Math. Z. 185, 93-104 (1984)

0 Springer-Verlag 1984

Extensions of Chain Rings

Hans-Heinrich Brungs ' and Giinter Torner ' Department of Mathematics, University of Alberta, Edmonton, Alberta, Canada

Fachbereich Mathematik der Universitat Duisburg, LotharstraBe 65, D-4100 Duisburg, Federal Republic of Germany

1. Introduction

It is the purpose of this paper to study the construction of integral domains R in which the lattice of right (left) ideals is linearly ordered. Such rings are called right (left) chain rings. The unique maximal ideal is denoted by J(R) and the group of units by U(R). If R contains exactly n prime ideals, not counting R and (O), we say that R has rank n. The question whether every right ideal in a right chain ring of rank one has to be two-sided was left unanswered in [3].

Using ideas of Mathiak and Rohlfing ([8, 91) we construct a localization R, of an Ore extension of a chain ring R, with a monomorphism a and a a- derivation 6. The ring R, is an extension of R, with J (R, )nR,=J(R, ) and is again a chain ring. However R, can turn out to be a rank one right chain ring without any two-sided ideals besides R,, J (R , ) and (0) even if R, is com- mutative or has infinite rank. This follows from an investigation of the re- lationship between right ideals (Sect. 3) twosided ideals, completely prime ideals (Sect. 4) and prime ideals in R, and R , (Sect. 5).

The construction discussed here fails to provide an answer to the question whether prime ideals exist in chain rings which are not completely prime. It will be proved (in the case 6=0) that such an ideal can exist in the extension R , only if one existed in R, already, see Theorem 4.

The value group of a rank one commutative or right invariant right chain ring is isomorphic to a subgroup of the real numbers under addition. Examples of rank one chain rings can be constructed with the above mentioned method whose value groups can be considered as generalizations of Chehatas simple ordered group (Sect. 6).

2. The Construction

Let R, be a right chain ring with monomorphism a and a-derivation 6. We construct the Ore skew polynomial ring R = R, [x, a , 61 = { ~ a i x i ; ai in R,) with multiplication in R defined by xa = o(a) x + 6(a) [ 6 ] .

It is assumed that a and 6 are compatible with J(R,) in the following sense:

94 H.H. Brungs and G. Torner

( V ) i) a(r ) is in J(R,) if and only if r is in J(R,) for r in R,. ii) 6 ( J ( R 0 ) ) is contained in J(R,).

Let S be the subset of polynomials x a i x i in R with at least one ai in U(R,). To show that S is multiplicatively closed we consider f ( x ) = x f k x k , g (x ) =Cg,x i in S with f, and gj the highest coefficient in U(R,). The coefficient hi+ of xi+' in h(x ) = f ( x ) g ( x ) has the form hi+ j= C f , o ~ ( ~ , ) + C fkgj. Here

k + f = i + i k + l > i + i

gi denotes an element obtained from a coefficient g, of g(x ) by applying a se- quence of mappings a or 6 with 6 appearing at least once. Since k+t >i+j in the second sum, either fk or g, is in J and hence either f , or gj is in J (Ro) using condition (V). Every summand in the first sum is in J(R,) except the term f ,ai(gj) which is a unit. This shows that h i+j is a unit and h ( x ) is in S. This argument also shows that f ( x ) g ( x ) can not be an element of S if one of the factors is not in S.

Since R , is a right chain ring, it is a right Ore-ring and the skew field of fractions Q = Q(Ro)= { a b - ' ; b +O, a in R,} exists. We need one further con- dition in order to make use of Q : given finitely many elements pi, i = 1, . . . , n in Q then there exists O+d in R, with dpi in R , for all i. However, the following lemma shows that this condition holds if and only if R , is also a left chain ring.

Lemma 1. Let R , be a right chain ring and Q its skew field of fractions. If for an arbitrary element q in Q an element O+d in R , exists with dq in R , then R , is a left chain ring.

Proof. Let a, b + 0 be elements in R , and q = a b ' in Q follows. Assume dab- ' =r is in R , for some d+O in R,. Then either r=dr, and a b - ' = d - ' r = r , or d = rd , and d- ' r = d; for elements d l , r, in R,. In the first case a b ' = r, , a =r, b and Rob?Roa follows, in the second case ab-' = d , I , b = d , a and R,a 2 Rob .

From now on it is assumed that R , is a right and left chain ring - chain ring for short. This implies that the elements of Q have the form a or a- ' where a is in R,.

Lemma 2. Let R , be a chain ring, a a monomorphism, 6 a a-derivation of R , satisfying condition (V) . Then there exists a natural extension of a and S respectively to Q defined through a(a- I ) = a(a)- and 6 (a - ') = - o ( a ) ' 6(a) a- I .

The proof of this lemma can be given by considering several cases and will be omitted.

We will denote the extension of a and 6 to Q by a , 6 again. The skew polynomial ring Q [x , a , 61 = {C c- ' aixi , 0 + c, ai in R,) can be constructed and is a left principal ideal domain. We used the fact that R , is a left chain ring to write the elements in Q in the form c - ' a with O+c, a in R,. The skew field D of quotients of the ring Q [ x , o, 61 exists and elements in this skewfield have the form

Extensions o f Chain Rings 95

where O+c, ai , b j are elements in R,. Hence D is the skew field of quotients of R , [x , a, 61. Given any two elements O+ f (x ) , g(x) in R, [x , a, 61. Then g(x) f ( x ) - = f; ( x )g l ( x ) and f , ( x ) g(x) =g, (x ) f ( x ) follows for certain elements f , ( x ) , g , ( x ) in R, [x , a, 61. If we cancel on the left the common factor of the coefficients of f , (x ) and g , ( x ) we can assume that one of these two elements is contained in S. Here we used the fact that R, is a right chain ring.

Lemma 3. For any two elements f (x ) , g(x) in R , [x , o, 61 = R there exist ele- ments f , (x) , g , (x) in R with f , ( x )g (x )=g , ( x ) f ( x ) and at least one of f , (x) and g,(x) is in S . The set S is an Ore system.

Proof. Only the last statement remains to be proved. If g(x) is in S and f ( x ) is in R then we are done if g,(x) is in S . If however f , (x ) is in S then f , ( x )g (x ) is in S which forces g,(x) into S by an earlier observation.

Theorem I. Let R , be a chain ring, o a monomorphism of R,, 6 a a-derivation satisfying condition (V). Then the ring of quotients R , = S ' R , [x , a ,6] exists and is a chain ring.

Proof. Lemma 3 states that S is an Ore-system and

R , = { f - ' ( x ) g ( x ) ; f ( x ) in S , g (x ) in R ) exists.

To prove that R , is a right chain ring we choose two arbitrary elements f - ' ( x ) g(x) and f - ' ( x ) h ( x ) in R , which we write with a common denominator f ( x ) in S . The elements g(x) and h(x) can be factored as follows: g(x)=gg,(x) , h(x)=hh,(x) with g, h in R,, g , (x) , h , (x) in S. Hence, f - ' ( x ) g ( x ) R , = f - ' ( x ) g R , and f - ' ( x ) h ( x ) R , = f - ' ( x ) hR, . The statement follows, since either g=i r or h=gr for some r in R,. To prove that R , is a left chain ring only principal left ideals of the form R , f (x) , R , g(x) with f (x) , g(x) in R have to be considered. Using Lemma 3 we know that there exist elements f , (x) , g,(x) in R , at least one of them in S with f , ( x )g ( x )=g , ( x ) f (x). If we assume that f , ( x ) is in S we obtain R,g (x )=R , f , (x )g(x) = R , g , ( x ) f ( x ) ~ R , f (x) . Similarly, R , f ( x ) z R , g ( x ) if g , ( x ) is in S . This proves the theorem.

The construction described above can be repeated with R , in place of R , if a suitable monomorphism o , and a a,-derivation 6 , are available satisfying condition ( V ) for R, . Assume 6=0 and that a second monomorphism a , exists for R,, satisfying (V ) and a commuting identity of the following form: o , a =orno, for some positive integer m. In that case extend a , to R by defining o , ( ~ a i x i ) = ~ o,(ai)iZim. It follows that o , ( s ( x ) ) is again in S if s (x ) is in S . We can therefore extend a , to R , , the condition (V) holds for R , , o , and 6 , = 0 and an extension R , of R , can be constructed. Finite and even transfinite chains of extensions of R, can be constructed in this way if suitable monomor- phisms and derivations are available.

3. One Sided Ideals

The relationship between right (left) ideals in R, and in R , is studied. It is one of the goals to decribe the set W,(R,) (W,(R,)) of all principal right (left) ideals

96 H.H. Brungs and G. Torner

in R , . This will be done in special cases, but not in general. Helpful in the following discussion if the introduction of an extension R, of R, which allows an extension & of the monomorphism a of R , such that 6 is an automorphism.

Lemma 4. Assume R , is a chain ring with a monomorphism a . Then there exists a chain ring R, containing R , and an extension 6 of a such that & is an automorphism of R, and for every a in R, there exists an n with c?'(cc) in R,. Condition ( V ) holds for k,, & i f it holds for R , and a.

Proof. We consider R,=R,[y, a ] = { x a i y i ; a' in R,} ; ya=a(a)y. The set M ={yn, n=0,1,2, ...} is an Ore-system in R, and the ring M 1 R 2 = { x - " x a i x i ; a, in R,} exists. R , = M p ' R 2 contains the subset R,= U y - " ~ o y", n =0,1,2, ... . We have R ~ ~ = y - n R o y n ~ y - n a ( R o ) y n = y - ~ n - l ' R o y ' n - ' ~ = R("-') o and it follows that R, is a subring and again a chain ring. If one defines &(y-"ayn)=y-"a(a)yn it follows that this is the mapping induced on R, by the inner automorphism of R , that sends u to yuy-I . The same is true for the inverse of these mappings and this shows that 6 is an automorphism of R, and an extension of a. Further, J(R,)= y " J ( R o ) y n . Hence R, and & satisfy (V) if R, and a do and this proves the lemma.

Can the a-derivation 6 also be extended from R, to the ring R,: This is possible if 6 and a commute i.e. if 06 = 60. In that case one defines 6(yyna yn) = y-" 6(a)yn where y-"ayn is in R, and a in R,. It must be checked that this is well defined and is a &-derivation of R,. This means that y-"ayn=y-"bym must imply y-" 6(a) yn = y-"6(b) ym for elements a, b in R,. We can assume that n>m and

y - ( n - m ) ayn-"= b or a = on-"(b) follows.

Hence y - n 6 ( a ) y n = y n 6 ( a n ~ m ( b ) ) y n = y ~ n a n ~ m ( 6 ( b ) ) y n ~ y m 6 ( b ) _ y m and the mapping $ is well defined. It follows very easily that 6(a+ P ) = 6(a) + $(P) and $(ag) = 8(a) 0) + ŝ (a for elements a, ,O in I?,. Finally, if the condition ( V ) holds for R,, a and 6 it also holds for R,, C? and 6:

The ring R, plays a crucial role in the proof of the following theorem.

Theorem 2. Let R , be a chain ring, a a monomorphism of R , and 6 a a- derivation satisfying condition (V). Then W,(R ,) = {x-"a R , , a in R,) and W;(R ,) = { R , axn, a in R,) for non-negative integers n, provided a is an automorphism or 6 is equal to zero. Further, x - " a R , = x - " b R , i f and only if om(a)Ro =an(b)Ro and x -"aR, ~ x - " b R , if and only if am(a)Roco"(b) R,. Proof. If a is an automorphism or 6=0 it is clear that the ring R , constructed above can be considered as a subring of R , and in fact that R , = S- Ro [x, a, 61 = 3- ' R^, [x , &, $1 = R , . To prove the last statement we observe that R , contains the subring R,[x, 6, $1 with the subset S^= { C a i x i ; ai in R,, at least one ai in u(R,)). The set Ŝ is an Ore system. The element x in R , remains algebraically independent over I?, , since 1 ai xi = 0 implies ~ x n a i x - " x i = O with xnaix-" in R, for all i and suitable n. Hence xnaix-" =0 and ai = 0 for all i.

Similarly there exists an integer n for a given element f ( x ) in S ̂ such that xn f ( x ) x - " is in R , and hence in S. This shows that every element in Ŝ is invertible in R , and R , = R , follows. The ring Rocx, 6, $1 is a right and left

Extensions of Chain Rings 97

Ore ring, since 6 is an automorphism. Further, 9- R, [x , 6 , = R, [x , 6 ,6 j 5- ' = R , since $={Xx ia i , a, in R,, at least one ai in U(R,)) using again the fact that 6 is an automorphism.

A principal right ideal in R , has the form f - ' ( x ) g (x )R , which is equal to g ^ , ( x ) f ; ' ( x ) R , for elements f , ( x ) in 9 and g*, (x) in R, [x , 6 , 8. Therefore f - l ( x ) g ( x ) R , =x-"axnR,=x-"aR, for a certain element a in R, and a certain non-negative integer n.

The principal left ideals in R , have the form R , f - ' ( x ) g(x) = R , g(x) with g(x) in R, [x , o, 61, g(x) = 1 a, xi = xi a, for suitable elements ai in R, . There- fore, R , g ( x ) = R l a = R l b x n for a certain element a=x-"bxn in R,, b in R,.

To prove the last part of the theorem we consider an element a in R,nU(R, ) . This implies a = f P ' ( x ) g ( x ) with f ( x ) = X a i x i , g(x)=Cb,xj in S . Hence, C aioi(a)xi = 1 bjxj with at least one bi in U (R,). This implies that oi(a) is a unit and because of (V) that a is in U(R,). Assume x n a R , = x m b R , . We multiply this equation from the left with x " + ~ and om(a)Rl = on(b)R, follows. Either om(a)r=on(b) or om(a)=on(b)r for some element r in R,. We saw that om(a)e=P(b) for a unit F in R , and r in U(R,) follows, using the above observation. The remaining statement is proved in the same fashion.

We single out the following facts:

Remark. Let R,, o, 6 and R , be as in Theorem 1.

Then U ( R , ) n R o = U(R,); J ( R , ) n R , = J(R,).

Corollary I. Assume R , is a chain ring, o a monomorphism of R,, 6 a a- derivation satisfying condition (V). If in addition 6 is an inner a-derivation, i.e. G(a) = za - o(a) z for some z in R , then every principal right ideal in R , has the form ( x - z ) " a R , for some a in R,, some nonnegative integer n.

Proof. We have x a = o(a )x + z a - o(a)z which implies ( x - z ) a = o(a) ( x - z). The corollary follows from Theorem 2 if x is replaced by x-z .

Corollary 2. Assume R,, o, 6 are as in Theorem 1, with 06 = 6 0 and o(a)R, 2 6(a) R , for all a in R , as additional condition. Then W,(R,) = { x - " a R , , a in R,, n a nonnegative integer).

Proof. We consider the ring R,. We saw (Sect. 3) that both o and 6 can be extended to R, with 66^=$6, since o6=6a. The principal right ideals of R , =S^- 'R,[X, 6, 8 are of the form a ~ , with a in R,, since 6 is an automor- phism (Theorem 2). We will show that a principal ideal U R , of R , can be written in the form x - " b k , for some b in R,. This is done by induction on n with 2 ( a ) in R,; the statement being true for n=O. Assuming a in R, with $(a) in R,, we have xa=;(a)x+$(a). However, a=&-"(a) for some a in R, and o(a)r=d(a) for a certain r in R , by assumption. Hence $(a)=$;-"(a)

A- - = o n (6 (a ) )=6-n (a (a ) r )=6- (n ')(a)&-"(r). We obtain: xcc=6(a)x +6- (n - ' ) (a )6 -n ( r )=6(a) (x+6-n(r ) ) and a=x- '6(a) (x+o-"(r ) ) follows. The element ( x + 6-"(r)) is in 9 and 6(a) R , =x-"6 R, , b in R, by induction. Therefore a R , = x-("+ l ) b R , and the above statement is proved.

98 H.H. Brungs and G. Torner

We complete the proof by showing that cz R , = v R , = w R , for v, w in R , is true only if v R , = w R , . This will follow if we show that an element f - ( x ) g(x) in R, is a unit in R , only if it is already a unit in R, . We can assume that f ( x ) = 1 and write g(x )=C c,xt=(C i i x i ) - ' (1 6,xj) with c, in R,, if, hj in R, for all t, i, j . This implies ~ l i i x i x c , x ' = x b ; . x j and at least one of the 6;s is a unit in R, . Using condition (V) we see that at least one of the ct7s has to be in u(R,) and hence in U(R,), (Remark after Theorem 2).

4. Two-sided Ideals

Let R,, o, 6 and R, be as in Theorem 1. The two-sided ideals of R , are described in terms of certain two-sided ideals of R,. To prepare the definition of these ideals in R, we consider the commuting rule for elements a in R, and a power xn of x. We have xna=A:(a)xn+A:-,(a)x "Lf...+ A;(a)xi+ ...

n

+ A",(a where A; is defined through (ot + 0)" = C A; ti if t is a commuting i = O

indeterminate. Hence A ; = a2 6 + 0 6 o + 6 02, A: = an and A", 6" for all n. Definition. A two-sided ideal I in R, is called (o, 6)-compatible if conditions i) and ii) are satisfied:

i) o ( 1 ) z I ; 6 ( 1 ) ~ I . ii) The element a in R, is contained in I if there exist f,, ..., fn in R,, not

n

all in J(R,), with 1 fjA{(a) in I for all i, 0 5 i 5 n. j t i

Remark I . The condition (V) is equivalent with the following condition: J(Ro) is (o, 6)-compatible.

It must be proved that ii) follows from (V) for the ideal J(Ro). Assume an n

element a exists in R,, not in J(R,), but x f,A:(a) in J for i=O, ..., n for j > i

certain elements f,, . . . , fn in R,, not all in J(R,). For m = max { j , f, a unit} we n

have f j AL(a) not in J, since f, A:(a) is not in J, but f, AL(a) is in J for j>m jbrn

-a contradiction.

Remark 2. An ideal I in R, is (a, 0)-compatible if and only if o ( 1 ) z l and o(a) in I implies a in I for any element a in R,.

We assume first that I is (o, 0)-compatible: If a is in I then o(a) is in I by definition. If a(a) is in I we choose n= 1, f,=O, f , = 1 and f,A;(a)+ f , Ah(a) = foa + f , 6(a) =0 is in I . Since f , A: (a) = o(a) is also in I we can conclude that a is in I. Conversely, let I be an ideal with o(a) in I if and only if a is I. We must show that condition ii) holds. We observe that A: = 0 for i < n, since 6 =O. We therefore assume fjA$a)=fjoj(a) in I , at least one J;, a unit and a in I follows.

Extensions of Chain Rings 99

Theorem 3. Let R,, R, be as in Theorem 1. There exists a one-to-one cor- respondence between ideals in R , and (a, 6)-compatible ideals in R, .

Proof. Let I be an ideal in R , . The intersection I, = 1 n R, is an ideal in R,. We prove that it is (o, 6)-compatible. Let a be in I, and xa=o(a )x+6(a ) follows. Either a(a)r=d(a) or a(a)=h(a)r for some r in R,. In the first case we have a(a)x+b(a)=o(a)(x+r) in I and hence o(a) in I and I,, since x + r is in S. Similarly, we conclude that 6(a) is in I, in the second case and o(a), d(a) in 1, follows for every a in I,. Let f,, ..., f, be elements in R,, not all in J(R,).

n

The element x f , x i is in S. Assume a is an element in R, such that ci n 0

= f, d!(a) is in 1, for i =0, . . . , n. It follows that cioRo z ciRo is true for a jzi

certain index i, and all i =0, . . . , n. That means that

is in I, S(x) is in S and a in I, follows. This proves the (a, 8)-compatibility of 10 .

We prove next that R , I, R , =I , and R, I,R, c 1 is obvious. Any element in R, has the form f ~ ' ( x ) g ( x ) = f ~ ' ( x ) a g , ( x ) with a in R,, f,(x), g,(x) in S. If such an element is also in I, then a is certainly in I,. Hence R, I, R , 2 I and I = R , (1 n R,)R , is proved.

To complete the proof of the theorem we must show that R, I,R, n R , =I, for any (a, 6)-compatible ideal I, in R, . Let a be in I,, f - g, h- ' k in R , , then it must be proved that b is in I, if b = f ' g a h l k is in R,. Using the (0,6)- compatibility of I, we see that g a = a l g , with a' in I, and g, in S. Using the Ore-property of S we obtain f l g a h l k = f L a ' g , h ' k = f l a ' h ; ' g , k for certain elements h,, g, in S. By the same property there exist elements u, w in S, c in R, with ua1=cwh1 and as before ua'=at'u, for u, in S, a" in I,. Since a"u,=cwh, and a" is in I,, u , ,w,h , are in S, the element c i s in I, and a'h;' =u- 'cw and finally b= f - ' u l c w g , k is in R, with c in I,. If we write t =u f and s = wg, k, we obtain t(x)b = cs(x) with t = t (x) and s = s(x) in S. Comparing

n n

coefficients we get: 1 tjAi(b) is an element in I, for all i if t(x)= 2 tixi. It j 2 i 1 = 0

follows that b is in I, and the theorem is proved. Using the last theorem it is easy to describe the completely prime ideals in

Rl.

Lemma 5. An ideal I in R , is completely prime if and only if I, = 1 n R, is a completely prime ideal in R,.

Proof. One direction is trivial and we assume that I, is completely prime in R,. It follows from the above theorem that I, is (a, 6)-compatible. It is sufficient to show, [4], that f - ' g f - ' g in I implies f - ' g in I for any element f - ' g in R , . We write f - ' g= f - ' a g l for a in R,, not in I,, f l , g , in S and will show that f - l a g , f - ' a g , is not an element in I. Using the (a,6)-

100 H.H. Brungs and G. Torner

compatibility of I , we know that for s (x ) in S there are elements a', a" not in I, with

s ( x ) a = a r s , ( x ) and a s l ( x ) = s ; ' ( x ) a " t ( x )

for certain elements s , (x) , s ,(x) and t(x) in S.

Hence : f - l g f - l g = f - l a g , f - j a g , = f - ' a f , - ' g , a g ,

= f - I f ; 1 a " w g 2 a g l = f ; ' f ~ ' a " a ' w , g , g ,

for f , f , , f,, g , , g,, g,, w, w , in S, a, a', a" in R,, not in I,. It follows from the above theorem that f - ' g f - g is not in I , since a"ar is not in I,.

5. Prime Ideals

The relationship between prime ideals in R , and R , is studied in this section.

Lemma 6. If I, is a (a , 6)-compatible prime ideal in R, then I = R , 1, R , is a prime ideal in R , .

Proof. Let f,- ' ( x ) a g , ( x ) = u and f; ' ( x ) b g , ( x ) = v be elements in R , not in I with f,(x), g,(x) in S, a, b in R,. This implies that a, b are not in I, and an element r exists in R , with a r b not in I,. Hence, ug; ' r f , v is an element in R , , not in I , and I is a prime ideal in R , .

Given a prime ideal 1 in R , , we cannot make a definite statement about I n R,. However, a related problem is discussed. Since it is not known whether prime ideals in chain rings must be completely prime ideals, it is of interest to decide whether a prime ideal which is not completely prime can appear in R , without such a prime ideal existing in R,. The following theorem answers this question in the case 6 = 0.

Theorem 4. Let R,, R , = S - R , [ x , o ] be as in Theorem 1 with 6 = O . I f there exist prime ideals in R , which are not completely prime then such prime ideals do also exist in R,.

We recall the following result from [4] (Lemma 2): Given elements a, r in a right chain ring then ra=uari for a unit u and an element r' in the ring. Using this we first prove the following technical result.

Lemma 7. Let R,, a and R , be as in Theorem 4, f ( x ) an element in S and a an element in R,. Then there exist integers i, j>= 0 and elements r, , r, in R,, u in U(R,) with a f ' ( x ) = s ( x ) - ' r , o i (a ) t ( x ) and f ( x )a=uoj (a ) r , f , ( x ) with s(x), t ( x ) , f l ( x ) in S.

Proof of Lemma 7. We have a f - ' ( x ) = s ( x ) ' b t ( x ) for elements s(x), t ( x ) in S, b in R,. Hence, s ( x ) a = b t ( x ) f ( x ) with s ( x ) = ~ c i x i . There exists an i with cioi(a)R,?cjoj(a)Ro for all j and b can be chosen equal to cioi(a) since t ( x ) f ( x ) is in S, proving the first part of the lemma with r, =c i . To prove the

Extensions of Chain Rings 101

second part let f ( x ) = di xi and f ( x ) a = di a i (a ) xi follows. As before, dja j (a)R,r>dio i (a)R, for all i and for a certain index j . Using the above cited result we have d jp j (a )= u aj(a)r2 for a unit u in R,, r2 in R , and f ( x ) a =d ja j (a ) f l ( x ) = u o J ( a ) r 2 f l ( x ) follows with f l ( x ) in S.

Proof of Theorem 4. Let Q + ( O ) be a prime ideal in R , , not completely prime. Let P be the smallest, completely prime ideal in R , containing Q. We know ( [ 4 ] , Theorem 3) that there are no two-sided ideals properly between P and Q. The intersections Po = P n R , and Q , = Q n R , are a-compatible ideals in R , with P, completely prime (Lemma 5), and Qo not completely prime. If Q , is prime we are done and we assume from now on that Q , is not prime. The proof of the theorem will be completed by exhibiting a prime ideal in R , which is not completely prime.

By assumption there exists an element a in R o with a not in Q,, but aR,a in Q,. Let P' be the smallest completely prime ideal in R , containing a, and P' G Po follows.

W; write I( ')= U v a R , ; I(")= U v a n ( a ) R o and 1'") is the smallest v e U ( R o ) v c U i R o )

two-sided ideal in R , containing on(a ) ( [ 4 ] , Lemma 2) for all nonnegative integers n .

The ideal and the element ( ~ " ( a ) ) ~ are contained in Q,.

Case 1 . I i l ) c I ( o ) . This implies o(a )= v a r for t. in U(R,) , r in R,. Therefore, on(a)= on- ' ( v ) on- ' ( a ) an - ' ( r ) is contained in I ("- ') and I i n ) c I ( " 'I follows for all n z 1. The product I ( i ' I ( J ) is contained in and hence in Q,, which implies that a i (a )vo j (a ) is an element of Qo for every v in U(R , ) and non- negative integers i , j . We know from Theorem 3, iv in [4] that there exists a unit u in U ( R , ) with (au)" not in Q for all positive n. With u = f - l ( x ) g ( x ) - f ( x ) , g (x) in S - we obtain

for elements r , , r, in R,, u , in U(R , ) , f , ( x ) , g , ( x ) in S using Lemma 6. This proves that ( a ~ ) ~ is in Q for every u in U ( R l ) - a contradiction.

Case 2. I ( ' ) . We assume first that 1'") s I("+ 'I for n = 0 , 1 , 2, . . . and want to show that (I("')2 G Q,. The element u on(a ) von (a )= an+ ' (a ) r von (a ) for some r in J(R,) where u, v are in U ( R o ) . As mentioned earlier, we can factor r =r , r, with r, in U(R,) , r, in J ( R o ) such that r2 va" (a )=von(a ) r ; for some r; in R,. We get on+'(a)r l von(a)r ;=(on+' (a) ) ' s r2 for some s in J (Ro) . This element is in Q,, since a2 is in Q,. The ideal I = U I(") satisfies I ' c Q , and a contradiction follows as in case 1.

We consider the last possibility: I(') $ I ( ' ) z . . . 5 Ii") = I("+ 'I. This implies I in) =I in+ ' ) for all nonnegative integers i. We obtain, as in the above argument, ( I ( k ) ) 2 ~ Q 0 for k = O , 1, ..., n - 1. Let Qb= u o n ( a ) R o . This is a two-sided

u € U ( R o ) ideal and there are no two-sided ideals properly between Qb and I(")= I ( [ 4 ] , Lemma 2).

Let z, y be in Qb and z=on(a ) r , r y = a n ( a ) s for r, s in R , follows. This implies that z = an(a)r y = ( ~ " ( a ) ) ~ s is contained in Q , and (Qb), G Q,.

102 H.H. Brungs and G. Torner

The ideal 1'") is idempotent, otherwise 1' 5 1 = 1'") and I 2 c Qb follows. This in turn implies 1 4 ~ Q 0 and ( ~ 2 4 ) ~ in Q, for every unit u in R, - a contradiction as in the previous cases. Hence, 12=1 is a complete prime ideal ([4], Theorem 1 (ii)).

The prime ideal I can not be finitely generated ([4], Theorem 2) as a right ideal and a n ( a ) R o ~ I. Therefore a unit u exists in R, with an(a)Ro ~ u o n ( a ) R o and u ' an(a)Ro E on(a) R,. This implies Qb 5 on(a) R, and o " ( ~ ) ~ contained in Qb shows that Qb is not completely prime. All conditions of Theorem 4 in [4] are satisfied for the pair I and Qb which proves that Qb is a prime ideal in R, which is not completely prime.

6. Further Remarks and Examples

We consider o-compatible (i.e. (o, 0)-compatible) ideals and begin with the following definition. Let 1 be a right ideal in R,. The largest o-compatible ideal f contained in I is called the o-kernel of 1.

Lemma 7. Let R be a chain ring, o an automorphism of R. i) Every ideal is o-compatible in R if a has finite order n.

ii) I f P is a completely prime ideal in R and not o-compatible then there exist non-finite strictly ascending and strictly descending chains of completely prime ideals.

iii) The o-kernel i o f an ideal I of R is completely prime if is n In. Proof. i) If a (a )=ar+O we obtain on(a)=ara(r) ... on- ' ( r )=a, and r is a unit. A similar argument shows that s is a unit if o (a )s=a .

ii) We assume o ( P ) s P , replacing o by a - ' if necessary. Then on+ ' (P )zan (P ) and o - " ( P ) ~ o - ( " + " ( P ) and all these ideals are completely prime.

iii) Let 1 be any ideal in R with o(1)c l . We show that D = r) an ( l )= i . Obviously we have o (D)sD . Assume that a(a) is in on(l) for all n, then a -.n- - I (b,) for some bn in I and all positive n. Hence, o(a) in D implies a in D and D is o-compatible. Let L be any o-compatible ideal between D and I . Then L c on(l) for all n and L c D, which shows L = D = L The same type of argu- ment shows that f = n a-"(I) in case a ( I ) z l .

We have the assumption i~ n In, and is a completely prime ideal if f = n In P >sing [4], Theorem 1 (iii). The containment o ( l ) c 1 implies o(P)c_ P and Ic P = n o n ( P ) s n o n ( l ) = i implies F=P is a completely prime ideal as the intersection of the on(P). Similarly, f = P = n o-"(P) in case I ~ o ( 1 ) which proves the lemma.

The group of values of a commutative rank one valuation ring is isomor- phic to a subgroup of the additive group of real numbers [7]. The construction discussed in the previous sections provides us with examples of rank one chain rings. The following generalization of the value group of a valuation ring was defined in [2] for an integral domain R. Let W={aR; a+O in R) and H(R) = {T; r $0 in R) where T is the mapping from W to W defined by T(a R) = ra R.

Extensions of Chain Rings 103

With f, . f2 = rF2 and f, 2 fz if r , a R E r,a R for all a in R we consider H(R) as a partially ordered semigroup. We are not able to determine H(R) for all rank one chain rings, but will compute this semigroup in a typical example.

Example 1 . Consider R o = K[[t]] the commutative power series ring in one variable over a commutative field K . Let a be the monomorphism of R, defined by a( t )= t2, o ( a ) = a for a in K. This monomorphism is J(R,)- compatible. One can therefore construct R , = S ' R, [x, a] and it follows that the principal right ideals of R, are of the form x " tk R, = x-" tkxn R , . However, x- ' t x is an element mapped onto t by the inner automorphism of R, induced by x. This inner automorphism is an extension of a . We therefore write X-"tk R, = tkJ2*Rl and this provides us with a description of all principal right ideals of R, , i.e. W(R,)= {tkJ2" R, , k, n nonnegative integers}. This set corre-

k sponds to the set W' of nonnegative rational numbers -, however

2" tq' R , 2 tq2 R, iff q, s q 2 in W'. Using this observation one can represent the elements in H(R,) by functions from W' to W' whose graphs consist of finitely many linear pieces. Let for example f (x) = t xZ + (t3 + t8) x + t1 O in R , .

Then t l + ~ Z R for 0 5 ~ 5 1

t 3 + "R, for 1 5 2 5 7 tlO+'R, for 7 5 2 , z i n W'.

The function can be represented by a function f defined in W' with f(z) = Min (1 +4z, 3 + 2z, 10+ z) for z in W'. In this example R,, t R, and (0) are the only o-compatible ideals of R, which implies that R , is a rank one right chain ring with no other two-sided ideals beside R, , J ( R , ) and (0). However, in this case rank R, =rank R , and the following (archimedian) property still holds: given u in J(R ,), O + v in R, then u" R, 5 vR, for some n.

Rohlfing in [9] has computed further examples of groups related to H(R, ) which can be considered as generalizations of the simple ordered groups constructed by Chehata in [5].

Example 2. We conclude with an example in which rank (R,) = co, rank (R ,) = 1 even though W(Ro) may be considered as a subset of W(R,) as it is always the case in the above construction. Let G =@ Zi, i = 1,2,3, . . . where Z i = (Z, +) is the additive group of the integers for every i. Then G is an ordered group under the lexicographical ordering and contains the subsemigroup H = {g in G, gzidentity = e l . Let K be a commutative field. The semigroup ring K [HI = { C a h h ; a, in K , h in H almost all a,=O) contains the multiplicatively closed subset M consisting of elements C a h h in K [HI with a,+O. We set R, =M- 'K[H] and H(R,)EH.

The mapping a of H with a (z , , z,, ...)=( c , , c,, ...) in H with c , =0, c i + = zi for i = 1,2, . . . induces a monomorphism of R, which is again called a . The only a-compatible ideals of R, are R,, J(R,) and (0). Hence, R, can be

104 H.H. Brungs and G. Torner

constructed and is a rank one chain ring. In this case however elements u in J(R,) , v+O in R , exist with u n R , z v R , for all n.

References

1. Boto-Mura, R.T., Brungs. H.H., Fischer, J.L.: Chain rings and valuation semigroups: Comm. Algebra 5, 1529-1547 (1977)

2. Brungs, H.H., Torner, G.: Right chain rings and the generalized semigroup of divisibility. Pacific J. Math. 97, 293-305 (1981)

3. Brungs, H.H., Torner, G.: Chain rings and prime ideals. Arch. Math. (Basel) 27, 253-260 (1976) 4. Brungs, H.H., Torner, G.: Prime ideals in right chain rings. Preprint 5. Chehata, C.G.: An algebraically simple ordered group. Proc. Lond. Math. Soc. 2, 183-197 (1952) 6. Cohn, P.M.: Free rings and their relations. London: Academic Press1971 7. Krull, W.: Allgemeine Bewertungstheorie. Journal Reine Angew. Math. 167, 160-196 (1932) 8. Mathiak, K.: Zur Bewertungstheorie nichtkommutativer Korper. J . Algebra 13, 586-600 (1981) 9. Rohlfing, U.: Wertegruppen nichtinvarianter Bewertungen. Dissertation, Braunschweig 1981

Received April 29, 1983