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Preprint submitted on 10 Oct 2020
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Good rings and homogeneous polynomials
J. Fresnel, Michel Matignon
To cite this version:
J. Fresnel, Michel Matignon. Good rings and homogeneous polynomials. 2020. �hal-02173007v3�
Good rings and homogeneous polynomials
J. Fresnel and M. Matignon October 9, 2020
Abstract
In 2011, Khurana, Lam and Wang define the following property. (*)A commutative unital ring A satisfies the property “power stable range one” if for all a, b∈ Awith aA+bA =A there is an integerN=N(a, b)≥1andλ=λ(a, b)∈Asuch thatbN+λa∈A×, the unit group ofA.
In 2019, Berman and Erman consider rings with the following property
(**)A commutative unital ring A has enough homogeneous polynomials if for anyk ≥1 and set S:={p1, p2, ..., pk}, of primitive points inAnand anyn≥2, there exists an homogeneous polynomial P(X1, X2, ..., Xn)∈A[X1, X2, ..., Xn]withdegP≥1andP(pi)∈A×for1≤i≤k.
We show in this article that the two properties (*) and (**) are equivalent and we shall call a commutative unital ring with these properties a good ring.
WhenAis a commutative unital ring of pictorsion as defined by Gabber, Lorenzini and Liu in 2015, we show thatAis a good ring. Using a Dedekind domain built by Goldman in 1963, we show that the converse is false.
1 Introduction
In this paper we consider only commutative and unital rings. As usualA× denotes the unit group ofA and ring homomorphisms send 1 to 1. In particular if A is the ring reduced to {0} then A× =A. In general we follow the notations in [L].
The main goal of this paper is to study a class of rings that we will call “good rings” and to analyse their relations with some classical or less classical properties.
A. Property “power stable range one”. Good points and good rings. In 2011, Khurana, Lam and Wang ([K.L.W], Definition 1.1 p. 123) were interested in the notion of “rings of square stable range one” which can be seen as an extension of the notion “nis the stable range of a ring” as defined by Bass in 1964 ([B], p. 498).
One says that a ring A satisfies the property “square stable range one” if for all a, b ∈ A with aA+bA=A, there is λ∈A such thatb2+λa∈A×, where A×.
In the epilogue of their paper ([K.L.W], p. 141) they give a generalization of the property “square stable range one”, namely
Definition 1.1. Property “power stable range one”. A ringA satisfies the property power stable range one, if for alla, b∈A with aA+bA=A, there is an integerN =N(a, b)≥1, λ=λ(a, b)∈A withbN +λa∈A×, the unit group ofA.
Let us re-interpret this notion in terms of primitive points.
Definition 1.2. Good points and good rings. LetA be a ring. Recall that a point p= (x1, x2, ..., xn)∈An isprimitiveifP
1≤j≤nxjA=A.
1) A primitive point(a, b)∈A2 is agood pointif there is an integer N =N(a, b)≥1 and λ=λ(a, b)∈A withbN+λa∈A×.
2) A ringA is agood ringif the primitive points in A2 are good points.
So, in other words, good rings are those satisfying the power stable range one property.
Remark 1.1. The ring Z of integers is a good ring. Namely, if(a, b)∈Z2 is a primitive point, b is a unit moduloaZand as the unit group aZ
Z
× is finite, then(a, b)is a good point.
B. Rings with enough homogeneous polynomials. Before going further, we remark that if A is a ring andp:= (x1, x2, ..., xn)∈An is such thatP(p) :=P(x1, x2, ..., xn)∈A× for some homogeneous polynomialP(X1, X2, ..., Xn)∈A[X1, X2, ..., Xn]ofdegP ≥1, thenpis a primitive point. Reciprocally ifp:= (x1, x2, ..., xn)∈An is primitive, then there isW(X1, X2, ..., Xn) :=P
1≤i≤nuiXi such that W(p)∈A×.
Now generalizing this equivalence, we define a new family of rings as suggested in [Be.E], namely Definition 1.3. Rings with enough homogeneous polynomials. A ring A has enough homo- geneous polynomials in two variables, (resp. enough homogeneous polynomials) if for all finite set S := {p1, p2, ..., pk} of primitive points in A2 with cardS := k ≥ 1, (resp. primitive points in An and any n ≥ 2), there is P(X1, X2) ∈ A[X1, X2] (resp. P(X1, X2, ..., Xn) ∈ A[X1, X2, ..., Xn]) with P homogeneous, degP ≥ 1 and P(pi) ∈ A× for 1 ≤ i ≤ k where P(pi) := P(p1,i, p2,i, ..., pn,i) and pi:= (p1,i, p2,i, ..., pn,i)∈An.
A main result is
Theorem 2.1. Let A be a ring. The following properties are equivalent.
i) The ring Ais a good ring,
ii) the ringAhas enough homogeneous polynomials.
iii) the ring Ahas enough homogeneous polynomials in two variables.
Remark 1.2. In the caseA=Z, we saw (Remark 1.1) thatZis a good ring. The implication i) implies ii) in Theorem 2.1 works by induction on the cardinality of the set of points we want to interpolate contrarily to the proof the ring Z has enough homogeneous polynomials in [Be.E], Theorem 0.1. Let us sketch the steps in their proof. The first point is to show that a field K has enough homogeneous polynomials. This follows from the classical avoiding lemma namely that if an ideal of a ring is included in a finite union of prime ideals then it is included in one of them. Then they show that the property “having enough polynomials” transfers to the direct product of two rings and deduce that fora∈Z− {0,±1}the ring aZ
Z has enough homogeneous polynomials. Now letS:={p1, p2, ..., pr}primitive points inZn. Again the avoiding lemma proves the existence of Pi ∈Z[X0, X1, ..., Xn] homogeneous and non constant such that Pi(pi)6= 0for1≤i≤r andPi(pj) = 0 forj6=i. Leta:=P1(p1)...Pr(pr)6= 0andpi the image of piin as aZ
Z
n. As aZ
Z has enough homogeneous polynomials there isH∈ aZ
Z[X0, X1, ..., Xn], homogeneous and non constant such thatH(pi)∈ aZ
Z
×. Then combining thePi with a lifting ofH inZ[X0, X1, ..., Xn], they get a non constant homogeneous polynomialQ withQ(pi) = 1 for alli.
C. Pictorsion rings. The last class is related with torsion in Picard groups, namely with ([G.L.L], Definition 0.3, p. 1191) we can define
Definition 1.4. Pictorsion rings. A ringA isa pictorsion ring if for all ringB which is finite over A, its Picard groupPic(B)is a torsion group.
Remark 1.3. We would like to comment on this notion. Namely, why is it usefull? Is the ring Z a pictorsion ring?
This notion is usefull in order to have a Noether normalization lemma for families (([C.MB.P.T], [G.L.L]).
If R is a pictorsion ring then Noether normalization lemma is valid for projective schemes over SpecR. Moreover a ring R is pictorsion if for all equidimensional projective schemes overSpecR there is a Noether normalization (([C.MB.P.T], [G.L.L]).
Now what’s aboutZ and pictorsion ?
A long time ago, Minkowski using the geometry of numbers proves the finiteness of the class group of the integral closure ofZ in a finite algebraic extension of Q and more generally the finiteness of the picard group of a ring which is finite over Z appears in ([MB], Theorem 2.3. p. 165) in his geometric proof of Rumely’s theorem on Skolem problem. So we can say that Zis a ring of pictorsion.
Note there is a probalistic approach in [Br.E] to Noether normalization lemma for the rings Z or Fq[T].
It is more subtle to express the property of good point with pictorsion, namely we prove the following caracterization of good points.
Theorem 3.1. Let A be a ring,(a, b)∈A2 a primitive point.
Let A[x, y] := X(aYA[X,Y−bX)A[X,Y] ] wherex(resp.y) is the image of X (resp. Y) by the natural epimor- phism. Moreover A[x, y] is endowed with the induced grading. LetS(a, b) := ProjA[x, y]. The following properties are equivalent.
i) The OS(a,b)(S(a, b))-moduleOS(a,b)(1)(S(a, b))is a torsion element in the Picard group of OS(a,b)(S(a, b)),
ii) there existsP(X, Y)∈A[X, Y] homogeneous of degree d≥1 with P(0,1), P(a, b)∈A×, iii) the point(a, b)∈A2 is a good point.
It follows that the ring Ais a good ring if and only if for all primitive point(a, b)∈A2, OS(a,b)(1)(S(a, b))is a torsion element in the Picard group ofOS(a,b)(S(a, b)).
More specifically, let Abe a pictorsion ring. AsOS(a,b)(S(a, b))is a free rank twoA-module (Propo- sition 3.1), it follows thatOS(a,b)(1)(S(a, b))is a torsion element inPic(OS(a,b)(S(a, b))). We get Corollary 3.1. Let Abe a pictorsion ring, then Ais a good ring.
A question is to know if there are good ringsA withPic(OS(a,b)(S(a, b))) is not a torsion group.
The answer uses ([G], Corollary 2 p. 118), a 1963 paper where Goldman shows the existence of a Dedekind domain A withZ[X] ⊂A ⊂Q[X], MA finite for all maximal idealM and such that its ideal class group isn’t a torsion group. Such a ring is a good ring but not a pictorsion ring (Proposition 4.13).
Acknowledgments.
We would like to thank Dino Lorenzini for indicating us the notion “n is the stable range of a ring”
and for attracting our attention on the first version of [Be.E]. We thank Qing Liu for showing us the subtleties of Picard group. Moreover his remarks on an early version of our paper gave us the opportunity to rewrite and simplify some proofs in using cohomological tools. We also warmly thank the referee for many useful suggestions which led to improvements in the exposition.
Outline of the paper.
In section 2, we prove in particular the equivalence of the two notions “good rings” and “having enough homogeneous polynomials” (Theorem (2.1)). Moreover we rephrase this in terms of sections of theA-schemePnA. We give also some examples of good rings.
In section 3, we give a geometric characterization of good points and good rings in terms of Picard group.
In section 4, in order to help the reader, we begin by a subsection in which we list the results concerning good rings and we postpone the proofs further. We discuss the stability of good rings by inductive limit, product, quotient, integral extension, localisation, transfert to a polynomial ring. Moreover we give many examples of good rings or not good rings. We also study the links between good rings and pictorsion (sous-section 4.3).
2 Invertible values of homogeneous polynomials and primitive points.
In the sequel we adopt the terminology introduced in the introduction.
Proposition 2.1. Let Abe a ring. The following properties are equivalent.
i) The ring Ais a good ring, ii) for alla∈A, the group (ρρa(A))×
a(A×) is a torsion group whereρa :A→ aAA is the natural epimorphism.
Proof. 1) We show thati)impliesii).
Leta∈Aandb∈Awithρa(b)∈(aAA)×, then there isa0, b0∈Awitha0a+b0b= 1and so(a, b)∈A2 is primitive point. Then by i) (a, b) is a good point and so there is an integer N ≥ 1, λ ∈ A with bN +λa∈A×, which means that(ρa(b))N ∈ρa(A×), i.e. ii)is satisfied.
2) We show that ii)impliesi).
Let (a, b)∈ A2 a primitive point and ρa :A → aAA, the natural epimorphism. Thenρa(b)∈(aAA)× and byii)there is an integerN ≥1with(ρa(b))N ∈ρa(A×). So there isλ∈AwithbN +λa∈A×, i.e.
(a, b)is a good point andi)is satisfied.
Remark 2.1. 1. Property ii) (Proposition 2.1) is trivially satisfied whena= 0ora∈A×. 2. IfA is a field, then partii)(Proposition 2.1) is trivially satisfied, so a field is a good ring.
3. LetAbe a ring. If for alla∈A− {0},(aAA)× is a finite or a torsion group, then part ii)(Proposition 2.1) is satisfied and soA is a good ring.
4. Let Abe a ring. If for all a∈A− {0}, aAA is finite, then Ais a good ring. In particular Zis a good ring.
5. LetAbe a ring. IfAis a local ring, one can easily show thatAis a good ring (Proposition 4.4). One can give an example of a good ringA anda∈A− {0} such that (aAA)× is not a torsion group (for exampleA:=Q[[T]], the formal power series ring with rational coefficients anda=T).
Proposition 2.2. On the section associated to a primitive point. Letp:= (a0, a1, ..., an)∈An+1 be a primitive point i.e. a0A+a1A+...+anA=A.
1. Let ρ:A[X0, X1, ..., Xn]→A[T] be the A-homomorphism with ρ(Xi) =aiT for 0≤i≤n. Then the homomorphism ρis an epimorphism and its kernel is
Pp:=P
0≤i<j≤n(aiXj−ajXi)A[X0, X1, ..., Xn].
2. The structural morphismProjA[T]→SpecAis an isomorphism. LetPnA:= ProjA[X0, X1, ..., Xn], and π : PnA →SpecA the structural morphism, then ρ induces a section σp : SpecA → PnA (i.e.
πσp=IdSpecA) and σp(SpecA) =V+(Pp).
More precisely if p∈SpecA, thenσp(p) =pA[X0, X1, ..., Xn] +Pp. We call σp,the section of π associated to the primitive pointp.
3. LetP ∈A[X0, X1, ..., Xn]homogeneous of degree d≥1. The following assertions are equivalent (a) V+(P)∩V+(Pp) =∅,
(b) V+(P)∩σp(SpecA) =∅, (c) P(a0, a1, ..., an)∈A×.
Proof. 1. Letλ0, λ1, ..., λn∈Abe such that λ0a0+λ1a1+...+λnan= 1, then ρ(λ0X0+λ1X1+...+λnXn) =T and soρis onto.
Asρ(aiXj−ajXi) = 0we havePp⊂Kerρ. LetP(X0, X1, ..., Xn)∈Kerρan homogenous polynomial of degreed, then we haveP(a0, a1, ..., an) = 0.
LetM⊂Aa maximal ideal. Asa0A+a1A+...+anA=Athere isai∈/ M. Thenai is invertible in AM and soP ∈P
0≤j≤n(aiXj−ajXi)AM[X0, X1, ..., Xn]. It follows that(Kerρ)M ⊂(Pp)M for allM and so(Kerρ)⊂Pp.
2. This is an exercise which is left to the reader.
3. As (b) rephrases (a) in the geometric language it is sufficient to prove (a) equivalent (c).
Let us assume that (a) is not verified. Then by 2. there is a prime idealp⊂A with
P(X0, X1, ..., Xn)∈pA[X0, X1, ..., Xn] +Pp. One hasP(a0, a1, ..., an)∈pand soP(a0, a1, ..., an)∈/A×. Let us assume now thatP(a0, a1, ..., an)∈/A×, then there isp∈SpecAsuch thatP(a0, a1, ..., an)∈p.
It follows thatP(X0, X1, ..., Xn)∈pA[X0, X1, ..., Xn] +Pp. Remark 2.2. On sections for the morphism π:PnA→SpecA.
We know that such a section is associated to an onto A-linear map f : An+1 → M where M is a locally free rank one A-module, ([G.D], Theorem 4.2.4, p. 74). The case of sections associated to a primitive point corresponds to the case whereM is a free rank oneA-module.
Theorem 2.1. Let Abe a ring. The following properties are equivalent.
i) The ring Ais a good ring,
ii) the ringAhas enough homogeneous polynomials,
iii) the ring Ahas enough homogeneous polynomials in two variables.
Proof. We show thati)impliesii)impliesiii) impliesi).
1) We showi)impliesii). The proof works by induction onk= cardS. Letn≥1.
1.1) If cardS = 1, then S ={p1 = (p1,1, p2,1, ..., pn,1)∈ An} and there are u1, u2, ..., un ∈ A with P
1≤j≤nujpj,1= 1.
ClearlyP(X1, X2, ..., Xn) :=P
1≤i≤nuiXi works.
1.2) Let k ≥ 1 and S0 := {p1, p2, ..., pk} ⊂ An, consisting in k primitives points. By induction hypothesis there is an homogeneous polynomialP(X1, X2, ..., Xn)∈A[X1, X2, ..., Xn]of degreed≥1 withP(pi) :=P(p1,i, p2,i, ..., pn,i)∈A×, where pi:= (p1,i, p2,i, ..., pn,i)for1≤i≤k.
Letq= (q1, q2, ..., qn)∈An be a primitive point withq /∈S0 . We want to find
R(X1, X2, ..., Xn)∈A[X1, X2, ..., Xn], an homogeneous polynomial of degreed0 ≥1, withR(p)∈A× for allp∈S0 and forp=q.
1.2.1) Letai,j,t:=pi,tqj−pj,tqi andAi,j,t(X1, X2, ..., Xn) :=pi,tXj−pj,tXi for1≤t≤k. We have Ai,j,t(q) =ai,j,t,Ai,j,t(pt) = 0.
LetAt:=P
1≤i,j≤nai,j,tA⊂A.
1.2.2)We show thatP(q)A+At=A, for all1≤t≤k.
Let us assume there is a maximal idealMin A withP(q)∈Mand At ⊂M, i.e. ai,j,t ∈Mfor all 1≤i, j≤n.
Let ρ: A → MA be the natural epimorphism, then ρ(pi,t)ρ(qj)−ρ(pj,t)ρ(qi) = ρ(ai,j,t) = 0 for all 1≤i, j≤n. This means that the matrix
ρ(p1,t) ρ(p2,t) ... ρ(pn,t) ρ(q1) ρ(q2) ... ρ(qn)
has rank≤1.
As pt is a primitive point we have (ρ(p1,t), ρ(p2,t), ..., ρ(pn,t)) 6= (0,0, ...,0) and so there is λt ∈ A with(ρ(q1), ρ(q2), ..., ρ(qn)) =ρ(λt)(ρ(p1,t), ρ(p2,t), ..., ρ(pn,t)).Now asqis a primitive point we have ρ(λt) 6= 0. Moreover ρ(P(q)) = ρ(λt)degPρ(P(pt)) and as P(pt) ∈ A× we get ρ(P(q)) 6= 0; a contradiction. It follows thatP(q)A+At=Afor all1≤t≤k.
1.2.3) It follows from 1.2.2) that1 =P(q)at+P
1≤i,j≤nui,j,tai,j,t for someat, ui,j,t∈A.
LetBt(X1, X2, ..., Xn) := P
1≤i,j≤nui,j,tAi,j,t(X1, X2, ..., Xn), then Bt(X1, X2, ..., Xn) is nul or ho- mogeneous of degree1.
Moreover, we have1 =P(q)at+Bt(q)andBt(pt) = 0. Then 1 =Q
1≤t≤k(P(q)at+Bt(q)) =P(q)a+Q
1≤t≤kBt(q), witha∈A.
It follows that(Q
1≤t≤kBt(q), P(q)) is a primitive point inA2 and as A is a good ring (Definition 1.2), there isN ≥1andλ∈AwithP(q)N+λQ
1≤t≤kBt(q) =∈A×.
1.2.4) Note that if α≥1, thenPα is homogeneous of degreeαdegP as P(pt)∈A× which prevent P to be a nilpotent element in A[X1, X2, ..., Xn]. It follows that up to changing P to Pα, we can assume thatNdegP ≥k.
Asq= (q1, q2, ..., qn)is a primitive point, there isu1, u2, ..., un∈AwithP
1≤s≤nusqs= 1.
LetW(X1, X2, ..., Xn) :=P
1≤s≤nusXs and R(X1, X2, ..., Xn) :=P(X1, X2, ..., Xn)N+λ(Q
1≤t≤kBt(X1, X2, ..., Xn))W(X1, X2, ..., Xn)NdegP−t. ThenR(pt) =P(pt)N ∈A×, in particularR(X1, X2, ..., Xn)is not0and with 1.2.3),
λ(Q
1≤t≤kBt(X1, X2, ..., Xn))W(X1, X2, ..., Xn)NdegP−tis nul or homogeneous of degreeNdegP, so R(X1, X2, ..., Xn)is homogeneous of degreeNdegP.
MoreoverR(q) =P(q)N +λQ
1≤t≤kBt(q) =∈A×. This showsii).
2) The implicationii) impliesiii) follows from the definition.
3) We showiii)impliesi).
Let us assume thati)isn’t satisfied, we show thatiii)isn’t satisfied.
So there is(a, b)∈A2 a primitive point which isn’t a good point, i.e. for all N ≥1 andλ∈A one hasbN−λa /∈A×.
Let assume there is an homogeneous polynomialP(X1, X2)∈A[X1, X2] of degree d ≥1 such that P(0,1) =:1∈A× andP(a, b) =:2∈A×. We write P(X1, X2) =a0X2d+a1X1X2d−1+...+adX1d thena0=P(0,1) =1∈A× and2=1bd+µawhere µ∈A.
It follows thatbd+ (1)−1µa=2(1)−1∈A×, which gives a contradiction.
Remark 2.3. In ([Be.E], Theorem 0.3), the authors show that PID (principal ideal domain) such that the quotients by maximal ideal are finite, have enough homogeneous polynomials.
Our Theorem 2.1 with Proposition 2.1, gives a characterization of good rings in terms of their quotient rings by principal ideals. With this tool we are able to give in Section 4, numerous examples of rings which are or aren’t good rings.
Now we can rephrase the fact that a ring has enough homogeneous polynomials in terms of sections associated to primitive points (Proposition 2.2 part 2.) and so we can characterize good rings in terms of an avoidance property (compare with [G.L.L], Theorem 5.1, p. 1188).
Theorem 2.2. Let A be a ring, PnA := Proj(A[X0, X1, ..., Xn]). Then the following assertions are equivalent.
1. The ringA is a good ring,
2. the ringAhas enough homogeneous polynomials,
3. for any finite family{p1, p2, ..., ps}of primitive points inAn+1there is an homogeneous polynomial P(X0, X1, ..., Xn)∈A[X0, X1, ..., Xn] withdegP ≥1 such that
V+(P)∩σpi(SpecA) =∅for1≤i≤sandσpi is the section associated topi.
3 Primitive points in A2 and Picard group
Proposition 3.1. Let A be a ring and (a, b) ∈ A2 a primitive point. Let A[x, y] := X(aYA[X,Y−bX)A[X,Y] ] wherex(resp. y) is the image ofX (resp. Y). MoreoverA[x, y] is endowed with the induced grading of A[X, Y]. LetS(a, b) := Proj(A[x, y]).
1. TheA-algebraOS(a,b)(S(a, b))is a free A-module of rank two.
More concretely, there is θ ∈ OS(a,b)(S(a, b))with θ|D+(x)= 0, θ|D+(y) = ay−bxy . One has θ2 =aθ, OS(a,b)(S(a, b))' T(a−TA[T)A[T]] and(1, θ)is a basis for the A-moduleOS(a,b)(S(a, b)).
Moreover the schemeS(a, b)is affine and isomorphic toSpec(OS(a,b)(S(a, b))).
2. Let d∈ N>0. When considering the OS(a,b)(S(a, b))-module OS(a,b)(d)(S(a, b)) as an A module, we have
OS(a,b)(d)(S(a, b)) = X
0≤k≤d
Axkyd−k.
3. Leta0, b0∈A withaa0+bb0= 1. There is an epimorphism u:A[x, y]→ (ay−bx)A[x,y]A[x,y] 'A[a0X+b0Y] which is defined byu(x) =a(a0X+b0Y) andu(y) =b(a0X+b0Y).
Let∆(a, b) := Proj(A[a0X+b0Y]),d≥1. Then uinduces an epimorphism
u0 : OS(a,b)(d)(S(a, b)) → O∆(a,b)(d)(∆(a, b)) such that u0(P(x, y)) = P(a, b)(a0X +b0Y)d where P(X, Y) ∈ A[X, Y] is homogeneous of degree d and {(a0X +b0Y)d} is a basis for the A-module O∆(a,b)(d)(∆(a, b)).
