Which Fields Have No Maximal Subrings?
A. AZARANG- O.A.S. KARAMZADEH
ABSTRACT- Fields which have no maximal subrings are completely determined.
We observe that the quotient fields of non-field domains have maximal subrings. It is shown that for each non-maximal prime idealP in a commu- tative ringR, the ringRPhas a maxi mal subri ng. It i s also observed that i fRis a commutative ring with jMax(R)j>2@0 or jR=J(R)j>22@0, then R has a maximal subring. It is proved that the well-known and interesting property of the field of the real numbers R (i.e., R has only one nonzero ring en- domorphism) is preserved by its maximal subrings. Finally, we characterize submaximal ideals (an idealIof a ringRis called submaximal if the ringR=I has a maximal subring) in the rings of polynomials in finitely many variables over any ring. Consequently, we give a slight generalization of Hilbert's Nullstellensatz.
Introduction
All rings in this article are commutative with 160. IfSi s a subri ng of a ri ngR, then 1R2S. The characteristic of a ringRis denoted by c(R) and the algebraic closure of a fieldKis denoted byK. A proper subring S of a ring R is said to be maximal if there is no subring of R properly betweenSandR. Unlike maximal ideals, whose existence is guaranteed by either Zorn's Lemma or Noetherianity of rings, maximal subrings need not always exist. We denote the set of maximal subrings (maximal ideals) of a ring by RgMax(R) (Max(R)). Maximal subrings are also studied in [17], [5], [6], [14], [13], [16], [2], [3] and [4]. In [2], maximal subrings of commutative rings are investigated and some useful criter- ions for the existence of maximal subrings are given. Using these cri- terions one can easily see that some important rings have maximal
(*) Indirizzo degli A.: Department of Mathematics, Chamran University, Ahvaz, Iran.
E-mail: a_azarang@scu.ac.ir karamzadeh@ipm.ir
subrings, see [2]. In [3], we have characterized exactly which Artinian rings have maximal subrings. In particular, in [3], it is shown that, ifFi s a field which has either zero characteristic or is not algebraic over some subfields (in particular every uncountable field) then F has a maximal subring. It is also interesting to note that ifRis an integral domain with the quotient field K, then R is a valuation ring with a unique nonzero prime ideal if and only if R is a maximal subring ofK, see [12, P. 43].
Consequently, ifF=Kis a function field of one variable (i.e., an extension field KF such that F is a finite algebraic extension of K(x), where x2Fis a transcendental element overK) then the setRgMax(F)6 ;and it coincides with the set of discrete valuation subrings ofF (note, by a subringRofF, we meanK4R4F), see [19, Theorem 1.1.13]. The latter fact is also an immediate consequence of a more general result, namely, if KF is a non-algebraic field extension, thenF has a maximal subring containingK, which is not a field, see [3, Corollary 1.2]. We also recall that ifV is a variety over an algebraically closed fieldk, then its coordinate ring k[V]k[x1;. . .;xn]
I(V) is a domain, hence it has a field of fractions Kk(V), which is called the function field ofV. In particular, ifV P1k, whereP1kis the projective line overk, then one can easily show thatP1kis bijective with the set of maximal subrings ofk(x) contai ni ngk, see [19, Theorem 1.2.2, Corollary 1.2.3] or [9, Problem 4.8]. More generally, ifVis a hyperelliptic curve, see [10, P. 298], then by an observation which ap- peared recently in [21], see also [19, Appendix B] and [20], the set V is bijective withRgMax(k(V)).
Now let us sketch a brief outline of this paper. Our aim in Section 1, is to determine exactly which fields have no maximal subrings. In fact, we show that the only fields E which have no maximal subrings are those of the form E S
n2TFpn, where p is a prime number, Fpn i s the unique subfield, with pn elements, of Fp, the algebraic closure of Fp Z
(p), and Ti s a subset ofN, the set of positive integers, such that T fn2N : every prime divisor of nis in a fixed set of prime num- bers, sayPg [ f1g. Finally, in Section 1, we show that ifAi s the set of fields without maximal subrings, up to isomorphism, then jAj 2@0. In Section 2, we find more ringsR withRgMax(R)6 ;. Also in Section 2, we investigate some hereditary properties between a ring and its maximal subrings. Finally, in Section 3, we determine exactly which ideals of R[x] are not submaximal, and as a consequence we give a generalization of Hilbert's Nullstellensatz.
1. Fields without maximal subrings
Before presenting our new results, let us cite the following facts from [3]. The next result is in [3, Proposition 1.1] and for a more general result, see [4, Proposition 2.1].
PROPOSITION1.1. LetF Kbe an algebraic field extension andSbe a maximal subring of F which is not a field. Then for any non-unit elementx2S, there exists a maximal subringRxofKsuchthatxis not a unit element inRx,Rx\F S,Rx[x 1]KandRxis not a field, but it contains the integral closure ofSinK.
COROLLARY1.2 [3, Corollary 1.2]. LetFbe a field andF Kbe a non- algebraic field extension of F. Then K has a maximal subring which containsF and is not a field.
COROLLARY 1.3 [3, Corollary 1.3]. Let K be a field. If K is of characteristic zero or uncountable thenKhas a maximal subring which is not a field.
The following is now immediate.
COROLLARY1.4. LetEbe a field without maximal subrings, thenEis algebraic overFpfor some prime numberp.
Next, we recall that ifKis a field andRis a domain containingKwhich is also algebraic over K, thenR is a field. This immediately implies that wheneverK is an algebraic field extension of a fieldF, then any subring betweenFandKis a field. In particular, any subring ofFpis a field (more generally, it is also well-known that any vector subspaceEofFpoverFpis a field, if and only if for anya2Ewe havean2E, for alln).
In what follows, in this section we are going to present exactly the form of fields which have no maximal subrings.
We need the next definitions.
DEFINITION1.5. LetNbe the set of positive integers andTN. Then T is said to be a field generating set (briefly FG-set) ifE S
n2TFpn i s a subfield ofFp, whereFpnis the unique subfield ofFpwithpnelements andp is a prime number. Moreover, T must be such that if TT0N and E S
n2TFpn S
n2T0Fpn, thenTT0.
DEFINITION1.6. LetT1T2 be twoFG-sets. ThenT1 is said to be a maximalFG-subset ofT2if there is noFG-set properly betweenT1andT2. REMARK 1.7. One can easily see that there is a one-one order pre- serving correspondence between theFG-subsets ofNand subfields ofFp, see also the Steinitz's numbers and their properties in either [7] or [18].
Using the previous remark and Corollary 1.4, we immediately have the following characterization of the fields without maximal subrings.
THEOREM1.8. A field F has no maximal subrings if and only if there is a prime number p such that either FFpor F is an infinite subfield of Fpsuchthat F S
n2TFpn, where T is a FG-set which has no maximal FG- subsets.
We know that the multiplicative groupFpnof nonzero elements ofFpnis cyclic of orderpn 1. This and the fact that whenever an abelian groupG contains two elements of ordersmandn, then it has an element of order [m;n], the least common multiple ofm;n, immediately yield the following characterization ofFG-sets.
PROPOSITION 1.9. TN is a FG-set if and only if it satisfies the following conditions.
(1) 12T
(2) If n2T, djn, then d2T.
(3) If m;n2T, then[m;n]2T.
PROOF. Let T be a FG-set, then (1) and (2) are evident. Hence let E S
n2TFpnbe the subfield ofFpcorresponding toT. Ifm;n2T, then let a2Fpn,b2Fpmbe elements inEwhich are of orderpn 1;pm 1 respec- tively. ThenEhas an elementgof order [pn 1;pm 1]. Butgis a root of the equationxpk x0, wherekis the degree off(x), the minimal polynomial of goverFp(note, the roots off(x) areg;gp;. . .;gpk 1). Clearly,gpk 11. Hence [pn 1;pm 1]jpk 1, consequently we havepn 1jpk 1,pm 1jpk 1.
Thusnjk,mjkimply that [m;n]jk. SinceFp(g)FpkEwe infer thatk2T and therefore [m;n]2T. The converse is evident. p By the previous proposition we have the following trivial examples of FG-sets.
EXAMPLE1.10. TNis a finite FG-set if and only if T consists of divisors of an integer n2N. Let P be a set of prime numbers, then T fn2N :prime divisors of n are in Pg [ f1gis a FG-set.
Theorem 1.8 shows that in order to know exactly which fields are without maximal subrings we must know the FG-sets which have no maximalFG-subsets. The next result precisely determines theseFG-sets.
PROPOSITION1.11. LetT be a FG-set. Then T has no maximalFG- subsets if and only if wheneverqis a prime number inT, thenqn2T for alln2N. Moreover,T has no maximalFG-subsets if and only ifT has the formT fn2N :everyprime divisor ofnis in a fixed set of prime numbers, sayPg [ f1g.
PROOF. LetTbe aFG-set with the above property. We claim thatThas no maximalFG-subsets. LetS4Tbe a maximalFG-subset ofTand obtain a contradiction. Taken2TnSand letnpm11pm22 pmkk be the factoriza- tion ofn. Sincen2=S, we infer that there exists 1rkwithpmrr2=S. Now put S1S[ f[m;ptr]:m2S;tmrg. Clearly, S1 i s a FG-set and S4S14T(note,pmr 2Tfor allm) which is the desired contradiction.
Conversely, let Thave no maximal FG-subsets. Suppose that q2T is a prime number such that qn2=T for some n2 and n is the least positive integer with this property and seek a contradiction. Since T i s a FG-set, we i nfer that qm2=T for all mn. Now put S fx2T: x6qn 1y; y2T; (q;y)1g and note that S is a maximal FG-subset of T. To see this, it is clear that S i s a FG-set. Let SS1T, where S1 i s a FG-set. If S6S1, then there exists qn 1y2S1nS, where y2T, (y;q)1. Hence qn 12S1 and therefore qn 1z2S1 for all z2T, (z;q)1 (note, z2SS1). Thus S1T and
we are done. The last part is now evident. p
COROLLARY1.12. aFG-setThas no maximalFG-subsets if and only if whenevert2T, th entn2T for alln2N.
REMARK1.13. Trivially by the above corollary,Ni s aFG-set without maximal FG-subsets, hence Fp has no maximal subrings for any prime numberp.
Next, we observe that the collection of allFG-sets which are without maximalFG-subsets is uncountable.
PROPOSITION 1.14. Let A be the collection of all FG-sets which are without maximalFG-subsets. ThenAis uncountable.
PROOF. Letp1;p2;. . .;pn;. . .be the sequence of prime numbers, i.e., p12; p23; p35, etc. We define the map f:A ! Q1
i1Xi, where Xi f0;1g for all i, as follows. For each A2 A we put f(A)5an >, wherean0 i fpn2=Aandan 1 i fpn2A. It is manifest thatfis one-one
and onto. HencejAj 2@0. p
COROLLARY1.15. LetfFi : i2Ig be the family of fields, which are without maximal subrings and have the same characteristic (up to isomorphism). ThenjIj 2@0. In particular, the set of all fields without maximal subrings(up to isomorphism)is of cardinality2@0.
2. More rings with maximal subrings and some hereditary properties In this section, we begin with the following interesting result, which is a generalization of the fact that each Artinian ring with zero characteristic has a maximal subring, see [3].
PROPOSITION 2.1. Every zero dimensional ring R of characteristic zero has a maximal subring.
PROOF. IfSZn f0g, then there exists a maximal idealMsuch that S\M ;, henceR=Mis a field with zero characteristic. ThereforeR=M has a maximal subring, by Corollary 1.3. ThusR has a maximal subring
too. p
COROLLARY2.2. LetfRig1i1be a family of rings with c(Ri)ni60 and(ni; nj)1for alli6j. ThenRQ1
i1Rihas a maximal subring.
PROOF. LetMQ1
i1Mi, where eachMiis a maximal ideal ofRi. Then R=Mis a zero dimensional ring (in fact, a regular ring) whose characteristic is clearly zero. Hence, by the previous proposition,R=M, a fortioriR, have
maximal subrings. p
PROPOSITION2.3. LetRbe a domain andF 6Rbe its quotient field.
ThenF has a maximal subring.
PROOF. If c(R)0, thenc(F)0 and we are done by Corollary 1.3.
Thus we may suppose thatc(R)p60 andRcontainsFp. SinceRis not field, we infer thatRis not algebraic overFp. HenceFis not algebraic over Fpand therefore by Corollary 1.2 we are done. p
The following is interesting.
COROLLARY2.4. LetRbe a ring andP be a prime ideal ofRwhich is not maximal. ThenRP has a maximal subring.
PROOF. LetFbe the quotient field ofR=P. ClearlyF6R=P. Hence by the previous propositionF has a maximal subring. But we haveRP
Pe F, wherePe is the extension ofPinRP. This means thatRP
Pe has a maximal subring and thereforeRPhas a maximal subring too. p PROPOSITION2.5. LetRbe a ring which contains a fieldF satisfying at least one of the following conditions:
(1) c(F)0.
(2) F is uncountable.
(3) F is not algebraic over a finite field.
Then R has a maximal subring.
PROOF. LetMbe a maximal ideal inR. Then the fieldR=Mcontains a copy ofF. Therefore ifFsatisfies in one of the above conditions, thenR=M clearly satisfies it too. Hence by Corollaries 1.2, 1.3 and 1.4,R=M has a maximal subring which implies thatRhas a maximal subring too. p PROPOSITION2.6. LetRbe a ring withjMax(R)j>2@0. ThenRhas a maximal subring.
PROOF. PutT fR=M : M2Max(R)g. IfR=M6R=M0for any two maximal ideals M,M0, thenjTj>2@0 and we are done by Corollary 1.15.
Hence we may assume thatR=MR=M0for two maximal idealsM;M0.
Now R
M\M0R M R
M0and by [2, Theorem 2.2], we see that R
M\M0has a maximal subring, a fortiori,Rhas a maximal subring. p COROLLARY2.7. LetfRigi2Ibe a family of rings withjIj>2@0. Then RQ
i2IRihas a maximal subring.
COROLLARY2.8. LetRbe a Hilbert ring. Then eitherRhas a maximal subring orjSpec(R)j 22@0.
PROOF. IfjMax(R)j>2@0, then we are done, by Proposition 2.6. Hence suppose thatjMax(R)j 2@0. SinceRis a Hilbert ring, every prime ideal of Ris an intersection of maximal ideals, thereforejSpec(R)j 22@0. p
The following is also in order.
PROPOSITION 2.9. Let R be a ring with jJ(R)j5jRj (or with jU(R)j5jRj, whereU(R)is the set of units ofR). IfjRj>22@0, thenRhas a maximal subring. In particular, if jR=J(R)j>22@0, then R has a maximal subring.
PROOF. First we note thatjJ(R)j jU(R)j. If eitherjMax(R)j>2@0or R=Mis uncountable for some maximal idealMofR, then we are done, by Proposition 2.6 or Corollary 1.3. Hence we may assume thatjMax(R)j 2@0, R=M is countable for all M2Max(R) and seek a contradiction. Clearly, R=J(R) embeds in Q
M2Max(R)
R
M and thereforejR=J(R)j 22@0 which is ab-
surd. The final part is now evident. p
The next interesting proposition shows that maximal subrings ofRare similar to R, regarding the existence of the unique nonzero ring en- domorphism.
PROPOSITION2.10. LetSbe a maximal subring ofR. ThenSis dense inR. Moreover, the only nonzero ring homomorphism fromSintoSis the identity.
PROOF. Clearly,S is uncountable. It is well-known that any additive subgroup ofRis either cyclic or dense inR. But clearly,Scannot be cyclic, hence it is dense. Now for the final part, let f:S !S be a nonzero homomorphism and consider the following two cases.
Case(I): Q6S. In this case, we first note that S is not a field and S\QZ(p), wherepis a prime number. To see this, sinceQ6S, we infer that there exists a prime number p such that 1
p2=S. But in view of [2, Theorem 3.3],Shas only one nonzero prime ideal, sayM. Thusp2Mand for every prime number q6p, we have q2=M. Consequently, 1
q2S and thereforeZ(p)Swhich implies thatS\QZ(p)(note,Z(p)is a maximal
subring ofQ). We also observe thatc(S=M)p, forp2M. Now we claim that f is an injective homomorphism. To this end, we show that Pkerf60, leads us to a contradiction. SinceS=Pembeds in the domain S, we infer thatPis a prime ideal and therefore we must havePM. But c(S=P)pwhich is absurd, forS=PImfR. Next, we show thatfis order preserving. Leta2S,a>0 and thereforeax2,x2R. We claim thatx2Sand we are done, for in that casef(a)f(x)2>0. To see this, x2aimplies thatxis integral overSand therefore ifx2=S, thenRS[x]
is integral overS. Consequently,Sis a field which is impossible. Hence we must have x2S and f(a)>0 which means that f is order preserving.
Finally, let f(a)b, where a;b2S,a6b and seek a contradiction. We may assume thata5b(note,f( a) b). SinceZ(p)is not a cyclic group, we infer that Z(p) is dense inR and therefore there exists q2Z(p) with a5q5b. Hence f(a)5f(q)5f(b). But trivially, f is the identity on Z(p) which means thatb5q5f(b), a contradiction.
Case(II):QS. First, let us get rid of the case whenSmight be a field (note, we shall see shortly in the next remark thatScan never be a field, but let us ignore this for a moment). For everya2RnS, we haveRS[a]
andRis algebraic overS(note,a 12S[a]). Hence we may naturally ex- tend f to f:R !R by definingfjS f and f(a)a. But it is well- known and easy to prove that f is the identity (in fact, we have already proved this in the last part of the proof of the previous case) and we are done. Hence we may suppose thatSis not a field. We now claim thatfis also order preserving in this case. It suffices to show that a>0 implies f(a)>0 for all a2S. We haveax2, where x2R. We now claim that x2S, for similar to the previous case,x2=Simplies thatSis a field which is absurd. Hence f(a)f(x)20. Now let q2Q be an element with 05q5a, then 0f(q)f(a). But triviallyfis the identity onQ, Hence 05qf(a), i.e.,f(a)>0. Finally, by applying the final part of the proof of the previous case and replacingZ(p) byQwe are done. p
We give the following remark for the sake of the reader.
REMARK 2.11. We should clarify our ignoring, intentionally, an im- portant fact concerning the maximal subringSofRin the previous proof.
Let us first recall an amazing and well-known theorem of Artin-Schreier, which says, whenever a fieldFis not algebraically closed but its algebraic closure, sayL, is a finite extension ofF, thenF must be real closed and LF(i), wherei2 1, see [11, P. 316, Theorem 17]. It follows easily that no maximal subring of a real closed field (note, the algebraic closure of a
real close field is a finite extension over it), let aloneR, can be a field (note, ifS is a maximal subring ofRwhich is a field, then clearly Ris a finite extension ofS, see [2, Theorem 3.3] and thereforeC, the field of complex numbers, is a finite extension ofS). We can also easily infer from the Artin- Schreier Theorem, that no maximal subring of an algebraically closed field with a nonzero characteristic, can be a field. Hence in the proof of Pro- position 2.10, although we know that,Sis actually not a field, but we prefer not to make use of this in the proof, on purpose, in order, to avoid invoking the sophisticated theorem of Artin-Schreier, to just get rid of the trivial case whenSmight be a field.
Now, we digress for a moment to record the following interesting fact.
PROPOSITION 2.12. Let R be a domain. Then there is no principal maximal ideal in the following rings.
(1) R[x], where R is an infinite domain whose set of units has smaller cardinality thanjRj.
(2) R[x1;x2;. . .;xn], n2.
PROOF. First we prove (1). LetM(f(x)) be a maximal ideal inR[x]
and seek a contradiction. Clearly, the setAof all elementsa2Rsuch that f(a) is a unit, has smaller cardinality thanjRj(note, if the set of units is finite thenAis finite too, i.e.,jAj5jRj, if the set of units is infinite and has car- dinalitya, thenjAj a5jRj). Thus there isb2Rsuch thatf(b)c60 i s not a unit (note,f(x)0 cannot have a root inR, for otherwiseR[x]=MR, which is absurd; we should also emphasize that we may assumec60 by just using the fact thatjAj5jRj). Now we note that (f(x))4(c;f(x))6R[x], for if 1cg(x)h(x)f(x), then 1c(g(b)h(b)), a contradiction. Hence (f(x)) is contained in a proper ideal (c;f(x)) which is a contradiction.
Finally we prove (2). This part must be well-known (at least whenRi s a field), but we give a quick proof. It suffices to prove it forn2. Hence let (f) be a maximal ideal inR[x1;x2] and obtain a contradiction. Clearly,f is not a constant element in R. Put f f0f1x2 fmxm2, where fi2R[x1], i0;1;. . .;m. It is clear that f2=R[x1], f2=R[x2] and f060.
Now we note that f F(x2)2R[x1][x2] and there is some k such that F(xk1)2=R, for otherwise for allr6swe havexr1 xs1jF(xr1) F(xs1) which is absurd. Finally, we claim that (f)4(F(xk1);f)6R[x1;x2], which is the de- sired contradiction. To this end, it is evident that (f)6(F(xk1);f). Hence we must show that (F(xk1);f)6R[x1;x2]. If (F(xk1);f)R[x1;x2], then
1F(xk1)gfh implies that 1F(xk1)(g(x1;xk1)h(x1;xk1)), where g;h2R[x1;x2]. Clearly, the latter equality is absurd. p REMARK2.13. We recall that a domainRi s aG-domain if and only if there is a principal maximal idealMinR[x] withM\R(0), see the proof of Theorem 24 in [12]. This and part (2) of the previous result immediately gives another proof of the well-known fact thatR[x] is neverG-domain.
Part (1) of Proposition 2.12, immediately yields the following corollary.
COROLLARY2.14. IfRis an infiniteG-domain andU(R)is the set of units ofR, thenjU(R)j jRj.
REMARK2.15. It is interesting to notice that if Ris a domain whose Jacobson radicalJis nonzero, thenjU(R)j jRj(note,jJj jRjand 1 x is a unit for allx2J). This immediately gives another quick proof to the previous corollary.
In [5], it is proved that ifRis a finite maximal subring of a ringT, thenT must be finite too. This is a trivial consequence of the fact that wheneverR is a maximal subring ofT, thenRis Artinian if and only ifTis Artinian and integral overR, see [3, the proof of Theorem 2.9] or [4, Theorem 3.8].
COROLLARY2.16. IfRis a maximal subring of an infiniteG-domain T, thenjU(R)j jRj.
PROOF. We note that Tis algebraic over R, for if t2T, then either t22R ort2R[t2]. Clearly,Tis also finitely generated as a ring over R (note, for eacht2TnR, we haveR[t]T). Now, by [12, Theorem 22],Ris a G-domain and by the preceding comment it must be infinite too and
therefore by Corollary 2.14, we are done. p
REMARK2.17. By the proof of the above corollary, and [12, Theorem 22], in fact ifRis a maximal subring of an integral domainT, thenRi s aG- domain if and only if T i s a G-domain. In particular, if R is a maximal subring of a ringT(not necessarily an integral domain), then an idealQin Ti s aG-ideal if and only ifQ\Ri s aG-ideal inR.
We remind the reader that wheneverRis a maximal subring of a ring T, then eitherTis integral overRorRis integrally closed inT. Also one
can easily see that ifRis a maximal subring ofT, then the conductor ideal (R:T) is a prime ideal inR, see [8] or [17, Theorem 1 and Theorem 7], see also [4, Remark 3.1]. Moreover, in [8] or [17] it is proved that if R i s a maximal subring of a ring T, then T is integral over R if and only if (R:T)2Max(R); it is also shown that ifRis integrally closed inT, then (R:T)2Spec(T). The following is now in order.
PROPOSITION2.18. LetRbe a maximal subring of a ringT. Then the following statements hold.
(1) If R is a Hilbert ring then so is T.
(2) If T is integral over R, then R is a Hilbert ring if and only if T is a Hilbert ring.
(3) If T is not integral over R and(R:T)2Max(T), then R is never a Hilbert ring.
PROOF. We remind the reader that, wheneverRis a Hilbert ring, then the rings of polynomials in finitely many variables overRand epimorphic images of R are also Hilbert, see [12]. This proves (1) (note, for any x2TnR, we haveR[x]T). Therefore for (2), it suffices to show thatRis Hilbert wheneverTis Hilbert. LetPbe aG-ideal ofR, sinceTis integral overR, there exists a prime idealQofTsuch thatQ\RP. Now, we have two cases, if Q6R, then RQT. Hence we infer that R=PT=Q.
ThereforeQis also aG-ideal ofT. ThusQis a maximal ideal ofT, which implies that P is a maximal ideal in R, too. Now, assume that QR.
ThereforeQPandR=Pis a maximal subring ofT=P. Inasmuch asR=Pis aG-domain, we infer thatT=Pis also aG-domain, by Remark 2.17. HenceP is a maximal ideal ofTand sinceT=Pis integral overR=P, we infer thatR=P is a field too, and we are done. For (3), we note that by Remark 2.17, M(R:T) i s aG-ideal ofR. Now, sinceTis not integral overR,Mis not a maximal ideal inR, by the above comment. Hence we are done. p Finally, we conclude this article with the following short section on submaximal ideals.
3. Submaximal Ideals
Let us for the sake of brevity, in this section, call a ringRsubmaximal, ifRhas a maximal subring and call an idealIofR, a submaximal ideal if R=Iis a submaximal ring (note,I6R). It is manifest that whenever R
contains a field which is either of zero characteristic or uncountable, then every ideal of R is submaximal. We also note that if R is a Noe- therian ring withjRj>2@0, then the nilradical ofRis submaximal, by [4, Theorem 2.9]. It is easy to see that a ringRis submaximal if and only if there exist a proper subringSofRandx2RnSsuch thatS[x]R, see [2, Theorem 2.5].
The following is interesting.
LEMMA 3.1. Let R be a ring and I be an ideal of R[x]. Then I is sub- maximal if and only if at least one of the following conditions holds,
(1) For any r2R, x r2=I.
(2) There exists an element r2R suchthat x r2I and I\R is submaximal in R. In particular in this case I(I\R;x r).
PROOF. Let f : R[x]!R[x]
I be the natural epimorphism and put Sf(R). We consider two cases.
(1) Suppose for each r2R, we have x r2=I which is equivalent to xI2=S. HenceR[x]
I S[xI] and we are done by the above comment.
(2) Suppose that x r2I for some r2R or equivalently, SR[x]
I . But clearly,R[x]
I S R
I\Rimplies thatIis submaximal in R[x] if and only if I\R is submaximal in R. It remains to be shown that I(x r;I\R). To this end, it is evident that (x r;I\R)I. Now let f(x)2Iand putf(x)(x r)g(x)b, whereg(x)2R[x] andb2R. Con- sequently,f(x)2(x r;I\R) and we are done. p COROLLARY3.2. The ringT is submaximal if and only if there exist a ring R and a ring epimorphism f:R[x]!T such that either Ker(f)\ fx r jr2Rg ;orR\Ker(f)is a submaximal ideal inR.
We recall that ifPi s a G-ideal of a ringR(i.e., R=Pi s a G-domain), thenPR\M, for some maximal idealMofR[x]. IfPis not maximal, then by the proof of [12, Theorem 24], R[x]=M is the quotient field of R=P.
The following is now immediate.
COROLLARY3.3. LetM be a maximal ideal in the ringR[x]. Then the following statements are equivalent.
(1) M is not submaximal in R[x].
(2) M(x r;m)for some r2R, where mM\R2Max(R)is not submaximal in R.
PROOF. We note that wheneverIis an ideal ofR[x] withx r2Ifor somer2R, thenI(x r;I\R). This and the previous lemma complete the proof if we show that whenMis not submaximal, thenmM\Ris maximal inR. To see this, we note thatmi s aG-ideal and if it is not maximal, thenR[x]
M R
m[t], for somet2R[x]
M nR
m noting thatR[x]
M is the quotient field of R
m
. Consequently, by the observation in the introduction of this section, we infer thatR[x]
M is submaximal, which is absurd. Thusmmust be
maximal. p
COROLLARY3.4. LetRbe a non-submaximal ring andIbe an ideal in R[x]. ThenIis submaximal inR[x]if and only ifI\ fx rjr2Rg ;.
COROLLARY3.5. Let R be a ring and I be an ideal in R[x1;. . .; xn].
ThenIis not submaximal if and only ifI(I\R; x1 r1;. . .; xn rn), whereri2RandI\Ris not submaximal inR.
In view of our main result in Section 1, the next corollary characterizes non-submaximal ideals in the rings of polynomials in finitely many vari- ables over a fieldK.
COROLLARY 3.6. Let K be a field and I be an ideal in R K[x1;. . .; xn]. ThenI is not submaximal if and only if I is a maximal ideal of the formI(x1 a1;. . .; xn an)andKis non-submaximal.
Using the above corollary and the Hilbert's Nullstellensatz, we next present a fact which can be considered as a slight generalization of Hil- bert's Nullstellensatz.
REMARK3.7. LetKbe an algebraically closed field andIbe an ideal in RK[x1;. . .;xn]. IfIis a non-submaximal ideal ofR, thenIis a maximal ideal of the form I(x1 a1;. . .;xn an) and KFp, for some prime
numberp(note, in this case there is no need forKto be algebraically closed to get the latter form) and ifIis a maximal ideal ofRwhich is submaximal, thenIhas the same form andKis a submaximal field. Consequently, in this case each proper ideal ofRis submaximal, too. We should also emphasize that in the former case proper non-maximal ideals are submaximal. Hence, when KFp andIis a proper non-maximal ideal inR, then there exist maximal subrings S of R such that IS and S=(S\M)Fp for all maximal idealsMofR.
Finally, in the following proposition, we characterize rings in which every maximal ideal ofR[x] is non-submaximal, which is a generalization of Corollary 3.6.
PROPOSITION3.8. LetRbe a ring withnonzero characteristic, sayn.
Then the following conditions are equivalent.
(1) Every maximal ideal of R[x]is non-submaximal
(2) R is a Hilbert ring and for eachmaximal ideal m of R, R=m is an algebraically closed field withnonzero characteristic.
In particular, if the number of prime divisors of n is k andjMax(R)j>k, then R is submaximal.
PROOF. If (1) holds, then by the proof of Corollary 3.3,Ris a Hilbert ring; and ifMis a maximal ideal inR[x], thenM(M\R;x r), where r2R. This shows thatR=(R\M) is algebraically closed field and we are done (note, ifm is a maximal ideal inR, and p(x) is an irreducible poly- nomial in (R=m)[x], thenM(m;q(x)), where q(x)Pn
i0aixi2R[x] with p(x)Pn
i0aixi, aiaim, is a maximal ideal of R[x] and M\Rm, henceR[x]=M(R=m)[x]
(p(x)) and since the latter field is non-submaximal we must have deg(p(x))1). The converse is similar, by applying Corollary 3.3. For the final part, since ifm is a maximal ideal of R, then we have R=mFp for some prime numberpjn, we infer that there exist distinct maximal ideals m1 and m2 such that R=m1R=m2, i.e., m1\m2 is submaximal, by [2, Theorem 2.2], and we are done. p Acknowledgments. We are very grateful to the referee for reading this paper carefully and giving helpful comments and corrections.
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Manoscritto pervenuto in redazione il 30 gennaio 2011.
