Good rings and homogeneous polynomials

TL;DR

This paper proves the equivalence of two properties related to rings, introduces the concept of good rings, and explores their characteristics through examples including rings of pictorsion and Dedekind domains.

Contribution

It establishes the equivalence of properties (*) and (**) defining good rings and analyzes their presence in rings of pictorsion and Dedekind domains.

Findings

Properties (*) and (**) are equivalent for commutative unital rings.

Rings of pictorsion are examples of good rings.

Dedekind domains can serve as counterexamples to the converse.

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 $\in$ A with aA + bA = A there are an integer N = N (a, b) $\ge$ 1 and $\lambda$ = $\lambda$(a, b) $\in$ A such that b N + $\lambda$a $\in$ A x , the unit group of A. In 2019, Berman and Erman consider rings with the following property (**) A commutative unital ring A has enough homogeneous polynomials if for any k $\ge$ 1 and set S := {p 1 , p 2 , ..., p k } , of primitive points in A n and any n $\ge$ 2, there exists an homogeneous polynomial P (X 1 , X 2 , ..., X n) $\in$ A[X 1 , X 2 , ..., X n ]) with deg P $\ge$ 1 and P (p i) $\in$ A x for 1 $\le$ i $\le$ 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. When A is a commutative unital ring of pictorsion as defined by Gabber, Lorenzini and Liu in 2015, we show that A is a good ring. Using a Dedekind domain we built by Goldman in 1963,we show that the converse is false.

Equations

$${\mathcal{O}}_{S(a,b)}(d)(S(a,b))=\sum _{0\le k\le d}A{x}^k{y}^{d-k}.$$

$$\{0\}\to \mathcal{J}\to {\mathcal{O}}_{{\mathbb{P}}^1}\to \frac{{\mathcal{O}}_{{\mathbb{P}}^1}}{\mathcal{J}}\to \{0\}.$$

$$\{0\}\to \mathcal{J}\to {\mathcal{O}}_{{\mathbb{P}}^1}\to {\mathcal{O}}_{S(a,b)}\to \{0\}.$$

$$\{0\}\to H^0({\mathbb{P}}^1,\mathcal{J})\to H^0({\mathbb{P}}^1,{\mathcal{O}}_{{\mathbb{P}}^1})\to \to H^0({\mathbb{P}}^1,{\mathcal{O}}_{S(a,b)})\stackrel{\partial }{\to }{H}^1({\mathbb{P}}^1,\mathcal{J})\to H^1({\mathbb{P}}^1,{\mathcal{O}}_{{\mathbb{P}}^1}).$$

$$\{0\}\to A\to H^0({\mathbb{P}}^1,{\mathcal{O}}_{S(a,b)})\stackrel{\partial }{\to }{H}^1({\mathbb{P}}^1,\mathcal{J})\to \{0\}.$$

$$\{0\}\to {\mathcal{O}}_{{\mathbb{P}}^1}(-2)\to {\mathcal{O}}_{{\mathbb{P}}^1}\to {\mathcal{O}}_{S(a,b)}\to \{0\}.$$

$$\{0\}\to {\mathcal{O}}_{{\mathbb{P}}^1}(d-2)\to {\mathcal{O}}_{{\mathbb{P}}^1}(d)\to {\mathcal{O}}_{S(a,b)}(d)\to \{0\}.$$

$$H^0({\mathbb{P}}^1,{\mathcal{O}}_{{\mathbb{P}}^1}(d))\stackrel{w}{\to }{H}^0({\mathbb{P}}^1,{\mathcal{O}}_{S(a,b)}(d))\to H^1({\mathbb{P}}^1,{\mathcal{O}}_{{\mathbb{P}}^1}(d-2))$$

$$H^0({\mathbb{P}}^1,{\mathcal{O}}_{S(a,b)}(d))={\mathcal{O}}_{S(a,b)}(d)(S(a,b))=\sum _{0\le k\le d}A{x}^k{y}^{d-k}.$$

$$\{0\}\to \mathcal{G}\to {\mathcal{O}}_{{\mathbb{P}}^1}\to {\mathcal{O}}_{\Delta (a,b)}\to \{0\},$$

$$\{0\}\to {\mathcal{O}}_{{\mathbb{P}}^1}(d-1)\to {\mathcal{O}}_{{\mathbb{P}}^1}(d)\to {\mathcal{O}}_{\Delta (a,b)}(d)\to \{0\}.$$

$$\{0\}\to {\mathcal{O}}_{{\mathbb{P}}^1}(d-1)({\mathbb{P}}^1)\to {\mathcal{O}}_{{\mathbb{P}}^1}(d)({\mathbb{P}}^1)\stackrel{u_d^{\prime \prime }}{\to }{\mathcal{O}}_{\Delta (a,b)}(d)(\Delta (a,b))\to H^1({\mathbb{P}}^1,{\mathcal{O}}_{{\mathbb{P}}^1}(d-1))$$

$$P(\mathbb{X})U(\mathbb{X})+Q(\mathbb{X})V(\mathbb{X})=R(\mathbb{X}).$$

$${\pi }_N\times {\pi }_N(C(P(z),Q(z))\subset \Theta (z).$$

$${\pi }_N\times {\pi }_N(C(P(z^{\prime }),Q(z^{\prime }))\subset \Theta (z).$$

$${\pi }_N\times {\pi }_N({\cup }_{z\in Z}C(P(z),Q(z))\subset {\cup }_{z\in Z_N}\{(x,y)\in (\frac{\mathbb{Z}}{p^N\mathbb{Z}}{)}^2|{\pi }_N(P(z))y-{\pi }_N(Q(z))x=0\}.$$

$$\alpha \le M.$$

$${\pi }_N(p^{\alpha })u(y-{\pi }_N(p^{\beta -\alpha })u^{-1}vx)=0.$$

$$\mathrm{card}\{(x,y)\in (\frac{\mathbb{Z}}{p^N\mathbb{Z}}{)}^2|{\pi }_N(P(z))y-{\pi }_N(Q(z))x=0\}\le p^{N+M}.$$

$$\mathrm{card}\{{\pi }_N\times {\pi }_N({\cup }_{z\in Z}C(P(z),Q(z))\}\le N^d{p}^{N+M}.$$

$${\pi }_N\times {\pi }_N(s,t)\notin {\pi }_N\times {\pi }_N({\cup }_{z\in Z}C(P(z),Q(z)).$$

$$(s,t)\notin {\cup }_{z\in Z}C(P(z),Q(z)).$$

$$B^{\times }=(X+YZ{)}^{\mathbb{Z}}\times \{\frac{P}{Q}|P,Q\in k[X,Y,Z],P(0,0,0)\ne 0,Q(0,0,0)\ne 0\}.$$

$$(\frac{B}{aB}{)}^{\times }=Y^{\mathbb{Z}}\times Z^{\mathbb{Z}}\times \{\frac{P}{R}|P,R\in k[Y,Z],R(0,0)\ne 0\}.$$

$${\rho }_a(B^{\times })=Y^{\mathbb{Z}}\times Z^{\mathbb{Z}}\times \{\frac{P}{R}|P,R\in k[Y,Z],R(0,0)\ne 0\}.$$

$$\{0\}\to \mathbb{Z}a\to \mathrm{Pic}(X)\to \mathrm{Pic}(U)\to \{0\}$$

$$\{1\}\to \frac{{\rho }_a(A{)}^{\times }}{{\rho }_a(A^{\times })}\to \mathrm{Pic}(B_a)\to \mathrm{Pic}(A{)}^2\to \mathrm{Pic}(\frac{A}{aA}).$$

$$\{1\}\to 1+\mathcal{I}t\to {\mathcal{F}}^{\times }\to {\mathcal{G}}^{\times }\to \{0\}$$

$$\{0\}\to 1+It\to B_a^{\times }\to C_a^{\times }\to \to H^1(X,\mathcal{I})\to \mathrm{Pic}(B_a)\to \mathrm{Pic}(C_a)\to H^2(X,\mathcal{I}).$$

$$\{1\}\to \frac{{\rho }_a(A{)}^{\times }}{{\rho }_a(A^{\times })}\to \mathrm{Pic}(C_a)\to \mathrm{Pic}(A{)}^2\to \mathrm{Pic}(\frac{A}{aA}),$$

Full text

In 2019, Berman and Erman consider rings with the following property

(**) A commutative unital ring $A$ has enough homogeneous polynomials if for any $k\geq 1$ and set $S:=\{p_{1},p_{2},...,p_{k}\}$, of primitive points in $A^{n}$ and any $n\geq 2$, there exists an homogeneous polynomial $P(X_{1},X_{2},...,X_{n})\in A[X_{1},X_{2},...,X_{n}]$ with $\deg P\geq 1$ and $P(p_{i})\in A^{\times}$ for $1\leq i\leq k$.

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.

When $A$ is a commutative unital ring of pictorsion as defined by Gabber, Lorenzini and Liu in 2015, we show that $A$ is 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 usual $A^{\times}$ denotes the unit group of $A$ and ring homomorphisms send $1$ to $1$. In particular if $A$ is the ring reduced to $\{0\}$ then $A^{\times}=A$. 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 “$n$ is 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 $a,b\in A$ with $aA+bA=A$, there is $\lambda\in A$ such that $b^{2}+\lambda a\in A^{\times}$, where $A^{\times}$.

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 ring $A$ satisfies the property power stable range one, if for all $a,b\in A$ with $aA+bA=A$, there is an integer $N=N(a,b)\geq 1,\ \lambda=\lambda(a,b)\in A$ with $b^{N}+\lambda a\in A^{\times}$, the unit group of $A$.*

Let us re-interpret this notion in terms of primitive points.

Definition 1.2**.**

Good points and good rings.* Let $A$ be a ring. Recall that a point

$p=(x_{1},x_{2},...,x_{n})\in A^{n}$ is primitive if $\sum_{1\leq j\leq n}x_{j}A=A$.*

*A primitive point $(a,b)\in A^{2}$ is a good point if there is an integer $N=N(a,b)\geq 1$ and *

$\lambda=\lambda(a,b)\in A$* with $b^{N}+\lambda a\in A^{\times}$.* 2. 2)

A ring $A$ is a good ring if the primitive points in $A^{2}$ are good points.

So, in other words, good rings are those satisfying the power stable range one property.

Remark 1.1**.**

The ring $\mathbb{Z}$ of integers is a good ring. Namely, if $(a,b)\in\mathbb{Z}^{2}$ is a primitive point, $b$ is a unit modulo $a\mathbb{Z}$ and as the unit group $\frac{\mathbb{Z}}{a\mathbb{Z}}^{\times}$ 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 and $p:=(x_{1},x_{2},...,x_{n})\in A^{n}$ is such that $P(p):=P(x_{1},x_{2},...,x_{n})\in A^{\times}$ for some homogeneous polynomial $P(X_{1},X_{2},...,X_{n})\in A[X_{1},X_{2},...,X_{n}]$ of $\deg P\geq 1$, then $p$ is a primitive point. Reciprocally if $p:=(x_{1},x_{2},...,x_{n})\in A^{n}$ is primitive, then there is $W(X_{1},X_{2},...,X_{n}):=\sum_{1\leq i\leq n}u_{i}X_{i}$ such that

$W(p)\in A^{\times}$.

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 homogeneous polynomials in two variables, (resp. enough homogeneous polynomials) if for all finite set $S:=\{p_{1},p_{2},...,p_{k}\}$ of primitive points in $A^{2}$ with $\operatorname{card}S:=k\geq 1$, (resp. primitive points in $A^{n}$ and any $n\geq 2$), there is $P(X_{1},X_{2})\in A[X_{1},X_{2}]$ (resp. $P(X_{1},X_{2},...,X_{n})\in A[X_{1},X_{2},...,X_{n}]$) with $P$ homogeneous, $\deg P\geq 1$ and $P(p_{i})\in A^{\times}$ for $1\leq i\leq k$ where $P(p_{i}):=P(p_{1,i},p_{2,i},...,p_{n,i})$ and $p_{i}:=(p_{1,i},p_{2,i},...,p_{n,i})\in A^{n}$.*

A main result is Theorem 2.1. Let $A$ be a ring. The following properties are equivalent.

1. i)

The ring $A$ is a good ring, 2. ii)

the ring $A$ has enough homogeneous polynomials. 3. iii)

the ring $A$ has enough homogeneous polynomials in two variables.

Remark 1.2**.**

In the case $A=\mathbb{Z}$, we saw (Remark 1.1) that $\mathbb{Z}$ is 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 $\mathbb{Z}$ 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 for $a\in\mathbb{Z}-\{0,\pm 1\}$ the ring $\frac{\mathbb{Z}}{a\mathbb{Z}}$ has enough homogeneous polynomials. Now let $S:=\{p_{1},p_{2},...,p_{r}\}$ primitive points in $\mathbb{Z}^{n}$. Again the avoiding lemma proves the existence of $P_{i}\in\mathbb{Z}[X_{0},X_{1},...,X_{n}]$ homogeneous and non constant such that $P_{i}(p_{i})\neq 0$ for $1\leq i\leq r$ and $P_{i}(p_{j})=0$ for $j\neq i$. Let $a:=P_{1}(p_{1})...P_{r}(p_{r})\neq 0$ and $\overline{p}_{i}$ the image of $p_{i}$ in as $\frac{\mathbb{Z}}{a\mathbb{Z}}^{n}$. As $\frac{\mathbb{Z}}{a\mathbb{Z}}$ has enough homogeneous polynomials there is $H\in\frac{\mathbb{Z}}{a\mathbb{Z}}[X_{0},X_{1},...,X_{n}]$, homogeneous and non constant such that $H(\overline{p}_{i})\in\frac{\mathbb{Z}}{a\mathbb{Z}}^{\times}$. Then combining the $P_{i}$ with a lifting of $H$ in $\mathbb{Z}[X_{0},X_{1},...,X_{n}],$ they get a non constant homogeneous polynomial $Q$ with $Q(p_{i})=1$ for all $i$.

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 ring $A$ is a pictorsion ring if for all ring $B$ which is finite over $A$, its Picard group $\operatorname{Pic}(B)$ is a torsion group.*

Remark 1.3**.**

We would like to comment on this notion. Namely, why is it usefull? Is the ring $\mathbb{Z}$ 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 $\operatorname{Spec}R$. Moreover a ring $R$ is pictorsion if for all equidimensional projective schemes over $\operatorname{Spec}R$ there is a Noether normalization (([C.MB.P.T], [G.L.L]).

Now what’s about $\mathbb{Z}$ and pictorsion ?

A long time ago, Minkowski using the geometry of numbers proves the finiteness of the class group of the integral closure of $\mathbb{Z}$ in a finite algebraic extension of $\mathbb{Q}$ and more generally the finiteness of the picard group of a ring which is finite over $\mathbb{Z}$ appears in ([MB], Theorem 2.3. p. 165) in his geometric proof of Rumely’s theorem on Skolem problem. So we can say that $\mathbb{Z}$ is a ring of pictorsion.

Note there is a probalistic approach in [Br.E] to Noether normalization lemma for the rings $\mathbb{Z}$ or $\mathbb{F}_{q}[T]$.

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, $(a,b)\in A^{2}$ a primitive point.

Let $A[x,y]:=\frac{A[X,Y]}{X(aY-bX)A[X,Y]}$ where $x$ (resp.$y$) is the image of $X$ (resp. $Y$) by the natural epimorphism. Moreover $A[x,y]$ is endowed with the induced grading. Let $S(a,b):=\operatorname{Proj}A[x,y]$. The following properties are equivalent.

1. i)

*The ${\cal O}_{S(a,b)}(S(a,b))$-module ${\cal O}_{S(a,b)}(1)(S(a,b))$ is a torsion element in the Picard group of *

${\cal O}_{S(a,b)}(S(a,b)),$** 2. ii)

there exists $P(X,Y)\in A[X,Y]$ homogeneous of degree $d\geq 1$ with $P(0,1),\ P(a,b)\in A^{\times},$ 3. iii)

the point $(a,b)\in A^{2}$ is a good point.

It follows that the ring $A$ is a good ring if and only if for all primitive point $(a,b)\in A^{2}$,

${\cal O}_{S(a,b)}(1)(S(a,b))$ is a torsion element in the Picard group of ${\cal O}_{S(a,b)}(S(a,b))$.

More specifically, let $A$ be a pictorsion ring. As ${\cal O}_{S(a,b)}(S(a,b))$ is a free rank two $A$-module (Proposition 3.1), it follows that ${\cal O}_{S(a,b)}(1)(S(a,b))$ is a torsion element in $\operatorname{Pic}({\cal O}_{S(a,b)}(S(a,b)))$. We get Corollary 3.1. *Let $A$ be a pictorsion ring, then $A$ is a good ring. * A question is to know if there are good rings $A$ with $\operatorname{Pic}({\cal O}_{S(a,b)}(S(a,b)))$ 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 $A$ with $\mathbb{Z}[X]\subset A\subset\mathbb{Q}[X]$, $\frac{A}{\mathfrak{M}}$ finite for all maximal ideal $\mathfrak{M}$ 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 the $A$-scheme $\mathbb{P}^{n}_{A}$. 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 $A$ be a ring. The following properties are equivalent.

1. i)

The ring $A$ is a good ring, 2. ii)

for all $a\in A$, the group $\frac{(\rho_{a}(A))^{\times}}{\rho_{a}(A^{\times})}$ is a torsion group where $\rho_{a}:A\to\frac{A}{aA}$ is the natural epimorphism.

Proof.

1. We show that i) implies ii).

Let $a\in A$ and $b\in A$ with $\rho_{a}(b)\in(\frac{A}{aA})^{\times}$, then there is $a^{\prime},b^{\prime}\in A$ with $a^{\prime}a+b^{\prime}b=1$ and so $(a,b)\in A^{2}$ is primitive point. Then by i) $(a,b)$ is a good point and so there is an integer $N\geq 1,\ \lambda\in A$ with $b^{N}+\lambda a\in A^{\times}$, which means that $(\rho_{a}(b))^{N}\in\rho_{a}(A^{\times})$, i.e. ii) is satisfied.

2. We show that ii) implies i).

Let $(a,b)\in A^{2}$ a primitive point and $\rho_{a}:A\to\frac{A}{aA}$, the natural epimorphism. Then $\rho_{a}(b)\in(\frac{A}{aA})^{\times}$ and by ii) there is an integer $N\geq 1$ with $(\rho_{a}(b))^{N}\in\rho_{a}(A^{\times})$. So there is $\lambda\in A$ with $b^{N}+\lambda a\in A^{\times}$, i.e. $(a,b)$ is a good point and i) is satisfied.

∎

Remark 2.1**.**

Property ii) (Proposition 2.1) is trivially satisfied when $a=0$ or $a\in A^{\times}$. 2. 2.

If $A$ is a field, then part ii) (Proposition 2.1) is trivially satisfied, so a field is a good ring. 3. 3.

Let $A$ be a ring. If for all $a\in A-\{0\}$, $(\frac{A}{aA})^{\times}$ is a finite or a torsion group, then part ii) (Proposition 2.1) is satisfied and so $A$ is a good ring. 4. 4.

Let $A$ be a ring. If for all $a\in A-\{0\}$, $\frac{A}{aA}$ is finite, then $A$ is a good ring. In particular $\mathbb{Z}$ is a good ring. 5. 5.

Let $A$ be a ring. If $A$ is a local ring, one can easily show that $A$ is a good ring (Proposition 4.4). One can give an example of a good ring $A$ and $a\in A-\{0\}$ such that $(\frac{A}{aA})^{\times}$ is not a torsion group (for example $A:=\mathbb{Q}[[T]]$, the formal power series ring with rational coefficients and $a=T$).

Proposition 2.2**.**

On the section associated to a primitive point.* Let $p:=(a_{0},a_{1},...,a_{n})\in A^{n+1}$ be a primitive point i.e. $a_{0}A+a_{1}A+...+a_{n}A=A$.*

*Let $\rho:A[X_{0},X_{1},...,X_{n}]\to A[T]$ be the $A$-homomorphism with $\rho(X_{i})=a_{i}T$ for $0\leq i\leq n$. Then the homomorphism $\rho$ is an epimorphism and its kernel is *

${\mathfrak{P}}_{p}:=\sum_{0\leq i<j\leq n}(a_{i}X_{j}-a_{j}X_{i})A[X_{0},X_{1},...,X_{n}]$. 2. 2.

The structural morphism $\operatorname{Proj}A[T]\to\operatorname{Spec}A$ is an isomorphism. Let $\mathbb{P}^{n}_{A}:=\operatorname{Proj}A[X_{0},X_{1},...,X_{n}]$, and $\pi:\mathbb{P}^{n}_{A}\to\operatorname{Spec}A$ the structural morphism, then $\rho$ induces a section $\sigma_{p}:\operatorname{Spec}A\to\mathbb{P}^{n}_{A}$ (i.e. $\pi\sigma_{p}=Id_{\operatorname{Spec}A}$) and $\sigma_{p}(\operatorname{Spec}A)=V_{+}({\mathfrak{P}}_{p})$.

More precisely if ${\mathfrak{p}}\in\operatorname{Spec}A$, then $\sigma_{p}({\mathfrak{p}})={\mathfrak{p}}A[X_{0},X_{1},...,X_{n}]+{\mathfrak{P}}_{p}$.

We call $\sigma_{p}$, the section of $\pi$ associated to the primitive point $p$. 3. 3.

Let $P\in A[X_{0},X_{1},...,X_{n}]$ homogeneous of degree $d\geq 1$. The following assertions are equivalent

1. (a)

$V_{+}(P)\cap V_{+}({\mathfrak{P}}_{p})=\emptyset$, 2. (b)

$V_{+}(P)\cap\sigma_{p}(\operatorname{Spec}A)=\emptyset$, 3. (c)

$P(a_{0},a_{1},...,a_{n})\in A^{\times}$.

Proof.

1. Let $\lambda_{0},\lambda_{1},...,\lambda_{n}\in A$ be such that $\lambda_{0}a_{0}+\lambda_{1}a_{1}+...+\lambda_{n}a_{n}=1$, then

$\rho(\lambda_{0}X_{0}+\lambda_{1}X_{1}+...+\lambda_{n}X_{n})=T$ and so $\rho$ is onto.

As $\rho(a_{i}X_{j}-a_{j}X_{i})=0$ we have ${\mathfrak{P}}_{p}\subset\operatorname{Ker}\rho$. Let $P(X_{0},X_{1},...,X_{n})\in\operatorname{Ker}\rho$ an homogenous polynomial of degree $d$, then we have $P(a_{0},a_{1},...,a_{n})=0$.

Let $\mathfrak{M}\subset A$ a maximal ideal. As $a_{0}A+a_{1}A+...+a_{n}A=A$ there is $a_{i}\notin\mathfrak{M}$. Then $a_{i}$ is invertible in $A_{\mathfrak{M}}$ and so $P\in\sum_{0\leq j\leq n}(a_{i}X_{j}-a_{j}X_{i})A_{\mathfrak{M}}[X_{0},X_{1},...,X_{n}]$. It follows that $(\operatorname{Ker}\rho)_{\mathfrak{M}}\subset({\mathfrak{P}}_{p})_{\mathfrak{M}}$ for all $\mathfrak{M}$ and so $(\operatorname{Ker}\rho)\subset{\mathfrak{P}}_{p}$.

2. This is an exercise which is left to the reader.

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 ideal ${\mathfrak{p}}\subset A$ with

$P(X_{0},X_{1},...,X_{n})\in{\mathfrak{p}}A[X_{0},X_{1},...,X_{n}]+{\mathfrak{P}}_{p}$. One has $P(a_{0},a_{1},...,a_{n})\in{\mathfrak{p}}$ and so $P(a_{0},a_{1},...,a_{n})\notin A^{\times}$.

Let us assume now that $P(a_{0},a_{1},...,a_{n})\notin A^{\times}$, then there is ${\mathfrak{p}}\in\operatorname{Spec}A$ such that $P(a_{0},a_{1},...,a_{n})\in{\mathfrak{p}}$. It follows that $P(X_{0},X_{1},...,X_{n})\in{\mathfrak{p}}A[X_{0},X_{1},...,X_{n}]+{\mathfrak{P}}_{p}$. ∎

Remark 2.2**.**

On sections for the morphism $\pi:\mathbb{P}^{n}_{A}\to\operatorname{Spec}A$.

We know that such a section is associated to an onto $A$-linear map $f:A^{n+1}\to M$ 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 where $M$ is a free rank one $A$-module.

Theorem 2.1**.**

Let $A$ be a ring. The following properties are equivalent.

1. i)

The ring $A$ is a good ring, 2. ii)

the ring $A$ has enough homogeneous polynomials, 3. iii)

the ring $A$ has enough homogeneous polynomials in two variables.

Proof.

We show that i) implies ii) implies iii) implies i).

We show i) implies ii). The proof works by induction on $k=\operatorname{card}S$. Let $n\geq 1$. 1.1) If $\operatorname{card}S=1$, then $S=\{p_{1}=(p_{1,1},p_{2,1},...,p_{n,1})\in A^{n}\}$ and there are $u_{1},u_{2},...,u_{n}\in A$ with $\sum_{1\leq j\leq n}u_{j}p_{j,1}=1$.

Clearly $P(X_{1},X_{2},...,X_{n}):=\sum_{1\leq i\leq n}u_{i}X_{i}$ works. 1.2) Let $k\geq 1$ and $S^{\prime}:=\{p_{1},p_{2},...,p_{k}\}\subset A^{n}$, consisting in $k$ primitives points. By induction hypothesis there is an homogeneous polynomial $P(X_{1},X_{2},...,X_{n})\in A[X_{1},X_{2},...,X_{n}]$ of degree $d\geq 1$ with $P(p_{i}):=P(p_{1,i},p_{2,i},...,p_{n,i})\in A^{\times}$, where $p_{i}:=(p_{1,i},p_{2,i},...,p_{n,i})$ for $1\leq i\leq k$.

Let $q=(q_{1},q_{2},...,q_{n})\in A^{n}$ be a primitive point with $q\notin S^{\prime}$ . We want to find

$R(X_{1},X_{2},...,X_{n})\in A[X_{1},X_{2},...,X_{n}]$, an homogeneous polynomial of degree $d^{\prime}\geq 1$, with $R(p)\in A^{\times}$ for all $p\in S^{\prime}$ and for $p=q$.

1.2.1) Let $a_{i,j,t}:=p_{i,t}q_{j}-p_{j,t}q_{i}$ and $A_{i,j,t}(X_{1},X_{2},...,X_{n}):=p_{i,t}X_{j}-p_{j,t}X_{i}$ for $1\leq t\leq k$. We have $A_{i,j,t}(q)=a_{i,j,t}$, $A_{i,j,t}(p_{t})=0$.

Let $\mathfrak{A}_{t}:=\sum_{1\leq i,j\leq n}a_{i,j,t}A\subset A$. 1.2.2) We show that $P(q)A+\mathfrak{A}_{t}=A$, for all $1\leq t\leq k$.

Let us assume there is a maximal ideal $\mathfrak{M}$ in $A$ with $P(q)\in\mathfrak{M}$ and $\mathfrak{A}_{t}\subset\mathfrak{M},$ i.e. $a_{i,j,t}\in\mathfrak{M}$ for all $1\leq i,j\leq n$.

Let $\rho:A\to\frac{A}{\mathfrak{M}}$ be the natural epimorphism, then $\rho(p_{i,t})\rho(q_{j})-\rho(p_{j,t})\rho(q_{i})=\rho(a_{i,j,t})=0$ for all $1\leq i,j\leq n$. This means that the matrix \left[\begin{array}[]{cccc}\rho(p_{1,t})&\rho(p_{2,t})&...&\rho(p_{n,t})\\ \rho(q_{1})&\rho(q_{2})&...&\rho(q_{n})\end{array}\right] has rank $\leq 1$.

As $p_{t}$ is a primitive point we have $(\rho(p_{1,t}),\rho(p_{2,t}),...,\rho(p_{n,t}))\neq(0,0,...,0)$ and so there is $\lambda_{t}\in A$ with $(\rho(q_{1}),\rho(q_{2}),...,\rho(q_{n}))=\rho(\lambda_{t})(\rho(p_{1,t}),\rho(p_{2,t}),...,\rho(p_{n,t})).$ Now as $q$ is a primitive point we have $\rho(\lambda_{t})\neq 0$. Moreover $\rho(P(q))=\rho(\lambda_{t})^{\deg P}\rho(P(p_{t}))$ and as $P(p_{t})\in A^{\times}$ we get $\rho(P(q))\neq 0$; a contradiction. It follows that $P(q)A+\mathfrak{A}_{t}=A$ for all $1\leq t\leq k$. 1.2.3) It follows from 1.2.2) that $1=P(q)a_{t}+\sum_{1\leq i,j\leq n}u_{i,j,t}a_{i,j,t}$ for some $a_{t},u_{i,j,t}\in A$.

Let $B_{t}(X_{1},X_{2},...,X_{n}):=\sum_{1\leq i,j\leq n}u_{i,j,t}A_{i,j,t}(X_{1},X_{2},...,X_{n})$, then $B_{t}(X_{1},X_{2},...,X_{n})$ is nul or homogeneous of degree $1$.

Moreover, we have $1=P(q)a_{t}+B_{t}(q)$ and $B_{t}(p_{t})=0$. Then

$1=\prod_{1\leq t\leq k}(P(q)a_{t}+B_{t}(q))=P(q)a+\prod_{1\leq t\leq k}B_{t}(q)$, with $a\in A$.

It follows that $(\prod_{1\leq t\leq k}B_{t}(q),P(q))$ is a primitive point in $A^{2}$ and as $A$ is a good ring (Definition 1.2), there is $N\geq 1$ and $\lambda\in A$ with $P(q)^{N}+\lambda\prod_{1\leq t\leq k}B_{t}(q)=\epsilon\in A^{\times}$. 1.2.4) Note that if $\alpha\geq 1$, then $P^{\alpha}$ is homogeneous of degree $\alpha\deg P$ as $P(p_{t})\in A^{\times}$ which prevent $P$ to be a nilpotent element in $A[X_{1},X_{2},...,X_{n}]$. It follows that up to changing $P$ to $P^{\alpha}$, we can assume that $N\deg P\geq k$.

As $q=(q_{1},q_{2},...,q_{n})$ is a primitive point, there is $u_{1},u_{2},...,u_{n}\in A$ with $\sum_{1\leq s\leq n}u_{s}q_{s}=1$.

Let $W(X_{1},X_{2},...,X_{n}):=\sum_{1\leq s\leq n}u_{s}X_{s}$ and

$R(X_{1},X_{2},...,X_{n}):=P(X_{1},X_{2},...,X_{n})^{N}+\lambda(\prod_{1\leq t\leq k}B_{t}(X_{1},X_{2},...,X_{n}))W(X_{1},X_{2},...,X_{n})^{N\deg P-t}$. Then $R(p_{t})=P(p_{t})^{N}\in A^{\times}$, in particular $R(X_{1},X_{2},...,X_{n})$ is not [math] and with 1.2.3),

$\lambda(\prod_{1\leq t\leq k}B_{t}(X_{1},X_{2},...,X_{n}))W(X_{1},X_{2},...,X_{n})^{N\deg P-t}$ is nul or homogeneous of degree $N\deg P$, so $R(X_{1},X_{2},...,X_{n})$ is homogeneous of degree $N\deg P$.

Moreover $R(q)=P(q)^{N}+\lambda\prod_{1\leq t\leq k}B_{t}(q)=\epsilon\in A^{\times}$. This shows ii). 2. 2)

The implication ii) implies iii) follows from the definition. 3. 3)

We show iii) implies i).

Let us assume that i) isn’t satisfied, we show that iii) isn’t satisfied.

So there is $(a,b)\in A^{2}$ a primitive point which isn’t a good point, i.e. for all $N\geq 1$ and $\lambda\in A$ one has $b^{N}-\lambda a\notin A^{\times}$.

Let assume there is an homogeneous polynomial $P(X_{1},X_{2})\in A[X_{1},X_{2}]$ of degree $d\geq 1$ such that $P(0,1)=:\epsilon_{1}\in A^{\times}$ and $P(a,b)=:\epsilon_{2}\in A^{\times}$. We write $P(X_{1},X_{2})=a_{0}X_{2}^{d}+a_{1}X_{1}X_{2}^{d-1}+...+a_{d}X_{1}^{d}$ then $a_{0}=P(0,1)=\epsilon_{1}\in A^{\times}$ and $\epsilon_{2}=\epsilon_{1}b^{d}+\mu a$ where $\mu\in A$.

It follows that $b^{d}+(\epsilon_{1})^{-1}\mu a=\epsilon_{2}(\epsilon_{1})^{-1}\in A^{\times}$, 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, $\mathbb{P}^{n}_{A}:=\operatorname{Proj}(A[X_{0},X_{1},...,X_{n}])$. Then the following assertions are equivalent.

The ring $A$ is a good ring, 2. 2.

the ring $A$ has enough homogeneous polynomials, 3. 3.

*for any finite family $\{p_{1},p_{2},...,p_{s}\}$ of primitive points in $A^{n+1}$ there is an homogeneous polynomial $P(X_{0},X_{1},...,X_{n})\in A[X_{0},X_{1},...,X_{n}]$ with $\deg P\geq 1$ such that *

$V_{+}(P)\cap\sigma_{p_{i}}(\operatorname{Spec}A)=\emptyset$* for $1\leq i\leq s$ and $\sigma_{p_{i}}$ is the section associated to $p_{i}$.*

3 Primitive points in $A^{2}$ and Picard group

Proposition 3.1**.**

Let $A$ be a ring and $(a,b)\in A^{2}$ a primitive point. Let $A[x,y]:=\frac{A[X,Y]}{X(aY-bX)A[X,Y]}$ where $x$ (resp. $y$) is the image of $X$ (resp. $Y$). Moreover $A[x,y]$ is endowed with the induced grading of $A[X,Y]$. Let $S(a,b):=\operatorname{Proj}(A[x,y])$.

The $A$-algebra ${\cal O}_{S(a,b)}(S(a,b))$ is a free $A$-module of rank two.

More concretely, there is $\theta\in{\cal O}_{S(a,b)}(S(a,b))$ with $\theta_{|D_{+}(x)}=0$, $\theta_{|D_{+}(y)}=\frac{ay-bx}{y}$. One has $\theta^{2}=a\theta$, ${\cal O}_{S(a,b)}(S(a,b))\simeq\frac{A[T]}{T(a-T)A[T]}$ and $(1,\theta)$ is a basis for the $A$-module ${\cal O}_{S(a,b)}(S(a,b))$.

Moreover the scheme $S(a,b)$ is affine and isomorphic to $\operatorname{Spec}({\cal O}_{S(a,b)}(S(a,b)))$. 2. 2.

Let $d\in\mathbb{N}^{>0}$. When considering the ${\cal O}_{S(a,b)}(S(a,b))$-module ${\cal O}_{S(a,b)}(d)(S(a,b))$ as an $A$ module, we have

[TABLE] 3. 3.

Let $a^{\prime},b^{\prime}\in A$ with $aa^{\prime}+bb^{\prime}=1$. There is an epimorphism $u:A[x,y]\to\frac{A[x,y]}{(ay-bx)A[x,y]}\simeq A[a^{\prime}X+b^{\prime}Y]$ which is defined by $u(x)=a(a^{\prime}X+b^{\prime}Y)$ and $u(y)=b(a^{\prime}X+b^{\prime}Y)$.

*Let $\Delta(a,b):=\operatorname{Proj}(A[a^{\prime}X+b^{\prime}Y])$, $d\geq 1$. Then $u$ induces an epimorphism *

$u^{\prime}:{\cal O}_{S(a,b)}(d)(S(a,b))\to{\cal O}_{\Delta(a,b)}(d)(\Delta(a,b))$* such that $u^{\prime}(P(x,y))=P(a,b)(a^{\prime}X+b^{\prime}Y)^{d}$ where $P(X,Y)\in A[X,Y]$ is homogeneous of degree $d$ and $\{(a^{\prime}X+b^{\prime}Y)^{d}\}$ is a basis for the $A$-module ${\cal O}_{\Delta(a,b)}(d)(\Delta(a,b))$.*

*Analogously, let $v:A[x,y]\to A[Y]$ with $v(x)=0$ and $v(y)=Y$ and if $\Delta(0,1):=\operatorname{Proj}(A[Y])$ and $d\geq 1$ one has an epimorphism $v^{\prime}:{\cal O}_{S(a,b)}(d)(S(a,b))\to{\cal O}_{\Delta(0,1)}(d)(\Delta(0,1))$ with *

$v^{\prime}(P(x,y))=P(0,1)Y^{d}$* where $P(X,Y)\in A[X,Y]$ is homogeneous of degree $d$ and $\{Y^{d}\}$ is a basis for the $A$-module ${\cal O}_{\Delta(0,1)}(d)(\Delta(0,1))$.*

Proof.

Let $\mathbb{P}^{1}=\mathbb{P}^{1}_{A}:=\operatorname{Proj}(A[X,Y])$, and ${\cal J}$ be the coherent sheaf of ideals on $\mathbb{P}^{1}$ defined by the homogeneous polynomial $X(aY-bX)$.

Easily we see that ${\cal J}$ is isomorphic to the twisted sheaf ${\cal O}_{\mathbb{P}^{1}}(-2)$.

Now, we have the exact sequence of sheaves on $\mathbb{P}^{1}$,

[TABLE]

and so

[TABLE]

Then (2) gives the long exact cohomological sequence

[TABLE]

As ${\cal J}\simeq{\cal O}_{\mathbb{P}^{1}}(-2)$, it follows from [L], Lemma 3.1 p. 195, that

$H^{0}(\mathbb{P}^{1},{\cal J})\simeq H^{0}(\mathbb{P}^{1},{\cal O}_{\mathbb{P}^{1}}(-2))=\{0\},\ H^{0}(\mathbb{P}^{1},{\cal O}_{\mathbb{P}^{1}})=A,$ and $H^{0}(\mathbb{P}^{1},{\cal J})\simeq H^{0}(\mathbb{P}^{1},{\cal O}_{\mathbb{P}^{1}}(-2))^{\vee}\simeq A$.

So we get the exact sequence of $A$-modules,

[TABLE]

In particular it follows that $H^{0}(\mathbb{P}^{1},{\cal O}_{S(a,b)})={\cal O}_{S(a,b)}(S(a,b))$ is a free $A$-module of rank two.

Moreover looking more precisely at (4) on the cover $\{D_{+}(X),\ D_{+}(Y)\}$ of $\mathbb{P}^{1}$, one shows that $\partial(\theta)$ is the generator $\frac{X(aY-bX)}{XY}$ of the $A$-module $H^{1}(\mathbb{P}^{1},{\cal J})$. This shows that $(1,\theta)$ is a basis for the $A$-module ${\cal O}_{S(a,b)}(S(a,b))$.

We show that $S(a,b)$ is finite over $\operatorname{Spec}A$ and so $S(a,b)\simeq\operatorname{Spec}{\cal O}_{S(a,b)}(S(a,b)).$ It suffices to remark that the canonical morphism $S(a,b)\to\operatorname{Spec}A$ is quasi-finite, more precisely for all prime ideal $\mathfrak{p}\subset A$, there is two prime homogeneous ideal $\mathfrak{P}_{1},\mathfrak{P}_{2}\subset A[x,y]$ with $\mathfrak{P}_{i}\cap A=\mathfrak{p}$ and $xA[x,y]+yA[x,y]\nsubseteq\mathfrak{P}_{i}$ for $i=1,2$. Namely we use [L], ex. 4.2. p. 155 in order to show that $S(a,b)\to\operatorname{Spec}A$ is finite. Moreover we know by part 1. that ${\cal O}_{S(a,b)}(S(a,b))$ is a finite $A$-module and so $S(a,b)\simeq\operatorname{Spec}{\cal O}_{S(a,b)}(S(a,b)).$

Note that we could as well remark that if $\pi:S(a,b)\to\operatorname{Spec}A$ is the structural morphism then $\pi^{-1}(D(a))\to D(a)$ (resp. $\pi^{-1}(D(b))\to D(b)$) is finite and as $\operatorname{Spec}A=D(a)\cup D(b)$ the result follows. 2) As ${\cal J}\simeq{\cal O}_{\mathbb{P}^{1}}(-2)$,  (2) gives

[TABLE]

Let $d\geq 1$, then after tensorizing by ${\cal O}_{\mathbb{P}^{1}}(d)$ we get

[TABLE]

The corresponding long cohomological exact sequence gives in particular the following small exact sequence

[TABLE]

where $w(P(X,Y))=P(x,y)$.

As $d\geq 1$, it follows from [L], Lemma 3.1 p. 195 that $H^{1}(\mathbb{P}^{1},{\cal O}_{\mathbb{P}^{1}}(d-2))=\{0\}$, which shows that the homomorphism $w:H^{0}(\mathbb{P}^{1},{\cal O}_{\mathbb{P}^{1}}(d))\to H^{0}(\mathbb{P}^{1},{\cal O}_{S(a,b)}(d))$ is an epimorphism; it follows that

[TABLE]

Let $X^{\prime}:=aY-bX$ and $Y^{\prime}:=a^{\prime}X+b^{\prime}Y$. As $aa^{\prime}+bb^{\prime}=1$, it follows that $\varphi:A[X,Y]\to A[X,Y]$ with $\varphi(P(X,Y))=P(X^{\prime},Y^{\prime})$ is an $A$-automorphism. In particular $X^{\prime},Y^{\prime}$ are $A$-algebraically independant and $A[X,Y]=A[X^{\prime},Y^{\prime}]$. Moreover if $P(X,Y)=Q(X^{\prime},Y^{\prime})\in A[X,Y]=A[X^{\prime},Y^{\prime}]$ then $P(X,Y)$ is homogeneous of degree $d$ if and only if $Q(X^{\prime},Y^{\prime})$ is homogeneous of degree $d$.

Let $u^{\prime\prime}:A[X,Y]=A[X^{\prime},Y^{\prime}]\to A[Y^{\prime}]$ be the $A$-homomorphism defined by $u^{\prime\prime}(X^{\prime})=0$ and

$u^{\prime\prime}(Y^{\prime})=Y^{\prime}$, then $u^{\prime\prime}$ induces an epimorphism still denoted $u^{\prime\prime}:{\cal O}_{\mathbb{P}^{1}}\to{\cal O}_{\Delta(a,b)}.$

Let ${\cal G}=\operatorname{Ker}u^{\prime\prime}$ be the ideal sheaf on $\mathbb{P}^{1}$ associated to $(aY-bX)A[X,Y]$. We have so the exact sequence

[TABLE]

where ${\cal G}(D_{+}(X^{\prime}))=\frac{X^{\prime}}{X^{\prime}}{\cal O}_{\mathbb{P}^{1}}(D_{+}(X^{\prime}))={\cal O}_{\mathbb{P}^{1}}(D_{+}(X^{\prime})$ and ${\cal G}(D_{+}(Y^{\prime})=\frac{X^{\prime}}{Y^{\prime}}{\cal O}_{\mathbb{P}^{1}}(D_{+}(Y^{\prime})).$

We see that ${\cal G}$ is isomorphic to ${\cal O}_{\mathbb{P}^{1}}(-1)$ and tensorizing (6) by ${\cal O}_{\mathbb{P}^{1}}(d)$, we get the exact sequence

[TABLE]

Hence we have the long exact sequence

[TABLE]

and as for $d\geq 1$, $H^{1}(\mathbb{P}^{1},{\cal O}_{\mathbb{P}^{1}}(d-1))=\{0\}$ by [L], Lemma 3.1 p. 195, we get that

$u^{\prime\prime}_{d}:{\cal O}_{\mathbb{P}^{1}}(d)(\mathbb{P}^{1})\to{\cal O}_{\Delta(a,b)}(d)(\Delta(a,b))$ is an epimorphism.

We describe the map $u^{\prime\prime}_{d}$.

We have the cover $\mathbb{P}^{1}=D_{+}(X^{\prime})\cup D_{+}(Y^{\prime})$, $\Delta(a,b)=V_{+}(X^{\prime})$ and so $D_{+}(X^{\prime})\cap\Delta(a,b)=\emptyset$ and $D_{+}(Y^{\prime})=\Delta(a,b)$.

It is sufficient to describe the map $u^{\prime\prime}_{d}:{\cal O}_{\mathbb{P}^{1}}(d)(D_{+}(Y^{\prime}))\to{\cal O}_{\Delta(a,b)}(d)(\Delta(a,b))=AY^{\prime d},$ where $\{Y^{\prime d}\}$ is a basis of ${\cal O}_{\Delta(a,b)}(d)(\Delta(a,b))$.

Let $\frac{Q(X^{\prime},Y^{\prime})}{Y^{\prime m}}\in{\cal O}_{\mathbb{P}^{1}}(D_{+}(Y^{\prime}))$ with $Q(X^{\prime},Y^{\prime})$ homogeneous of degree $m+d$.

By definition of $u^{\prime\prime}$, we have $u^{\prime\prime}_{d}(\frac{Q(X^{\prime},Y^{\prime})}{Y^{\prime m}})=Q(0,1)Y^{\prime d}$ and if $P(X,Y):=Q(X^{\prime},Y^{\prime})$, we have $Q(0,1)=P(a,b)$.

It follows that for $P(X,Y)\in{\cal O}_{\mathbb{P}^{1}}(d)(\mathbb{P}^{1})$ we get $u^{\prime\prime}_{d}(P(X,Y))=P(a,b)Y^{\prime d}=P(a,b)(a^{\prime}X+b^{\prime}Y)^{d}.$

Now we describe the map $u^{\prime}$.

Let $w:{\cal O}_{\mathbb{P}^{1}}(d)(\mathbb{P}^{1}){\rightarrow}{\cal O}_{S(a,b)}(d)(S(a,b))$ defined in (5). As $\operatorname{Ker}w\subset\operatorname{Ker}u^{\prime\prime}_{d}$, there is

$u^{\prime}:{\cal O}_{S(a,b)}(d)(S(a,b)){\rightarrow}{\cal O}_{\Delta(a,b)}(d)(\Delta(a,b))$ such that $u^{\prime\prime}_{d}=u^{\prime}w.$ Moreover, as $w$ is an epimorphism by (5), we deduce that $u^{\prime}(P(x,y))=P(a,b)(a^{\prime}X+b^{\prime}Y)^{d}$ for $P(X,Y)\in A[X,Y]$ homogeneous of degree $d$.

Now let us consider $\Delta(0,1):=\operatorname{Proj}(A[Y])$ and $v^{\prime\prime}:A[X,Y]\to A[Y]$ the $A$-morphism defined by $v^{\prime\prime}(X)=0$ and $v^{\prime\prime}(Y)=Y$. By an analogous method to the preceeding proof, the morphism $v^{\prime\prime}$ for $d\geq 1$ induces an epimorphism $v^{\prime\prime}_{d}:{\cal O}_{\mathbb{P}^{1}}(d)(\mathbb{P}^{1})\to{\cal O}_{\Delta(0,1)}(d)(\Delta(0,1))$ defined by $v^{\prime\prime}_{d}(P(X,Y))=P(0,1)Y^{d}$. It follows that $v^{\prime}:{\cal O}_{S(a,b)}(d)(S(a,b))\to{\cal O}_{\Delta(0,1)}(d)(\Delta(0,1))$ is onto and defined by $v^{\prime}(P(x,y))=P(0,1)Y^{d}$. ∎

Theorem 3.1**.**

Let $A$ be a ring, $(a,b)\in A^{2}$ a primitive point.

Let $A[x,y]:=\frac{A[X,Y]}{X(aY-bX)A[X,Y]}$ where $x$ (resp.$y$) is the image of $X$ (resp. $Y$) by the natural epimorphism. Moreover, $A[x,y]$ is endowed with the induced grading. Let $S(a,b):=\operatorname{Proj}A[x,y]$. The following properties are equivalent.

1. i)

*The ${\cal O}_{S(a,b)}(S(a,b))$-module ${\cal O}_{S(a,b)}(1)(S(a,b))$ is a torsion element in the Picard group of *

${\cal O}_{S(a,b)}(S(a,b)),$** 2. ii)

there exists $P(X,Y)\in A[X,Y]$ homogeneous of degree $d\geq 1$ with $P(0,1),\ P(a,b)\in A^{\times},$ 3. iii)

the point $(a,b)\in A^{2}$ is a good point.

Proof.

In order to simplify the notations we shall write $S$ for $S(a,b)$.

1)We show i) implies ii).

It follows from i) there is $d\geq 1$ such that ${\cal O}_{S}(d)(S)$ is a free rank one ${\cal O}_{S}(S)$-module and by Proposition 3.1 part 2. there is $P(X,Y)\in A[X,Y]$ homogeneous of degree $d$ such that $\{P(x,y)\}$ is a basis.

Then by Proposition 3.1 part 3, we have $u^{\prime}(P(x,y))=P(a,b)(a^{\prime}X+b^{\prime}Y)^{d}$

(resp. $v^{\prime}(P(x,y))=P(0,1)(Y)^{d}$) is a basis for $A(a^{\prime}X+b^{\prime}Y)^{d}$ (resp.for $A(Y)^{d}$), in other words

$P(a,b)\in A^{\times}$ (resp. $P(0,1)\in A^{\times}$).

2)We show ii) implies i).

Let $P(X,Y)\in A[X,Y]$ homogeneous of degree $d\geq 1$ with $P(0,1),\ P(a,b)\in A^{\times}$, we show that in $\operatorname{Proj}(A[X,Y])$ we have $V_{+}(P(X,Y))\cap V_{+}(X(aY-bX))=\emptyset$.

First we consider $V_{+}(P(X,Y))\cap V_{+}(X)$. Let $\mathfrak{P}\subset A[X,Y]$ be an homogeneous ideal with $X\in\mathfrak{P}$ and $XA[X,Y]+YA[X,Y]\nsubseteq\mathfrak{P}.$ We have $P(X,Y)=a_{0}X^{d}+a_{1}X^{d-1}Y+...+a_{d}Y^{d}$ with $d\geq 1$ and $a_{d}=P(0,1)\in A^{\times}$. If $P\in\mathfrak{P}$, then $a_{d}Y^{d}\in\mathfrak{P}$ and so $Y\in\mathfrak{P}$, a contradiction. So $V_{+}(P(X,Y))\cap V_{+}(X)=\emptyset.$

Now we consider $V_{+}(P(X,Y))\cap V_{+}(aY-bX)$. Let $a^{\prime},b^{\prime}\in A$ with $aa^{\prime}+bb^{\prime}=1$, we can write $P(X,Y)=b_{0}(a^{\prime}X+b^{\prime}Y)^{d}+b_{1}(a^{\prime}X+b^{\prime}Y)^{d-1}(aY-bX)+...+b_{d}(aY-bX)^{d}$ with $b_{i}\in A$. As $P(a,b)\in A^{\times}$ it follows that $b_{0}\in A^{\times}$. Let $\mathfrak{P}\subset A[X,Y]$ be an homogeneous ideal with $aY-bX\in\mathfrak{P}$ and $XA[X,Y]+YA[X,Y]\nsubseteq\mathfrak{P}$ and $P(X,Y)\in\mathfrak{P}$, then $a^{\prime}X+b^{\prime}Y\in\mathfrak{P}$, but

$(aY-bX)A[X,Y]+(a^{\prime}X+b^{\prime}Y)A[X,Y]=XA[X,Y]+YA[X,Y]$, a contradiction.

So $V_{+}(P(X,Y))\cap V_{+}(aY-bX)=\emptyset.$

Now we show that ${\cal O}_{S}(d)(S)$ is a free ${\cal O}_{S}(S)$-module of rank one. 2. 2.

We show that $P(X,Y)$ doesn’t divide zero in $A[X,Y]$.

We have $P(X,Y)=a_{0}X^{d}+a_{1}X^{d-1}Y+...+a_{d}Y^{d}$ with $a_{d}=P(0,1)\in A^{\times}$.

Let us assume that $Q(X,Y)=q_{0}X^{m}+q_{1}X^{m-1}Y+...+q_{k}X^{m-k}Y^{k}\in A[X,Y]$ with $q_{k}\neq 0$ and $P(X,Y)Q(X,Y)=0$, then the coefficient of $X^{m-k}Y^{d+k}$ in $P(X,Y)Q(X,Y)$ is nul, so $a_{d}q_{k}=0$, a contradiction. 3. 3.

As $P(X,Y)$ isn’t a zero divisor in $A[X,Y]$, we can consider the Cartier divisor in $\mathbb{P}^{1}_{A}$ defined by $\{(D_{+}(X),\frac{P(X,Y)}{X^{d}}),(D_{+}(Y),\frac{P(X,Y)}{Y^{d}})\}$. Let $\cal L$, the invertible sheaf on $\mathbb{P}^{1}_{A}$ associated to this divisor. Then ${\cal L}(D_{+}(X))$ (resp. ${\cal L}(D_{+}(Y))$) is the ${\cal O}_{\mathbb{P}^{1}}(D_{+}(X))$-free module with basis $\frac{X^{d}}{P(X,Y)}$ (resp. ${\cal O}_{\mathbb{P}^{1}}(D_{+}(Y))$-free module with basis $\frac{Y^{d}}{P(X,Y)}$) and $\frac{Y^{d}}{P(X,Y)}=(\frac{Y}{X})^{d}\frac{X^{d}}{P(X,Y)}$ with

$(\frac{Y}{X})^{d}\in{\cal O}_{\mathbb{P}^{1}}(D_{+}(XY))^{\times}.$

As the invertible sheaf ${\cal O}_{\mathbb{P}^{1}}(d)$ is such that ${\cal O}_{\mathbb{P}^{1}}(D_{+}(X))$ (resp. ${\cal O}_{\mathbb{P}^{1}}(D_{+}(Y))$) is an ${\cal O}_{\mathbb{P}^{1}}(D_{+}(X))$-free module of basis $X^{d}$ (resp. the ${\cal O}_{\mathbb{P}^{1}}(D_{+}(Y))$-free module of basis $Y^{d}$) and as

$(\frac{Y}{X})^{d}\in{\cal O}_{\mathbb{P}^{1}}(D_{+}(XY))^{\times}$ it follows that the two sheaves $\cal L$ and ${\cal O}_{\mathbb{P}^{1}}(d)$ are isomorphic. It follows that ${\cal L}\otimes_{{\cal O}_{\mathbb{P}^{1}}}{\cal O}_{S}\simeq{\cal O}_{S}(d)$. 4. 4.

We show that ${\cal L}\otimes_{{\cal O}_{\mathbb{P}^{1}}}{\cal O}_{S}\simeq{\cal O}_{S}$.

We know that ${\cal L}\otimes_{{\cal O}_{\mathbb{P}^{1}}}{\cal O}_{S}(D_{+}(x))$ (resp. ${\cal L}\otimes_{{\cal O}_{\mathbb{P}^{1}}}{\cal O}_{S}(D_{+}(y))$ is generated by $\frac{x^{d}}{P(x,y)}$ (resp. $\frac{y^{d}}{P(x,y)}$). It is sufficient to prove that $\frac{P(x,y)}{x^{d}}\in{\cal O}_{S}(D_{+}(x))^{\times}$ (resp. $\frac{P(x,y)}{y^{d}}\in{\cal O}_{S}(D_{+}(y))^{\times}$). As

$V_{+}(P(X,Y))\cap V_{+}(X(aY-bX))=\emptyset$, it follows that the set $V_{+}(P(x,y))\subset S$ is empty.

So $V(\frac{P(x,y)}{x^{d}})\cap D_{+}(x)=\emptyset$, $V(\frac{P(x,y)}{x^{d}})\cap D_{+}(y)=\emptyset$. It follows that ${\cal L}\otimes_{{\cal O}_{\mathbb{P}^{1}}}{\cal O}_{S}\simeq{\cal O}_{S}$.

We have shown that ${\cal O}_{S}(d)={\cal O}_{S}(1)^{\otimes d}\simeq{\cal O}_{S}$. Now as $S$ is affine by 3.1, it follows that $({\cal O}_{S}(1)({\cal O}_{S}(S)))^{\otimes d}\simeq{\cal O}_{S}(S)$.

3. We show ii) implies iii). Let $P(X,Y)=a_{0}Y^{d}+a_{1}XY^{d-1}+...+a_{d}X^{d}$, $d\geq 1$ with $P(0,1)=a_{0}\in A^{\times}$,

$P(a,b)=a_{0}b^{d}+a(a_{1}b^{d-1}+a_{2}b^{d-2}a+...+a_{d}a^{d-1})\in A^{\times}.$

Let $\lambda:=(a_{0})^{-1}(a_{1}b^{d-1}+a_{2}b^{d-2}a+...+a_{d}a^{d-1})\in A$, then $b^{d}+\lambda a\in A^{\times}$, i.e. $(a,b)\in A^{2}$ is a good point. 4)We show iii) implies ii).

As $(a,b)\in A^{2}$ is a good point, there is $N\geq 1$ and $\lambda\in A$, with $b^{N}+\lambda a\in A^{\times}$. Let $a^{\prime},b^{\prime}\in A$ with $aa^{\prime}+bb^{\prime}=1$ and $P(X,Y):=Y^{N}-\lambda X(a^{\prime}X+b^{\prime}Y)^{N-1}$.

Then $P(0,1)=1$ and $P(a,b)=b^{N}-\lambda a\in A^{\times}$. ∎

Remark 3.1**.**

If for any primitive point $(a,b)\in A^{2}$, iii) in Theorem 3.1 is satisfied, then $A$ is a good ring and conversely.

*As well if for all primitive point $(a,b)\in A^{2}$, i) in Theorem 3.1 is satisfied, i.e. the

${\cal O}_{S(a,b)}(S(a,b))$-module ${\cal O}_{S(a,b)}(1)(S(a,b))$ is a torsion element in the Picard group of ${\cal O}_{S(a,b)}(S(a,b))$, then $A$ is a good ring and conversely.*

Corollary 3.1**.**

Let $A$ be a pictorsion ring, then $A$ is a good ring.

Proof.

As ${\cal O}_{S(a,b)}(S(a,b))$ is a finite $A$-algebra, it follows that $\operatorname{Pic}({\cal O}_{S(a,b)}(S(a,b)))$ is a torsion group and so part i) in Theorem 3.1 is satisfied. Therefore pictorsion rings are good rings. ∎

Note that the condition to be of pictorsion for $A$ is strong : it would suffice that $\operatorname{Pic}({\cal O}_{S(a,b)}(S(a,b)))$ be a torsion group. This will be study in paragraph 4.3.

4 Examples of good rings and of not good rings.

4.1 Examples of good rings

In this section we first give a list of good rings (Definition 1.2) and most of proofs are postponed to the end of the section.

4.1.1 Properties of the family of good rings

Stability with inductive limits

A ring $A$ which is the inductive limit of good rings $(A_{i})_{i\in I}$ is a good ring. 2. 2.

Stability with finite products

Let $A=A_{1}\times A_{2}\times...\times A_{r}$ with $A_{i}$ a good ring for $1\leq i\leq r$ then $A$ is a good ring. Namely let $a=(a_{1},...,a_{r}),b:=(b_{1},...,b_{r}),u=(u_{1},...,u_{r}),v=(v_{1},...,v_{r})\in A$ with $au+bv=1$. Then $a_{i}u_{i}+b_{i}v_{i}=1$ for $1\leq i\leq r$ and there is $N_{i}\geq 1$ and $\lambda_{i}\in A_{i}$ with $b_{i}^{N_{i}}+\lambda_{i}a_{i}=\epsilon_{i}\in A_{i}^{\times}$, then for $N:=\prod_{1\leq i\leq r}N_{i}$ there is $\mu_{i}\in A_{i}$ and $\epsilon^{\prime}_{i}\in A_{i}^{\times}$ with $b_{i}^{N}+\mu_{i}a_{i}=\epsilon^{\prime}_{i}$, and so $b^{N}+\mu a=\epsilon^{\prime}$ where $\mu=(\mu_{1},...,\mu_{r})$ and $\epsilon^{\prime}=(\epsilon^{\prime}_{1},...,\epsilon^{\prime}_{r})$. 3. 3.

Stability by quotients

Let $A$ be a good ring and $I\subset A$ an ideal then $\frac{A}{I}$ is a good ring. Namely let $\rho:A\to\frac{A}{I}$ be the natural epimorphism and $(\rho(a),\rho(b))\in(\frac{A}{I})^{2}$, a primitive point, then there is $(a^{\prime},b^{\prime})\in A^{2}$ such that $aa^{\prime}+bb^{\prime}-1=c\in I$ and so $((aa^{\prime}-c),b)\in A^{2}$ is a primitive point. Now as $A$ is a good ring then $((aa^{\prime}-c),b)$ is a good point (Definition 1.2) and so there is $N\geq 1$ and $\lambda\in A$ with $b^{N}+\lambda(aa^{\prime}-c)=\epsilon\in A^{\times}$. It follows that $(\rho(b))^{N}+\rho(\lambda a^{\prime})\in\rho(A^{\times})\subset(\frac{A}{I})^{\times}$ and so $\frac{A}{I}$ is a good ring.

The reverse is false in general. For example, let $A:=\mathbb{Q}[T]$ and $I=T\mathbb{Q}[T]$, then $A$ isn’t a good ring (Proposition 4.8) and $\frac{A}{I}=\mathbb{Q}$ is a good ring. 4. 4.

Good rings and the Jacobson radical

1. (a)

Let $A$ be a ring and $\mathfrak{R}$ its Jacobson radical i.e. the intersection of the maximal ideals. Let $\mathfrak{A}$ be an ideal with $\mathfrak{A}\subset\mathfrak{R}$. Then $A$ is a good ring iff $\frac{A}{\mathfrak{A}}$ is a good ring, (Proposition 4.1). 2. (b)

Let $A$ be a ring and $\mathfrak{R}$ its Jacobson radical. If for all $x\in A-\mathfrak{R}$, $(\frac{A}{xA})^{\times}$ is a torsion group, then $A$ is a good ring, (Proposition 4.2).

In conclusion, on one side, in order to show that a ring is good ring, after a quotient we can assume that the Jacobson radical is trivial. On the other side, from a given good ring $B$, one can built new good rings such as $A:=\frac{B[X]}{X^{n}B[X]}$. Namely let $\mathfrak{R}$ be the Jacobson radical of $A$ and $x$ the image of $X$ in $A$. Then as $xA\subset\mathfrak{R}$ and $\frac{A}{xA}\simeq B$ is a good ring, it follows that $A$ is a good ring.

4.1.2 Classical examples of good rings

If $A$ is a field it follows from (Definition 1.2) that $A$ is a good ring. More generally let $A$ be a semi-local ring (i.e. $A$ has a finite number of maximal ideals) then $A$ is a good ring (Proposition 4.4). 2. 2.

A PIR (principal ideal ring) such that $\frac{A}{aA}$ is finite for all $a$ not a zero divisor, is a good ring (Proposition 4.5).

This generalizes the particular case of PID (principal ideal domain) which is considered in ([Be.E], Theorem 0.3). Namely, they show that if $A$ is a PID such that $\frac{A}{\mathfrak{M}}$ is finite for all maximal, then $A$ has enough polynomial and so $A$ is a good ring by Theorem 2.1. 3. 3.

In particular if $A$ is a Dedekind domain such that the residue fields $\frac{A}{\mathfrak{M}}$ are finite for the maximal ideals $\mathfrak{M}$, then for all $x\in A-\{0\}$, $\frac{A}{xA}$ is finite and so $A$ is a good ring (Proposition 4.2).

Remark 4.1**.**

We note the following

Let $A$ be a good ring and $S$ be a multiplicative subset. A natural question is if $S^{-1}A$ is a good ring.

The answer is yes for the classical examples listed above, but we give a counterexample (Proposition 4.7). 2. 2.

Let $A$ be a Dedekind domain with zero Jacobson radical. Is $A$ a good ring iff for all maximal ideal ${\mathfrak{M}}$ the residue field $\frac{A}{\mathfrak{M}}$ is finite?

Proposition 4.3 with $d=1$ gives a negative answer.

4.1.3 New examples

Good rings with Krull dimension $d\geq 1$ and infinite residue fields

Let $p\in\mathbb{Z}$, be a prime, $F:=\{\{0\}\cup\{p^{k}\ ,k\geq 1\}$, $d\geq 1$ and $Z:=F^{d}$.

Then $S:=\{P\in\mathbb{Q}[X_{1},X_{2},...,X_{d}]\ |\ \forall z\in Z,\ P(z)\neq 0\}$ is a multiplicative set in $\mathbb{Q}[X_{1},X_{2},...,X_{d}]$ and $A:=S^{-1}\mathbb{Q}[X_{1},X_{2},...,X_{d}]$ is a good ring. Moreover $A$ is noetherian, factorial, with zero Jacobson radical, Krull dimension $d$ and the residue fields $\frac{A}{\mathfrak{M}}$ with $\mathfrak{M}$ a maximal ideal, are infinite.

Note that in particular when $d=1$, the good ring $A$ is a PID with zero Jacobson radical and the residue fields $\frac{A}{\mathfrak{M}}$ with $\mathfrak{M}$ a maximal ideal, are infinite.

For the proof, see Proposition 4.3 2. 2.

Good rings and integral extensions

1. (a)

Let $A$ be a ring and $\rho:\mathbb{Z}\to A$ be the natural homomorphism. If $A$ is integral over $\rho(\mathbb{Z})$, then $A$ is a good ring (Proposition 4.6). 2. (b)

Let $A$ be a ring and $L$ be a sub-field of an algebraic closure of a finite field and $\rho:L[T]\to A$ be an homomorphism with $A$ integral over $\rho(L[T])$, then $A$ is a good ring (Proposition 4.6). 3. 3.

Good rings inside infinite products of rings

1. (a)

Let $(K_{i})_{i\in I}$, a family of fields indexed by the set $I$, then $\prod_{i\in I}K_{i}$ is a good ring. This follows from Proposition 2.1. 2. (b)

Let $A$ be the subring of $\mathbb{Z}^{\mathbb{N}}$ of stationary sequences. Let $x\in A$ be the sequence $(x_{k})_{k\geq 0}$ then $\frac{A}{xA}\simeq\prod_{k\geq 0}\frac{\mathbb{Z}}{x_{k}\mathbb{Z}}$. Then $(\frac{A}{xA})^{\times}$ is a torsion group and $A$ is a good ring (Remark 2.1 and Proposition 2.1). 4. 4.

Let $X$ be a compact topological space, $A:={\cal C}(X,\mathbb{R})$ be the ring of continuous functions on $X$ with real values, then $A$ is a good ring and more precisely if $f,g,u,v\in A$ with $fu+gv=1$, then $f^{2}+g^{2}\in A^{\times}$. 5. 5.

Let $X$ be an algebraic non singular curve over $\mathbb{Q}$; let us assume that $X(\mathbb{Q})\neq\emptyset$. Let $a\in X(\mathbb{Q})$ and $Y:=X(\mathbb{Q})-\{a\}$.

Let $A:=\{f\in K(X)\ |\ v_{y}(f)\geq 0,\ \forall y\in Y\}$ where $K(X)$ is the field of rational functions on $X$ and $v_{x}$ is the valuation at $x$. Then $A$ is a good ring and more precisely if $u,v,u^{\prime},v^{\prime}\in A$ with $uu^{\prime}+vv^{\prime}=1$; then $\forall y\in Y$, one has $v_{y}(u)=0$ or $v_{y}(v)$=0 and as the residue field $\frac{{\cal O}_{X,y}}{{\mathfrak{M}}_{y}}=\mathbb{Q}$, it follows that $v_{y}(u^{2}+v^{2})=0$ and so $u^{2}+v^{2}\in A^{\times}$.

4.1.4 Proofs

Proposition 4.1**.**

Let $A$ be a ring, $\mathfrak{R}$ its Jacobson radical i.e. the intersection of the maximal ideals and $\mathfrak{A}\subset\mathfrak{R}$ an ideal.

The following properties are equivalent.

1. i)

The ring $A$ is a good ring, 2. ii)

the ring $\frac{A}{\mathfrak{A}}$ is a good ring.

Proof.

As i) implies ii) is always satisfied (see 4.1.1 part 3), we show that ii) implies i).

Let $\rho:A\to\frac{A}{\mathfrak{A}}$ be the natural epimorphism and $a,b,a^{\prime},b^{\prime}\in A$ with $aa^{\prime}+bb^{\prime}=1$.

Then $\rho(a)\rho(a^{\prime})+\rho(b)\rho(b^{\prime})=1$. As $\frac{A}{\mathfrak{A}}$ is a good ring, there is $N\geq 1$, $\mu\in A$ with

$\rho(b)^{N}+\rho(\mu)\rho(a)=\epsilon\in(\frac{A}{\mathfrak{A}})^{\times}$. Let $e\in A$ with $\rho(e)=\epsilon$, then $b^{N}+\mu a=e+\alpha$ with $\alpha\in\mathfrak{A}$. If $e+\alpha\notin A^{\times}$ there is a maximal ideal $\mathfrak{M}\subset A$ with $e+\alpha\in\mathfrak{M}$. As $\alpha\in\mathfrak{R}$ then $e\in\mathfrak{M}$. As $\operatorname{Ker}\rho\subset\mathfrak{R}\subset\mathfrak{M}$, it follows that $\rho(\mathfrak{M})$ is a maximal ideal of $\frac{A}{\mathfrak{A}}$, but $\epsilon=\rho(e)\in\rho(\mathfrak{M})$ which is in contradiction with $\epsilon\in(\frac{A}{\mathfrak{A}})^{\times}$. Finally $A$ is a good ring. ∎

Proposition 4.2**.**

Let $A$ be a ring, $\mathfrak{R}$ its Jacobson radical. If for all $x\in A-\mathfrak{R}$, $(\frac{A}{xA})^{\times}$ is a torsion group, then $A$ is a good ring.

In particular if $A$ is a Dedekind domain such that the residue fields $\frac{A}{\mathfrak{M}}$ are finite for the maximal ideals $\mathfrak{M}$, then for all $x\in A-\{0\}$, $\frac{A}{xA}$ is finite and so $A$ is a good ring.

Proof.

Let $a,b,u,v\in A$ with $au+bv=1$.

If $a\in A-\mathfrak{R}$, then $(\frac{A}{aA})^{\times}$ is a torsion group by hypothesis and so there is $N\geq 1$ and $\lambda\in A$ with $b^{N}+\lambda a=1$, i.e. $(a,b)$ is a good point (Definition 1.2).

If $a\in\mathfrak{R}$, then $1-av\in A^{\times}$ and so $b\in A^{\times}$ and $b^{1}+0a=b\in A^{\times}$; again $(a,b)$ is a good point.

It follows that $A$ is a good ring (Definition 1.2).

Let us assume now that $A$ is a Dedekind domain such that the residue fields $\frac{A}{\mathfrak{M}}$ are finite for the maximal ideals $\mathfrak{M}$.

If $x\in A^{\times}$, then $\frac{A}{xA}=\{0\}$ and $(\frac{A}{xA})^{\times}$ is a torsion group.

If $x\notin A^{\times}$ and $x\neq 0$, then $xA=\mathfrak{M}_{1}^{\alpha_{1}}\mathfrak{M}_{2}^{\alpha_{2}}...\mathfrak{M}_{r}^{\alpha_{r}}$ where $r\geq 1$, $\mathfrak{M}_{i}$ a maximal ideal and $\alpha_{i}\geq 1$ for $1\leq i\leq r$. Then $\frac{A}{xA}\simeq\frac{A}{\mathfrak{M}_{1}^{\alpha_{1}}}\times\frac{A}{\mathfrak{M}_{2}^{\alpha_{2}}}\times...\times\frac{A}{\mathfrak{M}_{r}^{\alpha_{r}}}$ and as $|\frac{A}{\mathfrak{M}_{i}^{\alpha_{i}}}|=|\frac{A}{\mathfrak{M}_{i}}|^{\alpha_{i}}$ (to see this one can replace $A$ by its localisation at $\mathfrak{M}_{i}$ which is a principal local ring and $\mathfrak{M}_{i}$ by its maximal ideal) the ring $\frac{A}{xA}$ is finite and $(\frac{A}{xA})^{\times}$ is a torsion group and so $A$ is a good ring (Remark 2.1 and Proposition 2.1).

∎

Proposition 4.3**.**

*Let $p\in\mathbb{Z}$, be a prime, $F:=\{\{0\}\cup\{p^{k}\ ,k\geq 1\}$, $d\geq 1$ and $Z:=F^{d}$. Then

$S:=\{P\in\mathbb{Q}[X_{1},X_{2},...,X_{d}]\ |\ \forall z\in Z,\ P(z)\neq 0\}$ is a multiplicative set in $\mathbb{Q}[X_{1},X_{2},...,X_{d}]$ and

$A:=S^{-1}\mathbb{Q}[X_{1},X_{2},...,X_{d}]$ is a good ring. Moreover $A$ is noetherian, factorial, with zero Jacobson radical, Krull dimension $d$ and the residue fields $\frac{A}{\mathfrak{M}}$ with $\mathfrak{M}$ a maximal ideal, are infinite.*

Proof.

We shall write $\mathbb{X}$ for $X_{1},X_{2},...,X_{d}$.

0. Let $a\in A$ and $\rho_{a}:A\to\frac{A}{aA}$ be the natural epimorphism. Let $b\in A$ such that $(a,b)\in A^{2}$ is a primitive point i.e. $\rho_{a}(b)\in\rho_{a}(A)^{\times}$.

We prove that $\rho_{a}(b)\in\rho_{a}(A^{\times})$ i.e. $(a,b)\in A^{2}$ is a good point and so $A$ is a good ring.

1. One can write $a=\frac{P(\mathbb{X})}{T(\mathbb{X})}$, $b=\frac{Q(\mathbb{X})}{T(\mathbb{X})}$ with $P,Q,T\in\mathbb{Z}[\mathbb{X}]$ and $T\in S$. There is $U,V\in\mathbb{Z}[\mathbb{X}]$ and $R\in S$ with

[TABLE]

In particular $\forall z\in Z$, one has $R(z)\neq 0$. 2) There is $M\geq 1$ such that for all $z\in Z$, one has $(P(z),Q(z))\not\equiv 0$ modulo $p^{M}\mathbb{Z}^{2}$.

Let $\mathbb{Z}_{p}$ be the completion of $\mathbb{Z}$ for the $p$-adique absolute $|.|_{p}$. Then $\mathbb{Z}_{p}$ is compact and if $F\subset\mathbb{Z}_{p}$ is closed, it is compact. Moreover $Z=F^{d}$ is compact. The map $z\in Z\to|f(z)|_{p}\in\mathbb{R}$ is continuous and as $R(z)\neq 0$ for $z\in Z$, there is $M\geq 1$ with $R(z)\not\equiv 0\mod p^{M}\mathbb{Z}.$

From relation (8) we deduce that if $P(z)\equiv 0\mod p^{M}\mathbb{Z}$, (resp. $Q(z)\equiv 0\mod p^{M}\mathbb{Z}$ ) then

$Q(z)\not\equiv 0\mod p^{M}\mathbb{Z}$, (resp. $P(z)\not\equiv 0\mod p^{M}\mathbb{Z}$). 3) We show that there is $(s,t)\in\mathbb{Z}^{2}$ such that $\forall z\in Z$, one has $P(z)t-Q(z)s\neq 0$. 3.1) Let $z\in Z$ and $C(P(z),Q(z)):=\{(u,v)\in\mathbb{Z}^{2}\ |\ P(z)v-Q(z)u=0\}$.

Let $N\geq 1$ and $\pi_{N}:\mathbb{Z}\to\frac{\mathbb{Z}}{p^{N}\mathbb{Z}}$ be the natural epimorphism and

$\Theta(z):=\{(x,y)\in(\frac{\mathbb{Z}}{p^{N}\mathbb{Z}})^{2}\ |\ \pi_{N}(P(z))y-\pi_{N}(Q(z))x=0\}$, then

[TABLE]

We remark that if $z^{\prime}\equiv z$ modulo $p^{N}\mathbb{Z}^{d}$, then

[TABLE]

Let $Z_{N}:=\{0\}\cup\{p,p^{2},...,p^{N-1}\}$, it follows from (9) and (10) that

[TABLE]

3.2) We assume that $N\geq M$ where $M$ is defined in 2). Then $\forall z\in Z$, we have

$\operatorname{card}\{(x,y)\in(\frac{\mathbb{Z}}{p^{N}\mathbb{Z}})^{2}\ |\ \pi_{N}(P(z))y-\pi_{N}(Q(z))x=0\}\leq p^{N+M}$, and so there is $(a,b)\in\mathbb{Z}^{2}$, such that $\forall z\in Z$, we have $P(z)b-Q(z)a\neq 0$.

We assume that $\pi_{N}(P(z))y-\pi_{N}(Q(z))x=0$. We can write $\pi_{N}(P(z))=\pi_{N}(p^{\alpha})u$ and

$\pi_{N}(Q(z))=\pi_{N}(p^{\beta})v$ with $u,v\in\frac{\mathbb{Z}}{p^{N}\mathbb{Z}}^{\times}$. Moreover we can assume that $0\leq\alpha\leq\beta\leq N$.

Now we remark that 2) implies

[TABLE]

So we can write

[TABLE]

It follows that for $x\in\frac{\mathbb{Z}}{p^{N}\mathbb{Z}}$ there are $p^{\alpha}$ solutions $y$ to equation (13). This shows

[TABLE]

Now it follows from (11), (12), (13), (14) that

[TABLE]

Now for $N$ big enough, we have $N^{d}p^{N+M}<p^{2N}$ and so there is $(s,t)\in\mathbb{Z}^{2}$ such that

[TABLE]

and so

[TABLE]

i.e. $\forall z\in Z$, we have $P(z)t-Q(z)s\neq 0.$ 4) We prove that $\rho_{a}(b)\in\rho_{a}(A^{\times})$ i.e. $(a,b)\in A^{2}$ is a good point and so $A$ is a good ring.

We have $a=\frac{P(\mathbb{X})}{T(\mathbb{X})}$, $b=\frac{Q(\mathbb{X})}{T(\mathbb{X})}$ with $P,Q,T\in\mathbb{Z}[\mathbb{X}]$ and $T\in S$.

By 3), $\forall z\in Z$, one has $P(z)t-Q(z)s\neq 0$ and so $a(z)t-a(z)s\neq 0$ i.e. $a(\mathbb{X})t-b(\mathbb{X})s\in S\subset A^{\times}$.

If $s\neq 0$ then $b-\frac{t}{s}a\in A^{\times}$ and so $\rho_{a}(b)\in\rho_{a}(A^{\times})$.

If $s=0$, then $a\in S$ and $\rho_{a}(A)=\{0\}$ and again $\rho_{a}(b)\in\rho_{a}(A^{\times})$. 5) As $\mathbb{Q}[\mathbb{X}]$ is integral, noetherian and factorial , this is the same for $A=S^{-1}\mathbb{Q}[\mathbb{X}]$. 6) We show that $\dim A=d$.

As $\dim\mathbb{Q}[\mathbb{X}]=d$, it follows that $\dim A\leq d$.

We show that $\{0\}\subset X_{1}A\subset X_{1}A+X_{2}A\subset...\subset X_{1}A+X_{2}A+...+X_{d}A$ is a chain of prime ideals of length $d$.

Let $k\geq 1$ and $r_{k}:\mathbb{Q}[\mathbb{X}]\to\mathbb{Q}[X_{k+1},X_{k+2},...,X_{d}]$ the homomorphism with

$r_{k}(P(\mathbb{X})):=P(0,...,0,X_{k+1},X_{k+2},...,X_{d})$. If $T\in S$, $r_{k}(T)=r_{k}(0,...,0,X_{k+1},X_{k+2},...,X_{d})$.

If $z_{k+1},z_{k+2},...,z_{d}\in F$, then $(0,...,0,z_{k+1},z_{k+2},...,z_{d})\in Z$, so $r_{k}(0,...,0,z_{k+1},z_{k+2},...,z_{d})\neq 0$, and $r_{k}(T)\neq 0$.

It follows that $r_{k}$ extends to an homomorphism

$r^{\prime}_{k}:S^{-1}\mathbb{Q}[\mathbb{X}]\to\mathbb{Q}(X_{k+1},X_{k+2},...,X_{d})$, so $\operatorname{Ker}r^{\prime}_{k}=X_{1}A+X_{2}A+...+X_{k}A$ is a prime ideal.

As $r_{k}(X_{k+1})=X_{k+1}\neq 0$, we get that $\operatorname{Ker}r^{\prime}_{k}\neq\operatorname{Ker}r^{\prime}_{k+1}$. 7) We show that the Jacobson radical of the ring $A$ is zero.

Let $z\in Z$ and $f_{z}:\mathbb{Q}[\mathbb{X}]\to\mathbb{Q}$, defined by $f_{z}(P):=P(z)$. Then $f_{z}$ extends to an homomorphism $f^{\prime}_{z}:S^{-1}\mathbb{Q}[\mathbb{X}]\to\mathbb{Q}$ and ${\mathfrak{M}}_{z}:=\operatorname{Ker}f^{\prime}_{z}\subset A$ is a maximal ideal. Let $\frac{P(\mathbb{X})}{T(\mathbb{X})}\in A$ with $P(\mathbb{X})\in\mathbb{Q}[\mathbb{X}]$ and $T(\mathbb{X})\in S$, then $\frac{P(\mathbb{X})}{T(\mathbb{X})}\in{\mathfrak{M}}_{z}$ iff $P(z)=0$. As $F$ is infinite it is well known that this implies that $P=0$. 8) Let ${\mathfrak{M}}\subset A$, be a maximal ideal; then $\mathbb{Q}\subset\frac{A}{{\mathfrak{M}}}$ and so is infinite.

∎

Proposition 4.4**.**

Let $A$ be a field or a semi-local ring (i.e. the set of maximal ideals in $A$ is finite), then $A$ is a good ring.

Proof.

If $A$ is a field this is immediate from the definition. Now let us assume that the set of maximal ideals in $A$ is $\{\mathfrak{M}_{1},\mathfrak{M}_{2},...,\mathfrak{M}_{r}\}$ then the diagonal morphism $d:A\to\prod_{1\leq i\leq r}\frac{A}{\mathfrak{M}_{i}}$ induces an isomorphism of rings $\frac{A}{\mathfrak{R}}\simeq\prod_{1\leq i\leq r}\frac{A}{\mathfrak{M}_{i}}$. As $\frac{A}{\mathfrak{M}_{i}}$ is a good ring we deduce from example 2. in 4.1. that its the same for $\frac{A}{\mathfrak{R}}$ and so for $A$ (Proposition 4.2). ∎

Proposition 4.5**.**

Let $A$ be a principal ideal ring (PIR). We assume that for all $x\in A$ which is not a zero divisor, the quotient ring $\frac{A}{xA}$ is finite, then $A$ is a good ring.

Proof.

By ([Z.S] Theorem 33 page 245), we know that a principal ideal ring $A$ is a finite product of rings $A=A_{1}\times A_{2}\times...\times A_{r}$ where $A_{i}$ is a PID or a local ring whose the maximal ideal $\mathfrak{M}_{i}$ is generated by a nilpotent element.

If $A_{i}$ is a PID and $\mathfrak{M}_{i}=z_{i}A_{i}$ a maximal ideal, then $(1,..,1,z_{i},1,...,1)\in A$ is not a zero divisor. As $\frac{A_{i}}{z_{i}A_{i}}\simeq\frac{A}{(1,..,1,z_{i},1,...,1)A}$ is finite, then if $x_{i}\in A_{i}-\{0\}$, $\frac{A_{i}}{x_{i}A_{i}}$ is a finite ring and so $\frac{A_{i}}{x_{i}A_{i}}$ is finite for any $x_{i}\in A_{i}-\{0\}$. Now with Remark 2.1 and Proposition 2.1, $A_{i}$ is a good ring.

If $A_{i}$ is a local ring, then $A_{i}$ is a good ring (cf. example 5 in 4.1 ). Finally $A$ is a finite product of good rings and so $A$ is a good ring (cf. example 2 in 4.1 ).

∎

Proposition 4.6**.**

1. Let $A$ be a ring.
2. A.

Let $\rho:\mathbb{Z}\to A$ be the natural homomorphism. We assume that any element in $A$ is integral over $\rho(\mathbb{Z})$, then $A$ is a good ring. 3. B.

Let $L$ be a subfield of an algebraic closure of a finite field, $L[T]$ the polynomial ring in the variable $T$ over $L$. We assume there is $\rho:L[T]\to A$ an homomorphism with $A$ integral over $\rho(L[T])$, i.e. every element in $A$ is zero of a unitary polynomial with coefficient in $L[T]$, then $A$ is a good ring.

Proof.

As $\mathbb{Z}$ (resp. $L[T]$) is a pictorsion ring ([MB], Theorem 2.3. p. 165), it follows by Corollary 3.1 that $\mathbb{Z}$ (resp. $L[T]$) is a good ring.

Note that there is a direct algebraic proof of this result, due to its technical nature we refer to

[F.M], Proposition 4.6. ∎

4.2 Examples of rings $A$ which aren’t good rings.

4.2.1 Good rings and localization

In general a localisation of a good ring isn’t a good ring as shown with the following example.

Proposition 4.7**.**

Let $k$ be a field and $k[X,Y,Z]$, the polynomial ring. Let ${\mathfrak{M}}=(X,Y,Z)\subset k[X,Y,Z]$ be a maximal ideal and $A:=k[X,Y,Z]_{\mathfrak{M}}$, then $A$ is a good ring.

Let $B:=A[\frac{1}{X+YZ}]$ be the localized ring $S^{-1}A$, with $S:=\{(X+YZ)^{n}\ |\ n\in\mathbb{N}\}$, then $B$ isn’t a good ring.

Proof.

As $A$ is a local ring it is a good ring (Proposition 4.4).

Note that $A:=\{\frac{P}{R}\ |\ P\in k[X,Y,Z],\ R\in k[X,Y,Z],\ R(0,0,0)\neq 0\}$ and

$B=\{\frac{P}{(X+YZ)^{n}R}\ |\ P,R\in k[X,Y,Z],\ n\geq 0,\ R(0,0,0)\neq 0\}$.

As $X+YZ$ is irreducible in the factorial ring $k[X,Y,Z]$, we have

[TABLE]

Let $a:=X\in B$, then $\frac{B}{aB}\simeq\{\frac{P}{(YZ)^{n}R}\ |\ P,R\in k[Y,Z],\ n\geq 0,\ R(0,0)\neq 0.\}$ It follows that

[TABLE]

Let $\rho_{a}:B\to\frac{B}{aB}$, the natural epimorphism. It follows from (18) that

[TABLE]

Now with (19) and (20) we have $\frac{\rho_{a}(B)^{\times}}{\rho_{a}(B^{\times})}\simeq\frac{Y^{\mathbb{Z}}\times Z^{\mathbb{Z}}}{(YZ)^{\mathbb{Z}}}\simeq\mathbb{Z}.$

Then $\frac{\rho_{a}(B)^{\times}}{\rho_{a}(B^{\times})}$ is not torsion and so $B$ isn’t a good ring. ∎

4.2.2 Good rings and transfert to polynomial rings

A. The case of polynomial rings $k[T]$ with coefficients in a field is answered in the following proposition.

Proposition 4.8**.**

Let $k$ be a field, $A:=k[T]$ the polynomial ring of the variable $T$ and coefficients in $k$.

Then the following properties are equivalent.

1. i)

The ring $A$ is a good ring, 2. ii)

the field $k$ is algebraic over a finite field, 3. iii)

the group $k^{\times}$ is a torsion group.

Proof.

1. We show that ii) implies iii).

We remark that $\mathbb{F}_{p}^{alg}=\cup_{n\geq 1}\mathbb{F}_{p^{n}}$, it follows that $(\mathbb{F}_{p}^{alg})^{\times}$ and so $k^{\times}$ is a torsion group. 2) We show that iii) implies ii).

Necessarily $k$ has characteristic $\neq 0$, otherwise $\mathbb{Q}\subset k$, this contradict the fact that $k^{\times}$ is a torsion group. One can assume that $\mathbb{F}_{p}\subset k$. Let $x\in k^{\times}$, there is $n_{x}\geq 1$ with $x^{n_{x}}=1$ and so $k$ is algebraic over $\mathbb{F}_{p}$. 3) We show that i) implies iii).

One can assume that $k\neq\mathbb{F}_{2}$. Let $\theta\in k-\{0,1\}$, then $1=\operatorname{gcd}(T-\theta,T(T-1))$ so there is $u,v\in k[T]$ with $u(T-\theta)+vT(T-1)=1$. Let $b=T-\theta$, $a:=T(T-1)$, as $k[T]$ is a good ring there is $m\geq 1$ which depends on $\theta$ and $\lambda(T)\in k[T]$ with $b^{m}+\lambda(T)a=\epsilon\in k^{\times}=A^{\times}$ which a specialisation of $T$ by [math] and $1$ gives $(-\theta)^{m}=\epsilon,\ (1-\theta)^{m}=\epsilon$ and so $(\frac{1-\theta}{-\theta})^{m}=1$. Now we remark that $\theta\to\frac{1-\theta}{-\theta}$ is a permutation of $k-\{0,1\}$, it follows that $k^{\times}$ is a torsion group. 4) We show that ii) implies i).

One can assume that $k\subset\mathbb{F}_{p}^{alg}$. Let $z\in k[T]$, we show that $(\frac{k[T]}{zk[T]})^{\times}$ is a torsion group.

If $z=0$ as $k[T]^{\times}=k^{\times}$ then $(\frac{k[T]}{zk[T]})^{\times}=k^{\times}$ is a torsion group.

We now assume that $\deg z=d\geq 0$. If $d=0$ then $(\frac{k[T]}{zk[T]})^{\times}=\{1\}$.

If $d\geq 1$, then $\frac{k[T]}{zk[T]}$ is a $k$-vector space $V$ of dimension $d\geq 1$. Let $f\in(\frac{k[T]}{zk[T]})^{\times}$, then the map

$\mu_{f}:x\in V\to fx\in V$ belongs to $\operatorname{Gl}(V)$ and the map $f\in(\frac{k[T]}{zk[T]})^{\times}\to\mu_{f}\in\operatorname{Gl}(V)$ is an into homomorphism. After fixing a basis of the $k$-vector space $V$, the element $\mu_{f}$ and so $f$ can be identified with an element of $\operatorname{Gl}_{d}(k)$ and so for $n$ big enough with an element of $\operatorname{Gl}_{d}(\mathbb{F}_{p^{n}})$. It follows that $f$ is a torsion element.

∎

Remark 4.2**.**

It follows from Proposition 4.8 that if $k$ is a field of characteristic [math], then $k[T]$ isn’t a good ring.

B. The case of polynomial rings $A[T]$ with coefficients in an integral domain which is not a field is answered in the following proposition.

Proposition 4.9**.**

Let $A$ be an integral domain which is not a field, $B:=A[T]$ the polynomial ring of the variable $T$ and coefficients in $A$. Then $B$ isn’t a good ring.

Proof.

As $A$ isn’t a field, there is $\pi\neq 0$ and $\pi\notin A^{\times}$. Let $a:=1-\pi T$, and $S:=\{\pi^{k}\ |\ k\in\mathbb{N}\}$, then $\operatorname{card}S$ is infinite.

Let $\rho_{a}:A[T]\to\frac{A[T]}{aA[T]}$, the natural epimorphism. Then $\rho_{a}$ induces an isomorphism $f:S^{-1}A\to\frac{A[T]}{aA[T]}$, which is defined by $f(\frac{\lambda}{\pi^{n}})=\rho_{a}(\lambda T^{n})$.

We have $S^{-1}A\supseteq\pi^{\mathbb{Z}}A^{\times}=\pi^{\mathbb{Z}}\times A^{\times}$ and so $\rho_{a}(A[T])^{\times}\supseteq\rho_{a}(\pi)^{\mathbb{Z}}\times f(A^{\times})=\rho_{a}(T)^{\mathbb{Z}}\times f(A^{\times})$.

On the other hand, $A[T]^{\times}=A^{\times}$ and $\rho_{a}(A[T]^{\times})=f(A^{\times})$. It follows that $\rho_{a}(T)^{\mathbb{Z}}\subset\frac{\rho_{a}(A[T])^{\times}}{\rho_{a}(A[T]^{\times})}.$

Now as $\pi^{\mathbb{Z}}=\rho_{a}(T)^{\mathbb{Z}}\simeq\mathbb{Z}$, we see that $B=A[T]$ isn’t a good ring.

∎

C. The case of polynomial rings $A[T]$ with coefficients in a ring with zero Jacobson radical is answered in the following proposition.

Proposition 4.10**.**

Let $A$ be a ring with Jacobson radical nul. Then

If $\dim A\geq 1$, then $A[T]$ isn’t a good ring. 2. 2.

If $\dim A=0$ and if there is a maximal ideal ${\mathfrak{M}}\subset A$ such that $\frac{A}{\mathfrak{M}}$ isn’t contained in the algebraic closure of a finite field, then $A[T]$ isn’t a good ring.

Proof.

1. There is a prime ideal $\mathfrak{p}\subset A$ which isn’t maximal, i.e. $\frac{A}{\mathfrak{p}}$ is an integral domain and not a field. Then by 4.9 we know that $\frac{A}{\mathfrak{p}}[T]$ isn’t a good ring and then from stability by quotients (4.1.1. part 3.), the ring $A$ isn’t a good ring.

If $\frac{A}{\mathfrak{M}}$ isn’t contained in the algebraic closure of a finite field, by proposition 4.8 that $\frac{A}{\mathfrak{M}}[T]$ isn’t a good ring and from stability by quotients (4.1.1. part 3.), the ring $A[T]$ isn’t a good ring. ∎

Remark 4.3**.**

The preceding proposition doesn’t give an answer when $\dim A=0$ and for any maximal ideal ${\mathfrak{M}}$ of $A$, the residue field $\frac{A}{\mathfrak{M}}$ is contained in the algebraic closure of a finite field. In fact one can give many examples of such a ring $A$ with $A[T]$ a good ring.

Namely, let $A:=K_{1}\times K_{2}\times...\times K_{s}$, where $K_{i}$ is a field which is algebraic over a finite field for $1\leq i\leq s$. Then $A[T]\simeq\ K_{1}[T]\times K_{2}[T]\times...\times K_{s}[T]$ and it follows from Proposition 4.8 that $K_{i}[T]$ is a good ring and by 4.1.1. part 2. that $A[T]$ is a good ring.

Another example is the ring $A:=\mathbb{F}_{p}^{\mathbb{N}}$.

4.3 Good rings and pictorsion

See Definition1.2 and Definition 1.4

In the sequel we discuss the relation between good rings and torsion in Picard groups. A. Rings which aren’t of pictorsion.

We saw in Corollary 3.1 that a pictorsion ring is a good ring and rings which aren’t good rings aren’t of pictorsion.

Remark 4.4**.**

For example the ring $A=\mathbb{Q}[y]$ isn’t a good ring (Proposition 4.8) and so isn’a pictorsion ring. In this case $\operatorname{Pic}(A)$ is trivial and so there is a finite extension $B$ over $A$ such that the group $\operatorname{Pic}(B)$ is not a torsion group. One can give an explicit example of such an extension $B/A$ (compare with [G.L.L], Example 8.15 p. 1263).

Namely, Let $A:=\mathbb{Q}[y]$ be the ring of polynomial in the variable $y$ with coefficients in $\mathbb{Q}$. Then $\operatorname{Pic}(A)$ is trivial.

Let $B:=\frac{\mathbb{Q}[y,X]}{(X^{3}-X-y^{2}-y)\mathbb{Q}[y,X]}=\mathbb{Q}[y,x]$ where $x$ is the $X$ image. The ring $B$ is finite over $A$. We show that $\operatorname{Pic}(B)$ is not a torsion group and so $A$ is not of pictorsion.

Let $\mathbb{Q}[x,y,z]:=\frac{\mathbb{Q}[X,Y,Z]}{(X^{3}-XZ^{2}-Y^{2}Z-YZ^{2})\mathbb{Q}[X,Y,Z]}$, where $x$ (resp. $y$, $z$) is the $X$ (resp. $Y$,$Z$) image and the grading is induced by that of $\mathbb{Q}[X,Y,Z]$.

Let $S:=\operatorname{Proj}(\mathbb{Q}[x,y,z])$, $\infty:=(0:1:0)$. We have $S-\{\infty\}=D_{+}(z)$ and ${\cal O}_{S}(S-\{\infty\})=B$.

Let $p:=(0:0:1)$, the divisor $(p-\infty)$ has infinite order ([H], Example 4.3.8. p. 335). It follows that the ideal $\mathfrak{M}_{p}=Bx+By$ induces an element of infinite order in $\operatorname{Pic}(B)$.

B. Good rings with a finite Picard group.

i) Let $A$ be the ring of integers of a number fields or the ring of regular functions on a smooth affine curve over a finite field. Such a ring is a good ring (see 4.1.2. example 3) and $\operatorname{Pic}(A)$ is finite. Moreover for the same reason such a ring is of pictorsion.

ii) Let us consider the good rings in examples 5 in paragraph 4.1.3. In contrary to the examples in i), the residue fields aren’t finite and nevertheless one can give examples with a finite Picard group.

Namely, let $X$ be an algebraic non singular curve over $\mathbb{Q}$; let us assume that $X(\mathbb{Q})\neq\emptyset$. Let $a\in X(\mathbb{Q})$ and $Y:=X(\mathbb{Q})-\{a\}$. Let $A:=\{f\in K(X)\ |\ v_{y}(f)\geq 0,\ \forall y\in Y\}$, where $K(X)$ is the field of rational functions on $X$ and $v_{x}$ is the valuation at $x$. Let us assume that $\operatorname{Jac}(X)(\mathbb{Q})$ is finite (see [G.G], Theorem 3, p. 822 for a genus $2$ example and [B.S] for the existence of infinitely many $\mathbb{Q}$-elliptic curves).

Let $U:=X-\{a\}$, then $B:={\cal O}_{X}(U)$ is a Dedekind domain. Moreover we have the following exact sequence

[TABLE]

and as $\operatorname{Pic}(X)=\operatorname{Pic}^{0}(X)\oplus\mathbb{Z}a$ we get that $\operatorname{Pic}(U)\simeq\operatorname{Pic}^{0}(X)$. Now as $\operatorname{Pic}^{0}(X)\subset\operatorname{Jac}(X)(\mathbb{Q})$ it follows that $\operatorname{Pic}(U)$ is finite and then by [G], Corollary 1, p. 114, there is $S\subset B$, a multiplicative set such that $A=S^{-1}B$. In particular $\operatorname{Pic}(A)$ is finite. Note that we don’t know if the ring $A$ is a pictorsion ring. C. Good rings with a non torsion Picard group and so which aren’t of pictorsion.

Let us recall (Corollary 3.1) that a pictorsion ring is a good ring. There is still the question to find a good ring $A$ such that the Picard group of $A$ is not a torsion group. This is what we intend to do now.

Proposition 4.11**.**

Let $A$ be a ring, $a\in A$, $B_{a}:=\frac{A[T]}{T(T-a)A[T]}=A\oplus At$ where $t$ is the $T$ image. Let $\rho_{a}:A\to\frac{A}{aA}$ the natural epimorphism. Then we have the following exact sequence

[TABLE]

In particular, if $\frac{\rho_{a}(A)^{\times}}{\rho_{a}(A^{\times})}$ and $\operatorname{Pic}(A)$ are torsion groups then $\operatorname{Pic}(B_{a})$ is a torsion group.

Proof.

1. Let $I:=\{b\in A\ |\ ba=0\}$ and $C_{a}:=\frac{B_{a}}{It}$. We show that $\operatorname{Pic}(B_{a})\simeq\operatorname{Pic}(C_{a})$.

Let $X:=\operatorname{Spec}A$, $Y:=\operatorname{Spec}B_{a}$, $Z:=\operatorname{Spec}C_{a}$. The canonical morphisms $A\to B_{a}$ and $A\to B_{a}\to C_{a}$ induce continuous maps $f:Y\to X$ and $g:Z\to X$.

We have the exact sequence of sheaves $\{0\}\to{\cal I}t\to{\cal F}\to{\cal G}\to\{0\}$ where $\cal F$ is the sheaf defined over $X$ by the ideal $I$, and for all open $U\subset X$ we have ${\cal F}(U):={\cal O}(f^{-1}(U))$, ${\cal G}(U):={\cal O}(g^{-1}(U))$.

It follows the following exact sequence of sheaves

[TABLE]

where ${\cal F}^{\times}(U):={\cal O}(f^{-1}(U))^{\times}$, ${\cal G}^{\times}(U):={\cal O}(g^{-1}(U))^{\times}$.

Then (21) gives the long exact sequence of sheaves

[TABLE]

As $1+{\cal I}t\simeq{\cal I}$, we have $H^{1}(X,{\cal I})=H^{2}(X,{\cal I})=\{0\}$ ([L], Theorem 2.18, p. 186).

It follows from (22) that $\operatorname{Pic}(B_{a})\simeq\operatorname{Pic}(C_{a})$. 2) We show the exact sequence

[TABLE]

where the map $\operatorname{Pic}(C_{a})\to\operatorname{Pic}(A)^{2}$ is induced by

$u:B_{a}=A\oplus At\to A\times A$ with $u(\alpha+\beta t)=(\alpha,\alpha+\beta t).$

We have $\operatorname{Ker}u=\{\beta t\ |\ \beta a=0\}=It$. So $u$ induces an into homomorphism $v:C_{a}\to A\times A$. So $v$ induces a morphism ${\cal O}_{Z}\to{\cal O}_{X}\times{\cal O}_{X}$ and so a morphism ${\cal O}_{Z}^{\times}\to{\cal O}_{X}^{\times}\times{\cal O}_{X}^{\times}.$

Let now $W:=\operatorname{Spec}(\frac{A}{aA})$; then the homomorphism $(\ell,m)\to\rho_{a}(\ell)\rho_{a}(m^{-1})$ induces a morphism ${\cal O}_{X}^{\times}\times{\cal O}_{X}^{\times}\to{\cal O}_{W}^{\times}.$ We have the exact sequence

[TABLE]

from which we get the long exact sequence

[TABLE]

It induces an into homomorphism $\frac{\rho_{a}(A)^{\times}}{\rho_{a}(A^{\times})}\to\operatorname{Pic}C_{a}.$

It follows from (23) the exact sequence

[TABLE]

3. It follows from 1) and 2) the exact sequence

[TABLE]

If the groups $\frac{\rho_{a}(A)^{\times}}{\rho_{a}(A^{\times})}$ and $\operatorname{Pic}(A)$ are torsion groups then (24) implies that $\operatorname{Pic}B_{a}$ is also a torsion group. ∎

Proposition 4.12**.**

([G], Corollary 2) Let $\mathbb{Z}[T]$ the ring of polynomials of the variable $T$ with coefficients in $\mathbb{Z}$. Then there is a Dedekind domain $A$ with $\mathbb{Z}[T]\subset A\subset\mathbb{Q}[T]$ such that for each maximal ideal $\mathfrak{M}\subset A$, the field $\frac{A}{\mathfrak{M}}$ is finite and the group $A^{\times}$ is of finite type. Moreover the class group of $A$ is not a torsion group, i.e. $\operatorname{Pic}(A)$ is not a torsion group.

Proposition 4.13**.**

Let $A$ be the ring defined in 4.12. Then $A$ is a good ring.

Let $(a,b)\in A^{2}$ a primitive point, then the Picard group of ${\cal O}_{S(a,b)}(S(a,b))$ is not a torsion group and $A$ is not a pictorsion ring.

Proof.

1. As $A$ is a Dedekind domain and that for each maximal ideal $\mathfrak{M}\subset A$, the ring $\frac{A}{\mathfrak{M}}$ is finite, it follows that $\rho_{a}(A)$ is finite for all $a\neq 0$ (Proposition 4.2), and so $\frac{\rho_{a}(A)^{\times}}{\rho_{a}(A^{\times})}$ is finite for all $a\neq 0$ and trivially $\frac{\rho_{a}(A)^{\times}}{\rho_{a}(A^{\times})}=\{1\}$ if $a=0$. In particular $A$ is a good ring.
2. We have ${\cal O}_{S(a,b)}(S(a,b))=B_{a}:=\frac{A[T]}{T(T-a)A[T]}=A\oplus At$ (Proposition 3.1, part 1.).

It follows from Proposition 4.11 the following exact sequence

[TABLE]

2.1) Let us assume that $a\neq 0$. It follows that $\frac{A}{aA}$ is finite and so that $\operatorname{Pic}(\frac{A}{aA})$ is finite. Moreover as $\operatorname{Pic}(A)$ is not a torsion group it follows from (25) that $\operatorname{Pic}(B_{a})$ is not a torsion group, i.e. the Picard group of ${\cal O}_{S(a,b)}(S(a,b))$ is not a torsion group.

2.2) Let us assume that $a=0$. This means that

[TABLE]

We show that $\operatorname{Pic}(A)\simeq\operatorname{Pic}(B_{0})$.

Let $X:=\operatorname{Spec}A$, $Y:=\operatorname{Spec}B_{0}$. From the exact sequence

[TABLE]

we get the following exact sequence of sheaves

${1}\to 1+tA\to B_{0}^{\times}\to A^{\times}\to H^{1}(X,1+t{\cal O}_{X})\to\to\operatorname{Pic}(B_{0})\to\operatorname{Pic}(A)\to H^{2}(X,1+t{\cal O}_{X}).$

As $t{\cal O}_{X}$ is coherent, $H^{1}(X,1+t{\cal O}_{X})\simeq H^{1}(X,t{\cal O}_{X})=\{0\}$ and $H^{2}(X,1+t{\cal O}_{X})\simeq H^{2}(X,t{\cal O}_{X})=\{0\}$.

It follows that $\operatorname{Pic}(A)\simeq\operatorname{Pic}(B_{0})$ and so that the Picard group of ${\cal O}_{S}(S(0,b))$ is not a torsion group. ∎