + g(s))s
= Ef(s)s + Eg(s)s= nB(f) + 1B(g).
1g(fg) = 1Bg(m) = Em(s)s
= sES3)
SsES } () rts u(r)u2(t))s
sESe(0 ( s f(r)g(t))s
rt=s
= ( Sf(s)s) (Eg(s)s).
So TB is a ring homomorphism.
IB is onetoone, since if B (f) = I (g), then
Ef(s)s = Zg(s)s; f(s) = g(s) for all s, and thus f = g.
TIis onto, for if bs s E B[[CS]], there is a ring homo
morphism f:A>B such that f(s) = b for all s. Then
TI (f) = Eb s.
B s
Thus, IB is an isomorphism for each B. Up until this
point, we didn't really know [A,B] is a ring, but since
a,m
[A,Ba,m has two operations on it and with these operations
is isomorphic to a ring, it must also be a ring, and so
(A,a,m) is a coring.
To see that T is natural, we must show that the follow
ing diagram is commutative for all a:B>C
B
[A, B]am  ) T (B)
[A,] TS(
[AC a,m TS(C)
But if f E [A,B a,m then
TS(a) B (f) = TS(a)(Ef(s)s) = Zaf(s)s;
nC[A, a] (f) = C(af) = Zef(s)s.
Thus, the two functors are naturally equivalent.
We will show later (2.9) that if S is an infinite semi
group, then a functor of the type FS cannot be representable.
If S is the semigroup (0,1l under multiplication, then
it is easy to see that R0[S] 2 R for any ring R, and that T
is thus naturally equivalent to the identity functor. Thus,
the identity functor on R is representable, and, using the
construction for the coring described above, we get:
(1.3) Proposition: Let G = W(([x), the free ring on
one generator, and let a, m be defined by:
a(x) = ul(x) + u2(x)
m(x) = u (x)u2(x).
Then (G,a,m) is a coring and [G,] is naturally equivalent
a,m
to the identity functor on R.
Functors on the category of rings other than semigroup
functors can be represented in a similar manner. For example,
let T be the functor which takes a ring R to the additive
group R S R with multiplication (r,s) (r',s') =
(rr' ss', sr' + rs') (complexnumberlike multiplication).
9
Then T can be represented as C,] a,m where C = W(xl,x2),
the free ring on two generators, and coaddition and comul
tiplication are given by:
a(xi) = u (xi) + u2(xi), i = 1,2.
m(x ) = u1(x )u2(x1) u (x2)u2(x2)
m(x2) = ul(xl)u2(x2) + u (x2)u2(xl) .
The functor which sends a ring to its ring of quaternions
can also be represented in a similar manner. This will be
generalized later.
It would be of interest to characterize categorically
some of the more common functors on the category of rings,
for example, the functor which takes a ring R into R This
functor is of the form TS (or FS) where S is the semigroup
Ce..) U (0) of matrix units. The next proposition shows
that this functor is cokernel preserving.
(1.4) Proposition: Let S be a semigroup with 0. Then
FS is cokernel preserving if and only if S = S.
Proof: If f:R>R' is a ring homomorphism, then
coker(f) can be realized as R'/I, where I = id, the
ideal of R' generated by the image of f.
Suppose FS is cokernel preserving, and suppose x is in
Consider i:Z Q, t inclusion map from the
S S Consider i:Z>Q, the inclusion map from the
integers into the rational numbers. Coker(i) = 0, so
coker (T (i)) = TS(0) = 0. If I is the ideal of Q0SI]
generated by im(TS(i)), then Q0[SI/I = 0, so Q0[S] = I.
x E I (otherwise x would have to be in S2), and we have a
contradiction. Thus S = S2
Conversely, suppose S = S TS is cokernel preserving
if R'[S]/id (R'/id)[S] for any f:R>R'.
Since (R'/id)[S] V R'ES]/id s], it is suffi
cient to show id = (id)[S]. Suppose
x E id. Then
x = T(f)(Zrisi) + (Et!s.)T(f)(Et.si) +
+ T(f) (Zu.s.) Zu s.) + (Zr!s.)T(f) (v.si) (Ev's.),
where each r.,t.,u.,v. ER, and r',ti,u!,v. E R'. Then
n n
x = s S, f(u )u)s. + (s 2 r f(v )v)s..
ial s.sS. sf 'j)k) + i=l(SjSsi 3 k i
SK i ks 1
Noting that each coefficient is in id, we see that
x E id[S]
n
Now suppose x E (id)[S]. Then x = ilwis where
w. E ( ) Then
i id
w. = f(ai) + b!f(b.) + f(c.)c! + a!f(d.)d!
1 1 1 1 1 1 1 1 1
2
where a.,b.,c.,d E R and a!,b!,c!,d! E R'. Since S = S,
1 1 t 1 1 1 1 1
s. = t.u. for each i, and u. v.z.
1 1 1 1 1 1
n n n n
x = . f(a)si + S bjf(b.)si + i C f(ci)c. + .2 a.f(d.)d!
1=1 1 1 = 1 1 1=1 1 = 1
n n
= f(a.)si + ((bt.)(f(bi.)u) +
n n
+ 2 (f(c.)t.)(c!u.) + .2 (ati )f(d.)v.idz.
n n
= iiT(f)(a.iS ) + .il(b!i) (T(f)(bui)) +
J1 1 (l I i ()) +b1u1)
n n
+i2 T(f) (c.it)c!u. + i (a!ti)T(f) (dvi) (dzi.)
But each term of each summation is in id, so
x E id.
The functor R>R has the property that it takes simple
rings into simple rings. It follows easily from the next
proposition that TS cannot have this property if S is infi
nite.
(1.5) Proposition: Let S be an infinite nontrivial
semigroup with 0 with the property that any nonzero element
has at most a finite number of factors. Then S is not
simple.
Proof: Suppose S is simple. Since S is nontrivial,
S2 ; (0O. S is an ideal, so S = S, and SSS = S. For any
x / 0, consider T = (s E SISsS = ([0}. T is clearly an ideal
of S. T / S, for then SSS = (0], so T = 0. Thus SxS 0 [O)
for any x / 0, and SxS is an ideal, so SxS = S for any x / 0.
Now consider y in S. y E SxS for all x r 0, so for each x,
12
there exist s t such that y = s (xt ). But y has only a
x x x x
finite number of factors, so only a finite number of these
factorizations can be distinct. But then there must be x
such that xtx = yt for infinitely many y, a contradiction.
Corollary: If S is as in the above proposition, TS
does not take simple rings into simple rings.
Proof: If I is an ideal of S, then for any ring R,
RO[[CI3 is an ideal of ROCIs]].
CHAPTER II
ADJOINT FUNCTORS
One of the most fruitful areas of study in category
theory has been the theory of adjoint functors. In this
section we will investigate those functors on the category
of rings which have left adjoints.
If T:G9 and S:a>G are covariant functors, then S
is the left adjoint for T if there exists a natural equiva
lence of setvalued bifunctors
nB,A:[S(B),A] [B,T(A)].
We will make use of the following characterization:
(2.1) Theorem: A functor T:G>R has a left adjoint if
and only if for any object B in 3 there exists S(B) in G and
v :BTS(B) such that for any cp:BT(A), there exists a
unique '' :S(B)A such that
B B TS(B)
}T(CP'
T(A)
is commutative.
For a proof of this result, the reader is referred to
[1, p. 119].
14
The functors on the category of rings which have left
adjoints are precisely those which are representable. In
fact, Freyd [2] has stated the following more general result:
(2.2) Let G be a complete category, I/ an algebraic
variety, and T:G7 a covariant functor. T is representable
if and only if T has a left adjoint.
A proof discovered independently of Freyd is included
here for the special case of a functor from the category of
rings to itself. The proof is in two parts.
(2.3) Theorem: Suppose T:R> has a left adjoint S.
If (C,a,m) is a coring in R, then (S(C),S(a),S(m)) is also a
coring in R. In fact, [S(C),]S(a),S(m) [C,T()]a,m
Proof: Since S has a right adjoint, S preserves co
products, so S(a),S(m):S(C)>S(C*C) are actually coopera
tions. From the adjoint situation there is a natural equiva
lence n:[S(C),] [C,T()] of set valued functors. We will
show that for any ring R, R_ is a ring homomorphism. It
is known from the theory of adjoint functors that there is a
natural transformation Y:ST>1 such that if f:CT(R),
then R (f) S(f). Now suppose f,g E [C,T(R)]. Then
1R (f + g) = R 1(a) = YRS(a) = YRS(a)
1
and similarly for multiplication. Thus, n is a ring iso
morphism, [S(C),R]S(a) ,Sm) is a ring for any R, and
(S(C),S(a),S(m)) is a coring. T:[S(C)]>[C,T()] is a
natural equivalence in R.
15
Corollary: If T:R>R has a left adjoint S, then T is
representable.
Proof: Let (G,a,m) be the coring described in Proposi
tion (1.3) representing the identity functor. Then
[S(G),]S(a),S(m) [G,T()]a,m T.
(2.4) Theorem: Let (C,a,m) be a coring in 2. Then
[C,] :a,  has a left adjoint.
a,m
Proof: We will verify the conditions for Theorem (2.1)
and for any ring R we will construct S(R). Let R be any
ring, and as the first candidate for S(R) consider LC,
rER
w uth
with u :C> C the rt injection into the coproduct.
rER
There is a natural mapping from R>[C, CC defined by r u
rER
but this will not be a ring homomorphism in general. There
fore, consider the ideal I of \ \C generated by all elements
rER
of the form:
ur+s(c) a(c)
or urs(c) m(c)
where r,s E R and c E C. Let P:UC>_C / I be the canon
rER rER
ical quotient map, and define S(R) = C/I and define
rER /
VR:R>C,S(R)] by vR(r) = Pu To see that v is a ring
homomorphi, note that if c C, then
homomorphism, note that if c E C, then
(vR(r) + VR(s))(c) = a(c) =
= a(c) = Pa(c) =
= Pu+s(c) = vR(r + s)(c)
so vR(r) + v R(s) = v R(r + s) and similarly for multiplica
tion.
We must now show that if :R>C,B] is a ring homo
a,m
morphism, then there is a unique ':S(R)B such that
VR
R R )[c,S(R)]
[c,B]
commutes.
Since for each r E R, (r):C>B, there is a mapping
T.: C>B such that u = (r) for each r. If I c keri,
rER r
then T induces :L JC >B. Consider a generator for I,
rR /R
say ur+s(c) a(c).
(u r+(c) a(c)) = uu (c) Ta(c)
r+sr s r+s r
= (r + s)(c) a(c)
= (r + s) (c) < (r), (s)>a(c)
= 0
since t is a homomoruhism. Thus, I c kerg and we can
define 5 :L C  B b '(p(x)) T(x). To see that $'
rER / I
makes the appropriate diagram commute, note that if r E R,
[C, ]vR(r) = 'Pur = ur = (r)
If *:S(R)>B also makes the diagram commute, then
5*Pu = 'Pu for all r, so 5*P = P, and since P is onto,
r r
=* = Thus, is unique, and, by Theorem (2.1), [C,1
a,m
has a left adjoint.
We now know that a representable functor has a left
adjoint, but, in general, it is quite difficult to determine
precisely what the left adjoint of a particular functor
looks like. The following are a few examples where the left
adjoint can be rather easily found.
(2.5) Example: Lez G be a category with products and
coproducts, and let I be a set. Then the functor T on G de
fined by T(A) = TTA and T(f) = jTf has a left adjoint,
iEI iEI
namely S, where S(A) = LIA and S(f) = \f.
iE I iE I
Proof: For any objects A,B in G, define
TB A:S(B),A1>[B,T(A)1 as follows: If f:LjB>A, then
B iE I
fu :BA for each i. These morphisms induce a unique mor
phism (fui):B>TTA. Then define B,A (f) = (fui). That ri
iEI
is a natural equivalence of bifunctors is an easy exercise
in category theory.
18
If G = R, the category of rings, then the functor T de
fined above is naturally equivalent to the contracted semi
group functor TS, where S is a semigroup (with zero) of
orthogonal idempotents indexed by the set I.
(2.6) Example: Define the functor T on R by T(R) = Ro,
where R0has the same additive group as R, but has trivial
multiplication. Then T has a left adjoint S where S(A) =
F(A/A2) and F(G) denotes the tensor ring on the abelian
group G.
Proof: The category of abelian groups, Ab, can be con
sidered as a subcategory of R, namely the full subcategory
consisting of all rings with trivial multiplication. The
functor T can then be factored as
U I
where U is the forgetful functor and I is the inclusion
functor. F, the functor which sends an abelian group to its
tensor ring, is the left adjoint of U (See [3]). We will
show that I has a left adjoint J where J(R) = R/R2 and if
f:RR' then J(f):R/R2R'/R'2 is defined by J(f)(r + R2)
2
f(r) + R'2. Let R be a ring and G an abelian group. Define
G,R :CJ(R),G]>[R,I(G)] by G,R(a)(r) = a(r + R2) for any
Ea [J(R),G]. TG,R is clearly onetoone. It is also onto,
G,R
for suppose 3:R>G. Since G has trivial multiplication,
R2 c ker3, so 8:R/R2>G defined by 3(r + R2) = S(r) is well
defined and G,R (3) = 8. It is easily verified that is
G,R
19
natural in both variables so that J is the left adjoint of I.
The left adjoint of T = IU is FJ, which is the functor S
defined above.
If we note that the functor T is naturally equivalent
to the contracted semigroup functor TS where S is the two
element semigroup with zero multiplication, then we see that
the coring representing T is (W(x),a,m) where W(x) is the
free ring on one generator x with a(x) = ul(x) + u2(x) and
m(x) = 0.
(2.7) Example: For each fixed natural number n, define
a functor Tn on 2 by letting T (R) denote the subring of R
of all elements of additive order n. If f:R>R', then
T (f) = fT (R). Then T has a left adjoint S where
n n n n
S (R) = R/nR and if f:R>R', then S (f)(r + nR) = f(r)+nR'
Proof: If R and R' are rings, define DR',R:[Sn(R),R' ]9
[R,Tn(R')] by i R,R(f) (r) = f(r + nR). It is clear that
1R',R is onetoone. It is also onto, for suppose
h:RTn(R'). h(nr) = nh(r) = 0, so nR c kerh, so define
f:R/nR>R' by f(r + nR) = h(r). Then R,R (f) = h. That
R' ,R
T is natural in both variables is easily verified.
The coring which represents T will be discussed later
(3.12).
(2.8) Example: Let S be an infinite semigroup with 0
and let FS be the functor such that FS(R) = R [S] and
F (f)(Zr s) = Zf(rs)s. Then FS does not have a left adjoint.
Proof: We will show that FS does not preserve products.
Suppose pi:.ifZ>Z is the product of a countable collection
CO
of copies of the integers, and qi:il zC0[S]>ZO[S] is the
product of copies of Z0[S]. If FS preserves products, then
FS(Pi) :(i Z) 0[S] >[S] is also a product of copies of
ZO[s] and there exists an isomorphism :i l0[S](i Z)0[s]
such that F (p.i) = q. for all i. Let s1,s'2... be a count
able subset of S. Then x = (s1,s2,s3 E = rS], and
FS (pi)(x) = qi(x) = s. for each i. If (x) =
n n
S(n ,n ,.)t, then F (p x) = n..t. = s. for each
i1 j2 )j 1 1s1 1 ] 1
i. So t. = s. and n.. = 6.. for each i. But this is im
1 1 ]l ]l
possible since there is only a finite number of the t. and
1
an infinite number of the s.. Thus, F cannot preserve
products.
products.
CHAPTER III
STRUCTURE OF CORINGS
Kan [4] showed that the only comonoids in the category
of groups are free groups with a more or less trivial co
multiplication defined as follows: There exists a free
basis X for the comonoid C such that m(x) = ul(x)u2(x) for
each x E X.
Things are not so simple for corings in 2, however. In
the last chapter we saw examples of corings which are free
rings with nontrivial comultiplications. There are also
examples of corings in R which are not free. For example,
if (C,a,m) is a coring where C is free, and if S is the
n
functor defined in Example (2.8), then by Theorem (2.4),
(Sn (C),Sn (a),S (m)) is also a coring. But S (C) = C/nC
which cannot be a free ring since its additive group is
torsion.
We now have a large number of examples of corings which
are not free rings. However, if we let GK denote the
category of Kalgebras where K is a field, then it is not
known if there exist examples of nonfree corings in GK. In
fact, even though we do not know if a coabelian group in GK
must be free, we will show that a coabelian group in GK
must satisfy many of the properties of a free algebra.
First we need to know something about coproducts in GK'
(3.1) Construction of coproducts in G : This construc
tion is alluded to by Cohn [3] and outlined in more detail
by Bergman [5]. This construction can be used to construct
coproducts in R if the rings involved have free additive
groups.
Let Al and A2 be Kalgebras, and let X. be a basis for
A. as a Kvector space. Assuming A1 and A2 are disjoint,
let X1 U X2 = (x ,vEN}. For each finite sequence
J = (v ,...,v ) of elements of N, define the monomial
1 n
xj = x x V...x Two monomials xI, xj will be equal if
and only if the sequences I and J are equal. We will say
that a monomial x = x x ...x is proper if no two suc
1 2 n
cessive x belong to the same Xk, k=l,2. Let A be the K
1
vector space whose basis consists of all proper monomials.
We will define a multiplication on basis elements of A
as follows: Let x = x x ...x, x = x x ...xV be
S 1 42 m 1 V2 n
proper monomials.
Case 1: If x and x belong to different X., then
m 1
x 1= x x x n ... X
1 2 m 1 2 n
Case 2: If x and x belong to the same X., then
m 1
xm 1 i R i so x and x can be multiplied to give
m 1
s
Sx = j k.x.. Then define
4 M j=l jj
m 1 s
s
xx = k x x ..x xx x ...x
xI J j=l jPi1 X2 '' mlXX3 2 xV3 V n
With this multiplication A will be a Kalgebra and will
be the coproduct of Al and A2 in GKC
(3.2) Lemma: Let G be a category with zero maps, and
let (C,m) be a cogroup in G. Then if A is any object in G,
the zero map from C to A is the identity element of [C,A] m
Proof: Let 0:CA be the zero map, and let e:CA be
the identity element of [C,A] Since e is the identity of
m
[C,A] we know 0Oe = 0. But 00 = <0,0>m = 0 (since <0,0>
is the zero map from CC to A). Thus, e = 00, e = 0.
is the zero map from C*C to A). Thus, 0e = 00, so e = 0.
(3.3) Lemma: If (C,m) is a comonoid in
then m must be a monomorphism.
Proof: Since (C,m) is a comonoid in G,
monoid and must have an identity e such that
any f E [C,C]. In particular, if 1:C)C is
then m = 1. Since 1 is a monomorphism,
a monomorphism.
a category G,
[C,c]m is a
m = f for
the identity map
m must also be
(3.4) Lemma: If m:C>C*C is a cooperation in a
category G, then m is commutative if and only if Tm = m
where 7 = :C*C>C*C.
24
Proof: If m is commutative, then [C,C*C] is commuta
tive, so ul'u2 = u2'u1 or m = m. But
is the identity map on C*C, so m = Tm.
Conversely, suppose m = Tm. Let X be any object in G
and let f,g be elements of [C,Xi. Then fg = m =
= Tm = m = <u2 u>m =
= m = gf.
Thus m is commutative.
A free ring has a degree function deg defined on it
satisfying the following properties.
1. deg(x) is a positive integer if x 4 0; deg(0) = (
2. deg(xy) < max(deg(x),deg(y)}.
3. deg(xy) = deg(x) + deg(y).
A ring with such a degree function is called a valuation
ring.
(3.5) Proposition: Let (C,a) be a coabelian group in
GK where K is a field. Then C is a valuation ring with no
elements of degree zero.
Proof: Let [x ,vEN] be a Kvector space basis for C.
C*C contains two disjoint copies of C, so let [xl,v,vEN3 be
a basis for the first copy, and let (x2 ,vvENJ be a basis
for the second. Then u.(x ) = x. i, i = 1,2. Then C*C has
as a basis the set of all proper monomials in Y =
(x1J J\EN] U (x2,VJvN
[xl~ !VN UX~ "xIE3
25
A degree function deg can be defined on C as follows:
If c is a nonzero element of C, since a is a monomorphism
n
(Lemma (3.6)), a(c) A 0. a(c) = i.k.x where each x is
i 1
is a proper monomial from Y. Define deg(c) to be the maxi
mal length of the monomials x where the length of a
1
monomial x x ...x is just p. It is clear from the
1 2 p
vector space addition in C*C that deg (cd) r max(deg(c),
deg(d)}.
It remains to be shown that deg (cd) = deg(c) + deg(d).
Let c,d be nonzero elements of C, and suppose deg(c) = p,
n m
deg(d) = q. a(c) = k.x a(d) = .k.'x. Then
i= j=1 i J 3
one of the x say x = xi v xi ...xi must have length
i 1 1 2 pp
p, and one of the x say x = x jllx22..qq must have
Jj 2
length q. If i 1 jl, then x. Vand x. come from dif
p p1 1p p il
ferent copies of C in C*C, so x xJ = XilV ...x. Vx p* ..x .P
11 pp 1 1 Jqq
and has length p+q. On the other hand, suppose i = j .
Then xi and x are from the same copy of C and, when
multiplied, x x would not give a monomial of length p+q.
I J
However, a is commutative and by Lemma (3.4), Ta = a.
7(Xj) = X, = 1 2... where j' / ji for i=l,...q.
(In other words, to get x replace a factor of x of the
J j
form xlP by x2y and vice versa.) Since Ta = a, a(c) must
contain xj, as one of its summands. xIx =
x x. .. x.i x. x x. which has length
1 1 1 2 pp 11 22 333" qq
p + q. In either case (ip J jl, ip = jl) a(c)a(d) yields
a monomial of length p + q. It is easy to see that xIxJ
could not be obtained by multiplying two monomials XKXL of
lengths < p,q respectively unless xK = xI and xj = xL.
Thus, xIxJ (or xIx in case i = j1) will not be cancelled
out by another monomial of the same length. Thus,
deg(cd) = p + q, and deg is a valuation. By the way deg is
defined, there can be no elements of C of degree 0.
Corollary: A coabelian group in GK, K a field, can
have no zero divisors or idempotents.
Note: The above result holds for corings C in the
category of rings as long as the additive group of C is a
free abelian group.
Let (C,a,m) be a coring in R where C is the free ring
on some set X. We say that the coaddition is standard if
a(x) = ul(x) + u2(x) for each x in X. Note that all the
examples previously mentioned of corings on free rings
(corings representing semigroup functors, etc.) have had
standard coaddition. In fact, it is conjectured that any
27
coaddition on a free ring must be standard (for some suitable
free basis of the ring).
If C is the free ring on the set X = (x ,vEN), then C*C
can be thought of as the free ring on X U Y where Y =
([y,vEN} and x 1 Y = 0. The coproduct injections are de
fined by ul(x ) = x, u2 (x ) = y for each v E N.
The following proposition shows the structure of co
multiplication in standard corings.
(3.6) Proposition: If (C,a,m) is a coring in R where C
is free on X = (x ,vEN} and a is standard, then m must have
the following form for each v E N:
k
m(x) = i n np
V i=l niPi
where either p. = x y, or pi = y Vx and n. E Z.
1 1 1 i
Proof: m(x ) is in the free algebra on X U Y, so m(x )
is a "word" in the x 's and y 's. We will denote m(x ) by
w (x~,yV). If A is any ring and f,g:CA, then
(f + g) (x) = a(x ) = f(x ) + g(x )
(fg)(xU) = m(x ) = w (f(x ),g(x )).
If we consider the ring [C,C]a,m and the fact that 1 C0 = 0,
then we see that wP( (C(x ),0(x )) = w (x ,0) = 0. But
w (x ,0) consists of precisely those terms of w (x ,y )
which do not contain any elements from Y as factors. Since
28
w (x ,0) = 0, we must conclude that each term in the "word"
w (x ,y ) must contain at least one of the y (Or else
w = 0 which satisfies the conclusion of the proposition.)
A similar argument shows that each term must contain at
least one of the x .
We have now shown that each term of w (x ,y ) contains
at least one of the x. and one of the y All that remains
is to show that each term does not contain more than one.
Suppose that a term of w (x ,y,) contains more than one of
the x for example. Let V = v ,VEN), and let R be the free
ring on X U Y U V. Define flf2 ,f3:C>R by fl(x ) = x ,
f2(x ) = y, f3(x ) = v for each v. Since C,R a,m is a
ring, the distributive laws hold, and for each 4 E N,
((fl + f2.f 3)(x) = (flf3 + f2f3) (x)
w((fl + f2) (x),f3(x )) = w (fl(x ),f3 (x)) + w(f2(x) ,f3(x))
w (x + y V) = w (x ,v ) + w (y%,v)). *
To get the left hand side of *, replace each x in w (x ,yv)
by x + y and replace each yV by v But if a term in
w (x ,y ) contains more than one of the x when they are
replaced by the appropriate xV + y. and multiplied, some
terms would result which would contain at least one x one
y and one v (For example, suppose w (x ,yv) contains
the term xv y x Substituting, we get
1 2 3
(xl + Y )v, (x3 + y ), and expanding results in the term
1 1 2 3 3
xv V 2Y 3 among others.) It is easily seen that two terms
containing an xv, a y and a v cannot cancel each other
out. The right hand side cannot contain any terms with an
x y and v Thus is impossible and we must conclude
that no term in w (x ,y ) can contain more than one x and,
by a similar argument, could contain no more than one of the
y.. Thus, either w (x ,y ) = 0 or each term contains
exactly one of the x, and one of the y and w (x ,y ) must
be of the required form.
The functor A plays an important role in the theory
of adjoint functors in the category of abelian groups. The
tensor product of rings does not have the same role in the
category of rings, but it does have some relation to adjoint
functors. In the following we will investigate the relation
of tensor product of rings to adjoint functors on the
category of rings.
(3.7) Lenrna: Let (C,a,m) be a coring where C is free
on (xl...x n and a is standard. Let Z be the ring of
integers, and for i = 1,...,n, define f.:C>Z by
1
30
f.(x.) = 6.. Then, as an abelian group, [C,Z] is free
1 J 3 a,m
on [flf2 ... fn
Proof: Let G be the free abelian group on the genera
tors v1,v2 ...,v Define :CC,Z] >G by 5(h) =
1 2 n a,m
n
i=h(x.)v.. That 5 is onetoone and onto is clear from the
i=l 1 1
fact that C is free. We claim that is an additive iso
morphism:
n
(h + g) = s(a) = i. a(xi)v =
n n
= El(u (x.) + u2(xi))v = l (h(x ) + g(x.))v
n n
= i*h(x )vi + i g(x)v. = 5(h) + 5(g).
i=l i i i=l i 1
We now know that [C,z] is a free abelian group, and
a,m
a set of generators is (v.)}, i=1,2,...,n. But
n n
E = f.v = V.
(f) = 1f.(xv. j=l ij j i
i
So f (vi ), and [C,Z]a is free on fl f2'"" f n
Note: In all that follows, all tensor products are
over Z, the ring of integers.
(3.8) Prooosition: Let (C,a,m) be a coring where C is
free on xlX2 ". xn a is standard, and
n n
m(x ) = j k = zijkl(x )u2 (Yk) for each i, where zijk Z.
Then the functor :C,] is naturally equivalent to
a,m
c,zroof: From Lemma (3.7) we know that an element of
a,m
Proof: From Lemma (3.7) we know that any element of
[C,Z] can be written as:
a,m
m n m n
[ z..f.) r.]= .E (z..f. r.)]
j=1 1=1 j1i 3 =1 1=1 31 i
n m n m
= i l j Z (z.j.f. r.) = [E f. z.j r.]
Sj= 31 1 3 i=1 l 1 31 3
n m n
= f. = z r i r. '
i=1 i 3=1 31 =
We need to show that for any ring R there is a natural
isomorphismR $_:c z] R[C,R] If x E [c,z] R,
isomorphism iR :C,Z] a,m R>CR a,m If x CZ a,m R,
R a,m a,m a,m
then R(x) is a function in CC,R] and it is sufficient to
define it for the basis elements of C. Define ~ by
R
n
~ (.i zf. r)(x.) = z.r. Is is routine to verify that
Ri i 1 1 3 3 R
is well defined and is an additive homomorphism.
n
Note that f.f = iz. f. since
j k i=l ij3k i
(fj.fk (xi) = m(xi) =
n n
= ( z (x )u2(x))
n n
= v F1 =i zi f (x )fk(x) = Zijk
n
= ( E z.. f. )(x.).
i=l ijk i i
Also note that if y is an arbitrary element of [C,Z] a R,
a,m
n
then y can be written as .1 f. 9 r., and R is the function
11 1 1 R
from C to R defined by R (y)(xi) = r..
We now show that 5R is a multiplicative homomorphism.
If x is one of the basis elements of C, then
P
n n
R (i fi ri)(Zif r.')] (x) =
n n
=R[j (fj~j r)(fk rk') (Xp
n n
R j=1 k=l (fjk rjk') (Xp)
n n n
E E ( z r r rjr
R[j=l k=li zijk i k') (p
n n n
= [ (f z zr. r) (X)
R 1= i j=l k=1 zijkl rk'i (p
n n
= 2 Z rr
j=l k=l Zpjkrjrk
n n
SR(i fi 0 ri) R(i l f ti r.') (x) =
Ri=1 1 1 Ri=l 1 p
n n
= < R(i2 f. r.)A(. f. 9 r.')>m(xp)
= < (i ifr ), f ( if r )>(. I clz j(x )u2
R i=l i R i=l i i k=l zpjkU1 (Xj2xk))
n n n n
r.2 f 2
j=l k=l ZpjkR Ri 1 if i=li)r xk)(
n n
Sz r.r '.
j=1 k=1 Zpjk k'
Thus, R is a ring homomorphism.
R
n
R 1=1 i= i
SisAll that remainsfor is to show:CR, then Rat( ils natural. Supposefi h(xi)) = h.
nh
E (i= f S r )(x) = 0 for all j, so r. = 0 for all j, and
R il1 1 j j
n
E fi ri = 0. Thus I is an isomorphism.
i=l 1 R
All that remains is to show that is natural. Suppose
h:R>R'. We must show.J that the following diagram i.s comrnu
tative:
'[C,Z] R
a,m
1 h
[* c,Z]
a,m
R'
n
EC,h] R (. f r.) (x) =
R =h(r) .
= h(r.).
[C,R]a
a,m
[C,h]
) [C,R']
a,m
n
h IR (il fi ri) (xj)
R =
n n
S (1 h) ( E f. r) (x.) = hR (= f h(r.))(x.).
R'i=1 1 1 R 1=l 1 1
= h(rj).
Thus, 7 is a natural equivalence.
The following lemma uses some wellknown results about
Ab, the category of abelian groups. For complete statements
and proofs of these results, the reader is referred to [6].
(3.9) Lemma: Let G be an abelian group such that the
functor G 9 :AbAb has a left adjoint. Then G is free
and of finite rank.
Proof: G Z has a right adjoint for any group G.
Suppose it also has a left adjoint. Any functor on Ab with
a left adjoint must be naturally equivalent to LH,] for
some abalian group H. Thus, G 0 [H,]. G is an
>
34
exact functor, but [H,] is exact if and only if H is pro
jective, so H must be a free abelian group. Suppose H is
free on an infinite set X. G has a right adjoint, so
[H,] has a right adjoint and must preserve colimits, direct
sums in particular. If we let Gi be the cyclic group of
order i, i=1,2,..., then, since EH,] preserves direct sums,
00o o
[H'i, G ] i [H'Gi] Since H is free on X, T CHG.] 
=1 1 i=l 1 i=l 1
il( x G a torsion group. On the other hand,
i=l xEX i'
o ~o
CH, G.] ( G.) which contains some torsion free
i=1 i xEX i=1 1
elements if X is infinite, a contradiction. Thus, H is free
on a finite set. Since G G 0 Z = [H,Z] f H, G must also
be free and of finite rank.
(3.10) Theorem: If R is a ring, then the functor
R :29 has a left adjoint if and only if the additive
group of R is free and of finite rank.
Proof: Suppose R S has a left adjoint. The category
Ab can be considered as a subcategory of F, namely the full
subcategory of all rings with trivial multiplication. If G
is a ring with trivial multiplication, then R 0 G will also
have trivial multiplication, so R S can be considered as a
functor from Ab to Ab. Also, since limits in Ab are the
same as the corresponding limits in R, and since R S is
limit preserving on R, it follows that R :Ab>Ab is a
limitpreserving functor. Since Ab is complete, locally
35
small, and has a cogenerator, by the Special Adjoint Functor
Theorem, R Z :AbAb has a left adjoint. From Lemma (3.10),
R, as an abelian group, must be free and of finite rank.
Conversely, suppose R is free and of finite rank as an
abelian group, and let bl,b2 ...,bn be a basis for the
additive group of R. Then there exist integers n. such
ijk
n
that b.b = 7: n. b.
Sk i=l ijkbi
Let C be the free ring on x1,x2,.. .x and define co
operations a,m:C>C*C by
a(xi) = u (xi) + u2(xi)
n n
m(x ) = jE1 k) ijk U(x (x
We claim (C,a,m) is a coring and that [C,]a,m R . By
a,m
Proposition (3.8) it is sufficient to show FC,z] = R.
a,m
n
Define :C[C,Z] >R by (f) = i f(x )b..
a,m i=1 i 1
n n
t(f + g) = (f + g)(xi)b. = a(xi)b
= i=1 i i
n
= (u(xi) + u2(x))b
n n n
= i f(xi) + g(xi)]b. = i~f(x.)b. + )ig(x)b.
i=1 = 1 1 i=l(xi
= O(f) + (g)
n n
$(fg) = (fg)(x)b =. < f,g>m(x.)bi
=1 1 1 i=l i
n n n
E n)U (xk))bi
=i~ j=l k= nijkul(j )u2(k
n n n
7 1 Z ni f(x )g(xk)b.
i=l J=1 k=l ijk j K1
n n
(f) (g) = ( f(x.)b.) ( g(x )b )
1=1 1 1 i=l i 1
n n n
Z Z [f(x)g( b bb
j=l k=l xf(xj)g(k) ( i=lnjkbi)]
n n n
Sn(x )g(xk k)b
i=l j=l k=l ijkf(x )b
Since is clearly onetoone and onto, is an isomorphism.
Thus, [C,Z] is a ring and [C,] m [C,Z] R 0 .
a,m a,m a,m
Since R 0 is representable, it has a left adjoint.
Corollary: Let T:R> have a left adjoint. Then
T a R 0 if and only if T is representable by a coring
(C,a,m) where C is free on xX2,... ,xn, a is standard, and
n n
m(x ) = Z n. l(x )u2(Xk)
1 j=l k=l nijkUjl (
Proof: This follows from Proposition (3.8) and the
proof of the above theorem where it was found that the
coring representing R 0 is as stated in the corallary.
In light of the construction mentioned in (1.2), it is
seen that if S is a finite semigroup with zero, then the
semigroup ring functor TS is naturally equivalent to
ZO [S 0 . Not every functor representable by a finitely
generated free coring is naturally equivalent to a tensor
product functor, however. An example of such a functor is
the opposite functor T which sends a ring into its oppo
op
site ring. T C [C,] where (C,a,m) is a standard coring
op a,m
37
on one generator x, and m(x) = u2(x)ul(x). If Top R 0 ,
then T (Z) 3 R 0 Z 2 R. But T (Z) = Z, and Z 0 is not
op op
naturally equivalent to Top
In the category of abelian groups, the functor A 0 
always has a right adjoint. There are rings such that the
functor R 0 on the category of rings has a right adjoint,
but this is not true for every ring R.
(3.11) Example: Denote by Z the ring of integers
modulo n. Then Z 0 :R>S has a right adjoint.
Proof: We will show that Zn is naturally equiva
lent to the functor Sn described in Example (2.8), where
Sn(R) = R/nR. If R is a ring, define R:R/nR)Zn 0 R by
R(r + nR) = 1 S r. It is easily verified that R is a well
defined ring homomorphism. To show that R is onetoone
and onto, we can define YR:Zn 0 RR/nR by
SR(z + (n) 0 r) = zr + nR. It can be easily shown that
R /nR and that R 1 R It is also a routine
R'R R/nR R R ZnOR*
calculation to show that r is a natural transformation and
hence that S Z 8 .
n n
From Theorem (2.4) and the above, we know that if
(C,a,m) is a coring in R, then (Zn 0 C,l 8 a,l 0 m) will
38
also be a coring. In particular, if (G,a,m) is the coring
representing the identity functor, then (Zn S G,l a,i m)
is the coring representing the functor T described in
Example (2.8)
(3.12) Example: Z x Z S :S) does not have a right
adjoint.
Proof: The functor T on 2 defined by T(R) = R x R and
T(f) = f x f, discussed in Example (2.7), has a coring rep
resentation of the form needed for Proposition (3.8). It
follows that T is naturally equivalent to Z x Z , and we
will show that T cannot have a right adjoint. In particular,
we will show that it is not coproduct preserving.
If T is coproduct preserving, then, since u.:Z>Z Z,
1
i = 1,2, defines a coproduct, then u. x u.:Z x Z
(Z Z) x (Z Z) will also define a coproduct.
(Z Z) x (Z Z) will also define a coproduct.
(Z Z) x (Z Z)
u1 xu u x u2
Z x Z h Z x Z
1 /2
Z
39
If pl1P2 represent projections from the product, there must
exist a ring homomorphism, h, making the above diagram com
mute. Consider the element (ul(l),u2(1)) E (Z Z) x (Z Z).
(ul(l),u2(1)) = (ul x u )(1,0) + (u2 x u2)(0,1). So
h(ul(l),u2(l)) = h(ul x u1) (1,0) + h(u2 x u2) (0,1)
= P1(1,0) + P2(0,1) = 1 + 1 = 2.
But (ul(1),u2(1))2 = (ul(l),u2(1)); but
h(u(l),u2(l))2 [h(ul(l),u2(1)]2, so h cannot be a ring
homomorphism. Thus, T cannot be coproduct preserving and
cannot have a right adjoint.
CHAPTER IV
THE CATEGORY OF CORINGS
Let G be a category with coproducts. Let A1,A2,BIB2
be objects in C and let f.:ABB, i=1,2, be morphisms in G.
Define f *f :A *A21B *B2 as the unique morphism such that
the diagram
A*A2
A1 22
I fl fl*f2 f2 2
ull /2
B *B
1 ,2
is commutative.
The following lemma is an easy exercise in category
theory and the proof is omitted.
(4.1) Lemma: Suppose fi:A>Bi, gi:BC, and
h.:B.D, i=1,2 are morphisms in a category G. Then
1 1
i. (gl*g2) (fl*f2) = g1fl*g2f2.
2. (f *f2 =
If (Cl,al,ml) and (C2,a2,m2) are corings in a category
G, then a morphism f:C>C2 is called a coring homomorphism
if the diagram
lx x2
x 1 x2
C1*C1 ) C2*C2
f*f
is commutative with x = a and x = m. This definition is
dual to the notion of ring homomorphism if the ring
operations +, are considered as functions from the product
of the ring with itself to the ring.
If G is a category, then we shall denote by CR(G) the
category whose objects are corings in G and whose morphisms
are coring homomorphisms. That CR(G) is actually a category
is easily verified with the help of Lemma (4.1). We will be
concerned in particular with CR(R), the category of corings
in the category cf rings.
Let LA(2) denote the category whose objects are
functors from 2 to 2 which have left adjoints and whose
morphisms are natural transformations of functors. (Actually,
we don't know yet that LA(R) is a category since we don't
know that [F,G] is a set. This will be made clear in the
proof of the following theorem.)
(4.2) Theorem: CR(R) is equivalent to the dual cate
gory of LA(R).
Proof: Define a functor G:CR(R)LA(R) by
G(C,a,m) = [C,]a,m and if o:(C1,a,,m1)>(C2 a2,m2) is a
coring homomorphism, then G() :[C ,] 2 [C ,]
L a22m2 1 alml
is the natural transformation [_,]. (Recall that if
f:C2R, then [r,R](f) = f5.) We will show that G is
a contravariant equivalence.
We know from Theorem (2.4) that G(C,a,m) is an object
of IA(R). We must now show that G(f) is a natural trans
formation.
First we must show that r5,R] as defined above is a
ring homomorphism. If f,g E [C2R] a2,m then
[,R](f + g) = [,R](a2) = a2 =
= (5*9)al = al = <[,R](f),[ ,R](g)>a
= [,R] (f) + [ ,R] (g).
A similar argument holds for multiplication.
We now show that [*,] is a natural transformation. We
must show that if C(:R>S, then the diagram
[C Ra2, m2
2' a ,m2
[c;
[c ,s]
a2,r
EP,R]
> [CR R]
1 al ,mI
a]
[c1,a]
[a,s] 1
t2 [1, salm
r2 1 a^_,m^
is commutative. If f E [C2,R], then
[c1,ia] [ ,R] (f) = af = [ ,S] [C2,,] (f).
Suppose :(C alm )>(C2,a2,m2) and
Y:(C2,a2'm2)(C3,a3,m3) are coring homomorphisms. For G
to be a contravariant functor, we must show that G(Y) =
G(U)G(Y), or that [Yi,] = [D,] [Y,]. But if a E [C1 R],
then [~i,R] (c) = a'7 = [i,R] [Y,R] (a).
We have shown that G is a contravariant functor. We
would like to show that G is an equivalencethat G is faith
ful, full, and representative.
Faithful: Suppose ',Y:(Cl,al,ml)(C2,a2,m2) and
G(i) = G(Y). Then [?,] = [Y,], and [D,C2](1C)
2
[Y,C 1(12). Then 1 C2 = C2Y, and = Y.
Full: Suppose n:[C2,]>[C1,] is a natural trans
formation. Define = C2 (1 C). If R is any ring, and
C2 C2
) I
f:C2R is a morphism in R, then from the diagram
*C
Cc2c2 '2 > [cc 3
[C2'f] [c1,f]
[C2,R] > [C1,R]
we see that
SR(f) = R[C2'f](lC) = [Clf]1 (1 ) = f. *
If 9 is a coring homomorphism, then 1R(f) = f = [ ,R](f) =
G() (f), and G will be full. Note that from *,
TC2C2 (ui) = u i, i=1,2, and that C2*C2(a2) = a2
a2= TC2*C2(a2) = 2C2(a2) = C2*C2 ( + u2)
= T2*C(u1) + C2*C2(u2) = a
22C2"2 22*C2
= (i*,)al = ( *i)al.
Thus,
C1 > C2
C1lC1  C2*C2
45
is commutative, and a similar argument holds for comultipli
cation. Thus, 5 is a coring homomorphism and G is full.
That G is representative follows from Theorem (2.4).
Thus, G is a contravariant equivalence.
The remainder of this chapter will be devoted to in
vestigating some properties of the category of corings.
(4.3) Proposition: If the category G is cocomplete,
then so is CR(C), the category of corings in G.
Proof: It is sufficient to show that CR(G) has co
products and coequalizers.
CR(G) has coproducts: Let (Ci,a.,m.), iEI, be a family
of corings in G. Since C is cocomplete, LC. exists. Let
iEI
C = _jC.. Let a:CC*C be the unique morphism <(u.*u.)a.>
iI 1 1 1
which makes the following diagram commutative for each i:
C
C.
Ci*CC
C*C
Define m:C,C*C in a similar manner.
46
We claim (C,a,m) is a coring. Let A be any object in
G. Then [C.,A] is a ring for each i E I. Since the
category of rings has products, let P = R [C.,A] We
iEI 1 ai',mi
claim P [C,A]a,m. Define a function :P>[C,A]a,m in the
a,m a,m
following manner: Let (f.) be an element of P where each
1
f.:C A. These f. induce a map :CA such that
1 1 1 1
u. = f. for each i. Let ((f.)) = . is oneto
1 1 1 1 1
one, for if ((f.)) = ((g.)), then = , and for
each i, u. u. and f. = g.. Thus (f.) = (g.). is
1 1 1 1 1 1 1 1
also onto, for if h:C>A, then hu.:C.A for each i, and
1 1
(hu.) is an element of P such that 5((hu)) = h. Q also
preserves addition:
S((fi) + (gi))ui = $((fi + gi))ui = ui
= f. + g. = a..
C[((fi)) + ((gi))]ui = ( + )ui
= <,>au = <,>(ui*ui)a
= <ui,ui>a. = ai.
Since these are equal for each i, ((fi) + (gi))
((fi)) + ((gi)). A similar argument holds for multiplica
tion and is an isomorphism. Since [C,A]a is isomorphic
to a ring, it must itself be a ring for any A, and (C,a,m)
is a coring.
47
It must still be shown that (C,a,m) is the coproduct of
the (Ci,a.,mi) in CR(G). By the way a and m are defined, it
is clear that each ui is a coring homomorphism. Now suppose
f.:(C.,a.,m.)(D,a',m') is a coring homomorphism for each
i. Since C is the coproduct of the C. in G, there is a
1
unique map :C>D such that u. = f. for each i. But
1 1 1 1
is also a coring homomorphism since for each i,
a'u. = a'f. = (fi*fi)ai.
1 1 1 1 1 1
(*)au = (*)(ui*ui)ai
= (u.*u.)a = (f.*f.)a..
1 1 1 1 1 1 1 1
Thus, a' = (*)a, and
1 1 1
1
C >D
a a'
.1 * v
C*C > D*D
is commutative. Thus is a coring homomorphism, and
(C,a,m) is the coproduct of the (C.,a.,m.) in CR(G).
CR(G) has coeuualizers: Suppose f,g:AB are coring
homomorphisms where (A,al,ml) and (B,a2,m2) are corings in
G. Let a:B>C be the coequalizer of f and g in G. In the
diagram
B*B
f
A B C ~ ) C*C
g a
we see that
( a*c)a2f = (*a) (f*f)al = (af*af) a = (cg*ag)al
= (a*) (g*g)al = (a*a)a2g.
Thus, there is a unique morphism a:C>C*C making the above
diagram commute. m:C>C*C is defined similarly.
We will show that (C,a,m) is a coring. Let X be any
object in G. It is easily verified that
[f,x],[g,X]:[B,x] [A,x] is a pair of ring homomorphisms.
Since the category of rings has equalizers, this pair of
morphisms will have an equalizer E. E can be thought of as
(h E [B,XlIhf = hgl. We claim E : [C,X] Define
a,m
:E>[C,X] as follows: If h E E, then hf = hg, so there
exists h':C>X making the following diagram commute:
X
f h'
A B B C
g a
Define (h) = h'. If (h) = ; (k), then h'a = k'a, and h = k.
So 3 is onetoone. If Y:CX, then Ya:BX, and
Yaf = Yag. Then ya E E and (YC) = Y. Thus, 5 is onto.
is additive, for if h,k E E, then
49
[. (h + k)] = e(a2)a = a2.
[ (h) + (k)]a = aa = < (h), (k)>( *)a2
= < (h)., (k)a>a2 = a2.
Since S is an epimorphism, (h + k) = (h) + (k). A
similar argument holds for multiplication, and 5 must be a
homomorphism. Since [C,X] is isomorphic to a ring, it
a,m
must itself be a ring for any X, and (C,a,m) is a coring.
We must now show that a:BC is the coequalizer of f
and g in CR(G). It is clear from the way a and m were de
fined that a is a coring homomorphism. Suppose h:B>D is
a coring homomorphism where (D,a',m') is a coring in G, and
h is such that hf = hg. Since a:BC is the coequalizer of
f and g in G, there is a unique morphism k:C>D making
D
f k
f
A z B C
g a
commute. We would like to show that k is a coring homomor
phism. Since a'k' = a'h, and
(k*k)az = (k*k)(a*a)a2 = (ka*ka)a = (h*h)a2 = a'h,
and since 2 is an epimorphism, we see that
k
C 
a a'
\I k*k I
C*C )D*D
50
is commutative. A similar argument holds for comultiplica
tion, so k is a coring homomorphism.
Corollary: CR(R) is cocomplete.
Not many other properties of CR(R) are known. Robert
Davis has shown [7] that if R1 is the category of commutative
rings with identity, then the forgetful functor from CR( 1)
to R, has a right adjoint. However, it is not known if this
result holds for CR(R). The following result shows that if
K is a field, then CR(GK), with the obvious forgetful
functor to the category of sets, does not form an algebraic
variety (in the sense of Cohn).
(4.4) Proposition: If K is a field, then CR(G K) cannot
have both products and free objects (relative to the
"obvious" forgetful functor).
Proof: Let G be the free Kalgebra on one generator x,
and let a(x) = ul(x) + u2(x) and m(x) = ul(x)u2(x).
(G,a,m) is then a coring in GK. Suppose the product of
(G,a,m) with itself exists in CR(GK), and denote it by
(H,al,ml). Define (0,1):G>H as the unique coring homomor
phism making
H
Pl P2
G G
G
51
commutative, and define (1,0) similarly. (0,1)(x) and
(1,0) (x) are nonzero elements of H. Suppose that (F,a2,m2)
is the free object in CR(G ) on one generator y. Define
f:FH by f(y) = (0,1) (x) (1,0) (x).
plf(y) = p (0,1) (x)pl(l,0) (x) = Ox = 0.
p2f(y) = 2(0,1) (x)p2(l,0)(x) = x0 = 0.
Thus, plf = p2f = 0, so f = 0. But then (0,1)(x)(1,0)(x) = 0,
and H has zero divisors, a contradiction to the corollary of
Proposition (3.5).
CHAPTER V
THE CATEGORY OF SEMIGROUPS
Many of the methods used to study functors on R which
have left adjoints can also be used to study functors on 8,
the category of semigroups, which have left adjoints. In
particular, Freyd's Theorem (2.3) holds in *, and a functor
T:8gS will have a left adjoint if and only if T is repre
sentable as [C,] where (C,m) is a cosemigroup in 8.
Not every ring admits a coring structure. The note
following Proposition (3.5) shows that Z, for example, can
not be made into a coring in R. However, every semigroup
can be made into a cosemigroup by defining the comultiplica
tion to be one of the injections into the coproduct. If S
is a semigroup, then [S,X]ul is the set [S,X] with left
trivial multiplication. For some semigroups a "trivial"
comultiplication of this type is the only kind possible.
(5.1) Construction of Coproducts in S: Let S1 and S2
be semigroups. Suppose S1 n S2= 0, and let S1 U S =
(sl E A). An element of S1*S2 is a product of the form
s, s 2...s where no two successive sk belong to the same
A1 2 n i
53
Sk, k=l,2.= s s x= s ...s if and only if n = m
1 2 n 1 2 m
and sL = si for each i. If x = Sk s, ...s and
i 1 2 n
y = s s .2 .sm are two elements of S *S2, then xy will be
defined as follows:
Case 1: If s and s belong to different Sk, then
n 1
xy= s ...ss s ...s
1 2" n 11 2 Pm
Case 2: If s and s belong to the same Sk, then
n 1 k
sk su = s E Sk Then xy = Xsk ... ss s u S3 .
n 1 2 n 2 3 m
For a proof that this construction actually yields the
coproduct of S1 and S2' the reader is referred to [8].
A semigroup S is said to be periodic if for each s E S
there are integers n and m, n < m, such that sn = sm. In
particular, every finite semigroup is periodic.
(5.2) Pronosition: Let S be a periodic semigroup.
Then (S,m) is a cosemigroup in & if and only if m = ul1 or
u20 where 5 is an endomorphism of S such that 2 = .
Proof: Suppose m = ul, or u2 where = 5. If X is a
2
semigroup and f,g E [S,X], then fg = ul, = f.
(fg).h = (f ) .h = f2 f(g.h) = f3, so the operation is
associative and (S,m) is a cosemigroup.
54
Conversely, suppose (S,m) is a cosemigroup. If x E S,
then m(x) E 'S*S, so m(x) = s1s2...sk, a finite sequence of
elements of S*S, where each si E ul(S) or u2(S) and no two
successive s. are from the same u.(S), j=l,2.
1 J
Case 1. If the length k of m(x) is even, then the
length of m(xn) is nk. But since S is periodic, xp = xn
for p < n. But then nk = pk, an impossibility, so the
length of m(x) cannot be even.
Case 2: If the length k of m(x) is odd, then the
length of m(xn) is nk n + 1. But if x = xn, p < n, then
nk n + 1 = pk p + 1; n(k 1) = p (k 1); and k = 1.
Thus, the length of m(x) = 1, and m(x) = u (x') or
m(x) = u2(x') for some x' in S.
If there exist x,y in S such that m(x) = ul(x') and
m(y) = u2 (y'), then m(xy) = u (x')u2(y'). But this is im
possible since we showed above that the length of m(s) is 1
for all s in S. Thus, if m(x) = u.(x') for some x,x' in S,
then for every y in S there is a y' such that m(y) = u.(y').
If we define :SS by (y) = y' where m(y) = ui(y'), then
is a homomorphism and it is clear that m = u.$. Since m
1
is associative, [S,S] must be associative, so (11)1 =
m
(l(). (11)l = <m,l>m = <,l>m = 2
1(l1.) = 1~ = ; so 92 = .
55
The remainder of this chapter is devoted to a proof of
the fact that 8 has exactly two autoequivalences, I and I p
the identity and opposite functors. Two lemmas are needed,
the first of which is apparently due to P. A. Grillet.
We shall define a semigroup P to be projective if, when
ever f:P>Q and g:T>Q are semigroup homomorphisms with
g onto, then there is a homomorphism h:P>T such that the
following diagram is commutative:
P
T >Q
(5.3) Lemma: Every projective in & is free.
Proof: Let P be projective and let X be the set of
elements in P which are not in P Let SX be the free semi
group on X. There is a :SXP defined by (x) = x for all
x in X. To show that is onto, note first that there is at
least one free semigroup, say S, which maps onto P, say
g:S>P. Since P is projective, we have h:P>S such that
P
S h 1
56
is commutative. It follows that h is onetoone and so P is
isomorphic to a subsemigroup of a free semigroup. Then
every element of P can be factored into a product of un
factorable elements. Hence, ::SXP is onto. And since P
is projective, there is 5' such that
P
_ _ 1
SX >P
is commutative. Thus I' = 1. Hence ' is onetoone. To
show ' is onto it suffices to prove that 5'(x) = x for all
x E X. Let (x) = x 2...x n; then x = 5' (x) =
S(x )I(x2) ... (xn) = X1X2...xn. This shows that n = 1, so
x = x, = (x). Hence :PSX is an isomorphism.
If is the free semigroup on one generator, then
* can be realized as the free semigroup on two gene
rators, y and z, with injections ul(x) = y; u2 (x) = z.
(5.4) Lemma: If (,m) is a cosemigroup in g, then m
must have one of the following four forms:
(1) m(x) = yz
(2) m(x) = zy
(3) m(x) = y
(4) m(x) = z.
57
Proof: That these four forms yield associative comulti
plications is easily verified. In fact, the four forms pro
duce the following functors:
(1) I, the identity functor
(2) Iop, the opposite functor
(3) T the left trivial multiplication functor (i.e.,
it takes a semigroup to the same set but with left trivial
multiplication.)
(4) Tr, the right trivial multiplication functor.
To prove the converse, let m(x) be an element of the
free semigroup on y and z. Then m(x) = w(y,z) where w(y,z)
is some "word" or monomial in y and z. If X is any semi
group and f,g:>X, then (fg)(x) = m(x) =
w(f(x),g(x)). Since x can be mapped uniquely to any element
of X, associativity of [,X]m is equivalent to
w(w(a,b),c) = w(a,w(b,c)). This will be true for all semi
groups only if
w(x,w(y,z)) = w(w(x,y),z) *
in the free semigroup on x,y,z. If the number of y's ap
pearing in w(y,z) is n, then the number of x's appearing in
w(x,w(y,z)) is also n, since each y is replaced by one x.
In w(w(x,y),z), however, each y in w(y,z) is replaced by
w(x,y), so the number of x's appearing in w(w(x,y),z) is
n n
n implies n = n, so n = 0 or n = 1. Thus, w(y,z) con
tains either one y or no y's. The same holds for z, so we
see m(x) must have one of the four listed forms.
58
The following theorem and the proof thereof were sug
gested by a similar result obtained by Clark in [9] for the
category of rings.
(5.5) Theorem: If T:8> is an equivalence, then
either T a I, or T z I the identity and opposite functors,
op
respectively.
Proof: An equivalence has a left adjoint. Theorem (2.4)
and its corollary hold in S as well as R, so T [C,]
m
where (C,m) is a cosemigroup in 9. We see that C must be
projective. Suppose f:AB is onto, and g:C>B. Since
an equivalence takes extremal epimorphisms to extremal epi
morphisms, [C,fl:[C,A][C,B] is also onto. Since
g E [C,B], there is h E [C,A] such that C,f](h) = g. Since
[C,f] (h) = fh = g, C is projective.
Since projectives in 8 are free, C must be free on some
set X. If U:SSet is the forgetful functor, then
U[C,] : n (). Since [C,] is representative, every semi
m
X
group A is isomorphic to [C,A'] for some A'. Then the
underlying set of A must be the product of X copies of A',
so the order of A must either be infinite or ncard(X) for
some n. Since any set can be made into a semigroup, the
above statement can be true for all sets A if and only if
card (X) = 1. So C is free on one generator and C = .
From Lemma (5.4), T must then be one of the functors I, I ,
opa
59
TV, or T That I and I are equivalences is readily veri
1 r op
fied. T1 and Tr cannot be equivalences, for, if so, they
would be representative, and every semigroup would have to
be isomorphic to a semigroup with left trivial or right
trivial multiplication, which we know is not the case.
Thus, the only automorphisms on are I and I.
op
A property of semigroups is said to be categorical if
whenever S has the property, T(S) will also have the prop
erty for every autoequivalence T of 8. The above theorem
shows that the only semigroup properties which are not cate
gorical are those which are not preserved under the functor
I those properties which are not leftright symmetric.
op
For example, the property of having a left zero or a left
identity would not be categorical.
BIBLIOGRAPHY
[11 Mitchell, Barry. Theory of Categories. New York:
Academic Press, 1965.
[2] Freyd, P. "Algebra Valued Functors in General and
Tensor Products in Paricular," Colloquium
Mathematicum XIV, 1966.
[31 Cohn, P. M. Free Rings. New Haven, 1962.
:4] Kan, D. M. "On Monoids and Their Duals," Bol. Soc.
Math. Mex. 3, 52, 1958.
[5] Bergman, G. M. Cormuting Ele7rnts in Free Algebras and
Related Topics in Ring Theory. Dissertation,
Harvard University, 1967.
[6] Cohn, P. M. Morita Equivalence and Duality. Queen
Mary College Mathematics Notes.
[7] Davis, Robert. "Free Coalgebras in a Category of Rings,"
Proc. Amer. Math. Soc. 25, May, 1970.
[8] Clifford, A. H., and G. B. Preston. The Algebraic
Theory of Semigroups, v. II. Providence: American
Mathematical Society, 1967.
[9] Clark, W. E. "Which RingTheoretical Properties Are
Categorical?" To appear.
BIOGRAPHICAL SKETCH
Burrow Penn Brooks, Jr., was born in Starkville,
Mississippi, on May 4, 1943. He graduated from Starkville
High School in 1961 and received the degree of Bachelor of
Arts with a major in mathematics from Mississippi State
University in 1964. He received the degree of Master of
Science from Case Institute of Technology in 1966. Since
September, 1966, he has attended the University of Florida.
This dissertation was prepared under the direction of
the chairman of the candidate's supervisory committee and
has been approved by all members of that committee. It was
submitted to the Dean of the College of Arts and Sciences
and to the Graduate Council, and was approved as partial
fulfillment of the requirements for the degree of Doctor of
Philosophy.
August, 1970
Dean, College oJArts & Sciences
Dean, Graduate School
Supervisory Committee:
Chai man
P/yZ67
nu. ' /414
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