- Email: [email protected]
Coloring of commutative rings
JOIlKNAL
OF ALGEBRA
116,
208-226 (1988)
Coloring
of Commutative
Rings
ISTVAN BECK ~epa~trne~t of mathematics and Compotes Science, University of Haifn, Haifa 31999, Israel Communicated by Kent R. Fz&er
Received October 29. 1986
1. INTRODUCTION
The purpose of this article is to present the idea of coloring of a commutative ring. This idea establishes a connection between graph theory and commutative ring theory which hopefully will turn out to be mutually beneficial for these two branches of mathemathics. In this introductory paper we shall mainly be interested in characterizing and discussing the rings which are finitely colorable, leaving aside, for the moment, possible applications to graph theory. Let R be a commutative ring. We consider R as a simple graph whose vertices are the elements of R, such that two different elements x and y are adjacent iff xy = 0. We let x(R) denote the chromatic number of the graph, i.e., the minimal number of colors which can be assigned to the elements of R in such a way that every two adjacent elements have different colors. A subset C = (xi, .... x,} is called a clique provided xixj = 0 for all i # j. If R contains a clique with n elements, and every clique has at most y1 elements, we say that the clique number of R is n and write clique R = n. If the sizes of the cliques in R are not bounded we define clique R = co. We shall show that clique R = co actually entails the existence of an infinite clique. Obviously x(R) 3 clique R and for general graph G we certainly may have x(G) > clique G. However, in the case of commutative rings we have not found any example where x(R) > clique R. The lack of such counterexamples together with the fact that we have been able to establish the equality x(R) = clique R for certain (rather wide) classes of rings like reduced and principal ideal rings motivates the following conjecture. Conjecture 1. X(R) = clique R.
We start the paper with a few examples to give the reader some understanding of the concepts and problems involved. 208 0021”8693/88$3.00 Copyright 0 1988 by Academic Press, Inc. All rights of reproduction in any form reserved.
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In Section 3 we prove the main characterization of rings of finite chromatic number and name these rings Cokorings. Theorem 3.9 a ring R is a Coloring provided its nilradical is finite an equals a finite intersection of prime ideals. The next section discusses some of the characteristics of Colorings. Theorem 4.2 we learn that Colorings have a.c.c. on annihilators and important theorem enables us to show that in a Coloring every mini rime ideal is an associated prime ideal. In the fifth Section we show that the family of Colorings is closed with respect to certain operations. The main purpose of introduction separating elements in Section 0 is to ical attack Conjecture 1. We succeedin reducing the are of the ring and manage to prove Conjecture reduced or principal ideal rings. In these rings it is even true that ~(1) = clique I for any ideal I. The invariant x(R) (or clique R) is a useful num er to attach to a Coloring. We illustrate this in Section 7 y describing the finite rin chromatic number < 3. In this last section we also prove that x( clique < 4 and thus conjecture 1 is also chromatic number. Some readers of the paper would perhaps miss a formula for x( in terms of x(R,) and x(R,). Such a formula does not exist. It seemsthat x(W, x R2) depends on x(R,) and x(R2) and and R,. In general x(R, x R2) &x(R,) + x reduced the inequality becomes an equality. oblems involved and clarified how the 2-nilpoten ings. aper only scratches the surface of the comb divisors in a commutative ring. Ultimately, will involve graph theory. hen proving that x(R) = clique R for rings of low cbromat~c num (Sect. 7) we lean heavily on a useful graph theoretic result. It sa graph is 2-colorable if and only if the graph does not c~~ta~~any (see for instance [2]). Unfortunately, such a c~aracteri~a~o~ has not been found for k-colorable graphs when k > 3. Since our proof of X(R) = clique R for low chromatic numbers depends terization of the 2-colorable graphs, one could ture 1 would give us some insight into k-colorable ng my work with this paper I was lucky to people that could advise me and I am grateful ilertsen, K. P. Villanger, and A. Zaks for their pa Tke terminology is standard. We emphasize:
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ISTVAN BECK
R denotes a commutative ring with an identity, J is its nilradical, and R is said to be reduced if J= (0). Given subsets I and K of R, I : K= {r~R(rKcl). Moreover, O:I=AnnZ and if I=(x) we simply write 0 : x = Ann x and these last ideals are called annihilators. Given an ideal K, rad K= {Y E R 1r” E K}. The set of zero divisors in R is denoted Z(R). A prime ideal P is an associated prime ideal if P = Ann x for some element x in R. We let Ass R denote the set of associated prime ideals. The Krull dimension of R is denoted dim R and we use the simple fact that dim R = 0 if R is a finite ring. Finally, the number of elements in the set Z is denoted by #I or simply )I/, and A\B is the set {x 1XE A and x6 B).
2. CHROMATIC NUMBER OF SOMERINGS The proof of our first proposition is straightforward. PROPOSITION2.1. x(R) = 1 iff R is the zero-ring. PROPOSITION2.2. x(R) = 2 iff R is an integral domain, R g Zq, or R z
z2 cJa(~2). ProoJ: Suppose that x(R) = 2 and R is not an integral domain. Let xy = 0 where x and y are non-zero. Then (0, x, JJ} is a clique and since clique R < x(R) = 2 it follows that x = y. Thus x # 0 and x2 = 0. The ideal Rx is a clique and we conclude that #Rx = 2. Assume that z E Ann x. Then { 0, x, z} is a clique and therefore z E Rx= (0, x}. Hence Ann x = Rx. From the exact sequence
O-+Annx-+R-+Rx-+O we may conclude that #R = 4. If char R=4 we have Rr Z, and if char R=2
we derive Rs
z2wll(~2).
It is easily seen that these two rings have chromatic- and clique number 2. PROPOSITION2.3. Let pl, .... pk, ql, .... q1 be different prime numbers N=p:“~... p~qfm’+1...q,2mr+1. Then x(Z,) = clique Z, = and “k ml , . . qy + r. P;"'.Pkqt Proof: Let y, = p;’ . . . p;kqyl+ i . . . qy + l. Then yi = 0 in Z,, thus Z, y, is a clique with p;i . . . p;kqy’ . . . qy elements.
ht yi = yo,/qlii, 1
I is easily seen that this coloring attaches di~ere~t colors to a vertices.
After these prelimina~es we start the ~re~arat~~~s to ~baracte~~~ethe riglgs of Unite ~bromatic number. First a de~nition= An element x ia R is said to be~~~~~provid Rx is a ~~jte set. The fo~l~wi~~ lemma is essential in this section.
Proof. Let x, ) ~..~x,, ... be different finite ~~~rne~~s in $2. The ~~~rne~~~ x1x2, ...) x1x,, ... belong to the finite ideal Rx,. Thus for some infinite subsequence {a,) of (2, .... 73,... > we have xi xai = x1 .xa2= . . . . e then consider the sequence xai, xn2, ... and repeat the ~~o~e~~re-~o~~~~~~g in this way we construct a subsequence y1 p‘.a7 y,, .“. of the seq~e~~~ XI) .*.,x,, ... such that yi yj = yi yk when j, k > i. In this s~bs~q~~~~e
zij = yi - yi. Ubviously zyzkr = 0 if i < j < k < r. e scald anow ~~~s~~~~tan infinite clique. CorI .Z1,2Z3,4 =Z1,2Z3,5=O. Since 23,4#Z3,5 at kaSt one Of ZJ,~an Es.5IS rent from z~,~. If for instance z~,~fz,,, then (21,29z 1 is clique
with
two elements.
We obserw
that z~,~, zhs, aad .z6,$ we
merest
and if for instance z~,~f#{z~,~, z~,~) then {.z~,~,z~,~~z~,~) is a ~~i.~~~with three ~~~rne~~s. Continuing in this way we get an i~~~~te ~~~q~e”
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ISTVAN BECK
ProoJ: If R has an infinite clique C the homomorphic image C of C is a clique in i?= R/I, and since I is finite C is still infinite. Conversely, let { Zi >;” be an infinite clique in R. Hence xixj EI for i # j. Since the set of products (x~x~}~+j is a finite set we may apply the same technique as in the proof of Lemma 3.1 to establish an infinite clique in R. LEMMA 3.3. If the ring R contains a nilpotent element which is not finite then R contains an infinite clique.
Proo$ Assume xn = 0. The proof goes by induction on n. If x2 = 0 and Rx is infinite then R contains an infinite clique since in this case Rx itself is a clique. Suppose now that the lemma is true for elements of nilpotence degree n - 1. Then let x” = 0, n 2 3, and assume that Rx is infinite. Put y = x2. Then y”- ’ = 0. If Ry is infinite we conclude that R has an infinite clique. Assume next that Ry is finite. Then Rx/Ry is infinite. We note that Rx = RxjRy is an infinite clique in R = R/Ry. Since Ry is finite we conclude from Lemma 3.2 that R has an infinite clique. Remark. Hakon B. Eilertsen has pointed out that Lemma 3.3 can be proven without using Lemma 3.2. Having shown that the nilpotent elements are finite we now get LEMMA
3.4. If the nilradical of R is infinite then R has an infinite clique.
ProoJ Assume that the nilradical J is infinite. If every element in J is finite Lemma 3.1 implies that R contains an infinite clique. And if an element in J is not finite Lemma 3.3 implies that R contains an infinite clique. Assume that R is a ring without an infinite clique. Then Lemma 3.4 implies that the nilradical J is finite. Furthermore, Lemma 3.2 implies that R/J does not have an infinite clique. We therefore limit ourselves for a moment to reduced rings. LEMMA 3.5. Let R be a reduced ring which does not contain an infinite clique. Then R has a.c.c. on ideals of the form Ann x.
ProoJ: Assume that Ann a, < Ann a2 < . . . . Let xi E Ann a,\Ann aip 1, i= 2, 3, .... The non-zero elements yn = x,a,- i, n = 2, 3, ... form a clique and we claim that yi # yj when i # j. We have that yi yj = 0. The equality yi = yj would yield y: = yJ?= 0. This proves our claim and completes the proof of the lemma. Remark. This simple proof of Lemma 3.5 is due to H. B. Eilertsen.
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3.6. Let x and y be elements in R such that Aralnx and Ann y are dtjjferent prime ideals. Then xy = 0. LEMMA
Proof. Assume that xy # 0. Then x $ Ann y an y $ Arm x. Since Arm x and Arm y are prime ideals we derive Ann x : y = Anny:x=Ann y. However, Annx: y=Ann y:x=Ann Arm x = Ann y. e are ready to prove THEOREM
3.7. For a reduced ring R the following are e~~~~aie~t:
(1) X(R) isfinite. (2) Clique R is finite. (3) The zero-ideal in R is a finite intersection ofprime ideals. (4) R does not contain an infinite clique. ProoJ: The implications (1) s (2), (1) * (4), and (2) * (4) are evident. als. see (3) a (1) let (0) = P, n . . . n P, where P, 7.... P, are pri for e a coloring f on R by putting f(0) = 0 and f(x) = mi x#Q. We note that X(R)
214 THEOREM
ISTVAN BECK
3.9. The following conditions are equivalent for a ring R:
(1) x(R) isfinite. (2) Clique R is finite. (3) The nilradical in R is finite and equals a finite intersection of prime ideals. (4) R does not contain an infinite clique. Proof The implications (1) * (2), (1) =P(4), and (2) + (4) are trivial. To see (3) =+=(1) let J= P, n ... n Pk where PI, .... P, are prime ideals. If x $ J let f (x) = min{ i 1x $ P,}. This is a coloring of the elements outside J. Since J is linite we need merely a finite number of additional colors to color all of R. To prove (4) + (3) assume that R does not have an infinite clique. By Lemmas 3.2 and 3.4 this implies that J is finite and R/J does not have an infinite clique. It follows then from Theorem 3.7 that J is a finite intersection of prime ideals. We close this section with a somewhat peculiar application Theorem 3.9.
of
THEOREM 3.10. Let R be a ring which contains a finite ideal which is a finite intersection of prime ideals. Then the radical of any Jinite ideal is finite and equals a finite intersection of prime ideals. Furthermore, the ring has only a finite number of finite ideals.
Proof If R contains a finite ideal which is a finite intersection of prime ideals then x(R) < cc by the argument used in proving the implication (3) A (1) in Theorem 3.9. Let K be any finite ideal in R. Then R/K does not have an infinite clique (Lemma 3.2), hence x(R/K) < co (Theorem 3.9). Theorem 3.9 further implies that rad K/K is finite and rad K is a finite intersection of prime ideals. Since rad K/K is finite and K is finite we conclude that rad K is finite. Let A = (x 1x is finite}. Since clique R < co it follows from Lemma 3.1 that A is a finite ideal. Since A contains every finite ideal, the number of such ideals is finite. Remark. Observe that the word “finite” is only applied a finite number of times in this theorem.
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RlNGS
4. COLORINGS
Having characterized the rings of finite chromatic comber, we now ~~t~~dto discuss some of the properties enjoyed by these rings. At this stage it seemsappropriate to make the fo~~ow~~~ A ring R is called a Coloring ~ro~i~~~ xf
~~INITION.
4.1. If I is a ,finite ideal in a ring
&EMMA
, then I : x/Ann x is a finite
~-~~d~~e.
PVOC$ Consider the exact sequence O-+Annx-+I:x
f +(6:x)x------+
where f(t) = tx. Since (I : x)x c f it follows that I : ~~A~~ x is finite.
We generalize Lemma 3.5 and get the important THEOREM
4.2.
A Cobring has a.c.c. on ideals of t~~f~~~
ProoJ Let R be a Coloring and assume that Ann x1 < Ann x2 < . . . ~ The nonradical J is finite and we may assume that xi 4 J for i = I, 2, I._. Theorem 3.7 let J=P, n .I- n P, where the Pi are prime ideals. For an element x f R we get J:x=(P,:x)n
.‘. n(P,:x).
This shows that the family {J : x 1x E R > is finite. ~~~s~~ue~tly t a subsequence (yj> of (xi} for which J : yI = J: 7, = ~.. . Consider Ann y1 < Ann yZ < . . . . 9: y,=JJ:y,=
.‘..
This implies that Ann y1 < Ann y2 < ~.. c 9 : yi + that J: ~~~A~n y1 is finite. ernark. It can be shown that there exists a number N sue k~ any chain Ann x1 < Ann x2 < . . . -CAnn xk.
at k d N
Let R he a Coloring. Then Ass any minimal prime ideal P is an associatedprime I?, is a field or a finite ring.
THEOREM 4.3,
P. Furthermore,
~~~0~~ Assume R is a Coloring. Then clique ~rn~~~~§ that Ass R is finite.
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ISTVAN BECK
Let x E Z(R). Then XE Ann y for some maximal Ann y (Theorem 4.2) and Ann y is an associated prime ideal. If P is a minimal prime ideal pick x q!P. Then Ann x c P. Choose Ann t maximal in the family {Ann y 1Ann y c P}. We claim that Ann t is a prime ideal and establish thus that Ann t = P. Assume that ab E Ann t, a $ Ann t, and b # Ann t. If a $ P we consider Ann ta. It is easily shown that Ann t < Ann ta c P and we have arrived at a contradiction. If a E P we still consider Ann ta. If Ann ta c P the contradiction is repeated. On the other hand, if CE Ann ta, c$ P we consider Ann tc and get the contradiction Ann t < Ann tc c P. Let now P be a minimal prime ideal. We just showed that P = Ann x for some x. If x $ P one concludes that PR, = (0) and R, is a field. Suppose that x E P. Let J= P n P, n . . . n Pk where P,, .... Pk are the remaining minimal prime ideals. Pick y E P, n ... n P, -P. Then yPc J and since y$ P we conclude that PR, = JR,. The ideal J is finite and therefore PR, is finite. Obviously R/P E Rx. Since x E P = Ann x it follows that x2 = 0. Hence Rx is finite. Hence RxOR R, E R/P@, R, z R,jPR, is finite. We have shown that R, is finite. THEOREM 4.4. Let P be an associated prime ideal in a Coloring. Then either R, is a field or P is a maximal ideal. Proof Let Ann x = P. Suppose first that x E P. Then x2 = 0 and hence Rx is finite. But then R/P( E Rx) is finite, so R/P is a field. Hence P is a
maximal ideal. If x $ P and Ann x = P, then PR, = (0) and R, is a field. From Theorem 4.4 we get COROLLARY 4.5. An associated prime ideal in a Coloring maximal ideal or a minimal prime ideal.
is either a
5. THE FAMILY OF COLORINGS
This short section is devoted to a study of the properties enjoyed by the family of Colorings. Obviously we have THEOREM 5.1. A subring of a Coloring is itserf a Coloring.
The next theorem follows from Lemma 3.2 and Theorem 3.9.
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5.2. Let I be a finite ideal in a Coloring R. Then
THEOREM
Coloring. LEMMA
5.3. Let x be an element in a C&wing
Coloring. Proc$ Let F,, .... Fn be a clique in I?= R/A urthermore r 1x, .... r,x are distinct elements in /Ann x
nd from Lemma 4.8 we y Lemma 5.3 R/Ann x is a Colorin at I: x/Ann x is a finite ideal in /Ann x. One can a Theorem 5.2 and conclude that R/I : x is a Coloring. Proof.
COdUde
THEOREM5.5. A finite product of Colorings is a Coloring. s&ices to consider the product = I?, x R, of the Colori Put clique R, = n and clique R, = m. It is easily seen I?, 6 nm and Theorem 3.9 yields that R is a Coloring. e now generalize Lemma 5.3. 5.6. Let I be a jiinitel~v generated ideal in a C&wing. n I is a Coloring.
EOREM
Pr54$ Let I= (x,, .... x,). Then Ann I= Ann x1 n ... n Ann x,. ave an injection RJAnn I --) RJAnn x1 x . . x R/Ann x,. Each sf the ri R/Ann xi is a Coloring and by Theorems 5.1 and 5.5 the proof is complete. CQROLLARY 5.7. Let R be a Noetherian ring whose nilradical is ,fiinite. Then rad(Ann I)/Ann I is finite for any ideal I.
Proof. We apply Theorem 3.9 and deduce that Theorem 5.6 we conclude that RJAnn I is a Coloring. Sin a Coloring is finite the proof is complete. THEOREM 5.8. Let S be a multiplicative Then s is a Coloring. Moreover, x(R,) < x
closed set in a Co10
Proof. Let x(R) = n. To show that the grap to show that every finite subset is n-colorable [ 11.
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ISTVAN BECK
Let x1 =rl/s, .... x, = r,/s ,be a finite subset of R,. For each pair (i, j), 1 G i, j< m for which xixj = 0 pick sij E S such that svrirj = 0. Let t = 17s, and define rl= tr,. Then xixj = 0 iff rirj’=O. Since every subset of R is n-colorable we conclude that the set (xi, .... x,} is n-colorable. The proof also shows that clique R, d clique R.
6. SEPARATING E~E~NT~
The study of Colorings is essentially a look at the combinatorial aspects of the zero divisors. It is therefore not a surprise that the prime ideals have interesting properties in this respect. Let P be a prime ideal in a Coloring. If the elements contained in P have been colored we need at most one additional color for the elements outside of P. This is seen from the simple fact that xy = 0 entails x E P or y E P. It would be of interest to study ideals which enjoy this property, but for our purpose it seems more convenient to introduce the notion of separating elements. DEFINITION
6.1. An element x in R is separating provided x #O and
ab = 0 implies xg = 0 or xb = 0.
We shall also need elements satisfying a more local property. DEFINITION 6.2. Let I be an ideal. An element x E I is I-separating provided xI# (0) and whenever ab = 0 for some elements a, bgI then xa=O or xb=O.
(1) Note that x is separating it? x is R-separating. (2) We do not require a # b in ~e~nitions 6.1 and 6.2. (3) An R-separating element x contained in I fails to be I-separating if x1= (0). If, however, xf# (0) then x is also I-separating. Remarks.
Before we reveal the purpose of introducing separating elements we wish to discuss their existence. PROPOSITION Proof:
6.3. If Ann x is a prime ideal then x is separatirzg.
Straightforward.
PROPOSITION 6.4.
A non-zero ideal I in a Coloring contains a separating
elemem. Pro&
Since R has a.c.c. on annihilators (Theorem 4.2) it is easily seen
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that I contains an element x such that Arm x is a prime ideal. Allay then Proposition 6.3. Using the same technique we derive T~~oR~~ 6.5. Let I be an ideal in a ~olor~~~ not co~ta~~~edin the ~ilrad~ca~.Then I contains an I-separating element. BrooJ: Let x E I be such that Ann x is a prime ideal P and P ;b I. Then x is R-separating and xI# (0) since Ann x = P ;S I. If I is an ideal for which I2 = (0) then I does not c~~ta~~any ~-separatism element. THEOREM 6.6. Let Z be a principal ideal in a Coloring. If 1’ # (0) then I contains arr I-separating element.
Prroof: Let I= Rx and suppose x2 # 0. Since R has a.c.c. on annihilators it is easily shown that 0 : x2t is a prime ideal for some 1E xl is I-separating. Let a, b E B and assume ab = 0. Write a = rx and b = sx. Then rs-sx2 = 0. ence YSis contained in the prime ideal 0 : x2t. If for instance r E 0 : x22 we rive that (p.~)(xt) = 0, i.e., afxt) = 0. ~urthermore~ (tx)x = tx’ # 0 proves that (tx)I# 0. e now reveal the purpose of introducing separating ~l~~e~ts. LEMMA 6.7. Let I be an ideal in a Coloring and assume XE d B ~-~e~ar~ti~l~.Put I’ = Ann x n I.
(1) If x2 = 0 then clique F = clique I and x(F) = ~(~)~ (2) rf x2 # 0 then clique T = clique I- 1 aBd x(T) = ~(1) - 4. Proof. Assume first that x2 = 0. Then x E I’. Let clique I= y1an a maximal clique C = { y, , .... y,> in I, If x E C, say x = yr p th yz, .... yn E 1’ and as we noted x E I’. Hence Cc F, thus clique I’ = n. x & C, XC # (0) since C is a maxima1 clique in I. Assume that xyi xy, = 0 for 2 < i < m since x is Z-separating. Hence {x, yz9 .... yn> contained in P. This proves that clique I’ = az. We still assume that x2 = 0 and consider x(I’) and x(I). observe again that x E I’. color first F. If y E I- I’ we assign to y the bake color as to x, This shows that x(1’) = x(I). Assume next that x2 # 0, i.e., x $ I’. Then certainly cli ue 1’ Q clique I- 1 since to any clique in I’ we may adjoin x and create a larger clique in I. the other hand, let C = (yr, .... y,,> be a maximal clique in I. If x E C9
220
ISTVAN BECK
x = y, say, then again { yz, .... yHf is a clique in Z’. This shows that clique Z = clique I- 1. We leave to the reader to verify that if x2 # 0 then x(Z) = x(Z) - 1. We can now prove THEOREM 6.8. Let Z be an ideal in a Coloring. Let (.x1, ,.., x,) be a clique
Z-separating elements. Define k = #{i ( x: # 0} Ann(x,, .... x,). Then clique r = clique Z-k and x(Z’) = x(Z) - k. of
and
I’ = Zn
Proof: Since a proof of this theorem can utilize the same technique as the proof of Lemma 6.7, we omit the proof,
A useful corollary of Theorem 6.8 is the following THEOREM 6.9. Let P, , ,.., P, be the minimal prime ideals in a Coloring R. Let E(R) = # (i 1 R, is afield). Then clique R = clique J-I- E(R) and x(R) = x(J) + E(R).
Proof, Let 0 : xi = Pi (Theorem 4.3). Then {x,, .... x,} is a clique of R-separating elements. Apply now Theorem 6.8 and observe that R, is a field iff x? # 0. Theorem 6.9 makes it clear that in order to prove (or disprove) Conjecture 1 one may concentrate on the nilradical. In reduced rings the conjecture has been verified and we can prove a stronger result. THEOREM 6.10.
Let R be a reduced Coloring. Then clique Z= x(Z) for any
idea/ Zc R. Proof: By Theorem 6.5 Z contains an Z-separating element x. Since x2 # 0 one completes the proof using induction on clique Z (Lemma 6.7). We can also verify Conjecture 1 for principal ideal rings. THEOREM 6.3 1. Let R be a Coloring which is a principaE ideal ring. Then x(Z) = clique Zfor any ideal I in R.
ProoJ: We make a reduction to the case Zc J. Let PI, .... P, be the minimal prime ideals in R. If Z (f P, there exists an element y, E Z such that Ann yr = Pi. Then y, is Z-separating. Define I, = In Ann y, = In P, and observe that by Lemma 6.7 x(Z) = clique Z iff x(Z) = clique I’. It should now be evident that we may assume that Zc J. Assume I= Rx c J. If Z2= (0) then clique Z= x(Z) = #I. If 1’ # (0) the ideal Z contains an Z-separating element y, [Proposition 6.61. Let Z, = Zn
221
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Ann y, < I It follows from Lemma 6.7 that ~(1) = cli will clique I,. This completes the proof since the process above lead to an ideal I, for which 1: = (0) in which case x(Z,,) = chque In. Theorems 6.10 and 6.11 suggest the following conjecture. Conjecture 2. The chromatic number equals t e clique number for any
ideal in a Coloring. hen proving Theorem 6.11 we noted that it sufkes to prove Conjecture 2 for ideals contained in the nilradical. Moreover, a look at the proof of Theorem 6.11 should also make the proof of the next theore straightforward. THEOREM 6.12. Let R be a Coloring with the property that any ideal 1 for which I2 # (0) contains an I-separating element. ThePzx(I) = clique Ifor any ideal 1 in R.
Unfortunately, not every Coloring satisfies the hy Theorem 6.42. The local ring R = Z,[x, y] with relations x2 maximal ideal; M= (x, y) does not contain any M-separating element even ’ # (0). Despite this clique R = X(R) = p2. verified Conjecture 1 and 2 for principal i may wonder whether Conjecture 2 is true such rings. By the aid of Theorem 6.12 we shall pro First an observation. Let I= I, 0 I,. If 4, contains and I, separatin e~~rne~tx1 then x1 is as well I-s rating. This servation imphes that if a Coloring pi is a finite product of rings satisfying the hypothesis of Theorem 6.12 then this roperty is inherited by We therefore state THEoREM
6.13. Let R be a Coloring which is a finite product of reduced
rings and principal
ideal rings. Then x(I) = clique Ifor
any ideai IC R.
e now wish to characterize the rings in Theorem 6.13. The next theorem is a little surprising. 6.14. Let R be a local Coloring whose maximal ideal. Then R is reduced or a finite principal idea/ ring.
THEQREM
principal
ideal is a
Proof If R is finite then R is a local Artinian ring whose maxima%i is principal. Thus R is a principal ideal ring. Assume that R is not reduced and not finite. Let I be t y the set of finite elements. The ideal 1 is finite and d i maximal finite ideal.
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ISTVAN BECK
Pick x E I such that Ann x = P, P being a prime ideal. Since R/P z Rx we observe that R/P is a finite integral domain. Hence P equals the maximal ideal M = Rt. Let B = f : t 2 1. Since Ic Rt it follows that I= Bt. The element x E 1 annihilates t, hence the map I--+’ I is not an injection. Since I is finite we conclude that t2-c 1, showing that B > f. The ideal 0 : t = Ann M is finite. By Lemma 4.1 we conclude that B > I is finite, contradicting the maximality of I. LEMMA 6.15. Let R be an indecomposable Coloring. Assume that every maximal ideal which equals Ann x for some x E:J is principal. Then R is reduced or a$nite local principal ideal ring. ProoJ: Assume that R is neither reduced nor finite. It is easily seen that a finite ideal not contained in the nilra~~al contains an idempotent e # 0. Assuming R is indecomposable we conclude that J is the unique maximal finite ideal. Let M = Ann x be a maximal ideal, x E J. Assume M = Rt. We claim that 0 : t = Ann MC J and this proves that 0 : t is finite. Since M is a maximal ideal, Ann M is contained in every prime ideal except possibly M. If AnnM& M then AnnM-tM=R and (Ann M) -M= (0). Hence RzM@Ann M contradicting that R is indecomposable. Hence 0 : t is finite and Lemma 4.1 yields that J : t is finite. But J : t > J since (J: t)t = J and tJ
We conclude the section with THEOREM 6.16. The following are equivalent for a Coloring R.
(1) Every maximal ideal which equals Ann x for some x E:J is principal. (2) R is isomorphic to a product of a reduced ring with a smite principal ideal ring. ProojI Since the clique number in R is finite, R is a finite product of indecomposable rings. On each of these indecomposable rings we apply Lemma 6.15 and this takes care of (1) z- (2). The implication (2) + (1) is easily verified. Remark. One can prove that a finite ideal in a reduced ring is generated by an idempotent. Hence any finite ideal in a ring satisfying the conditions in Theorem 6.16 is principal. In particular J is principal.
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7. RINGS OF Low CHROMATIC NUMBER We conclude this paper with a short discussion of rings with low numbers. The conjecture X(R) = clique R is veritie clique R < 4. Applying some of our earlier results we list the finite rings of chromatic number ~3. It is easily seen that clique R = 2 implies that x( ) = 2. In view of ropositions 2.1 and 2.2 we may state chromatic
PROPOSITION 7.1. Given a Coloring R, then x(R) = clique clique R < 2 or x(R) < 2.
PROPOSITION
provided
7.2. Let R be a Coloring. Then clique
Proof. Since x(R) 3 clique R it suffices by roposition 7.1 to show that x(R) > 3 implies clique R > 3. Let + = R - (0) and assume that X(R) > 3. Then x( graph R* is not 2-colorable it contains an odd cycle x1 ) . the minimal length of an odd cycle in R* and assume 123 5. We have xIxz =x2x3 = ... =x,-~x, =x,x! =O. Suppose that xlxk =O for some k # 1,2, n. Then x1, .... xk and xk, .~~,x,, x1 are cycles of He < FI and one of them has an odd length. It follows that xixj = 0 only xj are neighbours in the cycle. Put y = x1 xj. Then yx, = yx4 = yx, = 0. Since y is adjacent to three of the elements of the cycle x1, .... x, we conclude that y does not belong to it. We can then produce an odd cycle y, x4, .... x, of length that R* contains an odd cycle of length 3 and hence cli The next case requires a little more work. THEOREM
7.3. Let R be a Coloring and k an integer ~4. Then x(
iff clique R = k. Moreover, x(R) = 5 implies clique R = 5. PuoojI According to Proposition 7.2 it sufhces to show that x( implies that clique R > 4. If R is a reduced ring x(R) equals clique R. We assume t nilradical J is non-zero. y Theorem 6.9 clique R =clique -I-E(R) and x( suffices to show that in our case clique J= x(J) Let 4 = J n Ann J. Since J is nilpotent and non-zero /113 2, note that d is a clique in J. If I= J, J is a clique and trivially x(S) = clique J. Also if \I/> 4 there is nothing to prove.
224
ISTVAN
BECK
If (II = 4 let x E J- I. Then I u {x} is a clique with 5 elements. If 111= 3 and x(J) > 4 at least two different elements U, u in J-Z are adjacent. Thus 111u (u, u} is a clique with 5 elements. The only case which offers some difficulties is that in which 111= 2 and x(J) 3 5. Let I= (0, C). Since Z is an ideal C + C= 0. Since x(J) 3 5 the set J-I requires at least three distinct colors. Hence we can pick a minimal odd cycle A,, .... A, in J- I and assume IZ3 5. If A, A, = 0 for some k # 1, 2, n the cycle will decompose in two smaller cycles of which one is odd. Hence A,A, =0 only if Ai and Aj are neighbours. As in the proof of Proposition 7.2 we conclude that A,A, does not belong to the cycle since A,A, is adjacent to at least three of the members of the cycle. Let if 1, 2, IZ. If i is even A, A,, A,, .... Aj_ I is an odd cycle of length i- 1
-.+Ak--ZAk+Ak--lAk+A: A,+
... +AzkplAk
=A;+2(k-2)C=A;#O.
This proves that B$ I. Hence (0, A,- I 9 A,, B, C} is a clique in J. This completes the proof of Theorem 7.3. We shall now find the finite rings R for which x(R) < 3. By this restriction to finite rings Propositions 2.1 and 2.2 imply that x(R) = 1 iff R = (0) and x(R) = 2 iff R is a finite field, R g 2, or R z Z,[X]/(X*). A ring R has chromatic number 3 iff its clique number is 3. (Theorem 7.3). By Theorem 6.9 clique R = clique J+ E(R) where E(R) is the number of prime ideals P for which R, is a field. Since clique J3 1 the possible values of E(R) are 0, 1, and 2. Case c(R) = 2. Then clique J= 1, hence J= (0). Recall that dim R = 0.
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m 3.8 (0) = M, n M2 for some maximal ideals ct of two finite fields. E(R) = 1. Let P be a prime ideal for whi R = 0, P is both maximal and minimal. Since eorem 4.3). Let 0 :x= P. Since R, is a P + Ann B = R and we derive that R z R/P x where k is a finite field and it is easily seen t S = 2. Since R is not reduced we get that R z k Case
Case E(R)= 0. Then clique J= 3. Every belongs thus to Ass R. Suppose that Ann x = are different prime ideals. Then xy = 0 and since E(R) = 0, x2 = y* = 0. Th {O, x, y, x + y ) is a clique in J. Since cliq We conclude that R has al Rx is a clique, hence 2 d / and 0 : P is a clique. If Ann P strictly contain elements and we have a contradiction. We conclu Case A. lRx j = 3. Since P . Rx = (0) it exact sequence Q+Annx+R-+Rx-tO we now conclude that (R I= 9. If char R = 9 then R r Z, and if char en R r Z,[X]/(X*).
lRxl=2. Then Rx=(O,x) and P#Rs=AnnP. P - Rx. If y2 = 0 then (0, x, y, x + y j is a clique with 4 elements r, it is easily seen that y E P-
Y=
Pick y E P - Rx. Since (0, x) = Ann P it can be shown that there exists an element YE R such that ry = x. We note that YC$Rx. lain r(sy) = sx = 0. ves that Ann(sy) d Rx entailing that sy E for any element y E P - Rx. Consider the exact sequence
0-(0,x)---+
P
f b (0,x) -
e have shown that
0 where f(t) = ty.
It follows that /PJ= 4. Moreover R/P r Rx, hence lR\ = 8. (i) char R = 2. According to the exact sequence (I) the i consists of the elements (0, x, Y, x + r} where x2 = XT= 0 and ry = x.
(1)
226
ISTVAN BECK
If r2 = 0, clique P = 4. Hence r2 # 0. If r2 = r + x then r( 1 - r) = x and this entails that r E Rx. We conclude that r2 = x. Hence P= (0, r, r2, r + r’} wherer3=Oandtheunitsare 1, l+r, l+r’, 1+r+r2. It is easily seen that RrZ2[X](X3). (ii) char R = 4. We noted that P= (0, r, r2, r + r’} where r3 = 0. Since 22 = 4 = 0 it follows that 2 E P. If 2 = r we derive the contradiction r2 =0 and 2 = r + r2=r(l +I) leads to the same contradiction. Hence 2 = r*. This implies that 2r = r3 = 0. Define a homomorphism 40:Z,[T] + R by cp(T) = r. Then q is surjective. Moreover T2 - 2 E ker cp and 2Ts ker cp. Since # Z,[ T]/(2T, T2 - 2) = 8 we conclude that ker cp= (2T, T2 - 2). Hence R z Z,[X]/(2X, X2 - 2). (iii) If char R=8 then RgZZ,. We summarize: The following is a list of the finite rings R for which x(R) < 3. x(R) = 1. R = (0). x(R) = 2.
(i) R is a finite field (ii) RrZ, (iii) R z Z,[X]/(X’).
x(R) = 3.
(i) (ii) (iii) (iv)
R is a product of two finite fields
RrkxZ,,kalinitelield R r k x Z,[X]/(X*),
k a finite field
Rr 2,
(v) RrZ, (vi) R z Z,[X]/(X’)
(vii) R E Z2[X]/](X3) (viii) R s Z,[X]/(2X,
X2 - 2).
REFERENCES AND P. ERD~S, A colour problem for infinite graphs and a problem in theory of relations, Nederl. Akad. Wetensch. Proc. Ser. A 54 (1953), 371-373. 2. D. K~NIG, “Theorie der endlichen und unendlichen Graphen,” Leipzig, 1936. Reprinted Chelsea, New York, 1950. 3. H. B. EILERTSEN, Graphs of 0-semigroups, unpublished. 1. N. G. DE BRUIJN
OF ALGEBRA
116,
208-226 (1988)
Coloring
of Commutative
Rings
ISTVAN BECK ~epa~trne~t of mathematics and Compotes Science, University of Haifn, Haifa 31999, Israel Communicated by Kent R. Fz&er
Received October 29. 1986
1. INTRODUCTION
The purpose of this article is to present the idea of coloring of a commutative ring. This idea establishes a connection between graph theory and commutative ring theory which hopefully will turn out to be mutually beneficial for these two branches of mathemathics. In this introductory paper we shall mainly be interested in characterizing and discussing the rings which are finitely colorable, leaving aside, for the moment, possible applications to graph theory. Let R be a commutative ring. We consider R as a simple graph whose vertices are the elements of R, such that two different elements x and y are adjacent iff xy = 0. We let x(R) denote the chromatic number of the graph, i.e., the minimal number of colors which can be assigned to the elements of R in such a way that every two adjacent elements have different colors. A subset C = (xi, .... x,} is called a clique provided xixj = 0 for all i # j. If R contains a clique with n elements, and every clique has at most y1 elements, we say that the clique number of R is n and write clique R = n. If the sizes of the cliques in R are not bounded we define clique R = co. We shall show that clique R = co actually entails the existence of an infinite clique. Obviously x(R) 3 clique R and for general graph G we certainly may have x(G) > clique G. However, in the case of commutative rings we have not found any example where x(R) > clique R. The lack of such counterexamples together with the fact that we have been able to establish the equality x(R) = clique R for certain (rather wide) classes of rings like reduced and principal ideal rings motivates the following conjecture. Conjecture 1. X(R) = clique R.
We start the paper with a few examples to give the reader some understanding of the concepts and problems involved. 208 0021”8693/88$3.00 Copyright 0 1988 by Academic Press, Inc. All rights of reproduction in any form reserved.
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In Section 3 we prove the main characterization of rings of finite chromatic number and name these rings Cokorings. Theorem 3.9 a ring R is a Coloring provided its nilradical is finite an equals a finite intersection of prime ideals. The next section discusses some of the characteristics of Colorings. Theorem 4.2 we learn that Colorings have a.c.c. on annihilators and important theorem enables us to show that in a Coloring every mini rime ideal is an associated prime ideal. In the fifth Section we show that the family of Colorings is closed with respect to certain operations. The main purpose of introduction separating elements in Section 0 is to ical attack Conjecture 1. We succeedin reducing the are of the ring and manage to prove Conjecture reduced or principal ideal rings. In these rings it is even true that ~(1) = clique I for any ideal I. The invariant x(R) (or clique R) is a useful num er to attach to a Coloring. We illustrate this in Section 7 y describing the finite rin chromatic number < 3. In this last section we also prove that x( clique < 4 and thus conjecture 1 is also chromatic number. Some readers of the paper would perhaps miss a formula for x( in terms of x(R,) and x(R,). Such a formula does not exist. It seemsthat x(W, x R2) depends on x(R,) and x(R2) and and R,. In general x(R, x R2) &x(R,) + x reduced the inequality becomes an equality. oblems involved and clarified how the 2-nilpoten ings. aper only scratches the surface of the comb divisors in a commutative ring. Ultimately, will involve graph theory. hen proving that x(R) = clique R for rings of low cbromat~c num (Sect. 7) we lean heavily on a useful graph theoretic result. It sa graph is 2-colorable if and only if the graph does not c~~ta~~any (see for instance [2]). Unfortunately, such a c~aracteri~a~o~ has not been found for k-colorable graphs when k > 3. Since our proof of X(R) = clique R for low chromatic numbers depends terization of the 2-colorable graphs, one could ture 1 would give us some insight into k-colorable ng my work with this paper I was lucky to people that could advise me and I am grateful ilertsen, K. P. Villanger, and A. Zaks for their pa Tke terminology is standard. We emphasize:
210
ISTVAN BECK
R denotes a commutative ring with an identity, J is its nilradical, and R is said to be reduced if J= (0). Given subsets I and K of R, I : K= {r~R(rKcl). Moreover, O:I=AnnZ and if I=(x) we simply write 0 : x = Ann x and these last ideals are called annihilators. Given an ideal K, rad K= {Y E R 1r” E K}. The set of zero divisors in R is denoted Z(R). A prime ideal P is an associated prime ideal if P = Ann x for some element x in R. We let Ass R denote the set of associated prime ideals. The Krull dimension of R is denoted dim R and we use the simple fact that dim R = 0 if R is a finite ring. Finally, the number of elements in the set Z is denoted by #I or simply )I/, and A\B is the set {x 1XE A and x6 B).
2. CHROMATIC NUMBER OF SOMERINGS The proof of our first proposition is straightforward. PROPOSITION2.1. x(R) = 1 iff R is the zero-ring. PROPOSITION2.2. x(R) = 2 iff R is an integral domain, R g Zq, or R z
z2 cJa(~2). ProoJ: Suppose that x(R) = 2 and R is not an integral domain. Let xy = 0 where x and y are non-zero. Then (0, x, JJ} is a clique and since clique R < x(R) = 2 it follows that x = y. Thus x # 0 and x2 = 0. The ideal Rx is a clique and we conclude that #Rx = 2. Assume that z E Ann x. Then { 0, x, z} is a clique and therefore z E Rx= (0, x}. Hence Ann x = Rx. From the exact sequence
O-+Annx-+R-+Rx-+O we may conclude that #R = 4. If char R=4 we have Rr Z, and if char R=2
we derive Rs
z2wll(~2).
It is easily seen that these two rings have chromatic- and clique number 2. PROPOSITION2.3. Let pl, .... pk, ql, .... q1 be different prime numbers N=p:“~... p~qfm’+1...q,2mr+1. Then x(Z,) = clique Z, = and “k ml , . . qy + r. P;"'.Pkqt Proof: Let y, = p;’ . . . p;kqyl+ i . . . qy + l. Then yi = 0 in Z,, thus Z, y, is a clique with p;i . . . p;kqy’ . . . qy elements.
ht yi = yo,/qlii, 1
I is easily seen that this coloring attaches di~ere~t colors to a vertices.
After these prelimina~es we start the ~re~arat~~~s to ~baracte~~~ethe riglgs of Unite ~bromatic number. First a de~nition= An element x ia R is said to be~~~~~provid Rx is a ~~jte set. The fo~l~wi~~ lemma is essential in this section.
Proof. Let x, ) ~..~x,, ... be different finite ~~~rne~~s in $2. The ~~~rne~~~ x1x2, ...) x1x,, ... belong to the finite ideal Rx,. Thus for some infinite subsequence {a,) of (2, .... 73,... > we have xi xai = x1 .xa2= . . . . e then consider the sequence xai, xn2, ... and repeat the ~~o~e~~re-~o~~~~~~g in this way we construct a subsequence y1 p‘.a7 y,, .“. of the seq~e~~~ XI) .*.,x,, ... such that yi yj = yi yk when j, k > i. In this s~bs~q~~~~e
zij = yi - yi. Ubviously zyzkr = 0 if i < j < k < r. e scald anow ~~~s~~~~tan infinite clique. CorI .Z1,2Z3,4 =Z1,2Z3,5=O. Since 23,4#Z3,5 at kaSt one Of ZJ,~an Es.5IS rent from z~,~. If for instance z~,~fz,,, then (21,29z 1 is clique
with
two elements.
We obserw
that z~,~, zhs, aad .z6,$ we
merest
and if for instance z~,~f#{z~,~, z~,~) then {.z~,~,z~,~~z~,~) is a ~~i.~~~with three ~~~rne~~s. Continuing in this way we get an i~~~~te ~~~q~e”
212
ISTVAN BECK
ProoJ: If R has an infinite clique C the homomorphic image C of C is a clique in i?= R/I, and since I is finite C is still infinite. Conversely, let { Zi >;” be an infinite clique in R. Hence xixj EI for i # j. Since the set of products (x~x~}~+j is a finite set we may apply the same technique as in the proof of Lemma 3.1 to establish an infinite clique in R. LEMMA 3.3. If the ring R contains a nilpotent element which is not finite then R contains an infinite clique.
Proo$ Assume xn = 0. The proof goes by induction on n. If x2 = 0 and Rx is infinite then R contains an infinite clique since in this case Rx itself is a clique. Suppose now that the lemma is true for elements of nilpotence degree n - 1. Then let x” = 0, n 2 3, and assume that Rx is infinite. Put y = x2. Then y”- ’ = 0. If Ry is infinite we conclude that R has an infinite clique. Assume next that Ry is finite. Then Rx/Ry is infinite. We note that Rx = RxjRy is an infinite clique in R = R/Ry. Since Ry is finite we conclude from Lemma 3.2 that R has an infinite clique. Remark. Hakon B. Eilertsen has pointed out that Lemma 3.3 can be proven without using Lemma 3.2. Having shown that the nilpotent elements are finite we now get LEMMA
3.4. If the nilradical of R is infinite then R has an infinite clique.
ProoJ Assume that the nilradical J is infinite. If every element in J is finite Lemma 3.1 implies that R contains an infinite clique. And if an element in J is not finite Lemma 3.3 implies that R contains an infinite clique. Assume that R is a ring without an infinite clique. Then Lemma 3.4 implies that the nilradical J is finite. Furthermore, Lemma 3.2 implies that R/J does not have an infinite clique. We therefore limit ourselves for a moment to reduced rings. LEMMA 3.5. Let R be a reduced ring which does not contain an infinite clique. Then R has a.c.c. on ideals of the form Ann x.
ProoJ: Assume that Ann a, < Ann a2 < . . . . Let xi E Ann a,\Ann aip 1, i= 2, 3, .... The non-zero elements yn = x,a,- i, n = 2, 3, ... form a clique and we claim that yi # yj when i # j. We have that yi yj = 0. The equality yi = yj would yield y: = yJ?= 0. This proves our claim and completes the proof of the lemma. Remark. This simple proof of Lemma 3.5 is due to H. B. Eilertsen.
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3.6. Let x and y be elements in R such that Aralnx and Ann y are dtjjferent prime ideals. Then xy = 0. LEMMA
Proof. Assume that xy # 0. Then x $ Ann y an y $ Arm x. Since Arm x and Arm y are prime ideals we derive Ann x : y = Anny:x=Ann y. However, Annx: y=Ann y:x=Ann Arm x = Ann y. e are ready to prove THEOREM
3.7. For a reduced ring R the following are e~~~~aie~t:
(1) X(R) isfinite. (2) Clique R is finite. (3) The zero-ideal in R is a finite intersection ofprime ideals. (4) R does not contain an infinite clique. ProoJ: The implications (1) s (2), (1) * (4), and (2) * (4) are evident. als. see (3) a (1) let (0) = P, n . . . n P, where P, 7.... P, are pri for e a coloring f on R by putting f(0) = 0 and f(x) = mi x#Q. We note that X(R)
214 THEOREM
ISTVAN BECK
3.9. The following conditions are equivalent for a ring R:
(1) x(R) isfinite. (2) Clique R is finite. (3) The nilradical in R is finite and equals a finite intersection of prime ideals. (4) R does not contain an infinite clique. Proof The implications (1) * (2), (1) =P(4), and (2) + (4) are trivial. To see (3) =+=(1) let J= P, n ... n Pk where PI, .... P, are prime ideals. If x $ J let f (x) = min{ i 1x $ P,}. This is a coloring of the elements outside J. Since J is linite we need merely a finite number of additional colors to color all of R. To prove (4) + (3) assume that R does not have an infinite clique. By Lemmas 3.2 and 3.4 this implies that J is finite and R/J does not have an infinite clique. It follows then from Theorem 3.7 that J is a finite intersection of prime ideals. We close this section with a somewhat peculiar application Theorem 3.9.
of
THEOREM 3.10. Let R be a ring which contains a finite ideal which is a finite intersection of prime ideals. Then the radical of any Jinite ideal is finite and equals a finite intersection of prime ideals. Furthermore, the ring has only a finite number of finite ideals.
Proof If R contains a finite ideal which is a finite intersection of prime ideals then x(R) < cc by the argument used in proving the implication (3) A (1) in Theorem 3.9. Let K be any finite ideal in R. Then R/K does not have an infinite clique (Lemma 3.2), hence x(R/K) < co (Theorem 3.9). Theorem 3.9 further implies that rad K/K is finite and rad K is a finite intersection of prime ideals. Since rad K/K is finite and K is finite we conclude that rad K is finite. Let A = (x 1x is finite}. Since clique R < co it follows from Lemma 3.1 that A is a finite ideal. Since A contains every finite ideal, the number of such ideals is finite. Remark. Observe that the word “finite” is only applied a finite number of times in this theorem.
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RlNGS
4. COLORINGS
Having characterized the rings of finite chromatic comber, we now ~~t~~dto discuss some of the properties enjoyed by these rings. At this stage it seemsappropriate to make the fo~~ow~~~ A ring R is called a Coloring ~ro~i~~~ xf
~~INITION.
4.1. If I is a ,finite ideal in a ring
&EMMA
, then I : x/Ann x is a finite
~-~~d~~e.
PVOC$ Consider the exact sequence O-+Annx-+I:x
f +(6:x)x------+
where f(t) = tx. Since (I : x)x c f it follows that I : ~~A~~ x is finite.
We generalize Lemma 3.5 and get the important THEOREM
4.2.
A Cobring has a.c.c. on ideals of t~~f~~~
ProoJ Let R be a Coloring and assume that Ann x1 < Ann x2 < . . . ~ The nonradical J is finite and we may assume that xi 4 J for i = I, 2, I._. Theorem 3.7 let J=P, n .I- n P, where the Pi are prime ideals. For an element x f R we get J:x=(P,:x)n
.‘. n(P,:x).
This shows that the family {J : x 1x E R > is finite. ~~~s~~ue~tly t a subsequence (yj> of (xi} for which J : yI = J: 7, = ~.. . Consider Ann y1 < Ann yZ < . . . . 9: y,=JJ:y,=
.‘..
This implies that Ann y1 < Ann y2 < ~.. c 9 : yi + that J: ~~~A~n y1 is finite. ernark. It can be shown that there exists a number N sue k~ any chain Ann x1 < Ann x2 < . . . -CAnn xk.
at k d N
Let R he a Coloring. Then Ass any minimal prime ideal P is an associatedprime I?, is a field or a finite ring.
THEOREM 4.3,
P. Furthermore,
~~~0~~ Assume R is a Coloring. Then clique ~rn~~~~§ that Ass R is finite.
216
ISTVAN BECK
Let x E Z(R). Then XE Ann y for some maximal Ann y (Theorem 4.2) and Ann y is an associated prime ideal. If P is a minimal prime ideal pick x q!P. Then Ann x c P. Choose Ann t maximal in the family {Ann y 1Ann y c P}. We claim that Ann t is a prime ideal and establish thus that Ann t = P. Assume that ab E Ann t, a $ Ann t, and b # Ann t. If a $ P we consider Ann ta. It is easily shown that Ann t < Ann ta c P and we have arrived at a contradiction. If a E P we still consider Ann ta. If Ann ta c P the contradiction is repeated. On the other hand, if CE Ann ta, c$ P we consider Ann tc and get the contradiction Ann t < Ann tc c P. Let now P be a minimal prime ideal. We just showed that P = Ann x for some x. If x $ P one concludes that PR, = (0) and R, is a field. Suppose that x E P. Let J= P n P, n . . . n Pk where P,, .... Pk are the remaining minimal prime ideals. Pick y E P, n ... n P, -P. Then yPc J and since y$ P we conclude that PR, = JR,. The ideal J is finite and therefore PR, is finite. Obviously R/P E Rx. Since x E P = Ann x it follows that x2 = 0. Hence Rx is finite. Hence RxOR R, E R/P@, R, z R,jPR, is finite. We have shown that R, is finite. THEOREM 4.4. Let P be an associated prime ideal in a Coloring. Then either R, is a field or P is a maximal ideal. Proof Let Ann x = P. Suppose first that x E P. Then x2 = 0 and hence Rx is finite. But then R/P( E Rx) is finite, so R/P is a field. Hence P is a
maximal ideal. If x $ P and Ann x = P, then PR, = (0) and R, is a field. From Theorem 4.4 we get COROLLARY 4.5. An associated prime ideal in a Coloring maximal ideal or a minimal prime ideal.
is either a
5. THE FAMILY OF COLORINGS
This short section is devoted to a study of the properties enjoyed by the family of Colorings. Obviously we have THEOREM 5.1. A subring of a Coloring is itserf a Coloring.
The next theorem follows from Lemma 3.2 and Theorem 3.9.
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5.2. Let I be a finite ideal in a Coloring R. Then
THEOREM
Coloring. LEMMA
5.3. Let x be an element in a C&wing
Coloring. Proc$ Let F,, .... Fn be a clique in I?= R/A urthermore r 1x, .... r,x are distinct elements in /Ann x
nd from Lemma 4.8 we y Lemma 5.3 R/Ann x is a Colorin at I: x/Ann x is a finite ideal in /Ann x. One can a Theorem 5.2 and conclude that R/I : x is a Coloring. Proof.
COdUde
THEOREM5.5. A finite product of Colorings is a Coloring. s&ices to consider the product = I?, x R, of the Colori Put clique R, = n and clique R, = m. It is easily seen I?, 6 nm and Theorem 3.9 yields that R is a Coloring. e now generalize Lemma 5.3. 5.6. Let I be a jiinitel~v generated ideal in a C&wing. n I is a Coloring.
EOREM
Pr54$ Let I= (x,, .... x,). Then Ann I= Ann x1 n ... n Ann x,. ave an injection RJAnn I --) RJAnn x1 x . . x R/Ann x,. Each sf the ri R/Ann xi is a Coloring and by Theorems 5.1 and 5.5 the proof is complete. CQROLLARY 5.7. Let R be a Noetherian ring whose nilradical is ,fiinite. Then rad(Ann I)/Ann I is finite for any ideal I.
Proof. We apply Theorem 3.9 and deduce that Theorem 5.6 we conclude that RJAnn I is a Coloring. Sin a Coloring is finite the proof is complete. THEOREM 5.8. Let S be a multiplicative Then s is a Coloring. Moreover, x(R,) < x
closed set in a Co10
Proof. Let x(R) = n. To show that the grap to show that every finite subset is n-colorable [ 11.
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ISTVAN BECK
Let x1 =rl/s, .... x, = r,/s ,be a finite subset of R,. For each pair (i, j), 1 G i, j< m for which xixj = 0 pick sij E S such that svrirj = 0. Let t = 17s, and define rl= tr,. Then xixj = 0 iff rirj’=O. Since every subset of R is n-colorable we conclude that the set (xi, .... x,} is n-colorable. The proof also shows that clique R, d clique R.
6. SEPARATING E~E~NT~
The study of Colorings is essentially a look at the combinatorial aspects of the zero divisors. It is therefore not a surprise that the prime ideals have interesting properties in this respect. Let P be a prime ideal in a Coloring. If the elements contained in P have been colored we need at most one additional color for the elements outside of P. This is seen from the simple fact that xy = 0 entails x E P or y E P. It would be of interest to study ideals which enjoy this property, but for our purpose it seems more convenient to introduce the notion of separating elements. DEFINITION
6.1. An element x in R is separating provided x #O and
ab = 0 implies xg = 0 or xb = 0.
We shall also need elements satisfying a more local property. DEFINITION 6.2. Let I be an ideal. An element x E I is I-separating provided xI# (0) and whenever ab = 0 for some elements a, bgI then xa=O or xb=O.
(1) Note that x is separating it? x is R-separating. (2) We do not require a # b in ~e~nitions 6.1 and 6.2. (3) An R-separating element x contained in I fails to be I-separating if x1= (0). If, however, xf# (0) then x is also I-separating. Remarks.
Before we reveal the purpose of introducing separating elements we wish to discuss their existence. PROPOSITION Proof:
6.3. If Ann x is a prime ideal then x is separatirzg.
Straightforward.
PROPOSITION 6.4.
A non-zero ideal I in a Coloring contains a separating
elemem. Pro&
Since R has a.c.c. on annihilators (Theorem 4.2) it is easily seen
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219
that I contains an element x such that Arm x is a prime ideal. Allay then Proposition 6.3. Using the same technique we derive T~~oR~~ 6.5. Let I be an ideal in a ~olor~~~ not co~ta~~~edin the ~ilrad~ca~.Then I contains an I-separating element. BrooJ: Let x E I be such that Ann x is a prime ideal P and P ;b I. Then x is R-separating and xI# (0) since Ann x = P ;S I. If I is an ideal for which I2 = (0) then I does not c~~ta~~any ~-separatism element. THEOREM 6.6. Let Z be a principal ideal in a Coloring. If 1’ # (0) then I contains arr I-separating element.
Prroof: Let I= Rx and suppose x2 # 0. Since R has a.c.c. on annihilators it is easily shown that 0 : x2t is a prime ideal for some 1E xl is I-separating. Let a, b E B and assume ab = 0. Write a = rx and b = sx. Then rs-sx2 = 0. ence YSis contained in the prime ideal 0 : x2t. If for instance r E 0 : x22 we rive that (p.~)(xt) = 0, i.e., afxt) = 0. ~urthermore~ (tx)x = tx’ # 0 proves that (tx)I# 0. e now reveal the purpose of introducing separating ~l~~e~ts. LEMMA 6.7. Let I be an ideal in a Coloring and assume XE d B ~-~e~ar~ti~l~.Put I’ = Ann x n I.
(1) If x2 = 0 then clique F = clique I and x(F) = ~(~)~ (2) rf x2 # 0 then clique T = clique I- 1 aBd x(T) = ~(1) - 4. Proof. Assume first that x2 = 0. Then x E I’. Let clique I= y1an a maximal clique C = { y, , .... y,> in I, If x E C, say x = yr p th yz, .... yn E 1’ and as we noted x E I’. Hence Cc F, thus clique I’ = n. x & C, XC # (0) since C is a maxima1 clique in I. Assume that xyi xy, = 0 for 2 < i < m since x is Z-separating. Hence {x, yz9 .... yn> contained in P. This proves that clique I’ = az. We still assume that x2 = 0 and consider x(I’) and x(I). observe again that x E I’. color first F. If y E I- I’ we assign to y the bake color as to x, This shows that x(1’) = x(I). Assume next that x2 # 0, i.e., x $ I’. Then certainly cli ue 1’ Q clique I- 1 since to any clique in I’ we may adjoin x and create a larger clique in I. the other hand, let C = (yr, .... y,,> be a maximal clique in I. If x E C9
220
ISTVAN BECK
x = y, say, then again { yz, .... yHf is a clique in Z’. This shows that clique Z = clique I- 1. We leave to the reader to verify that if x2 # 0 then x(Z) = x(Z) - 1. We can now prove THEOREM 6.8. Let Z be an ideal in a Coloring. Let (.x1, ,.., x,) be a clique
Z-separating elements. Define k = #{i ( x: # 0} Ann(x,, .... x,). Then clique r = clique Z-k and x(Z’) = x(Z) - k. of
and
I’ = Zn
Proof: Since a proof of this theorem can utilize the same technique as the proof of Lemma 6.7, we omit the proof,
A useful corollary of Theorem 6.8 is the following THEOREM 6.9. Let P, , ,.., P, be the minimal prime ideals in a Coloring R. Let E(R) = # (i 1 R, is afield). Then clique R = clique J-I- E(R) and x(R) = x(J) + E(R).
Proof, Let 0 : xi = Pi (Theorem 4.3). Then {x,, .... x,} is a clique of R-separating elements. Apply now Theorem 6.8 and observe that R, is a field iff x? # 0. Theorem 6.9 makes it clear that in order to prove (or disprove) Conjecture 1 one may concentrate on the nilradical. In reduced rings the conjecture has been verified and we can prove a stronger result. THEOREM 6.10.
Let R be a reduced Coloring. Then clique Z= x(Z) for any
idea/ Zc R. Proof: By Theorem 6.5 Z contains an Z-separating element x. Since x2 # 0 one completes the proof using induction on clique Z (Lemma 6.7). We can also verify Conjecture 1 for principal ideal rings. THEOREM 6.3 1. Let R be a Coloring which is a principaE ideal ring. Then x(Z) = clique Zfor any ideal I in R.
ProoJ: We make a reduction to the case Zc J. Let PI, .... P, be the minimal prime ideals in R. If Z (f P, there exists an element y, E Z such that Ann yr = Pi. Then y, is Z-separating. Define I, = In Ann y, = In P, and observe that by Lemma 6.7 x(Z) = clique Z iff x(Z) = clique I’. It should now be evident that we may assume that Zc J. Assume I= Rx c J. If Z2= (0) then clique Z= x(Z) = #I. If 1’ # (0) the ideal Z contains an Z-separating element y, [Proposition 6.61. Let Z, = Zn
221
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Ann y, < I It follows from Lemma 6.7 that ~(1) = cli will clique I,. This completes the proof since the process above lead to an ideal I, for which 1: = (0) in which case x(Z,,) = chque In. Theorems 6.10 and 6.11 suggest the following conjecture. Conjecture 2. The chromatic number equals t e clique number for any
ideal in a Coloring. hen proving Theorem 6.11 we noted that it sufkes to prove Conjecture 2 for ideals contained in the nilradical. Moreover, a look at the proof of Theorem 6.11 should also make the proof of the next theore straightforward. THEOREM 6.12. Let R be a Coloring with the property that any ideal 1 for which I2 # (0) contains an I-separating element. ThePzx(I) = clique Ifor any ideal 1 in R.
Unfortunately, not every Coloring satisfies the hy Theorem 6.42. The local ring R = Z,[x, y] with relations x2 maximal ideal; M= (x, y) does not contain any M-separating element even ’ # (0). Despite this clique R = X(R) = p2. verified Conjecture 1 and 2 for principal i may wonder whether Conjecture 2 is true such rings. By the aid of Theorem 6.12 we shall pro First an observation. Let I= I, 0 I,. If 4, contains and I, separatin e~~rne~tx1 then x1 is as well I-s rating. This servation imphes that if a Coloring pi is a finite product of rings satisfying the hypothesis of Theorem 6.12 then this roperty is inherited by We therefore state THEoREM
6.13. Let R be a Coloring which is a finite product of reduced
rings and principal
ideal rings. Then x(I) = clique Ifor
any ideai IC R.
e now wish to characterize the rings in Theorem 6.13. The next theorem is a little surprising. 6.14. Let R be a local Coloring whose maximal ideal. Then R is reduced or a finite principal idea/ ring.
THEQREM
principal
ideal is a
Proof If R is finite then R is a local Artinian ring whose maxima%i is principal. Thus R is a principal ideal ring. Assume that R is not reduced and not finite. Let I be t y the set of finite elements. The ideal 1 is finite and d i maximal finite ideal.
222
ISTVAN BECK
Pick x E I such that Ann x = P, P being a prime ideal. Since R/P z Rx we observe that R/P is a finite integral domain. Hence P equals the maximal ideal M = Rt. Let B = f : t 2 1. Since Ic Rt it follows that I= Bt. The element x E 1 annihilates t, hence the map I--+’ I is not an injection. Since I is finite we conclude that t2-c 1, showing that B > f. The ideal 0 : t = Ann M is finite. By Lemma 4.1 we conclude that B > I is finite, contradicting the maximality of I. LEMMA 6.15. Let R be an indecomposable Coloring. Assume that every maximal ideal which equals Ann x for some x E:J is principal. Then R is reduced or a$nite local principal ideal ring. ProoJ: Assume that R is neither reduced nor finite. It is easily seen that a finite ideal not contained in the nilra~~al contains an idempotent e # 0. Assuming R is indecomposable we conclude that J is the unique maximal finite ideal. Let M = Ann x be a maximal ideal, x E J. Assume M = Rt. We claim that 0 : t = Ann MC J and this proves that 0 : t is finite. Since M is a maximal ideal, Ann M is contained in every prime ideal except possibly M. If AnnM& M then AnnM-tM=R and (Ann M) -M= (0). Hence RzM@Ann M contradicting that R is indecomposable. Hence 0 : t is finite and Lemma 4.1 yields that J : t is finite. But J : t > J since (J: t)t = J and tJ
We conclude the section with THEOREM 6.16. The following are equivalent for a Coloring R.
(1) Every maximal ideal which equals Ann x for some x E:J is principal. (2) R is isomorphic to a product of a reduced ring with a smite principal ideal ring. ProojI Since the clique number in R is finite, R is a finite product of indecomposable rings. On each of these indecomposable rings we apply Lemma 6.15 and this takes care of (1) z- (2). The implication (2) + (1) is easily verified. Remark. One can prove that a finite ideal in a reduced ring is generated by an idempotent. Hence any finite ideal in a ring satisfying the conditions in Theorem 6.16 is principal. In particular J is principal.
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7. RINGS OF Low CHROMATIC NUMBER We conclude this paper with a short discussion of rings with low numbers. The conjecture X(R) = clique R is veritie clique R < 4. Applying some of our earlier results we list the finite rings of chromatic number ~3. It is easily seen that clique R = 2 implies that x( ) = 2. In view of ropositions 2.1 and 2.2 we may state chromatic
PROPOSITION 7.1. Given a Coloring R, then x(R) = clique clique R < 2 or x(R) < 2.
PROPOSITION
provided
7.2. Let R be a Coloring. Then clique
Proof. Since x(R) 3 clique R it suffices by roposition 7.1 to show that x(R) > 3 implies clique R > 3. Let + = R - (0) and assume that X(R) > 3. Then x( graph R* is not 2-colorable it contains an odd cycle x1 ) . the minimal length of an odd cycle in R* and assume 123 5. We have xIxz =x2x3 = ... =x,-~x, =x,x! =O. Suppose that xlxk =O for some k # 1,2, n. Then x1, .... xk and xk, .~~,x,, x1 are cycles of He < FI and one of them has an odd length. It follows that xixj = 0 only xj are neighbours in the cycle. Put y = x1 xj. Then yx, = yx4 = yx, = 0. Since y is adjacent to three of the elements of the cycle x1, .... x, we conclude that y does not belong to it. We can then produce an odd cycle y, x4, .... x, of length that R* contains an odd cycle of length 3 and hence cli The next case requires a little more work. THEOREM
7.3. Let R be a Coloring and k an integer ~4. Then x(
iff clique R = k. Moreover, x(R) = 5 implies clique R = 5. PuoojI According to Proposition 7.2 it sufhces to show that x( implies that clique R > 4. If R is a reduced ring x(R) equals clique R. We assume t nilradical J is non-zero. y Theorem 6.9 clique R =clique -I-E(R) and x( suffices to show that in our case clique J= x(J) Let 4 = J n Ann J. Since J is nilpotent and non-zero /113 2, note that d is a clique in J. If I= J, J is a clique and trivially x(S) = clique J. Also if \I/> 4 there is nothing to prove.
224
ISTVAN
BECK
If (II = 4 let x E J- I. Then I u {x} is a clique with 5 elements. If 111= 3 and x(J) > 4 at least two different elements U, u in J-Z are adjacent. Thus 111u (u, u} is a clique with 5 elements. The only case which offers some difficulties is that in which 111= 2 and x(J) 3 5. Let I= (0, C). Since Z is an ideal C + C= 0. Since x(J) 3 5 the set J-I requires at least three distinct colors. Hence we can pick a minimal odd cycle A,, .... A, in J- I and assume IZ3 5. If A, A, = 0 for some k # 1, 2, n the cycle will decompose in two smaller cycles of which one is odd. Hence A,A, =0 only if Ai and Aj are neighbours. As in the proof of Proposition 7.2 we conclude that A,A, does not belong to the cycle since A,A, is adjacent to at least three of the members of the cycle. Let if 1, 2, IZ. If i is even A, A,, A,, .... Aj_ I is an odd cycle of length i- 1
-.+Ak--ZAk+Ak--lAk+A: A,+
... +AzkplAk
=A;+2(k-2)C=A;#O.
This proves that B$ I. Hence (0, A,- I 9 A,, B, C} is a clique in J. This completes the proof of Theorem 7.3. We shall now find the finite rings R for which x(R) < 3. By this restriction to finite rings Propositions 2.1 and 2.2 imply that x(R) = 1 iff R = (0) and x(R) = 2 iff R is a finite field, R g 2, or R z Z,[X]/(X*). A ring R has chromatic number 3 iff its clique number is 3. (Theorem 7.3). By Theorem 6.9 clique R = clique J+ E(R) where E(R) is the number of prime ideals P for which R, is a field. Since clique J3 1 the possible values of E(R) are 0, 1, and 2. Case c(R) = 2. Then clique J= 1, hence J= (0). Recall that dim R = 0.
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RINGS
225
m 3.8 (0) = M, n M2 for some maximal ideals ct of two finite fields. E(R) = 1. Let P be a prime ideal for whi R = 0, P is both maximal and minimal. Since eorem 4.3). Let 0 :x= P. Since R, is a P + Ann B = R and we derive that R z R/P x where k is a finite field and it is easily seen t S = 2. Since R is not reduced we get that R z k Case
Case E(R)= 0. Then clique J= 3. Every belongs thus to Ass R. Suppose that Ann x = are different prime ideals. Then xy = 0 and since E(R) = 0, x2 = y* = 0. Th {O, x, y, x + y ) is a clique in J. Since cliq We conclude that R has al Rx is a clique, hence 2 d / and 0 : P is a clique. If Ann P strictly contain elements and we have a contradiction. We conclu Case A. lRx j = 3. Since P . Rx = (0) it exact sequence Q+Annx+R-+Rx-tO we now conclude that (R I= 9. If char R = 9 then R r Z, and if char en R r Z,[X]/(X*).
lRxl=2. Then Rx=(O,x) and P#Rs=AnnP. P - Rx. If y2 = 0 then (0, x, y, x + y j is a clique with 4 elements r, it is easily seen that y E P-
Y=
Pick y E P - Rx. Since (0, x) = Ann P it can be shown that there exists an element YE R such that ry = x. We note that YC$Rx. lain r(sy) = sx = 0. ves that Ann(sy) d Rx entailing that sy E for any element y E P - Rx. Consider the exact sequence
0-(0,x)---+
P
f b (0,x) -
e have shown that
0 where f(t) = ty.
It follows that /PJ= 4. Moreover R/P r Rx, hence lR\ = 8. (i) char R = 2. According to the exact sequence (I) the i consists of the elements (0, x, Y, x + r} where x2 = XT= 0 and ry = x.
(1)
226
ISTVAN BECK
If r2 = 0, clique P = 4. Hence r2 # 0. If r2 = r + x then r( 1 - r) = x and this entails that r E Rx. We conclude that r2 = x. Hence P= (0, r, r2, r + r’} wherer3=Oandtheunitsare 1, l+r, l+r’, 1+r+r2. It is easily seen that RrZ2[X](X3). (ii) char R = 4. We noted that P= (0, r, r2, r + r’} where r3 = 0. Since 22 = 4 = 0 it follows that 2 E P. If 2 = r we derive the contradiction r2 =0 and 2 = r + r2=r(l +I) leads to the same contradiction. Hence 2 = r*. This implies that 2r = r3 = 0. Define a homomorphism 40:Z,[T] + R by cp(T) = r. Then q is surjective. Moreover T2 - 2 E ker cp and 2Ts ker cp. Since # Z,[ T]/(2T, T2 - 2) = 8 we conclude that ker cp= (2T, T2 - 2). Hence R z Z,[X]/(2X, X2 - 2). (iii) If char R=8 then RgZZ,. We summarize: The following is a list of the finite rings R for which x(R) < 3. x(R) = 1. R = (0). x(R) = 2.
(i) R is a finite field (ii) RrZ, (iii) R z Z,[X]/(X’).
x(R) = 3.
(i) (ii) (iii) (iv)
R is a product of two finite fields
RrkxZ,,kalinitelield R r k x Z,[X]/(X*),
k a finite field
Rr 2,
(v) RrZ, (vi) R z Z,[X]/(X’)
(vii) R E Z2[X]/](X3) (viii) R s Z,[X]/(2X,
X2 - 2).
REFERENCES AND P. ERD~S, A colour problem for infinite graphs and a problem in theory of relations, Nederl. Akad. Wetensch. Proc. Ser. A 54 (1953), 371-373. 2. D. K~NIG, “Theorie der endlichen und unendlichen Graphen,” Leipzig, 1936. Reprinted Chelsea, New York, 1950. 3. H. B. EILERTSEN, Graphs of 0-semigroups, unpublished. 1. N. G. DE BRUIJN