A note on a sheaf of real regular functions

TL;DR

This paper confirms that for real rings and integral domains, the global sections of a sheaf on the real Zariski spectrum coincide with a specific localization of the ring, answering a question posed in 1979.

Contribution

It provides a positive answer to Coste and Roy's 1979 question about the structure sheaf on the real Zariski spectrum for real rings and integral domains.

Findings

The global sections of the sheaf equal the localization of the ring at a certain multiplicative set.

The result applies specifically to real rings and integral domains.

It advances understanding of the structure sheaf in real algebraic geometry.

Abstract

Coste and Roy in 1979 defined a structural sheaf on the real Zariski spectrum of a semi-real ring $A$ and asked whether the ring of the global sections is $\sum^{-1}_1 A$ where $\sum_1$ is the multiplicative subset $\{1+\sum_{i=1}^n a_i^2|a_i\in A, n\in N\}$ of $A$. We give a positive answer to this question for real rings and integral domains.

Equations

$$\sum\nolimits_{f}=\big{\{}f^{2m}+\sum_{i=1}^{n}a_{i}^{2}\,|a_{i}\in A,m\in\mathbb{N}\cup\{0\},n\in\mathbb{N}\big{\}}.$$

$$\sqrt[R]{I}=\{a\in A|\exists m\in \mathbb{N}\exists b_1,\dots ,b_p\in A{a}^{2m}+b_1^2+\cdots +b_p^2\in I\}$$

$$D(f)\subseteq D(f^{2m}+\sum x^2)=D(a_1{f}_{i_1}+\cdots +a_k{f}_{i_k})\subseteq D(f_{i_1})\cup \cdots \cup D(f_{i_k}),$$

$$D(g_i)=D(h_i^2)\cap D(h_i^{2n_i}+{\sum }^i{x}^2)=D(h_i),$$

$$O_X(D(1))=O_X({\mathrm{Spec}}_{R}A)\cong {\sum }_1^{-1}A.$$

$$h\big{[}a(f^{2m}+\sum y^{2})-b(f^{2n}+\sum x^{2})\big{]}=0.$$

$$(f^{2k}+\sum z^{2})\big{[}a(f^{2m}+\sum y^{2})-b(f^{2n}+\sum x^{2})\big{]}=0.$$

$$\big{[}(g_{i}g_{j})^{2m_{ij}}+\sum\nolimits^{ij}x^{2}\big{]}(a_{i}g_{j}-a_{j}g_{i})=0.$$

$$(g_i{g}_j{)}^{n_{ij}}(a_i{g}_j-a_j{g}_i)=0.$$

$$g_j^{m+1}(g_i^m{a}_i)-g_i^{m+1}(g_j^m{a}_j)=0$$

$$g_{j}a=\sum _i{b}_i{a}_i{g}_j=\sum _j{b}_i{g}_i{a}_j=(f^{2k}+\sum x^2)a_j,$$

Full text

1. Introduction

By ring we mean a commutative with identity. The real Zariski spectrum of $A$ denoted by $\operatorname{\textrm{Spec}}_{R}A$ is a topological space whose underlying set is the set of all real prime ideals of $A$ and its topology is described as follows. Let $I$ be an ideal of $A$, we define $V(I)$ as the set all real prime ideals $\mathfrak{p}$ of $A$ with $\mathfrak{p}\supseteq I$. We have

Lemma 1.1**.**

(i) If $I$ and $J$ are two ideals of $A$, then $V(IJ)=V(I)\cup V(J)$.

(ii) If $\{I_{i}\}$ is any set of ideals of $A$, then $V(\sum I_{i})=\bigcap V(I_{i})$.

(iii) If $I$ and $J$ are two ideals, $V(I)\subseteq V(J)$ if and only if $\sqrt[R]{I}\supseteq\sqrt[R]{J}$.

Proof.

The same as ([3], Lemma 2.1) and see Theorem 1.3 below for the real radical $\sqrt[R]{I}$. ∎

Note that the ring $A$ is semi-real iff $\operatorname{\textrm{Spec}}_{R}A\neq\emptyset$. Now take the subsets of the form $V(I)$ to be the closed subsets. This defines a topology on $\operatorname{\textrm{Spec}}_{R}A$ which we call the real Zariski topology of $\operatorname{\textrm{Spec}}_{R}A$. Note that $V(A)=\emptyset$ and $V(\{0\})=\operatorname{\textrm{Spec}}_{R}A$. For any element $f\in A$, we denote by $D(f)$ the open complement of $V((f))$. It is easy to see that the open sets of the form $D(f)$ form a basis for the topology of $\operatorname{\textrm{Spec}}_{R}A$. An open set of the form $D(f)$ is called a principal open subset while its complement $V((f))$ is called a principal closed subset.

Now for a ring $A$, Coste and Roy defined a sheaf of rings $\mathcal{O}_{\operatorname{\textrm{Spec}}_{R}A}$ on the real Zariski spectrum $\operatorname{\textrm{Spec}}_{R}A$ as follows [2]. For an open $U\subseteq\operatorname{\textrm{Spec}}_{R}A$, $\mathcal{O}_{\operatorname{\textrm{Spec}}_{R}A}(U)$ consists of all functions $s\colon U\rightarrow\coprod_{\mathfrak{p}\in\operatorname{\textrm{Spec}}_{R}A}\!A_{\mathfrak{p}}$, satisfying the following two conditions. Firstly for all $\mathfrak{p}\in U$ we have $s(\mathfrak{p})\in A_{\operatorname{\mathfrak{p}}}$, secondly for each $\operatorname{\mathfrak{p}}\in U$ there is a neighborhood $U_{\operatorname{\mathfrak{p}}}$ of $\operatorname{\mathfrak{p}}$ in $U$ and also $a,f\in A$ such that for all $\mathfrak{q}\in U_{\operatorname{\mathfrak{p}}}$ we have $f\notin\mathfrak{q}$ and $s(\operatorname{\mathfrak{q}})=\frac{a}{f}\in A_{\operatorname{\mathfrak{q}}}$. For $V\subseteq U$ we define $\mathcal{O}_{\operatorname{\textrm{Spec}}_{R}A}(U)\rightarrow\mathcal{O}_{\operatorname{\textrm{Spec}}_{R}A}(V)$ by the restriction of functions. Let $X=\operatorname{\textrm{Spec}}_{R}A$ and $\operatorname{\mathfrak{p}}\in X$, it is easily seen that there is a canonical isomorphism $\mathcal{O}_{X,\operatorname{\mathfrak{p}}}\cong A_{\operatorname{\mathfrak{p}}}$. Though Coste and Roy had motivated from topos theoretic considerations, the sheaf $\mathcal{O}_{\operatorname{\textrm{Spec}}_{R}A}$ can be regarded as a generalization of affine irreducible real algebraic varieties in the sense of [1] in the same vein that affine schemes generalize affine algebraic varieties.

We denote by $\sum_{1}$ the multiplicative subset $\{1+\sum_{i=1}^{n}a_{i}^{2}\,|a_{i}\in A,n\in\mathbb{N}\}$ of $A$. Also for a nonzero $f\in A$ we define the following multiplicative subset of $A$.

[TABLE]

Obviously if $A$ is a semi-real ring, then $\sum_{1}\neq\emptyset$ and if $A$ a real ring, then $\sum_{f}\neq\emptyset$. Now we are able to state Coste and Roy’s question.

Question 1.2** ([2], Page 48).**

Do we have $\operatorname{\mathcal{O}}_{X}(D(f))\cong\sum_{f}^{-1}A$?

We show in Theorem 2.3 that the canonical homomorphism from $\sum_{f}^{-1}A$ into $\operatorname{\mathcal{O}}_{X}(D(f))$ is always injective and in the case of $A$ being an real ring or an integral domain is surjective. Hence an isomorphism. We don’t know whether this is the case for every semi-real ring. We shall make several use of the abstract version of the real Nullstellensatz.

Theorem 1.3** ([1], Proposition 4.1.7).**

Let $A$ be a commutative ring and $I$ an ideal of $A$. Then

[TABLE]

is the smallest real ideal of $A$ containing $I$. The ideal $\sqrt[R]{I}$, called the real radical of $I$, is the intersection of all real prime ideals containing $I$ (or is $A$ itself if there is no real prime ideal containing $I$).

Notation 1.4**.**

For the sake simplicity we often denote a sum of squares by $\sum x^{2}$,$\sum y^{2}$ and the like. When we say that $\sum x^{2}$ is a sum of square in $A$, we mean that there are $n\in\mathbb{N}$, $a_{1},\dots,a_{n}$ in $A$ such that $\sum x^{2}=\sum_{i=1}^{n}a_{i}^{2}$. In order to distinguish sums of squares we sometimes use superscripts, e.g., $\sum^{1}x^{2},\sum^{2}x^{2},\dots,\sum^{i}x^{2}$. Also when there is no danger of confusion we use expressions such as $f^{2}\sum x^{2}=\sum(xf)^{2}$.**

2. The ring of global sections

Lemma 2.1**.**

For any $f\in A$, $D(f)$ is quasi-compact.

Proof.

Let $D(f_{i}),i\in I$ be a open cover of $D(f)$, then $V((f))=\bigcap_{i\in I}V((f_{i}))=V(\sum_{i\in I}(f_{i}))$. So $f\in\sqrt[R]{\sum_{i\in I}(f_{i})}$ and it follows that there are $a_{1},\dots,a_{k}$ in $A$ and $f_{i_{1}},\dots,f_{i_{k}}$ such that $a_{1}f_{i_{1}}+\cdots+a_{k}f_{i_{k}}=f^{2m}+\sum x^{2}$ for some $m\in\mathbb{N}$. Therefore

[TABLE]

thus $D(f)$ has a finite subcover. ∎

Now we will need the following useful lemma.

Lemma 2.2** **(Basic Lemma).

Let $X=\operatorname{\textrm{Spec}}_{R}A$, $f\in A$ and $s\in\operatorname{\mathcal{O}}_{X}(D(f))$, then there are $g_{1},\dots,g_{n}\in A$ such that $D(f)=D(g_{1})\cup\cdots\cup D(g_{n})$ and $s\upharpoonright D(g_{i})=\frac{a_{i}}{g_{i}}$ for some $a_{i}\in A$.

Proof.

Let $D(f)=\{\operatorname{\mathfrak{p}}_{i}|i\in I\}$, then each $\operatorname{\mathfrak{p}}_{i}$ has a neighborhood $D(h_{i})\subseteq D(f)$ such that $s\upharpoonright D(h_{i})=\frac{b_{i}}{f_{i}}$ so that for all $\operatorname{\mathfrak{q}}\in D(h_{i})$ we have $f_{i}\notin\operatorname{\mathfrak{q}}$. This means that $D(h_{i})\subseteq D(f_{i})$ for $i\in I$ which implies that $h_{i}^{2n_{i}}+\sum^{i}x^{2}=u_{i}f_{i}$ for some $n_{i}\in\mathbb{N}$ and $\sum^{i}x^{2},u_{i}\in A$. Thus $s\upharpoonright D(h_{i})=\frac{b_{i}}{f_{i}}=\frac{u_{i}b_{i}}{h_{i}^{2n_{i}}+\sum^{i}x^{2}}$. Now set $a_{i}=u_{i}b_{i}h_{i}^{2}$ and $g_{i}=h_{i}^{2}(h_{i}^{2n_{i}}+\sum^{i}x^{2})$, so $s\upharpoonright D(h_{i})=\frac{a_{i}}{g_{i}}$. Observing that

[TABLE]

we get $s\upharpoonright D(g_{i})=\frac{a_{i}}{g_{i}}$. Also $\bigcup_{i\in I}D(g_{i})=\bigcup_{i\in I}D(h_{i})=D(f)$, then by quasi-compactness we can choose finitely many of $D(g_{i})$ covering $D(f)$. ∎

Theorem 2.3**.**

Let $A$ be an real ring or an integral domain, $f\in A$ and let $X=\operatorname{\textrm{Spec}}_{R}A$, then $\operatorname{\mathcal{O}}_{X}(D(f))\cong\sum_{f}^{-1}A$. So taking $f=1$ we get

[TABLE]

Proof.

We suppose that $A$ is an real ring. The case of $A$ being an integral domain is dealt with similarly or even simpler because of the fact that all the localizations are the subrings of the quotient filed of $A$. We define a map $\psi\colon\sum_{f}^{-1}A\rightarrow\operatorname{\mathcal{O}}_{X}(D(f))$ by sending $\frac{a}{f^{2n}+\sum x^{2}}$ to $s\in\operatorname{\mathcal{O}}_{X}(D(f))$ where $s(\operatorname{\mathfrak{p}})=\frac{a}{f^{2n}+\sum x^{2}}$ for all $\operatorname{\mathfrak{p}}\in D(f)$. Obviously $\psi$ is homomorphism. We first show that $\psi$ is injective. Note that for injectivity we do not need $A$ is a real ring or integral domain. Let $\psi(\frac{a}{f^{2n}+\sum x^{2}})=\psi(\frac{b}{f^{2m}+\sum y^{2}})=s\in\operatorname{\mathcal{O}}_{X}(D(f))$, then for any $\operatorname{\mathfrak{p}}\in D(f)$, both $\frac{a}{f^{2n}+\sum x^{2}}$ and $\frac{b}{f^{2m}+\sum y^{2}}$ are equal to $s(\operatorname{\mathfrak{p}})\in A_{\operatorname{\mathfrak{p}}}$. Hence there is $h\notin\operatorname{\mathfrak{p}}$ such that

[TABLE]

Let $I$ be the annihilator of $a(f^{2m}+\sum y^{2})-b(f^{2n}+\sum x^{2})$, then $h\in I$, $h\notin\operatorname{\mathfrak{p}}$ so $I\nsubseteq\operatorname{\mathfrak{p}}$. This holds for any $\operatorname{\mathfrak{p}}\in D(f)$, so we conclude that $V(I)\cap D(f)=\O$. Therefore $f\in\sqrt[R]{I}$, thus there are $k\in\mathbb{N}$ and $\sum z^{2}$ in $A$ such that

[TABLE]

Observing that $(f^{2k}+\sum z^{2})\in\sum_{f}$, we get $\frac{a}{f^{2n}+\sum x^{2}}=\frac{b}{f^{2m}+\sum y^{2}}$ in $\sum_{f}^{-1}A$. Hence $\psi$ is injective.

Let’s show that $\psi$ is surjective. Let $s\in\operatorname{\mathcal{O}}_{X}(D(f))$, by Basic Lemma there are $g_{1},\dots,g_{n}\in A$ such that $D(f)=D(g_{1})\cup\cdots\cup D(g_{n})$ and $s\upharpoonright D(g_{i})=\frac{a_{i}}{g_{i}}$ for some $a_{i}\in A$. So for any $\operatorname{\mathfrak{p}}\in D(g_{i}g_{j})=D(g_{i})\cap D(g_{j})$, there is $h_{\operatorname{\mathfrak{p}}}\notin\operatorname{\mathfrak{p}}$ such that $h_{\operatorname{\mathfrak{p}}}(a_{i}g_{j}-a_{j}g_{i})=0$ in $A$. Let $I$ be the annihilator of $a_{i}g_{j}-a_{j}g_{i}$, so $h_{\operatorname{\mathfrak{p}}}\in I$, $h_{\operatorname{\mathfrak{p}}}\notin\operatorname{\mathfrak{p}}$ thus $\operatorname{\mathfrak{p}}\nsupseteq I$ and this holds for any $\operatorname{\mathfrak{p}}\in D(g_{i}g_{j})$, therefore we have $V(I)\cap D(g_{i}g_{j})=\O$ from which it follows that $g_{i}g_{j}\in\sqrt[R]{I}$. So there are $m_{ij}\in\mathbb{N}$ and $\sum^{ij}x^{2}$ in $A$ such that

[TABLE]

since $A$ is a real ring there are $n_{ij}\in\mathbb{N}$ such that

[TABLE]

So taking $m$ larger than all $n_{ij}$ we get $(g_{i}g_{j})^{m}(a_{i}g_{j}-a_{j}g_{i})=0$ for all $i,j\in\{1,\dots,n\}$. Rewriting as

[TABLE]

and then replacing each $g_{i}$ by $g_{i}^{m+1}$, and $a_{i}$ by $g_{i}^{m}a_{i}$, we still have $s$ represented on on $D(g_{i})$ by $\frac{a_{i}}{g_{i}}$ and furthermore, we have $g_{i}a_{j}=g_{j}a_{i}$ for all $i,j\in\{1,\dots,n\}$. On the other hand having $V((f))=\bigcap^{n}_{i=1}V((g_{i}))=V(\sum^{n}_{i=1}(g_{i}))$ we get $f\in\sqrt[R]{\sum^{n}_{i=1}(g_{i})}$ and it follows that there are $b_{1},\dots,b_{k}$ in $A$ such that $b_{1}g_{1}+\cdots+b_{n}g_{n}=f^{2k}+\sum x^{2}$ for some $k$. Now let $a=a_{1}b_{1}+\cdots+a_{n}b_{n}$. Then for each $j$ we have

[TABLE]

which means that $\frac{a}{f^{2k}+\sum x^{2}}=\frac{a_{j}}{g_{j}}$ on $D(g_{j})$. So $\psi(\frac{a}{f^{2k}+\sum x^{2}})=s$ everywhere, which shows that $\psi$ is surjective, hence an isomorphism. ∎

Acknowledgment. The research of the author was in part supported by a grant from IPM (No. 99030403).