I nternat, J. Math. & Math. Si.
Vol. 2 No. 4 (1979) 627-650
627
A CLASS OF RINGS WHICH ARE ALGEBRAIC OVER THE INTEGERS
DOUGLAS F. RALL
Department of Mathematics
Furman University
Greenville, South Carolina 29613
U.S.A.
(Received June 23, 1978 and in Revised form October 30, 1978)
ABSTRACT.
A well-known theorem of N. Jacobson states that any periodic
associative ring is commutative.
Hersteir0 generalized
Several authors (most notably Kaplansky and
the "periodic polynomial" condition and were still able to
conclude that the rings under consideration were commutative.
this paper we develop
In particular, an associative ring R is said to be
a quasi-anti-integral (QAI) ring if for every a
+ nkak
direct sum
In
a structure theory for a class of rings which properly
contains the periodic rings.
integer k and integers
(See [3])
nl,n2,...,nk (all
# 0
in R there exist a positive
depending on a), so that 0
nla
n2a2+.
In the main theorems of this paper, we show that any QAl-ring is a subof prime QAl-rings, which in turn are shown to be left and right
orders in division algebras which are algebraic over their prime fields.
KEY WORDS AND PHRASES. Anti-integral, Quasi-anti-integral, Periodic, Prime.
1980 MATHEMATICS SUBJECT CLASSIFICATION CODES.
16AI 8, 16A40.
Py 16A48 Secondary 16A12,
D.F. RALL
628
I.
INTRODUCTION.
In this paper we give a characterization of an associative ring which satisfies
an algebraic condition over the integers.
This condition can be viewed as a natural
generalization of one studied by Osborn for a power-associative ring (i.e. a ring
in which the subring generated by each element is associative).
[5]
As in
R will
be called periodic if for every element a of R, there exists an integer n=n(a)
n
greater than 1, such that a
[5,
In
a.
l
p. 321
Osborn proves the following
theorem which relates the periodic property to a more general kind of algebraic
condition.
THEOREM i.
The following are equivalent for a power-associative ring R:
i.
R is a periodic ring.
2.
Every element a of R is anti-integral over the ring of integers.
there are integers
nka
k
3.
k,n2,n3,
+ nk_l ak-i +
n
k
+
n
2a
2
That is,
depending on a so that
+
0.
a
R is a discrete direct sum of ideals each of which is an algebraic algebra
over a periodic prime field and contains no nonzero nilpotent elements.
It follows that if R is a periodic ring, then R
T(R)
{x
E
Rlnx-
0
for some integer n
T(R), where
0} is the torsion ideal of R.
In
this paper we consider a class of associative rings, which are not necessarily
equal to their torsion ideals, but which satisfy a generalized antl-integral
relation.
In particular, if F is a commutative, associative ring with i, and
R is a power-associative F-algebra, then a nonzero element a of R is said to
be quasi-a, ti.i.ntegral
(QA.I).
over F
in F (depending on a) such that 0
if there are ’scalars’
@ nla
2
n2a +
+
nl,
k
nka
n 2,
n
k
R is a QA!-
algebra over F if every nonzero element of R is QAI over F.
After developing some of the basic properties of QAl-algebras over Z, the
ring of integers, we show that a prime associative ring which is QAI over Z is
CLASS OF RINGS WHICH ARE ALGEBRAIC OVER INTEGERS
629
a left and right order in a division algebra which is algebraic over its prime
field, and conversely.
We then show that an arbitrary associative QAl-algebra
over the integers is a subdirect sum of prime QAl-rings.
We present an example
to show that not every such subdirect sum yields a QAl-ring.
PROPERTIES OF QAI-ALGEBRAS.
2.
For the remainder of this paper, unless specifically stated otherwise, the
term ring will mean associative ring without any assumption being made about
the existence of a unity element.
If R is a ring which is a QAl-algebra over
If R is a ring, then a ring Q(R) is said
Z we will simply call R a QAl-ring.
to be a left quotient ring for R and R is called a left order in
Q(R) provided
(i) there is a monomorphism h of R into Q(R) such that (ii) if x is a regular
element (i.e. one which is neither a left nor a right zero divisor) in R, then
h(x) is invertible in Q(R), and (iii) Q(R) consists of all elements of the
form
h(a)-lh(b)
for a, b in R with a regular in R.
right order are defined in a similar manner.
Right quotient ring and
If R has a left quotient ring
Q(R), we will often assume that R is a subring of Q(R) and so dispense with
the monomorphism h.
We state the following well-known results for future reference.
[ 5 ], 3 ]
LEMMA 2.
for their proofs.
If R is a left order in Q(R) and xI,
Xn
x2,
trary elements of Q(R), then there are elements a, b I, b 2,
with a regular in R, such that x
LEMMA 3.
i
-ibi (i
a
i, 2,
@F
are arbi-
bn
in
R,
n).
Suppose that R is an F-algebra, where F is a commutative and
associative ring with i and with a left quotient ring Q(F).
x in Q(F)
See
R can be written as x
c
-i
Then every element
@ r for c regular in F and some r in R.
The following lemma is well-known in its more general form, but we state
it here and use it only for associative rings considered as algebras over Z.
630
D.F. RALL
The proof in the general case may be found in
LEMMA 4.
p. 189
]
Suppose R is an associative algebra over the ring of integers
Z, and that T(R)
f(r)
[ 5,
(0).
Then the map f from R into Q
Z R given by
(Q is the field of rationals)
1 8 r is a ring monomorphlsm.
We observe that if A is a prime periodic ring of characteristic p # 0,
Q(A).
then A is a periodic field and so A is a left order and in fact A
In fact, the class of periodic fields is precisely the class of prime periodic
rings which are left and right orders in fields of nonzero characteristic which
are algebraic over their prime fields.
It follows from Theorem i that any periodic ring is a QAl-rlng.
In the
characteristic zero case we will show that QAI rings play a role similar to
that of periodic rings in characteristic p.
LEMMA 5.
Let D be any division algebra of characteristic zero which is
algebraic over Q and suppose R is a subrlng of D.
PROOF.
Let a be a nonzero element of R.
is a polynomial p(x)
p(a)
0.
n
r x
n
+
+
r0a
0, I,
r
0
in
Q[x]
such that
Assume such a polynomial of minimal degree has been chosen.
-(r an+l +
n
+
n and 0
0
r2a3 + rla2).
the least common multiple of d
i
Since D is Q-algebralc, there
r2x2 + rlx +
a is not a zero divisor we may assume r
0
R is a QAl-rlng.
dr0a
O
O.
n+l
-dr a
n
But then
Suppose r i
dn.
d I,
Since
Then dr
cidi
i
-1
and let d be
is an integer for
dr2a3 drla2
so a is QAI,
and R is a QAl-ring.
The class of QAl-rings, although containing the periodic rings, can now
be seen to be quite different from the class of periodic rings.
integral
quaternions is a noncommutative QAl-ring (compare with
The ring of
[2,
Thm.
A periodic ring is Jacobson semisimple since every element generates an
idempotent, while the subring of Q consisting of all quotients of the form
Ii]).
CLASS OF RINGS WHICH ARE ALGEBRAIC OVER INTEGERS
2r/2s+l, (2r,2s
+ i)
I,
is a Jacobson radical ring which is also QAI by
Lemma 5.
Let R be an associative QAl-ring with torsion ideal T(R).
THEOREM 6.
Then
(a) R has no nonzero nilpotent elements, and thus every
idempotent of R is contained in the center of R, Z(R).
(b) T(R) is a QAl-ring in its own right with the following
decompositions:
(i)
T(R) is a periodic ring which is a direct sum of
algebraic Z -algebras each of which has no nilpotent
P
elements.
(ii)
T(R) is a periodic ring which is a subdirect sum of
primitive periodic rings (i.e. a subdirect sum of
periodic fields).
(c)
If S is any subring of R, then S is a QAl-ring.
(d)
Any homomorphic image of R which is torsion-free is a QAl-ring.
PROOF.
square to 0.
(a)
It suffices to prove that R has no nonzero elements which
Let r # 0 be an element of R.
Then there are integers
3
2
a such that 0
+ akrk Then
nr
n, k, a 2, a 3,
k
a2r + a3r +
2+i
2
0 implies r
0 for all positive integers i and this yields,
clearly, r
a2r2 + a3r3 +
+
akrk
0 a contradiction.
no nonzero nilpotent elements.
R, then (ex- exe)
2
0
(xe
If e
exe)
2
2.
Thus r
2
0 and hence R has
e and x is an arbitrary element of
Thus ex
exe
xe and so e is in
Z(R).
(b)
T(R) is an ideal of R and it is clear that the property of being
QAI is inherited by subrings, so T(R) is itself a QAl-ring.
631
D.F. RALL
632
{x in
Let R
(i)
P
Rlpmx
0 for some positive integer m},
R is a subset of T(R) and is in
P
fact an ideal of R and hence of T(R). T(R) is actually a
for each prime number p.
discrete direct sum of the collection of ideals {R
p a
P
Let x be an arbitrary element of R
P
prime.
m
a positive integer m such that p x
(px)
mm
p x
m
of R.
(pmx)xm- i
By (a), px
Also 0
nx
0.
There exists
This implies
0 so px is a nilpotent element
0 and hence R
is an algebra over Z
P
a2x2 a3x3
+
+
+
atxt
P
where n, a 2,
a
t
are integers which may be considered as coefficients from Z
P
since nx # 0. Therefore, each R is an algebraic Z -algebra
P
P
with no nonzero nilpotent elements and so a periodic ring by
Theorem i.
By (i) T(R) is a periodic ring and by
(ii)
commutative.
Thin.
ii]
T(R) is
If a is an element of T(R), then for some
positive integer m greater than
e
[2
m-I is an
a
idempotent.
m
I, a
a which implies
But the Jacobson radical J(T(R))
contains no Idempotents which yields a is not an element of
J(T(R)).
Thus J(T(R))
(0) and T(R) is a subdirect sum of
primitive rings which must also be commutative and periodic,
since they are homomorphic images of T(R).
Therefore T(R) is
a subdirect sum of periodic fields.
(c)
This is clear since the QAl-property is an element-wlse condition.
(d)
If I is an ideal of R for which R/I is torslon-free, then R/I is
a QAl-rlng follows from the fact that
na
n2 a2 +
n3a3 +
0 # na
n22 +
n
0
#
+
+
nk ak
+ nka
implies
if a
a
+ I @ 0
in
R/I.
633
CLASS OF RINGS WHICH ARE ALGEBRAIC OVER INTEGERS
0 then R is a QAl-ring if and only if every element of
If T(R)
LEMMA 7.
R satisfies some polynomial with integer coefficients and has no nonzero
nilpotent elements.
The necessity follows directly from Theorem 6.
PROOF.
Then there is some polynomial p(x)
be a nonzero element of R.
2
a Integers so
a2x + alx with a 1, a 2,
t
alw 0 and we are finished. Otherwise, let
set z
arwt-(k-l) + + ak+lw2 + akw. If
but z
k+l
Conversely, let w
0, a contradiction.
that p(w)
O.
a x
t
If a 1
k be minimal so that
0, w is QAI
z
t
+
+
0 then
0 and
akw
Otherwise, z
@
0
Therefore R is a QAl-ring.
It will sometimes be convenient to use the obviously equivalent form of
a in R is QAI if and only if there exists a polynomial
the definition:
p(x)
ala #
alX +
0.
a2x2 +
+
akxk
with integer coefficients so that p(a)
0 and
(p(x) will be called an integral polynomial.)
Then
Suppose R and S are both power-associative QAl-rings.
THEOREM 8.
so is their direct sum R @ S.
S.
Let (r, s) be an element of R
PROOF.
We may assume that both r and
s are nonzero for otherwise (r, s) is QAI trivially.
The argument will be
divided into four cases determined by the additive orders of r and s, which
will be denoted by o(r) and o(s), respectively.
Case i.
Then r e T(R)
m; n, m both finite.
n, o(s)
o(r)
and@
e T(S).
By Theorem 6 T(R) and T(S) are both periodic rings, and by the proof of Theorem
13.2
[5]
T(S).
so is T(R)
But T(R)
+ T(S)
T(R
(R)
S).
Thus (r, s)
T(R
a periodic ring and hence (r, s) is QAI.
Case 2.
o(r)
n, finite and o(x)
integral polynomial p(x)
Let f(x)
np(x)
polynomial and
nals
(nal)x
alx +
+
a2x2 +
(na2)x2 +
Since s is QAI there exists an
(R).
+
+
akxk
with p(s)
(nak)xk.
0 and
als
f(x) is an integral
0 since s has infinite additive order.
Thus
# 0.
S),
634
D.F. RALL
nal(r,s)
f((r,s))
(nalr, nals) (0, nals)
2
nal(r,s) + na2(r,s) +
(nalr, nals) +(na2r
2
(0, 0) in R
+
nak(r,s
na2s 2) +
S, yet
k
+
(nr
k
nsk)
(np(r), np(s))= (0, 0)
(f(r), f(s))
Therefore, (r,s) is QAI.
o(r)
Case 3.
-,
o(s)
n, finite.
-,
o(s)
-.
Same as Case 2
interchange the
roles of r and s.
o(r
Case 4.
alx +
polynomial p(x)
Now, p((r,s))
+
Otherwise, p(s)
h(p(s))
0 in S.
p(x) and h(x).
h(p(x))
@
0 in S.
blx + b2x2
integral polynomial h(x)
g(x)
(0, 0) and
0 in S, then p((r,s))
(r,s) is QAI.
alr
# 0
and p(r)
0.
aksk)
(0, p(s)).
(p(r), p(s))
If p(s)
+ ak xk with
+
+
ak(r,s)k
a2(r,s)2
+ (akrk
als) + (a2r2 a2s2) +
al(r,s)
(alr
a2x2 +
Since R is a QAI-rlng, there is an integral
+
al(r,s)
(0, 0) in R
S so
But S is a QAl-rlng so there is an
+
n
b x with
n
blP(S)
0 and
Define g(x) to be the composition of the two polynomials
That is,set
a2x2 +
+ b2(alX + a2x2 +
+ bn(alx + a2x2 +
bl(alx
+
xk)
+ akxk)2 +
+
+ akxk) n
Upon inspection of this polynomial one sees that the coefficient of the linear
term is
blaI
o(r)
blal(r,s)
0 and that g(x) is clearly an integral polynomial.
Since
o(s), we get
(blalr, blalS)
(0, 0) the zero element of R
S.
However,
CLASS OF RINGS WHICH ARE ALGEBRAIC OVER INTEGERS
g(<r,s)) --h(p((r,s)))
QAI.
h((0, p(s)))
(0, h(p(s)))
635
(0, 0), and <r,s) is
Since these four cases are exhaustive, the theorem is proved.
COROLLARY 9.
sum R
R
I
Suppose R
7.
R
Let
{Rt} t
e A
Then their direct
be any collection of QAl-rlngs, and suppose
Then R is a QAI-rlng.
is their weak direct sum.
t
PROOF.
Let a be any nonzero element of R.
of the components of a are zero; i.e. if a
finite number of elements
member of
are QAl-rlngs.
This can be proved easily by induction.
COROLLARY i0.
teA
R
R2,
Rn is also a QAl-rlng.
2
PROOF.
R
I
{tl, t2,
Rtl Rt2
@
tk}.
Rtk
t
tl, t2,
k
Then all but a finite number
(at) t
e A
then there exist a
in A such that a
s
0 if s is not a
Then a can be considered to be an element of
which is a QAI-rlng.
I
Thus a is QAI and R
R
teA
t
is a QAl-rlng.
As can be seen from Theorem 6 and Corollary i0, the class of QAl-rlngs is
closed under taking subrlngs, direct sums and torslon-free homomorphlc images.
However, Z/nZ is a homomorphlc image of Z which
is not a QAl-rlng whenever n
To see that the class of QAl-rlngs is not closed under
is not square-free.
complete direct products consider the following example.
EXAMPLE.
set R
W
i.
Let W
i
Let x
Suppose x is QAI.
that 0
nx
Z, the ring of integers for i
R be that element x
(xi)
That is, suppose there exist n,
a2x2 + a3x3 +
+
akxk.-
i, 2, 3,
such that x
a2, a3,
and
i
a
k
i.
integers so
Since two elements of R are equal if
and only if they are equal in each component, this equation yields the following
636
D.F. RALL
infinite system of equations:
a2 +
n
a3 +
2n
4a
2
+ 8a 3 +
3n
9a 2
+ 27a 3 +
mn
2
m a
2
3
+
m a
3
+ ak
2kak
+ 3kak+
+
+
k
which implies that 2 divides n
which implies that 3 divides n
which implies that m divides n
m a
k
This is clearly a contradiction since n can have only a finite number of
Therefore R is not a QAl-ring.
divisors.
3.
THE PRIME CASE
Let R be an arbitrary QAl-ring, and P(R) be the prime radical of R.
P(R)
is the intersection of all the prime ideals of R and is also a nil ideal of
R,
hence equals (0), since R has no nilpotent elements and thus certainly no nil
ideals.
This fact, together with examples of QAl-rings which are Jacobson
radical rings, led us to investigate QAl-rlngs via the prime-semi-prime route
instead of the primitlve-semisimple one.
LEMMA Ii.
PROOF.
Q
ZR
If R is a prime associative ring then Q
If R is prime, then it has a characteristic p.
(0) since m/n
(R)
mp/np
r
Suppose there exist elements x, y in Q
Lemma 3, x
0
(s
-1
m/np e pr
r
0, then by Lemma 4, R is embedded in Q
If p
8 r
R is also prime
8z
s
-i
I)(Q
r I, y
8 R)(t
t
-i
-l 8 r
2)
r
ez
If p
m/np 00
(R)Z
# 0, then
O.
R via f:r ---+ i
R such that x(Q
Z R)y
(0).
2
for some s, t e Z and r I, r e R.
2
(s
-1Q t- l)
8
(rlRr2).
In particular
r.
By
Then
CLASS OF RINGS WHICH ARE ALGEBRAIC OVER INTEGERS
0
(s-l(st)t-1)
and so
0
Q@ZR
But f:R
(rlRr2).
1 0
But R is prime and so r
I
0.
rlRr 2
(rlRr 2)
0 or r
637
is one-to-one
O.
2
Q@ZR
Therefore,
x(Q@zR)y
LEMMA 12.
Suppose R is a prime ring with no nonzero nilpotent elements.
(0) implies x
0 or y
0 and
is prime.
Then R has no nonzero, zero-divisors.
PROOF.
Now (yRx) 2
0.
(yRx)(yRx)
yR(xy)Rx
0 or y
now forces x
LEMMA 13.
S
Q@ZR
0 and thus yRx
Therefore xy
0.
The primeness of R
0.
0 or y
0 implies x
0.
Suppose R is a prime QAl-ring of characteristic zero.
Then
is also a QAl-rlng.
PROOF.
Q@ZR.
Let x
c-ler.
in R such that x
x
By hypothesis R has no nonzero nilpotent elements.
Suppose xy
By Lemma 3, there is an integer c and an element r
S is a Q-algebra, so we may rewrite x as
-I
c (1 e r).
R is a QAl-ring and r
that 0
# nr
n2r
2
R so there are integers n, n2, n 3,
n3r3 +
+
+
nkrk
n such
k
Since S is a Q-algebra
it is torsion-
free so
0
0
@ n(18r)
nx
l@(n2r2 + n3r3 + + nkrk)
(ln2r2) + (ln3r3) + + (lkrk)
n2(lr)2 + n3(lr)3 + + nk(l@r) k. Now
n@r
l@nr
n(c-lr)--
c
c-ln 2(18r)2
-1
(nr)
-1
c
that 0
nx
+
But if a i
a2x2 + a3x3 +
nlc
i-i
n3(lr)
3
+
+
nk(lr) k)
+
for i
+ akxk-
+
c-lnk(18r) k
+ nkck-l(c-lr) k,
c-ln3(18r) 3 +
n2c(c-lr) 2 + n3c2(c-1Or)3 +
is a Q-algebra.
2
(n2(lr)
k then we have shown
2, 3,
with a i
since S
Z.
Therefore since x
was arbitrary, S is a QAI-rlng which is also a Q-algebra.
S
D.F. RALL
638
If R is a prime QAl-ring of characteristic zero, then R
COROLLARY 14.
can be embedded as a subring in a Q-algebra S which is a prime QAl-ring.
PROOF.
Let S
Q
By Lemma II and Lemma 13, S is a prime QAl-ring
R.
@Z
By Lemma 4 R is embedded inside S by r-----+ l@r.
and is clearly a Q-algebra.
If we require, in addition to the hypothesis of the last corollary, that
our ring R have a unity element I, then we can also conclude that Q
In particular,
division algebra.
0
@
mx
There exist integers m,
0 =mx-
a2x2 + a3x3 +
(a2x2 +
+
+
atxt.
atxt)
We wish to show that x
a
a2, a3,
a2x +
-i
invertible, x
2
3
-i
m
a x
x(m-i
+
+
(a2.1
+
atx
t-i
a3x +
t
so that
Then
(a2x +
atxt-l)).
But by Lemma 12,
a3x2 +
+
+
(a2x +
(a2.1 + a3x +
R has no zero-divisors and then we get m-i
Thus m.l
R is a
let S be a Q-algebra which is a prime
QAl-ring with i and let x be a nonzero element of S.
is invertible.
@z
+
a x
t
t-2)
+
at xt-l)
atxt-2 )-x.
0.
So x is
and we have proven the
following.
THEOREM 15.
If R is a prime Q-algebra with i which is also a QAl-ring
then R is a division algebra.
If one assumes that R is a prime QAl-ring with i, then the resulting
Q-algebra
Q@ZR
satisfies the hypothesis of Theorem 15.
The following theorem
shows that we need not assume the existence of i.
THEOREM 16.
If R is a prime Q-algebra which is also a QAl-ring, then R
is a division algebra, algebraic over Q.
PROOF.
Let a be a nonzero element of R.
n, k and integers n 2,
may assume that nk
at least one of the n
such that 0
n3,
There exist positive integers
na
2
n2 a +
+
nka
k
We
0- for otherwise the polynomial is of smaller degree and
i
0.
Since R is a Q-algebra,
CLASS OF RINGS WHICH ARE ALGEBRAIC OVER INTEGERS
a
k
-i
n
(nak
(n2a
2
+ nk_ I ak-l) ).
+
639
A, the Q-subalgebra of
That is
R generated by a is finite dimensional.
A
A, the right multiplication map given by a r (x)
xa.
a is a Q-linear transformation which is one-to-one since R, being a prime
r
Consider a
r
Now A is finite dimensional so a r is
QAl-ring, has no nonzero zero-divisors.
also onto, and by a standard argument a is invertible and hence a is invertible
r
in A.
Similarly, every b in A which is nonzero is invertible, and since A is a
commutative subalgebra, it is a field.
e
2
e, and for any y # 0 in R, e(ey
nonzero zero-divisors, ey
y
Let e in A be the identity of A.
y)
0
(ye
y)e.
Since R has no
ye and e is an identity for R.
Thus every
nonzero element of R is invertible and R is a divisiom algebra.
The fact that
R is algebraic over Q follows from the definition of QAI-ring.
Having characterized prime Q-algebras which are QAl-rings as
bein
division
algebras which are algebraic over the rationals, we now wish to investigate the
nature of the embedding of a prime, characteristic zero QAl-ring R into
It is interesting that R turns out to be a left and right order in S
Q@ZR.
Q@ZR.
To demonstrate this, it is convenient to know the inverse of an arbitrary
c-l@r
nonzero element
LEMMA 17.
element of S.
tru
in S.
t-l@s
Suppose e
is the identity in S and
Then the multiplicative inverse of
c-l@r
c-l@r
is any nonzero
is given by
d-l@uwhere
cds- tur.
PROOF.
By Theorem 16,
c-l@r
is invertible since S is a division algebra.
(c-lr) -1 d-lu. Then (c-l@r) (d-lu) t-l(R)s (d-leu) (c-l@r),
t-l@s (dc)-l@ur. Since S is a Q-algebra, upon
only if (cd)-l@ru
Suppose
and
multiplying by tcd in Z, we get t@ru
is taken over Z, so l@tru
equivalent to tru
cds
l@cds
fur.
cd@s
l@tur.
t@ur.
if
But the tensor product
Applying Lepta 14 this is
640
D.F. RALL
We now proceed to characterize prime QAl-rings of characteristic zero.
THEOREM 18.
Let R be a prime QAl-ring of characteristic zero.
Then R is
a left and right order in a rational division algebra which is algebraic over Q.
PROOF.
By Corollary 14 and Theorem 16 the function f
given by f(r)
iOr is a ring monomorphlsm of R into the Q-divlsion algebra
QOzR which
S
QOzR
R
It remains then to show
is algebraic over the rational field.
that f actually embeds R as a left and right order in S.
By definition, we must show:
If x is regular in R, then f(x) is invertlble in S; and
(i)
f(x)-If(y)
Every element z in S can be written as z
(ll)
(and
f(u)f(v) -I) where x, y, u, v are in R with x and v regular.
z
To establish (1), it is necessary only to observe that since f is one-to-one
and S is a division algebra, f(x)
Let e
it follows that
lms =IOtm
From
be the identity for S.
t
t-18ms
18m
t-1Osm which
IOsm and hence, because f is one-to-one, ms
If m
0; then pm
10pm
By Lemma 17 (iOpm) -I
(10pm)-l(lm2)
0
by m in R, we get tpwm
torsion-free implies
2
pwm
lm
z
-lem where
qsm
qm.
q-18w where tpwm
q-18wm2. But tpwm
sqm
tqm.
tpmw
qs
Thus t(pwm
It follows that 10pwm
Since S is a Q-algebra, if we multiply by
p
p
sm.
tm
0 is an integer
p
since R is prime of characteristic zero.
(q-18w)(lm2)
2
(t-los)(lm)
lm
(1By) -i (100) for any y @ 0 in R.
0, then write z
assume m
0.
(lm)(t-los)
implies (multiplying by t)
Let z be an arbitrary element of S, say z
and m is in R.
0 in R.
iOx will be invertlble for any x
(pq)-I
2
2
qs.
So
Consider
Then
and upon multiplying
qm)
f(pwm
0 and R being
2)
then we see that
f(qm)
q-lwm
2
10qm.
CLASS OF RINGS WHICH ARE ALGEBRAIC OVER INTEGERS
Therefore z
(q-l@w)(l@m2)
p-l@m q-l@wm2
then seen to be the collection of all elements of
f(x)-if(y).
in
S’
Thus R is a left order in S.
641
(l@pm)-l(l@m2). S
the form (l@x)-l(l@y)
The proof for
’R
is
is a right order
is similar to the one above.
We now turn our attention to the remaining case in the prime situation-that of prime QAl-rings of nonzero characteristic.
Indeed it is here that we
find the overlap with periodic rings.
THEOREM 19.
Suppose R is a prime QAl-ring of characteristic p # 0.
Then
R is a periodic field.
PROOF.
T(R), by Theorem 6 R is a periodic ring and by Lemma
Because R
If x # 0 in R then there exists a positive
2
n-i
n
e # 0. We
e
x
x. Then if e
integer n greater than i such that x
12, R has no nonzero zero-divisors.
claim that e must be the identity element for R.
R such that ey # y (or ye # y).
(or ye- y # 0 and (ye
y)e
But then ey
0).
nonzero zero-divisors and hence e
R
k
w
If not, then there is y in
y # 0 and yet e(ey
y)
0
This yields a contradiction since R has no
n-i
x
is an identity for R.
For any w in
k-i
is an idempotent.
w for some k greater than i and as above w
k-i
the same argument as before w
Invertible and by
[2,
of characteristic p
THEOREM 20.
Thin.
# 0
Ii]
e.
By
Thus every nonzero element of R is
R is commutative.
Therefore a prime QAl-ring
is a periodic field.
For a ring R to be a prime QAl-ring it is necessary and
sufficient that it be a left and right order in a division algebra which is
algebraic over its prime field.
PROOF.
The necessity follows directly from Theorem 18 (characteristic
zero) and Theorem 19 (characteristic p).
For sufficiency, let R be a left
and right order in a division algebra D, where D is algebraic over its prime
D.F. RALL
642
field F.
a
0
0 in R; then there exist a positive integer n and
Smppose x
an in F such that anxn +
0, a I,
n+l
x gives a x
n
Case i.
+
+
If char F
alx2 + a0x
p
+
alX + a0
0.
Multiplying by
0.
0 then the ai can be considered as integers
(between 0 nd p).
Case 2.
If char F
0, then F
Q and a i
bic i
Set d equal to the least common multiple of c o
0, i,
integer for i
n and
n+l
(dan)X
-i
c I,
c
n.
+
2
+
+
In either case x is QAI so R is a QAl-ring.
(dal)x
n.
0, i,
i
Then da
(da0)x
is an
i
0.
Since a subring of a division
ring is always prime R is a prime QAl-ring.
In fact, we can drop the assumption that R be an order in Theorem 20.
A ring R is a prime QAl-ring if and only if R is a subring
COROLLARY 21.
of a division algebra which is algebraic over its prime field.
There are several instances where a QAl-ring R, being a special type of
prime ring, must not only be a left order in a division ring but must itself
be a division ring.
THEOREM 22.
Suppose R is a primitive QAl-ring.
Then R is a division
ring which is algebraic over its prime field.
PROOF.
A primitive ring is prime; so if R has nonzero characteristic, it
will be a periodic field by Theorem 19.
We thus assume that R is a primitive
By the Density Theorem R is isomorphic to a
QAl-rin of characteristic zero.
dense ring of linear transformations of a vector space V over a division ring D.
We claim that the D-dimension of V is one.
Suppose not; that is, suppose that
{v I, v 2} is a D-linearly independent subset of V.
elements x and y in R so that yv
I v 2, yv 2
QAl-ring so there exist integers m, m2, m 3,
that 0
#
mx
m2x2 + m3x3 +
+
mtx
t
and 0
Since R is dense, there exist
0, xv I
m
# ny
t
0, xv
2
v
2.
and n, n 2, n 3,
2
n2Y
+
n3y3 +
But R is a
n
+
nsY
s
s
so
CLASS OF RINGS WHICH ARE ALGEBRAIC OVER INTEGERS
Then mnxy
But 0
0.v
mx(n2y2 +
I
+
nsyS)
ay2)vI
(mnxy
ay
2
mx (n
where a
ay(yv I)
mnx(yvI)
HomD(V,V),
+
n3Y
+
+
nsys-2)
mnv2, which is a contra-
Thus the dimension of V over
diction since D has characteristic zero if R does.
D is one and so R is isomorphic to D--
2
643
and R is a division ring.
R
is algebraic over its prime field by the definition of QAI.
COROLLARY 23.
Suppose R is a QAl-ring which is an algebra over a field F.
Then R is a subdlrect sum of division algebras over F.
PROOF.
Since R is a QAl-ring, R is algebraic over F.
radical is a nil ideal of R and hence (0).
primitive F-algebras,
{Dr} t
A"
Each D
t
Thus the Jacobson
R is then a subdirect sum of
is a homomorphic image of R and has
the same characteristic as R so is a QAl-ring.
The corollary follows from
Theorem 22.
THEOREM 24.
PROOF.
A simple QAl-ring must be a division ring.
For each positive integer m, let mR
of R, and since R is simple, either mR
which mR
(0), then R
(0) or mR
{mrlr
R.
e R}.
mR is an ideal
If there is an m for
is a prime QAl-ring of nonzero characteristic.
Theorem 19 R is a periodic field.
Otherwise, mR
By
R for every positive integer
m.
In this case R is a Q-algebra and so a division ring by Theorem 16.
4.
THE GENERAL CASE.
In our study of prime QAl-rings we considered two separate cases
characteristic zero and another of prime characteristic.
one of
If R is any QAl-ring,
then R is semiprime; however, a semiprime ring may have characteristic zero and
still contain elements of finite order.
The dichotomy we will use in the general
case will be the torsion T(R) and the torsion-free R/T(R) cases, then
attempting to put these two extremes together to say something about R.
often the case, it is the latter which is the most difficult.
As is
D.F. RALL
644
The torsion ideal, T(R), is a subdirect
Let R be a QAl-ring.
THEOREM 25.
sum of periodic fields and
R/T(R)
is a subring of an algebra S over the
rationals Q, so that S is a subdirect sum of Q-division algebras which are
algebraic over Q.
PROOF.
Just
The conclusions about T(R) are
a torsion-free QAl-ring by Theorem 6.
If f: M
so is S by Lemma 13.
Since M is a QAl-ring,
Q@ZM.
Set S
S is given by f(m)
monomorphlsm since M is torsion-free.
R/T(R),
Let M
Theorem 6.
l@m then f is a ring
Thus we may consider M as a subring of
a Q-algebra S which is itself a QAl-ring.
We claim that S is a subdirect sum
of prime Q-algebras which are QAl-rings.
S has no nonzero nilpotent elements
and hence no nonzero nil ideals.
J
{Ill
a
(0)
E
J
a
.
{anln
is a Q-algebra ideal of S and I
so J
a
By
Zorn’s Lemma
pick I
a
is a positive
maximal in J
and the intersection of the collectlon {I 8
algebra ideal
(0).
Suppose a is a nonzero element of S, and set
Each S/I
a
Then I
a
la
E
$}.
integer}
is a prime
a
S} of ideals is
is a Q-algebra which forces it to be torsion-free and hence a
QAl-ring by Theorem 6.
Thus S is a subdlrect sum of the collection
{S/la} a
S
E
of prime Q-algebras which are QAI-rlngs, and we are done by applylng Theorem 16.
Suppose R is a torsion-free QAl-ring.
It follows from Theorem 25 that we
may embed R in a rational algebra which has a nice representation.
Can we then
say that R also has a similar representation?
THEOREM 26.
Suppose R is a torslon-free QAl-ring.
Then R can be written
as a subdirect sum of orders in Q-division algebras each of which is algebraic
over Q.
PROOF.
given by f(r)
1Or is a ring embedding.
By the proof of Theorem 25,
subdlrect sum of a collection of Q-dlvision algebras
D
t
QOzR
Since Q is the quotient field of Z, the homomorphlsm f: R
is algebraic over Q.
{Dr} t
Consider the following diagram:
e
A"
QSZR
is a
For each t in A,
645
CLASS OF RINGS WHICH ARE ALGEBRAIC OVER INTEGERS
h
QOzR
-’
’t
p
h is the embedding of
QOzR
A
1
D
t
Pw
into the complete direct product,
w
is a subring of D
w
and a rlng homomorphic image of R.
Dt
and
Pw
Pw(h(f(R))).
Set E
is just the projection homomorphism onto the wth coordinate.
E
A
t
In fact, E
is an order
For if d is an element of D and i is the unity of D__, then there are
w
in D
w
elements
n-10e
Pw (h (f (me))
and
m-lr
ran- i and
such that
Pw(h(n-le))
Pw (h (f (nr))
nmd.
(Pw(h(f(me))))-l(pw(h(f(nr)))).
clear that R is a
i and
Thus d
Pw(h(m-lr))
(ran- I)
-i
d.
Then
(rand)
By the way E t is defined for each t in A, it is
subdlrect sum of
_{Et} t
c
A"
(The proof that
Ew
is an order in
Dw is due to Professor F. Kosler.)
COROLLARY 27.
If R is a torslon-free QAl-rlng, then R is a subdlrect sum
of prime QAl-rlngs of characteristic zero.
Using the notation of the proof of Theorem 26, R is a subdlrect sum
PROOF.
of
{Et} t
Thus E
t
e
A"
E
t
is a subring
(order)of
a Q-division algebra and is a QAl-ring.
is prime of characteristic zero.
Reviewing What we have done up to this point in the general case, we see
that by Theorem 25 and Theorem 26 if R is either a torsion QAl-rlng or a torsionfree QAl-rlng, then R is a subdlrect sum of orders in division algebras, each
being algebraic over its prime field.
Our present goal is to get a similar
decomposition for an arbitrary QAl-rlng.
DEFINITION.
If R is a ring then a subset M of R is called a multlpllcatlve
system if
i)
0 is not in M, and
2)
whenever m and n are elements of M, so is their product mn.
D.F. RALL
646
M is a maximal multlplicatlve system if it is not a proper subset of another
multlpllcatlve system.
Observe that if R is a QAl-rlng, then R has no nonzero nilpotent elements;
so if r is a nonzero element of R, N
r
{rnln
} is a multlpllcatlve
i, 2, 3,
We state the following three lemmas without proof; see
system.
LEMMA 28.
If R is a QAl-rlng and a
[i].
0 in R, then there is a maximal
multlpllcatlve system containing a.
LEMMA 29.
If M is a maximal multiplicatlve system in a ring R without
nilpotent elements, then the set-theoretlc complement c(M) is an ideal.
LEMMA 30.
Let R be a ring without nilpotent elements and for each x
@
0
in R let M be a maximal multlpllcatlve system of R containing x with complement
x
C(Mx).
Then R is a subdlrect sum of the integral domains
{R/C(Mx)} x
4
0
R"
As we observed earlier, all homomorphlc images of a QAI-rlng need not
In Theorem 6 we showed that if I is an ideal of R such that R/I
be QAl-rlngs.
is torslon-free then
R/I
is a QAI-rlng.
However, in the decomposition of R, we
can make use of the following:
LEMMA 31.
Then
R/c(M)
PROOF.
Let R be a QAl-rlng and M a maximal multlpllcatlve system in R.
is a QAl-ring.
For notational convenience we denote R/c(M) by A and let x represent
the image of x under the natural homomorphism f: R
see that A is an integral domain.
(i)
# 0.
From Lemma 30 we
If A has characteristic zero, then we are
finished since it will then be torslon-free.
terlstlc a prime p
A.
Thus we may assume A has charac-
We make two observations:
Since A has no zero divisors, it suffices to show that for a in
R with a
0
a satisfies some relation of the form
k
k-I
nkx + nk_iX +
+
one--of the integers n
i
n2x2 + nix
0 with at least one--any
not divisible by p.
CLASS OF RINGS 41CH ARE ALGEBRAIC OVER INTEGERS
(ii)
647
If a is a nonzero element of R of finite additive order m,
mf(a)
then ma
f(ma)
f(0)
Thus f(a)
0.
a has finite
Therefore, since the
additive order k and k divides m.
characteristic of A is p, if a in R is of order q where q is
#
a prime, q
{x e R
T(R) --O7.R where R
P
P
0 in A.
f(a)
p, then a
lpkx--
{x e R Ipx
0 for some k}
0} (R has no nilpotent elements)
Let a # 0 be an element of R.
a e T(R)
Case i.
If a
a
0, we are done.
s such that i < s < t with a
collection of integers
and
mlas
a
# 0.
I + a2 +
s
Hence a
a
Ps
n
Rpi
and a
P
a
in A.
s
k
+
mk’ mk(as) +
ml’ m2’
0
we also have
2
+
mm(as)
+
m2()2
mla
mlas
I
But for some
m2(as)
2
+
mlas
0
and
+ mla s
+
ml
# 0
(p does not divide
ml).
is QAI.
a
T(R)
0 in A, we are done.
Suppose that a
+ rn_lan-i +
+
r2a2 + rla
If p does not divide r I, then
rla #
@
Since R is a QAl-ring,
0
rn such that
there exist integers n, r I, r 2,
rna
e
i
This implies that p does not divide m
s
Case 2.
If a
R
e R
a
0 and by (ii) there exists an integer
Otherwise a
ink(a--s )k + mk_l(aL )k-i +
mk()k + mk-i ()k-I +
Since a
+ at,
0
with
rla #
0.
0 in A and we are finished.
648
D.F. RALL
rna--n + rn_+/-a--n-i +
a
+
rna--n + r --n-i
If p divides r I, then 0
n_
--n-i
--n-2
+
(rna
If p does not divide
r2,
then
r 2, then as above we have r
n
@ 0
r2a
r2a
+
r2a
rna
+
+
r2a)a
+
+
ra=
0
bl, b2,
bn;
i
for i
I, 2,
for i
the highest power of p which divides all of the r
integers
r3 a2 + r^az
---O"
If p divides
Continuing in this manner,
it follows that either a is QAI by (1) or p divides r
i
rla
in A and we are finished.
()n-2 +
Therefore, assume p divides r
+
--2
+
rn_l a
But A has no zero divisors and a # 0 so
--2
+
i, 2,
n.
pt
n, and suppose
Write
i.
ptbj
rj
at least one of these integers, say
by,
is
for some
is not
divisible by p.
n
+
+
r2a2 + rla
Now, 0
r na
Set z
bn an +
+
i.
Otherwise z
(by,
p)
b2 a2 + bla.
By Case I, either z
# 0 and
t
p (b
nan
+
+
0 in R, then a is QAI by I, since
If z
t
since p z
# 0
Rp.
0, z e T(R); in fact, z
0 in A which implies
and we are finished again by l; or z
b2 a2 + bla).
in A.
bnn +
+
b2a2 + bl
a
0
If this is the case, (R is QAI)
then for some integers
u,
Cu; ClZ
Cl, c2,
ClZ #
0 and
Cu z
0 implies p does not divide c
and
ClZ #
0
c z)
u
0 in A.
u
+
+
u
I.
+
+
Then
2
+
0.
ClZ
Cu()u +
c2z
+
c2()2 + ClZ
This gives
+
n
+
c2() 2 + Cl(Z)_ Cu(b
n
c2(bnn
+
+
+ b2a +
ba 2 + ba) 2 + c l(bnn +
bfa) u +
+
ba 2 + bl)
CLASS OF RINGS WHICH ARE ALGEBRAIC OVER INTEGERS
_
-
Note that the coefficient of a in the last equation is
not divide b I, then p does not divide
If p divides b
w
z
n
b a
n
bla
Clb I which
I but does not divide
+
+
b2,
b3a3 + b2a2.
649
Clb I.
If p does
implies a is QAI by ().
let
Then
--w
--z bla --z,
b a
1
z
because A has characteristic p, which divides b
I.
Cu()u +
c2()2 + cl Cu(bn
+
Note that the coefficient
Clb 2,
a is QI by (i).
of--a 2
here is
+
Clb 2,
+
ba2)U +
and since p does not divide
If p divides b 1 and b but not b 3, let v-- z
2
and do as before, since again v- z
Since there is b
by p we eventually reach a stage where a is QAI by (1).
implies a is QAI.
THEOREM 32.
Hence A
R/c(M)
Y
bla- b2a
which is not divisible
In any case, a in R
is a QAl-rlng.
Suppose R is a QAl-ring.
Then R is a subdirect sum of prime
QAl-rings and hence a subdirect sum of periodic fields and orders in algebraic
rational division algebras.
PROOF.
Let F be the family of all maximal multiplicative systems in R.
By Lemma 30, R is a subdirect sum of the integral domains {R/c(M)}
Lemma 31 implies that each R/c(M) is a prime QAl-ring.
M
E
F’ and
Theorem 18 and
Theorem 19 yield the desired form for the subdlrect summands.
The example at the end of Section 2 shows that the converse of Theorem 32
fails to hold.
The reader is encouraged to compare Theorem 32 to
[5,
Theorem
2
D.F. RALL
650
ACKNOWLEDGMENT.
A majority of this paper appeared in the Ph.D. dissertation
written under the direction of Professor F. Kosler at the University of Iowa.
The author would llke to express his appreciation to Professor Kosler for
his help and guidance.
REFERENCES
i.
Hentzel, Irvln Roy. Alternative Rings Without Nilpotent Elements, Proc.
Amer. Math. Soc. 42 (1974), 373-376.
2.
Jacobson, N. Structure Theory for Algebraic Algebras of Bounded Degree,
Ann. Math. 46 (1945), 695-707.
3.
Structure of Rings,
Jacobson, N.
Provldence, R. I., 1964.
4.
Luh, Jiang and J. C. K. Wang. The Structure of a Certain Class of
Rings, Math. Japo.n. 20 (1975), 149-157.
5.
Osborn, J. Marshall.
163-369.
American Mathematical Society,
Varieties of Algebras, Advances in Math.
_8 (1972),