Algebraic characterizations of homeomorphisms between algebraic varieties

After some reminders on integral extensions and normalization, we recall the notions of subintegral extensions and the respective constructions of Traverso [31], Andreotti and Bombieri [1] of the seminormalization for ring extensions and morphisms between algebraic varieties.

Notation and terminology. Let A be a ring. The Zariski spectrum \({{\,\textrm{Spec}\,}}A\) of A is the set of prime ideals of A. It is a topological space for the topology whose closed sets are generated by the sets \(\mathcal {V}(f)=\{{{\,\mathrm{\mathfrak {p}}\,}}\in {{\,\textrm{Spec}\,}}A\mid f\in {{\,\mathrm{\mathfrak {p}}\,}}\}\) for \(f\in A\). We denote by \({{\,\mathrm{\mathrm Max}\,}}A\) the subspace of maximal ideals of A. For \({{\,\mathrm{\mathfrak {p}}\,}}\in {{\,\textrm{Spec}\,}}A\), we denote by \(k({{\,\mathrm{\mathfrak {p}}\,}})\) the residue field at \({{\,\mathrm{\mathfrak {p}}\,}}\).

Let \((X,\mathcal {O}_X)\) be a variety over k. For \(x\in X\), we denote by k(x) the residue field at x; for an affine neighborhood U of x, the element x corresponds to a prime ideal \({{\,\mathrm{\mathfrak {p}}\,}}_x\) of \(\mathcal {O}_X(U)\) and we have \(k(x)=k({{\,\mathrm{\mathfrak {p}}\,}}_x)\). In case X is affine, we denote by k[X] the coordinate ring of X i.e \(k[X]=\mathcal {O}_X(X)\). Let K be a field containing k. We denote by X(K) the set \({{\,\mathrm{\mathrm Mor}\,}}({{\,\textrm{Spec}\,}}K,X)\) of K-rational points. If \(K=k\), then X(k) is also the set of k-closed points of X, i.e the points of X with residue field equal to k. We have thus an inclusion \(X(k)\hookrightarrow X\) that makes X(k) a topological space for the Zariski topology. We denote by \(\mathcal {O}_{X(k)}\) the sheaf of regular functions on X(k), for \(x\in X(k)\) we have \(\mathcal {O}_{X,x}=\mathcal {O}_{X(k),x}\). In the case k algebraically closed, for an open subset U of X, we may identify \({{\,\mathrm{\mathrm Max}\,}}\mathcal {O}_X(U)\) with U(k) by the Nullstellensatz, and similarly we identify the regular functions on U with those on U(k), namely \(\mathcal {O}_X(U)=\mathcal {O}_{X(k)}(U(k))\). If T is a subset of X or X(k) or \({{\,\textrm{Spec}\,}}A\), then we will denote by \(\overline{T}\) the closure of T for the Zariski topology.

A ring extension \(i:A\rightarrow B\) induces a map \({{\,\textrm{Spec}\,}}(i):{{\,\textrm{Spec}\,}}B\rightarrow {{\,\textrm{Spec}\,}}A\), given by \({{\,\mathrm{\mathfrak {p}}\,}}\mapsto ({{\,\mathrm{\mathfrak {p}}\,}}\cap A)=i^{-1}({{\,\mathrm{\mathfrak {p}}\,}})\). Let \(\pi :Y\rightarrow X\) be a morphism between algebraic varieties over k and let \(\mathcal {O}_X\rightarrow \pi _*\mathcal {O}_Y\) be the associated morphism of sheaves of rings on X. For any affine open subsets \(U'\subset Y\), \(U\subset X\) such that \(\pi (U')\subset U\), the ring map \(\mathcal {O}_X(U)\rightarrow \mathcal {O}_Y(U')\) is an extension if \(\pi \) is dominant. For a field extension \(k\rightarrow K\), we denote by \(\pi _{K}:Y(K)\rightarrow X(K)\) the induced map. Remark that \(\pi _k\) is also the restriction of \(\pi \) to the k-closed points.

In the sequel, \(\mathcal K(A)\) (resp. \(\mathcal K\)) will denote the total ring of fractions of A (resp. the sheaf of total ring of fractions on X).

1.1 Reminder on integral extensions and normalization

A ring map \(A\rightarrow B\) is called of finite type (resp. finite) if it makes B a finitely generated A-algebra (resp. A-module). The ring map \(A\rightarrow B\) is birational if it induces an isomorphism between \(\mathcal K(A)\) and \(\mathcal K(B)\).

Let \(A\rightarrow B\) be a ring extension. An element \(b\in B\) is integral over A if b is the root of a monic polynomial with coefficients in A, which is equivalent for A[b] to be a finite A-module. As a consequence

is a ring called the integral closure of A in B. The extension \(A\rightarrow B\) is called integral if \(A_B'=B\). In case \(B=\mathcal K(A)\), the ring \(A_\mathcal {K(A)}'\) is denoted by \(A'\) and is simply called the integral closure of A. The ring A is called integrally closed (resp. in B) if \(A=A'\) (resp. \(A=A_B'\)).

We recall the classical definitions of morphisms considered in this paper. Let \(\pi :Y\rightarrow X\) be a morphism between algebraic varieties over k. We say that \(\pi \) is birational if \(\pi \) induces a bijection between the set of generic points of irreducible components of Y and the set of generic points of the irreducible components of X, and for every generic point \(\eta \) of an irreducible component of Y the local ring map \(\mathcal {O}_{X,\pi (\eta )}\rightarrow \mathcal {O}_{Y,\eta }\) is an isomorphism. We say that \(\pi \) is of finite type if \(\pi \) is quasi-compact and for all affine subsets \(U'\subset Y\), \(U\subset X\) with \(\pi (U')\subset U\), the ring map \(\mathcal {O}_X (U)\rightarrow \mathcal {O}_Y(U')\) is of finite type. We say that a morphism \(Y\rightarrow X\) between algebraic varieties over k is finite (resp. integral) if \(\pi \) is affine and for any open subset \(U\subset X\) the ring extension \(\mathcal {O}_X(U)\rightarrow \mathcal {O}_Y(\pi ^{-1}(U))\) is finite (resp. integral). In this paper, a morphism between algebraic varieties is always of finite type.

Let X be an algebraic variety over k. The normalization of X, denoted by \(X'\), is the algebraic variety over k with a finite birational morphism \(\pi ':X'\rightarrow X\), called the normalization morphism such that for any open subset \(U\subset X\) we have \(\mathcal {O}_{X'} (\pi '^{-1}(U))=\mathcal {O}_X(U)'\). We say that X is normal if \(\pi '\) is an isomorphism. A point \(x\in X\) is called normal if \(\mathcal {O}_{X,x}\) is integrally closed.

We will use frequently the following surjectivity results.

Proposition 1.1

Let \(A\rightarrow B\) be an integral extension of rings. The maps \({{\,\textrm{Spec}\,}}B\rightarrow {{\,\textrm{Spec}\,}}A\) and \({{\,\mathrm{\mathrm Max}\,}}B\rightarrow {{\,\mathrm{\mathrm Max}\,}}A\) are surjective and closed.

As a consequence, if \(\pi \) is a finite morphism between algebraic varieties over k, then Proposition 1.1 implies that \(\pi \) and \(\pi _{k}\) are surjective.

1.2 Subintegral extensions and seminormalization

We recall the concept of subintegral extensions introduced by Traverso [31].

Definition 1.2

Let \(A\rightarrow B\) be an extension of rings.

1. (1)

   For \({{\,\mathrm{\mathfrak {p}}\,}}\in {{\,\textrm{Spec}\,}}B\), we say that \({{\,\textrm{Spec}\,}}B\rightarrow {{\,\textrm{Spec}\,}}A\) is equiresidual at \({{\,\mathrm{\mathfrak {p}}\,}}\) if the extension \(k({{\,\mathrm{\mathfrak {p}}\,}}\cap A)\rightarrow k({{\,\mathrm{\mathfrak {p}}\,}})\) is an isomorphism. Let \(W\subset {{\,\textrm{Spec}\,}}B\), we say that \({{\,\textrm{Spec}\,}}B\rightarrow {{\,\textrm{Spec}\,}}A\) is equiresidual by restriction to W if for any \({{\,\mathrm{\mathfrak {p}}\,}}\in W\), \({{\,\textrm{Spec}\,}}B\rightarrow {{\,\textrm{Spec}\,}}A\) is equiresidual at \({{\,\mathrm{\mathfrak {p}}\,}}\). If \(W={{\,\textrm{Spec}\,}}B\), we simply say equiresidual. The extension \(A\rightarrow B\) is called equiresidual if \({{\,\textrm{Spec}\,}}B\rightarrow {{\,\textrm{Spec}\,}}A\) is.

2. (2)

   The extension \(A\rightarrow B\) and the map \({{\,\textrm{Spec}\,}}B\rightarrow {{\,\textrm{Spec}\,}}A\) are called subintegral if the extension is integral and \({{\,\textrm{Spec}\,}}B\rightarrow {{\,\textrm{Spec}\,}}A\) is bijective and equiresidual.

Note that a field extension is equiresidual if and only if it is an isomorphism. We extend these definitions to the geometric setting, adding moreover a notion of hereditarily birational morphism that have been introduced in the real setting in [14].

Definition 1.3

Let \(\pi :Y\rightarrow X\) be a dominant morphism between algebraic varieties over k.

1. (1)

   We say that \(\pi \) is equiresidual if, for any \(y\in Y\), the field extension \(k(\pi (y))\rightarrow k(y)\) is an isomorphism.

2. (2)

   We say that \(\pi \) is subintegral if \(\pi \) is integral, bijective and equiresidual.

3. (3)

   We say that \(\pi :Y\rightarrow X\) is hereditarily birational if for any affine open subsets \(U'\subset Y\), \(U\subset X\) with \(\pi (U')\subset U\) and for any irreducible algebraic subvariety \(V=\mathcal {V}({{\,\mathrm{\mathfrak {p}}\,}})\simeq {{\,\textrm{Spec}\,}}(\mathcal {O}_Y(U')/{{\,\mathrm{\mathfrak {p}}\,}})\) in \(U'\), the morphism

   $$\begin{aligned} \pi _{|V}: V\rightarrow W=\mathcal {V}({{\,\mathrm{\mathfrak {p}}\,}}\cap \mathcal {O}_X(U))\simeq {{\,\textrm{Spec}\,}}\big (\mathcal {O}_X(U)/({{\,\mathrm{\mathfrak {p}}\,}}\cap \mathcal {O}_X(U))\big ) \end{aligned}$$

   is birational.

Geometrically speaking, a dominant morphism \(\pi :Y\rightarrow X\) is equiresidual if and only if it is hereditarily birational. Indeed, for any affine open subsets \(U'\subset Y\), \(U\subset X\) with \(\pi (U')\subset U\) and for any irreducible algebraic subvariety \(V=\mathcal {V}({{\,\mathrm{\mathfrak {p}}\,}})\simeq {{\,\textrm{Spec}\,}}(\mathcal {O}_Y(U')/{{\,\mathrm{\mathfrak {p}}\,}})\) in \(U'\), the restricted morphism

is birational if and only if the extension \(k({{\,\mathrm{\mathfrak {p}}\,}}\cap \mathcal {O}_X(U))=\mathcal K(W)\rightarrow k({{\,\mathrm{\mathfrak {p}}\,}})=\mathcal K(V)\) is an isomorphism.

An hereditarily birational morphism is not necessarily bijective. However, adding an integrality assumption and using Proposition 1.1, we get the following characterization.

Lemma 1.4

Let \(\pi :Y\rightarrow X\) be an integral morphism between algebraic varieties over k. The following properties are equivalent:

1. (1)

   \(\pi \) is hereditarily birational and injective.

2. (2)

   \(\pi \) is subintegral.

The notion of subintegral extension leads to the notion of seminormalization, in a similar way that integral extensions lead to normalization.

In order to define the notion of seminormality, we need to consider sequences of ring extensions. A ring C is called intermediate between the rings A and B if there exists a sequence of extensions \(A\rightarrow C\rightarrow B\). In that case, we say that \(A\rightarrow C\) and \(C\rightarrow B\) are intermediate extensions of \(A\rightarrow B\) and we say in addition that \(A\rightarrow C\) is a subextension of \(A\rightarrow B\).

Seminormal extensions are maximal subintegral extensions.

Definition 1.5

Let \(A\rightarrow C\rightarrow B\) be a sequence of two extensions of rings with \(A\rightarrow C\) subintegral. We say that C is seminormal between A and B if for every intermediate ring D between C and B, with C different from D, then \(A\rightarrow D\) is not subintegral.

We say that A is seminormal in B if A is seminormal between A and B. We say that A is seminormal if A is seminormal between A and \(A'\).

Given an extension of rings \(A\rightarrow B\), Traverso [31] proved respectively that there exists a unique intermediate ring which is seminormal between A and B. To this purpose, he introduced the ring

where \(\mathrm{{Rad}}\) stands for the Jacobson radical, namely the intersection of all the maximal ideals. The idea to build \(A_B^+\) is, for all \({{\,\mathrm{\mathfrak {p}}\,}}\in {{\,\textrm{Spec}\,}}A\), to glue together all the prime ideals of \(A_B'\) lying over \({{\,\mathrm{\mathfrak {p}}\,}}\) (see [27]).

Theorem 1.6

[31]

Let \(A\rightarrow B\) be an extension of rings. Then \(A_B^+\) is the unique ring which is seminormal between A and B. Moreover, for any intermediate ring C between A and B, the extension \( A\rightarrow C\) is subintegral if and only if \(C\subset A^+_B\).

The ring \(A^+_B\) is called the seminormalization of the ring extension \( A\rightarrow B\) or the seminormalization of A in B. Note that

The ring \(A^+_{A'}\) is called the seminormalization of A and is simply denoted by \(A^+\). Note that if A and B are domains, then A and \(A_B^+\) have in particular the same fraction field.

1.3 Seminormalization of a morphism between algebraic varieties

Andreotti and Bombieri [1] have introduced and built the seminormalization of a scheme in another one. In this section, we provide a different and elementary construction of the seminormalization of an affine algebraic variety in another one. The seminormalization answers the following question. Let \(Y\rightarrow X\) be a dominant morphism between algebraic varieties over k. Does there exist a biggest algebraic variety Z such that \(Y\rightarrow X\) factorizes through Z and \(Z\rightarrow X\) is subintegral ?

We recall first the notion of normalization of a variety in another one. Let \(\pi :Y\rightarrow X\) be a dominant morphism between algebraic varieties over k. The integral closure \((\mathcal {O}_X)_{\pi _*(\mathcal {O}_Y)}'\) of \(\mathcal {O}_X\) in \(\pi _*(\mathcal {O}_Y)\) is a coherent sheaf [30, Lem. 29.53.15] and by [17, II Prop. 1.3.1] it is the structural sheaf of a variety over k.

Definition 1.7

Let \(\pi :Y\rightarrow X\) be a dominant morphism between algebraic varieties over k. The variety with structural sheaf equal to the integral closure of \(\mathcal {O}_X\) in \(\pi _*(\mathcal {O}_Y)\) is called the normalization of X in Y and is denoted by \(X_Y'\).

Note that the normalization of a variety in another one is not necessarily a normal variety, nor it admits a birational morphism onto the original variety.

For a dominant morphism \(Y\rightarrow X\) between algebraic varieties over k, we say that an algebraic variety Z over k is intermediate between X and Y if \(Y\rightarrow X\) factorizes through Z. For affine varieties, it is equivalent to say that k[Z] is an intermediate ring between k[X] and k[Y]. Recall the universal property satisfied by the normalization of a variety in another one:

Proposition 1.8

Let \(Y\rightarrow X\) be a dominant morphism between algebraic varieties over k. Let Z be an intermediate variety between X and Y. Then \(Z\rightarrow X\) is finite if and only if it factorizes \(X_Y'\rightarrow X\).

Proof

By [30, Lem. 29.44.4] and since we work with morphisms which are locally of finite type, then \(Z\rightarrow X\) is finite if and only if it is integral. The proof follows now from the definition of \(X'_Y\). \(\square \)

We describe now an elementary construction of the seminormalization of an affine algebraic variety in another one. Let \(Y\rightarrow X\) be a dominant morphism between affine algebraic varieties over k. We want to check that the ring \(A_1=k[X]^{+}_{k[Y]}\) is the coordinate ring of an algebraic variety. We know that the morphism \(X_Y'\rightarrow X\) is finite by Proposition 1.8, so we can apply Lemma 1.9 below to the extensions

to conclude.

Lemma 1.9

Let \(\pi :Y\rightarrow X\) be a finite morphism between affine algebraic varieties over k. Let A be a ring such that \(k[X]\subset A\subset k[Y]\). Then A is the coordinate ring of a unique affine algebraic variety over k and \(\pi \) factorizes through this variety.

Proof

Since k[Y] is a finite module over the Noetherian ring k[X] it is a Noetherian k[X]-module. Thus the ring A is a finite k[X]-module as a submodule of a Noetherian k[X]-module. It follows that A is a finitely generated algebra over k and the proof is done. \(\square \)

For a general construction of the seminormalization, one needs to check that the seminormalization of the local charts of an affine covering glue together to give a global variety. This is done by Andreotti and Bombieri [1] using Grothendieck criterion [17, II Prop. 1.3.1] concerning the quasi-coherence of sheaves. It leads to the following definitions.

Definition 1.10

Let \(\pi :Y\rightarrow X\) be a dominant morphism between algebraic varieties over k. The seminormalization of X in Y is the algebraic variety \(X^+_Y\) over k with structural sheaf equal to the seminormalization of \(\mathcal {O}_X\) in \(\pi _*(\mathcal {O}_Y)\).

We call \(X^{+}\) the seminormalization of X in its normalization \(Y=X'\). We say that X is seminormal in Y (resp. seminormal) if \(X=X^+_Y\) (resp. \(X=X^+\)).

Note that the seminormalization of X in Y is birational to X, even if \(Y\rightarrow X\) is not birational. It is not the case for the normalization of X in Y. We have in general

Theorem 1.6 implies the following universal properties for the seminormalization of a variety in another one:

Proposition 1.11

Let \(Y\rightarrow Z\rightarrow X\) be a sequence of dominant morphisms between algebraic varieties over k. Then \(Z\rightarrow X\) is subintegral if and only if \(X^+_Y\rightarrow X\) factorizes though Z.

In the classical study of the seminormalization, some basic properties such as the local nature happen to be not so straightforward to prove. A nice algebraic approach has been proposed by Manaresi [26], in the spirit of the relative Lipschitz saturation [25], via the saturation of a ring A in another ring B which is integral over A. The saturation coincides with the seminormalization when the ring extension is finite. We aim to study the properties of this saturation for more general extensions, and establish its universal properties.

2.1 Universal property of the saturation

We define the saturation of a ring extension analogously to [26], but for non-necessarily integral extensions.

Definition 2.1

Let \(A\rightarrow B\) be an extension of rings. The saturation of A in B, denoted by \(\widehat{A}_B\), is defined by

where the nil radical \(\mathrm{{NilRad}}\) denotes the ideal of nilpotent elements.

We say that A is saturated in B if \(\widehat{A}_B=A\). The saturation of A is its saturation in \(A'\) and it is simply denoted by \(\widehat{A}\). We say that A is saturated if \(\widehat{A}=A\).

Recall that the nilradical is the intersection of all prime ideals. In order to study the saturation, we need to understand better the relation between prime ideals in A and B and prime ideals in \({{\,\mathrm{B\otimes _{A} B}\,}}\). For a ring extension \(A\rightarrow B\), we introduce the notation \(\varphi _1\), \(\varphi _2\) for the ring morphisms \(\varphi _i:B\rightarrow {{\,\mathrm{B\otimes _{A} B}\,}}\) defined by

The data of a prime ideal \(\omega \) in \({{\,\mathrm{B\otimes _{A} B}\,}}\), or more precisely the data of a morphism \(g:{{\,\mathrm{B\otimes _{A} B}\,}}\rightarrow k(\omega )\) with kernel \(\omega \), is equivalent to the data of a 4-tuple of prime ideals

such that

and such that the composition

coincides with g.

The saturation is a ring, compatible with inclusion.

Lemma 2.2

Let \(A\rightarrow B\) be an extension of rings.

1. (1)

   \(\widehat{A}_B\) is a subring of B containing A.

2. (2)

   If \(A\subset C \subset B\), then \(\widehat{A}_B \subset \widehat{C}_B\).

Proof

1. (1)

   The set \(\widehat{A}_B\) is an A-module as the kernel of the A-module morphism

   $$\begin{aligned} B\xrightarrow {\varphi _1-\varphi _2} \dfrac{{{\,\mathrm{B\otimes _{A} B}\,}}}{\mathrm{{NilRad}}({{\,\mathrm{B\otimes _{A} B}\,}})}. \end{aligned}$$

   The stability under product comes from the identity

   $$\begin{aligned} b_1b_2\otimes _A 1-1\otimes _A b_1b_2=(b_1\otimes _A 1)(b_2\otimes _A 1-1\otimes _A b_2)+(1\otimes _A b_2)(b_1\otimes _A 1-1\otimes _A b_1) \end{aligned}$$

   and the fact that \(\mathrm{{NilRad}}({{\,\mathrm{B\otimes _{A} B}\,}})\) is an ideal.

2. (2)

   The image of a nilpotent element by the ring morphism \({{\,\mathrm{B\otimes _{A} B}\,}}\rightarrow B\otimes _C B\) remains nilpotent.

\(\square \)

In order to give a universal property of the saturation, we recall the notion of radicial extension introduced by Grothendieck [16, I, def. 3.7.2]. We also introduce a notion of radicial sequence of extensions similarly to [27], due to the lack of integrality of the ring extensions.

Definition 2.3

1. (1)

   An extension of rings \(A\rightarrow B\) is called radicial if \({{\,\textrm{Spec}\,}}B\rightarrow {{\,\textrm{Spec}\,}}A\) is injective and equiresidual.

2. (2)

   A sequence of extensions \(A\rightarrow C\rightarrow B\) of rings is called radicial if the restriction of \({{\,\textrm{Spec}\,}}C\rightarrow {{\,\textrm{Spec}\,}}A\) to the image of \({{\,\textrm{Spec}\,}}B\rightarrow {{\,\textrm{Spec}\,}}C\) is injective and equiresidual.

Note that an extension (resp. a sequence of extensions) of fields \(K\rightarrow K'\) (resp. \(K\rightarrow K'\rightarrow K''\)) is radicial if and only if \(K\rightarrow K'\) is an isomorphism.

The saturation provides radicial sequences of extensions.

Proposition 2.4

Let \(A\rightarrow B\) be a ring extension. For any \(C\subset \widehat{A}_B\), the sequence \(A\rightarrow C\rightarrow B\) is radicial.

Before entering into the proof, we state an elementary result about field extensions (giving a proof of (2) implies (3) of Theorem 2.6 in the case of a sequence of field extensions).

Lemma 2.5

Let \(K\rightarrow K'\rightarrow K''\) be a non radicial sequence of field extensions i.e such that \(K\rightarrow K'\) is not an isomorphism. There are two K-morphisms \(K''\rightarrow L\) into a (algebraically closed) field L whose compositions with \(K'\rightarrow K''\) are distinct.

Proof

Since \(K\not = K'\) then it follows from [16, Prop. I.3.7.1] that there are two distinct K-morphisms \(\psi _1, \psi _2: K'\rightarrow L'\) into a field \(L'\). The point is to extend them to \(K''\).

For \(i\in \{1,2\}\), one can embed the field extensions \(K'\rightarrow K''\) and \(\psi _i: K' \rightarrow L'\) into a common extension \(K'\rightarrow L'_i\) by amalgamation [8, Chap 5, 4,Prop. 2]. Denote by \(\psi _i':K''\rightarrow L'_i\) the induced extension. By amalgamation of \(L'_1\) and \(L_2'\) over \(K''\), one can assume that \(\psi _1'\) and \(\psi _2'\) take values in a common field L.

The morphisms \(\psi _1',\psi _2': K'' \rightarrow L\) fulfill the requirements since the restriction of \(\psi _i'\) to \(K'\) coincides with \(\psi _i\). \(\square \)

It is classical that one can choose \(L=L'\) in the proof of Lemma 2.5 if the extension \(K'\rightarrow K''\) is moreover algebraic (we don’t need it in the paper), and this is used in [25] to prove that the Lipschitz saturation is stable under contraction: in the setting of Proposition 2.4, if \(C\rightarrow B\) is integral, then the Lipschitz saturation of A in C is equal to the intersection of C with the Lipschitz saturation of A in B. In our context the extensions are not assumed to be integral, and this contraction property does not hold (cf. Example 3.5).

Proof of Proposition 2.4

Let \({{\,\mathrm{\mathfrak {p}}\,}}_1\), \({{\,\mathrm{\mathfrak {p}}\,}}_2\) be two prime ideals of B lying over the same ideal \({{\,\mathrm{{\mathfrak q}}\,}}\) of A. A first step is to prove that \({{\,\mathrm{\mathfrak {p}}\,}}_1\) and \({{\,\mathrm{\mathfrak {p}}\,}}_2\) lie over the same ideal of C.

Let \({{\,\mathrm{\mathfrak {p}}\,}}\) be a prime ideal of \(k({{\,\mathrm{\mathfrak {p}}\,}}_1)\otimes _{k({{\,\mathrm{{\mathfrak q}}\,}})}k({{\,\mathrm{\mathfrak {p}}\,}}_2)\), and \(\omega =({{\,\mathrm{\mathfrak {p}}\,}}_1,{{\,\mathrm{\mathfrak {p}}\,}}_2,{{\,\mathrm{{\mathfrak q}}\,}},{{\,\mathrm{\mathfrak {p}}\,}})\in {{\,\textrm{Spec}\,}}({{\,\mathrm{B\otimes _{A} B}\,}})\) be the corresponding element, coming with a morphism \(g:{{\,\mathrm{B\otimes _{A} B}\,}}\rightarrow k(\omega )\) with \(\ker g=\omega \). For \(c\in {{\,\mathrm{\mathfrak {p}}\,}}_1\cap C\), we have \(g\circ \varphi _1(c)=g(c\otimes _A 1)=0\) by construction of g. The element \({{\,\mathrm{1\otimes _{A} c - c\otimes _{A} 1}\,}}\) is nilpotent in \({{\,\mathrm{B\otimes _{A} B}\,}}\) by assumption, so that

As a consequence \(g\circ \varphi _2(c)=0\) and thus \(c\in {{\,\mathrm{\mathfrak {p}}\,}}_2\cap C\). By symmetry we obtain \({{\,\mathrm{\mathfrak {p}}\,}}_1\cap C= {{\,\mathrm{\mathfrak {p}}\,}}_2\cap C.\)

The second step is to prove that the extension \(\phi :k({{\,\mathrm{{\mathfrak q}}\,}})\rightarrow k({{\,\mathrm{\mathfrak {p}}\,}}_1\cap C)\) is an isomorphism, where \({{\,\mathrm{{\mathfrak q}}\,}}={{\,\mathrm{\mathfrak {p}}\,}}_1\cap A\). Assume by contradiction that \(\phi \) is not an isomorphism, and consider the composition

By Lemma 2.5, there are two distinct \(k({{\,\mathrm{{\mathfrak q}}\,}})\)-morphisms \(\psi _1,\psi _2:k({{\,\mathrm{\mathfrak {p}}\,}}_1)\rightarrow L\) into some field L, which remain distinct by restriction to \(k({{\,\mathrm{\mathfrak {p}}\,}}_1\cap C)\). Thus there exists \(c\in C\) such that \(\psi _1\circ \pi (c)\not =\psi _2\circ \pi (c)\), where \(\pi :B\rightarrow k({{\,\mathrm{\mathfrak {p}}\,}}_1)\) denotes the natural morphism.

The morphisms \(\psi _1\) and \(\psi _2\) induce a morphism \(\psi :k({{\,\mathrm{\mathfrak {p}}\,}}_1)\otimes _{k({{\,\mathrm{{\mathfrak q}}\,}})}k({{\,\mathrm{\mathfrak {p}}\,}}_1)\rightarrow L\) given by

for \(b_1,b_2\in B\). The kernel \({{\,\mathrm{\mathfrak {p}}\,}}\) of \(\psi \) gives rise to a prime ideal \(\omega =({{\,\mathrm{\mathfrak {p}}\,}}_1,{{\,\mathrm{\mathfrak {p}}\,}}_2,{{\,\mathrm{{\mathfrak q}}\,}},{{\,\mathrm{\mathfrak {p}}\,}})\) of \({{\,\mathrm{B\otimes _{A} B}\,}}\) coming with a morphism \(g:{{\,\mathrm{B\otimes _{A} B}\,}}\rightarrow k(\omega )\rightarrow L.\) By our choice of c, the element

is not zero, so that \({{\,\mathrm{1\otimes _{A} c - c\otimes _{A} 1}\,}}\) does not belong to \(\ker g=\omega \), contradicting the inclusion \(C\subset \widehat{A}_B\). \(\square \)

Actually the converse of the preceding result holds true, and it gives rise to universal properties of the saturation, in terms of radicial sequences of extensions. This result is, up to our knowledge, not present in the literature.

Theorem 2.6

Let \(A\xrightarrow {i} C\xrightarrow {j} B\) be a sequence of extensions of rings. The following properties are equivalent:

1. (1)

   For any field K, the following map is injective

   $$\begin{aligned} {{\,\textrm{Spec}\,}}(j)\circ ({{\,\mathrm{\mathrm Mor}\,}}({{\,\textrm{Spec}\,}}K, {{\,\textrm{Spec}\,}}B))\rightarrow {{\,\mathrm{\mathrm Mor}\,}}({{\,\textrm{Spec}\,}}K, {{\,\textrm{Spec}\,}}A) \end{aligned}$$
   $$\begin{aligned} ({{\,\textrm{Spec}\,}}(j)\circ \alpha )\mapsto {{\,\textrm{Spec}\,}}(i)\circ ({{\,\textrm{Spec}\,}}(j)\circ \alpha ) \end{aligned}$$
2. (2)

   For any field K, if \(\psi _1:B\rightarrow K\) and \(\psi _2:B\rightarrow K\) are two field morphisms distinct by composition with j, then they are distinct by composition with \(j\circ i\).

3. (3)

   The sequence \(A\rightarrow C\rightarrow B\) is radicial.

4. (4)

   \(j(C)\subset \widehat{A}_B\).

5. (5)

   The kernel of the morphism \({{\,\mathrm{C\otimes _{A} C}\,}}\rightarrow C\) defined by \(c_1\otimes _A c_2\mapsto c_1c_2\) is included in the nilradical of \({{\,\mathrm{B\otimes _{A} B}\,}}\).

Proof

The equivalence between (1) and (2) is straightforward. Since \(\ker ({{\,\mathrm{C\otimes _{A} C}\,}}\rightarrow C)\) is generated by the elements of the form \(c\otimes _A1-1\otimes _A c\) for \(c\in C\), then (4) \(\Leftrightarrow \) (5). Note that (4) implies (3) by Proposition 2.4.

Let us prove that (3) implies (2) by contraposition. Let \(\psi _1:B\rightarrow K\) and \(\psi _2:B\rightarrow K\) be two morphisms in a field K such that \(\psi _1\circ j\not =\psi _2\circ j\) and \(\psi _1\circ j\circ i=\psi _2\circ j \circ i\). Let \({{\,\mathrm{\mathfrak {p}}\,}}_1\), \({{\,\mathrm{\mathfrak {p}}\,}}_2\) and \({{\,\mathrm{{\mathfrak q}}\,}}\) denote respectively the kernels of \(\psi _1\), \(\psi _2\) and \(\psi _1\circ j\circ i:A\rightarrow K\). For \(i=1,2\) we get the following commutative diagram:

If \({{\,\mathrm{\mathfrak {p}}\,}}_1\cap C\) is not equal to \({{\,\mathrm{\mathfrak {p}}\,}}_2\cap C\), then \({{\,\textrm{Spec}\,}}C\rightarrow {{\,\textrm{Spec}\,}}A\) is not injective on the image of \({{\,\textrm{Spec}\,}}B\).

If \({{\,\mathrm{\mathfrak {p}}\,}}_1\cap C\) is equal to \({{\,\mathrm{\mathfrak {p}}\,}}_2\cap C\), then \(\psi _1\) and \(\psi _2\) induce two different \(k({{\,\mathrm{{\mathfrak q}}\,}})\)-morphisms \(\psi '_\iota : k({{\,\mathrm{\mathfrak {p}}\,}}_1\cap C) \rightarrow K\) since \(\psi _1\circ j\not =\psi _2\circ j\). Clearly, the extension \(k({{\,\mathrm{{\mathfrak q}}\,}})\rightarrow k({{\,\mathrm{\mathfrak {p}}\,}}_1\cap C)\) cannot be an isomorphism.

In both cases, the extension \(A\rightarrow C\rightarrow B\) is not radicial.

Finally we prove that (2) implies (4) by contraposition. By assumption there are \(c\in C\) and \(\omega \in {{\,\textrm{Spec}\,}}{{\,\mathrm{B\otimes _{A} B}\,}}\) such that \({{\,\mathrm{1\otimes _{A} c - c\otimes _{A} 1}\,}}\notin \omega \). The ideal \(\omega \) comes with a morphism \(g:{{\,\mathrm{B\otimes _{A} B}\,}}\rightarrow K\) with \(\ker g=\omega \). Consider the composition of g with the morphisms \(\varphi _1\) and \(\varphi _2\) defined in (a). By construction \(g\circ \varphi _1:B\rightarrow K\) coincides with \(g\circ \varphi _2:B\rightarrow K\) when composed with \(j\circ i\), but not when composed with j because \(g\circ \varphi _1 (j(c))\ne g\circ \varphi _2(j(c))\). This contradicts (2). \(\square \)

If we focus on the particular case of radicial extensions rather than sequences, we recover [16, Prop. I.3.7.1], with an additional condition using the saturation.

Proposition 2.7

Let \(i:A\rightarrow B\) be an extension of rings and \({{\,\textrm{Spec}\,}}(i):{{\,\textrm{Spec}\,}}B\rightarrow {{\,\textrm{Spec}\,}}A\) be the associated map. The following properties are equivalent:

1. (1)

   For any field K, the map

   $$\begin{aligned}{} & {} {{\,\mathrm{\mathrm Mor}\,}}({{\,\textrm{Spec}\,}}K, {{\,\textrm{Spec}\,}}B)\rightarrow {{\,\mathrm{\mathrm Mor}\,}}({{\,\textrm{Spec}\,}}K, {{\,\textrm{Spec}\,}}A) \\{} & {} \alpha \mapsto {{\,\textrm{Spec}\,}}(i)\circ \alpha \end{aligned}$$

   is injective.

2. (2)

   If \(\psi _1:B\rightarrow K\) and \(\psi _2:B\rightarrow K\) are two distinct morphisms then the compositions \(\psi _1\circ i\) and \(\psi _2\circ i\) are different.

3. (3)

   \(i:A\rightarrow B\) is radicial.

4. (4)

   \(B=\widehat{A}_B\).

5. (5)

   The kernel of the morphism \({{\,\mathrm{B\otimes _{A} B}\,}}\rightarrow B\) defined by \(b_1\otimes _A b_2\mapsto b_1b_2\) is included in the nilradical of \({{\,\mathrm{B\otimes _{A} B}\,}}\).

Proof

Direct consequence of Theorem 2.6, using the fact that an extension \(A\rightarrow B\) is radicial if and only if the sequence of extensions \(A\rightarrow B\rightarrow B\) is so. \(\square \)

2.2 Saturation for varieties

We begin with the definition of radiciality and saturation for morphisms [16, Chap. I,3.7.2]. Then, we extend these definitions to sequences of morphisms.

Definition 2.8

1. (1)

   Let \(\pi :Y\rightarrow X\) be a dominant morphism between algebraic varieties over k. We say that \(\pi \) is radicial if \(\pi \) is injective and equiresidual.

2. (2)

   Let \(Y{\mathop {\rightarrow }\limits ^{\phi }} Z{\mathop {\rightarrow }\limits ^{\psi }} X\) be a sequence of dominant morphisms between algebraic varieties over k. We say that \(Y\rightarrow Z\rightarrow X\) is radicial if \(\psi \) is injective and equiresidual by restriction to the image of \(\phi \).

3. (3)

   Let \(\pi :Y\rightarrow X\) be a dominant morphism between algebraic varieties over k. We say that X is saturated in Y if, for any affine subsets \(U'\subset Y\), \(U\subset X\) with \(\pi (U')\subset U\), the ring \(\mathcal {O}_X(U)\) is saturated in \(\mathcal {O}_Y(U')\).

Remark 2.9

1. (1)

   Let \(\pi :Y\rightarrow X\) be a dominant morphism between algebraic varieties over k. Then \(\pi \) is radicial if, for any affine subsets \(U'\subset Y\), \(U\subset X\) with \(\pi (U')\subset U\), the extension \(\mathcal {O}_X(U)\rightarrow \mathcal {O}_Y(U')\) is radicial.

2. (2)

   Let \(Y{\mathop {\rightarrow }\limits ^{\phi }} Z{\mathop {\rightarrow }\limits ^{\psi }} X\) be a sequence of dominant morphisms between algebraic varieties over k. Then \(Y\rightarrow Z\rightarrow X\) is radicial if, for any affine open subsets \(U'\subset Y\), \(U''\subset Z\), \(U\subset X\) with \(\phi (U')\subset U''\) and \(\psi (U'')\subset U\), the sequence of extensions \(\mathcal {O}_X (U)\rightarrow \mathcal {O}_Z (U'')\rightarrow \mathcal {O}_Y (U')\) is radicial.

In order to translate the universal property of the saturation in terms of varieties, we recall the notion of universal injectivity from [16, Chap. I, 3.4.3, 3.7.1].

Definition 2.10

1. (1)

   A morphism \(\pi :Y\rightarrow X\) between algebraic varieties over k is called universally injective if for any field extension \(k\rightarrow K\), the map \(\pi _K:Y(K)\rightarrow X(K)\) is injective.

2. (2)

   A sequence of morphisms \(Y\rightarrow Z\rightarrow X\) between algebraic varieties over k is called universally injective if for any field extension \(k\rightarrow K\), the map \(Z(K)\rightarrow X(K)\) is injective by restriction to the image of \(Y(K)\rightarrow Z(K)\).

Grothendieck [16, Prop. 3.7.1] proved that the notions of radicial and universally injective morphisms coincide. The universal property given in Theorem 2.6 implies that it is also the case if we consider sequences of morphisms rather than morphisms.

Proposition 2.11

Let \(Y{\mathop {\rightarrow }\limits ^{\phi }} Z{\mathop {\rightarrow }\limits ^{\psi }} X\) be a sequence of dominant morphisms between algebraic varieties over k. The following properties are equivalent:

1. (1)

   \(Y\rightarrow Z\rightarrow X\) is universally injective.

2. (2)

   \(Y\rightarrow Z\rightarrow X\) is radicial.

3. (3)

   For any affine open subsets \(U'\subset Y\), \(U''\subset Z\), \(U\subset X\) with \(\phi (U')\subset U''\) and \(\psi (U'')\subset U\) then \(\mathcal {O}_Z(U'')\subset \widehat{\mathcal {O}_X(U)}_{\mathcal {O}_Y(U')}\).

Let \(\pi :Y\rightarrow X\) be a dominant morphism between affine algebraic varieties over k. Contrarily to the seminormalization case it is not clear whether \(\widehat{k[X]}_{k[Y]}\) is a finitely generated algebra over k and thus is the coordinate ring of an affine algebraic variety.

As a consequence of the previous remark, the statement \(\pi \) is subintegral if and only if \(X^+_Y=Y\) has no equivalent when \(\pi \) is radicial. Nevertheless we get:

Corollary 2.12

Let \(\pi :Y\rightarrow X\) be a dominant morphism between algebraic varieties over k. Then \(\pi \) is radicial if and only if, for any affine subsets \(U'\subset Y\), \(U\subset X\) with \(\pi (U')\subset U\), we have \(\widehat{\mathcal {O}_X(U)}_{\mathcal {O}_Y(U')}=\mathcal {O}_Y(U')\).

In general, the seminormalization is only included in the saturation.

Lemma 3.1

Let \(A\rightarrow B\) be an extension of rings. Then \(A^+_B\subset \widehat{A}_B.\)

Proof

Since \(A\rightarrow A^+_B\) is subintegral then \(A\rightarrow A^+_B\rightarrow B\) is radicial. The inclusion \(A^+_B\subset \widehat{A}_B\) follows by Theorem 2.6. \(\square \)

A priori, there is no special relationship between the saturation and integrality. However saturation and seminormalization coincide when we restrict to integral extensions.

Proposition 3.2

Let \(A\rightarrow B\) be an integral extension of rings. Then \(A^+_B= \widehat{A}_B.\)

Proof

The inclusion \(A^+_B \subset \widehat{A}_B\) comes from Lemma 3.1. The sequence \(A\rightarrow \widehat{A}_B\rightarrow B\) is radicial by Theorem 2.6. Since \( \widehat{A}_B\rightarrow B\) is integral then \({{\,\textrm{Spec}\,}}B\rightarrow {{\,\textrm{Spec}\,}}\widehat{A}_B\) is surjective by Proposition 1.1. It follows that \(A\rightarrow \widehat{A}_B\) is radicial and integral and thus subintegral. This forces \( \widehat{A}_B\) to be equal to the seminormalization \(A^+_B\) of A in B by Theorem 1.6. \(\square \)

We state the relations between saturation and seminormalization for varieties induced by Lemma 3.1 and Proposition 3.2.

Proposition 3.3

Let \(\pi :Y\rightarrow X\) be a dominant morphism between varieties over k.

1. (1)

   If \(\pi \) is subintegral, then \(\pi \) is radicial.

2. (2)

   If X is saturated in Y, then X is seminormal in Y.

3. (3)

   If \(\pi \) is moreover integral, then \(\pi \) is subintegral if and only if \(\pi \) is radicial, and X is saturated in Y if and only if X is seminormal in Y.

We compare the notions of seminormality and relative seminormality.

Proposition 3.4

Let \(\pi :Y\rightarrow X\) be a dominant morphism between varieties over k. If X is seminormal then X is seminormal in Y.

Proof

From Proposition 1.11 the morphism \(\varphi :X_Y^+\rightarrow X\) is subintegral and factorizes \(\pi \). Since \(\varphi \) is birational and finite then it follows that the normalization map \(X'\rightarrow X\) factorizes through \(\varphi \). Then, by subintegrality of \(\varphi \) and by Proposition 1.11, the morphism \(X^+ \rightarrow X\) factorizes through \(\varphi \). Since the seminormality of X implies that \(X^+ \rightarrow X\) is an isomorphism, we get that \(\varphi \) is an isomorphism. \(\square \)

The converse of Proposition 3.4 does not hold (e.g. take \(Y=X\) with X not seminormal).

From [10, Rem. 1.4], we know that the notions of relative seminormalization and relative saturation differ when we do not consider integral extensions of rings and integral morphisms of varieties. We end this section by providing explicit examples to illustrate that the notions of relative saturation and seminormalization do not coincide for varieties over an algebraically closed field, in any dimension. The examples are built on Example 3.5, constructed from a nodal curve, for which we offer two arguments: a simple geometric one, and a direct computational one in order to construct the generalization in any dimension in Example 3.6.

Example 3.5

1. (1)

   Let X be the nodal plane curve with coordinate ring \(A=k[X]=k[x,y]/(y^2-x^2(x+1))\). Its normalization \(X'\) has coordinate ring \(A'=k[X']=k[x,z]/(z^2-(x+1))=A[y/x]\) and the inclusion \(A\rightarrow A'\) is given by \((x,y)\mapsto (x,xz)\). Let Y be defined by removing one of the two points \(p=(0,1)\) and \(q=(0,-1)\) of \(X'(k)\) lying above the singular point of X(k), say p. The coordinate ring of Y is

   $$\begin{aligned} B=k[Y]=k[x,z,s]/(z^2-(x+1),s(z-1)-1)=A'[1/(z-1)]=A'[s]=A[y/x,s], \end{aligned}$$

   and we have a sequence of inclusions \(A\rightarrow A'\rightarrow B\). Then \(A^+_B=A\) whereas \(\widehat{A}_B=B\). To see the first point, since the variety X is seminormal (see [15]) then X is seminormal in Y by Proposition 3.4. For the second point, note that \(A\rightarrow B\) is radicial because, for irreducible curves, the prime ideals correspond to the generic point and the closed points, and here \(Y\rightarrow X\) is birational with \(Y(k)\rightarrow X(k)\) bijective. As a consequence \(\widehat{A}_B=B\) by Proposition 2.7.

2. (2)

   We revisit the nodal curve example proving the equality \(\widehat{A}_B=B\) using the very definition of the saturation. Keeping previous notation, set \(\alpha =(z\otimes _A 1)-(1\otimes _A z)\) and \(\beta =(s\otimes _A 1)-(1\otimes _A s)\). It suffices to prove that \(\alpha \) and \(\beta \) are nilpotent elements of \({{\,\mathrm{B\otimes _{A} B}\,}}\). Indeed \(\widehat{A}_B\) is a ring containing x, z and s so that \(\widehat{A}_B=B\) in that case. Note that, using the equality \(y+x=x(z+1)\) in B, we have

   $$\begin{aligned} x \alpha =\big (x(z+1)\otimes _A 1\big ) - \big (1\otimes _A x(z+1)\big )= \big ((y+x)\otimes _A 1\big ) - \big (1\otimes _A (y+x)\big )=0 \end{aligned}$$

   hence, using the equality \(xs=z+1\) in B,

   $$\begin{aligned} \alpha ^2=\alpha \big ((z+1)\otimes _A 1-1\otimes _A(z+1)\big )=x\alpha (s\otimes _A 1 - 1\otimes _A s)=0. \end{aligned}$$

   Actually we even have \(\alpha =0\) in \({{\,\mathrm{B\otimes _{A} B}\,}}\), since a straightforward computation shows that \(\alpha =\frac{1}{4}\alpha ^3\). Finally, using the equality \(\alpha = (z-1)\otimes _A 1-1\otimes _A(z-1)\) and the relation \(s(z-1)=1\), we observe that \(0=(s\otimes _A 1)\alpha (1\otimes _A s) =-\beta .\)

Example 3.6

Consider the curves X and Y as in Example 3.5. For \(n\ge 1\), the variety \(X\times {{\,\mathrm{\mathbb A}\,}}_k^n\) is seminormal in \(Y\times {{\,\mathrm{\mathbb A}\,}}_k^n\), whereas the saturation of \(X\times {{\,\mathrm{\mathbb A}\,}}_k^n\) in \(Y\times {{\,\mathrm{\mathbb A}\,}}_k^n\) is \(Y\times {{\,\mathrm{\mathbb A}\,}}_k^n\).

To see this, note that X and \({{\,\mathrm{\mathbb A}\,}}_k^n\) are seminormal, so \(X\times {{\,\mathrm{\mathbb A}\,}}_k^n\) is also seminormal [15, Cor. 5.9] and thus \(X\times {{\,\mathrm{\mathbb A}\,}}_k^n\) is seminormal in \(Y\times {{\,\mathrm{\mathbb A}\,}}_k^n\) by Proposition 3.4.

For the saturation, although the radiciality of

is not so straightforward since we no longer work with curves as in Example 3.5 (1), the computations done in Example 3.5 (2) still prove that \(\widehat{A[t_1,\cdots ,t_n]}_{B[t_1,\cdots ,t_n]}\) contains x, z and s, and so is equal to \(B[t_1,\ldots ,t_n]\).

Continuous rational functions and regulous functions have been originally studied in real algebraic geometry [13, 22, 24], where the continuity is regarded with respect to the Euclidean topology, which can be studied algebraically via semialgebraic open sets [7]. For an algebraic variety X over \({{\,\mathrm{\mathbb C}\,}}\), we can consider the Euclidean (or strong) topology of the complex points \(X({{\,\mathrm{\mathbb C}\,}})\) seen as a topological variety. For example, if X is affine, then we have \(X\subset \mathbb A_{{{\,\mathrm{\mathbb C}\,}}}^n\) for some \(n\in {{\,\mathrm{\mathbb N}\,}}\) and the strong topology is induced by the natural inclusion \(X({{\,\mathrm{\mathbb C}\,}})\subset {{\,\mathrm{\mathbb R}\,}}^{2n}\). With this point of view Bernard has developed in [6] the theory of regulous functions for algebraic varieties over \({{\,\mathrm{\mathbb C}\,}}\).

Subintegral extensions and regulous functions are strongly related in real algebraic geometry as developed in [14]. Working with the field of complex numbers, we know from the work of Bernard that the same holds true in the geometric case. The purpose of this section is first to generalize the work of Bernard to varieties over any algebraically closed field k of characteristic zero, and then to relate it to the theory of relative seminormalization. In particular, we prove in this section Theorems A and D stated in the introduction.

4.1 Generalizing the strong topology of \({{\,\mathrm{\mathbb C}\,}}\)

Since there is a priori no natural strong topology on the k-rational points of a variety over k, we use the theory of real closed fields to define such a topology as in [18, 21] (see also [5] for a recent cohomological use of this approach).

From Artin Schreier theory [4], we know the existence of (possibly many) real closed subfields of k with algebraic closure equal to k. For instance for \(k=\mathbb {C}\), let us consider a transcendental basis B of \(\mathbb {R}\) over \(\mathbb {Q}\); the algebraic closure of \(\mathbb {Q}(B)\) is \(\mathbb {C}\). Let \(B'=B{\setminus }\{a\}\) with \(a\in B\); one has an isomorphism \(\mathbb {Q}(B)\simeq \mathbb {Q}(B')\). The ordering on \(\mathbb {R}\) induces an ordering on \(\mathbb {Q}(B')\) which one may arbitrarily extend to an ordering of \(\mathbb {Q}(B')(a)\) (namely of \(\mathbb {Q}(B)\)). Then, the real closure of \(\mathbb {Q}(B)\) with respect to this ordering cannot be isomorphic to \(\mathbb {R}\). Indeed, such an isomorphism should preserve all the rational numbers and also send all the elements of \(B'\) into those of B (since it preserves the inequalities between transcendant and rational numbers), giving no possible image for a. Playing likewise with \(B'\) and so on, one may produce infinitely many real closed fields whose algebraic closure is \(\mathbb {C}\).

Back to the general case, let \(R\subset k\) denote one of these real closed fields. Then \(k=R[\sqrt{-1}]\) and R comes with a unique ordering. The ordering on R gives rise to an order topology on the affine spaces \(R^n\), in a similar way as the Euclidean topology on \({{\,\mathrm{\mathbb R}\,}}^n\), even if the topological space R is not connected (except in the case \(R={{\,\mathrm{\mathbb R}\,}}\)) or the closed interval [0, 1] is in general not compact.

We use this choice of R to define a topology on the closed points of an algebraic variety over k. First, for an algebraic variety X over R, choose an affine covering of X by Zariski open subsets \(U_i\), and endow each affine sets \(U_i(R)\) with the order topology. These open sets glue together to define the order topology on X(R), and this topology does not depend on the choice of the covering. This topological space can be endowed additionally with the structure of a semialgebraic space by considering the sheaf of continuous semialgebraic functions [11, 12], or even of a real algebraic variety with the sheaf of regular functions on the R-points [7, 19].

Consider now the case of a quasi-projective algebraic variety X over k. By Weil restriction [33], we associate to X an algebraic variety \(X_R\) over R whose R-points are in bijection with the k-points of X. We endow X(k) with the topology induced by the order topology on \(X_R(R)\), and we call it the R-topology on X(k).

If X is no longer quasi-projective, then the Weil restriction does not necessarily exist. Anyway choose an affine open covering \((U_i)_{i\in I}\) of X, endow the R-points of the Weil restrictions \((U_i)_R\) with the order topology, and note that these open sets glue together to define a topology on X(k). This topology does not depend on the choice of the covering by [29, Lemma 5.6.1], and we call it the R-topology on X(k). Again one can consider X(k) as a semialgebraic space in the sense of [12] or as a real algebraic variety in the sense of [7].

The R-topology on X(k) has many good properties, for instance X(k) is semialgebraically connected and of pure dimension twice the dimension of X if X is irreducible [21]. For \(k={{\,\mathrm{\mathbb C}\,}}\) and \(R={{\,\mathrm{\mathbb R}\,}}\), the \({{\,\mathrm{\mathbb R}\,}}\)-topology is nothing more than the strong topology. The choice of a different real closed field R in k will lead to different topologies on X(k) (for instance the semialgebraic fundamental group does depend on the choice of R [21]). Already with \(k={{\,\mathrm{\mathbb C}\,}}\), one can choose a real closed field different from \({{\,\mathrm{\mathbb R}\,}}\), even for instance a non-Archimedean \(R\subset {{\,\mathrm{\mathbb C}\,}}\). We will see however that in our setting, the choice of the real closed field doesn’t matter.

4.1.1 Basics on the R-topology of k-varieties

In this section we fix a real closed field R with algebraic closure k.

Let X be a quasi-projective algebraic variety over k. Recall that by Weil restriction [29] :

1. (1)

   The variety \(X_R\) is nonsingular if X is nonsingular. More precisely, a k-point in X is singular if and only if its corresponding R-point in \(X_R\) is singular.

2. (2)

   A Zariski open subset \(U\subset X\) induces a Zariski open subset \(U_R\subset X_R\).

3. (3)

   A proper morphism \(Y\rightarrow X\) between quasi-projective algebraic varieties over k induces a proper morphism \(Y_R\rightarrow X_R\).

4. (4)

   A finite morphism \(Y\rightarrow X\) between quasi-projective algebraic varieties over k induces a finite morphism \(Y_R\rightarrow X_R\).

Let X be an affine algebraic variety over k. A regular function on X gives rise to a polynomial (and thus continuous) mapping \(X_R(R)\rightarrow R^2\). Indeed the regular function is polynomial, and by Weil restriction a polynomial function to k induces a polynomial mapping to \(R^2\) by taking the real and imaginary parts. Finally a polynomial function is continuous with respect to the R-topology.

The topological properties of R-varieties coming from k-varieties are much more moderate than for general R-varieties. For instance, although complex irreducible varieties are locally of equal dimension, irreducible algebraic subsets of \({{\,\mathrm{\mathbb R}\,}}^n\) may have isolated points. From [7], a real algebraic variety is called central if its subset of nonsingular points is dense with respect to the R-topology. For a subset \(A\subset X(k)\), we denote by \(\overline{A}^R\) the closure of A with respect to the R-topology. We denote by \({{\,\textrm{Reg}\,}}(X(k))\) the set of nonsingular points of X(k).

Proposition 4.1

Let X be an irreducible algebraic variety over k. Then X(k) is central:

Proof

The question being local, it suffices to assume X is affine, and in particular the Weil restriction of X exists. If X is nonsingular, then so is \(X_R\) by Weil restriction, and so \(X_R(R)\) is central. Otherwise, consider a resolution \(\sigma :\tilde{X} \rightarrow X\) of the singularities of X. Then \(\sigma _k\) is surjective since k is algebraically closed, and we have that \(\sigma _k\) induces a bijection

Let \(x\in X(k)\), and choose a preimage \(\tilde{x}\in \sigma _k^{-1}(x)\). The centrality of \(\tilde{X}(k)\) implies the existence of a continuous semialgebraic curve \(\tilde{\gamma }:[0,1]\rightarrow \tilde{X}(k)\) with \(\tilde{\gamma }(0)=\tilde{x}\) and \(\tilde{\gamma }(t)\in \tilde{U}\) for \(t\in (0,1]\subset R\) by the Curve Selection Lemma [7, Theorem 2.5.5]. Its composition \(\gamma =\sigma _k \circ \tilde{\gamma }\) is a continuous semialgebraic curve from [0, 1] to X(k) with \(\gamma (0)=x\) and \(\gamma (t)\in U\) for \(t\in (0,1]\subset R\). As a consequence x belongs to the closure with respect to the R-topology of \({{\,\textrm{Reg}\,}}(X(k))\) in X(k), and so X(k) is central. \(\square \)

In particular, if the irreducible algebraic variety X over k has dimension d, then the local semialgebraic dimension of X(k) at any point \(x\in X(k)\) is equal to 2d.

The following result is not valid in general for algebraic varieties over R, and even for \(R={{\,\mathrm{\mathbb R}\,}}\).

Lemma 4.2

Let X be an irreducible algebraic variety over k. A non-empty Zariski open subset of X(k) is dense with respect to the R-topology.

Proof

The question being local, it suffices to assume X is affine, and in particular the Weil restriction of X exists. A Zariski open subset remains Zariski open by Weil restriction. Combined with Proposition 4.1, it suffices to check that a non-empty Zariski open set U in a central irreducible algebraic variety over R is dense with respect to the R-topology. This last property is classical; for instance, the complement is an algebraic subset of strictly smaller dimension, and a semialgebraic triangulation of X(k) adapted to the complement shows that locally, a point in the complement is in the boundary of a semialgebraic simplex in U(k). \(\square \)

Over a general real closed field, the notion of compact sets is advantageously replaced by closed and bounded semialgebraic sets. For instance, the image of a closed and bounded semialgebraic set by a continuous semialgebraic map is again closed and bounded (and semialgebraic) [7, Theorem 2.5.8]. A semialgebraic map is called proper with respect to the R-topology if the preimage of a closed and bounded semialgebraic set is closed and bounded.

Lemma 4.3

Let \(\sigma :\tilde{X}\rightarrow X\) be a proper morphism between irreducible varieties over k. Then \(\sigma _k\) is proper with respect to the R-topology. If \(\sigma \) is moreover birational, then \(\sigma _k\) is surjective.

Proof

The notion of properness is local on the target, so that there is an affine covering of X such that for any open affine subset U in the covering, the restriction \(\sigma '=\sigma _{|\sigma ^{-1}(U)}\) of \(\sigma \) to \(\sigma ^{-1}(U)\) is proper. The properness of \(\sigma '\) is kept by Weil restriction, so that \(\sigma '_k\) is proper with respect to the R-topology by [11, Theorem 9.6]. Finally \(\sigma _k\) is proper with respect to the R-topology since that notion of properness is local on the target too by [12, Proposition 5.7].

If \(\sigma \) is birational, there are Zariski open sets \(\tilde{U}\subset \tilde{X}\) and \(U\subset X\) such that \(\sigma _{|\tilde{U}}\) is a bijection onto U. Then

the right hand side equality coming from Lemma 4.2. Finally \(\sigma _k(\tilde{X}(k))\) is closed for the R-topology by properness of \(\sigma _k\), so that taking the closure of U(k) with respect to the R-topology gives the result by Lemma 4.2. \(\square \)

Lemma 4.4

Let \(\sigma :\tilde{X}\rightarrow X\) be a finite morphism between algebraic varieties over k. Then \(\sigma _k\) is closed with respect to the R-topology.

Proof

By [11, Theorem 4.2], a finite morphism between algebraic varieties over R is closed with respect to the R-topology. The result follows since finiteness is local and Weil restriction preserves finite morphisms. Alternatively when X and \(\tilde{X}\) are irreducible, a finite morphism is proper, and apply Lemma 4.3. \(\square \)

4.2 Regulous functions on k-rational points

Fix a real closed field R with \(R[\sqrt{-1}]=k\).

4.2.1 Characterization of continuous rational functions

The choice of R induces a topology, hence a notion of continuity. We define continuous rational functions on an algebraic variety over k as follows.

Definition 4.5

Let X be an algebraic variety over k. Let \(U\subset X\) be an open subset of X. A continuous rational function on U(k) is a function from U(k) to k which is the continuous extension to U(k) of a rational function on X, when X is endowed with the R-topology.

This notion comes initially from real algebraic variety [13, 22, 24], and has been studied also in complex geometry [6]. In the setting of Definition 4.5, it depends a priori on the choice of R. A rational function that is continuous with respect to the R-topology can be characterized by the fact that it becomes regular after applying a relevant proper birational map.

Proposition 4.6

Let X be an algebraic variety over k. Let \(f:X(k)\rightarrow k\) be an everywhere defined function, and assume that f coincides with a regular function on a non-empty Zariski open subset of X(k). Then, f is continuous with respect to the R-topology if and only if there is a proper birational map \(\sigma :\tilde{X}\rightarrow X\) such that \(f\circ \sigma _k:\tilde{X}(k)\rightarrow k\) is regular.

Proof

Arguing similarly to [6, Lem. 4.4], we may assume X is irreducible. Assume f to be continuous, and denote by g the rational function on X that coincides with f on a non-empty Zariski open subset of X(k). One can resolve the indeterminacy of the rational map g by a sequence of blowings-up along nonsingular centers, giving rise to a proper birational morphism \(\sigma :\tilde{X}\rightarrow X\) such that \(g\circ \sigma _k:\tilde{X}(k)\rightarrow {{\,\mathrm{\mathbb P}\,}}^1(k)\) is regular. The functions \(f\circ \sigma _k\) and \(g\circ \sigma _k\) are equal on a Zariski dense subset of \(\tilde{X}(k)\), so they are equal on a subset dense with respect to the R-topology by Lemma 4.2. Therefore they coincide on \(\tilde{X}(k)\) by continuity. As a consequence the regular function \(g\circ \sigma _k\) takes its values in k rather than in \({{\,\mathrm{\mathbb P}\,}}^1(k)\).

Conversely, let \(C\subset k\) be a closed subset with respect to the R-topology. The set \((f\circ \sigma _k)^{-1}(C)\) is closed by continuity of \(f\circ \sigma _k\), and its image under \(\sigma _k\) is equal to \(f^{-1}(C)\) by surjectivity of \(\sigma _k\) via Lemma 4.3. As a consequence \(f^{-1}(C)\) is closed by properness of \(\sigma _k\) with respect to the R-topology, thanks to Lemma 4.3 again. \(\square \)

Note that the characterization of continuity given above, via a resolution of indeterminacy, does not refer to the choice of R. In particular, the continuity with respect to the R-topology of a rational function is independent of the choice of the real closed field R.

Let \(U\subset X\) be an open subset of X. The continuous rational functions on U(k) are the sections of a presheaf of k-algebras on X(k), denoted by \(\mathcal K_{X(k)}^0\) in the sequel. Since the presheaf of rational functions \(\mathcal K\) is a sheaf, and the presheaf of locally continuous functions on X(k) for the R-topology is also a sheaf for the Zariski topology, then \(\mathcal K_{X(k)}^0\) is a sheaf called the sheaf of continuous rational functions. It makes \((X(k),\mathcal K_{X(k)}^0)\) a ringed space. In case X is affine, we simply denote by \(\mathcal {K}^0(X(k))\) the global sections of \(\mathcal K_{X(k)}^0\) on X(k). A dominant morphism \(\pi :Y\rightarrow X\) between varieties over k induces an extension \(\mathcal K_{X(k)}^0\rightarrow (\pi _{k})_*\mathcal K_{Y(k)}^0\), hence a morphism \((Y(k),\mathcal K_{Y(k)}^0)\rightarrow (X(k),\mathcal K_{X(k)}^0)\) of ringed spaces.

An important fact is that continuous rational functions on a normal variety are regular. More precisely, next proposition says that continuous rational functions are integral on the regular ones and thus are already regular if the variety is normal. The set of indeterminacy points of a continuous rational function is related to the normal locus of the ambient variety.

Proposition 4.7

Let X be an algebraic variety over k. We have:

1. (1)

   \(\mathcal K_{X(k)}^0\subset (\pi _{k}^\prime )_* \mathcal {O}_{X'(k)}\) where \(\pi ':X'\rightarrow X\) is the normalization map.

2. (2)

   If x is a normal closed point of X(k) then \(\mathcal K_{X(k),x}^0=\mathcal {O}_{X,x}\).

Proof

The properties are local so we may assume X is affine. The original proof in [6, Proposition 4.7] uses only one argument related to the complex setting. It is the density with respect to the strong topology of a Zariski dense open set, that can be replaced by Lemma 4.2. Note that Hartogs Lemma used in the proof is valid over k: if X is normal then the restriction map \(\mathcal {O}(X(k))\rightarrow \mathcal {O}({{\,\textrm{Reg}\,}}(X(k)))\) is surjective since by [20, p. 124]:

\(\square \)

In the remaining of this section, we show how seminormalization and continuous rational functions are in relation for varieties over k. We begin with a description of regular functions on the relative seminormalization in terms of regular functions on the relative normalization.

Proposition 4.8

Let \(Y\rightarrow X\) be a dominant morphism between algebraic varieties over k. Let U be an open subset of X. Then

where \(\pi ':X'_Y\rightarrow X\) (resp. \(\pi ^+:X^+_Y\rightarrow Y\)) is the relative normalization (resp. seminormalization) morphism.

Proof

By [6, Cor. 3.7] (which is written in the case U is affine, but all of Section 3 there is valid for any open subset U), we have

The radical being an intersection of maximal ideals, we see that the functions in \(\mathcal {O}_{X^{+}_Y}((\pi ^+)^{-1}(U))\) correspond to the elements of \( \mathcal {O}_{X'_Y}((\pi ')^{-1}(U))\) constant on the fibers of \(\pi '_{k}\). \(\square \)

We give a characterization of the structural sheaf of the seminormalization of an algebraic variety over k in another one with continuous rational functions generalizing the main result in [6] to the relative seminormalization and over any algebraically closed field of characteristic zero. In order to state the result, we use the fiber product of two sheaf extensions.

Theorem 4.9

Let \(\pi :Y\rightarrow X\) be a dominant morphism between algebraic varieties over k. Then

where \(\pi ^+:X^+_Y\rightarrow X\) is the relative seminormalization morphism.

Proof

We may assume X and Y are affine and thus we want to prove that

where the right hand side stands for the fiber product of the rings. Let \(\pi ':X_Y^\prime \rightarrow X\) be the relative normalization map.

We consider the following diagram

where the horizontal maps from the top (resp. the bottom) are given by composition with respectively \(\pi '\) and \(Y\rightarrow X_Y^\prime \) (resp. \(\pi _{k}^\prime \) and \(Y(k)\rightarrow X_Y^\prime (k)\)).

We have \(k[X^{+}_Y]\subset k[Y]\) by definition. Since \(\pi ^+\) is subintegral then, proceeding similarly to Bernard’s proof of [6, Thm. 3.1], we show that \(\pi ^+_k\) is a homeomorphism with respect to the R-topology. Since \(\pi ^+\) is in addition birational, the composition by \(\pi _k^+\) gives an isomorphism between \(\mathcal K^0(X(k))\) and \(\mathcal K^0(X_Y^+(k))\). Therefore we get \(k[X^{+}_Y]\subset \mathcal K^0(X_Y^+(k))=\mathcal K^0(X(k))\). In particular \(k[X^{+}_Y]\subset {{\,\mathrm{\mathcal {K}^0}\,}}( X(k))\times _{{{\,\mathrm{\mathcal {K}^0}\,}}(Y(k))}k[Y]\).

Let us prove the converse inclusion. Let \(f\in {{\,\mathrm{\mathcal {K}^0}\,}}( X(k)))\times _{{{\,\mathrm{\mathcal {K}^0}\,}}(Y(k))}k[Y]\). The continuous rational function f is integral over k[X] by Proposition 4.7, therefore \(f\in k[X'_Y]\) since additionally \(f\in k[Y]\). As a function on \(X'_Y(k)\), the function f is constant on the fibers of \(X'_Y(k)\rightarrow X(k)\) since f induces a continuous function on X(k). By Proposition 4.8, we obtain \(f\in k[X^+_Y]\). It gives the reverse inclusion \({{\,\mathrm{\mathcal {K}^0}\,}}( X(k))\times _{{{\,\mathrm{\mathcal {K}^0}\,}}(Y(k))}k[Y]\subset k[X^+_Y]\). \(\square \)

A continuous rational function on a normal algebraic variety over k is a regular function by Proposition 4.7. The fact that the normalization of X in Y is not necessarily normal imposes to take the fiber product with \(\mathcal {O}_{Y(k)}\) in Theorem 4.9. We state as a theorem the particular case \(Y=X'\), the statement becoming much simpler by Proposition 4.7. It says that the ring of continuous rational functions is isomorphic to the ring of regular functions on the seminormalization. This generalizes the main result in [6] over any algebraically closed field of characteristic zero.

Theorem 4.10

Let X be an algebraic variety over k. The ringed space \((X^+(k),\mathcal {O}_{X^{+}(k)})\) is isomorphic to \((X(k),\mathcal K_{X(k)}^0)\).

Since \(X^+\) doesn’t depend of the choice of the real closed field, Theorem 4.10 gives another way to check that the continuity property of a rational function does not depend on the chosen real closed field.

4.2.2 Regulous functions, subintegrality and homeomorphisms

A regulous function f is a continuous rational function that satisfies the additional property that f remains rational by restriction to any subvariety. The first author proved that it is always the case [6, Prop. 4.14] for complex varieties, contrarily to the real case [22].

The following result asserts that continuous rational functions are always regulous.

Corollary 4.11

Let X be an algebraic variety over k and let \(f\in \mathcal K^0(X(k))\). For any Zariski closed subset V of X, the restriction \(f_{\mid V(k)}\) belongs to \(\mathcal K^0(V(k))\).

Proof

The proof of [6, Proposition 4.14] works verbatim using Theorem 4.10. \(\square \)

Definition 4.12

Let X and Y be algebraic varieties over k. Let \(E\subset X(k)\) and \(F\subset Y(k)\) be two closed subsets for the R-topology. We say that a map \(h:E\rightarrow F\) is biregulous if it is a homeomorphism for the R-topology which is birational i.e there exist two dense Zariski open subsets \(U_1\subset \overline{E}\), \(U_2\subset \overline{F}\), a birational map \(g:\overline{E}\rightarrow \overline{F}\) with \(g_{\mid U_1}:U_1\rightarrow U_2\) an isomorphism such that \(g=h\) by restriction to \(U_1\cap E\). In this situation, we say that E and F are biregulous or biregulously equivalent.

In this situation and if in addition \(E=Y(k)\), \(F=X(k)\) and h is a homeomorphism for the Zariski topology, so that the morphism \(\mathcal K_{X(k)}^0\rightarrow (h)_*\mathcal K_{Y(k)}^0\) is well-defined, then this morphism is an isomorphism and thus the ringed spaces \((Y(k),\mathcal K_{Y(k)}^0)\) and \((X(k),\mathcal K_{X(k)}^0)\) are isomorphic.

We are now in position to generalize the characterization of subintegrality via homeomorphisms, as in [6, Thm. 3.1], over any algebraically closed field of characteristic 0. We also add a radicial property in the equivalences. It explains how subintegral extensions, continuous rational functions and biregulous morphisms are related. It gives an extended version of Theorem A stated in the introduction.

Theorem 4.13

Let \(\pi :Y\rightarrow X\) be a finite morphism between algebraic varieties over k. The following properties are equivalent:

1. (1)

   \(\pi \) is subintegral.

2. (2)

   \(\pi _{k}\) is bijective.

3. (3)

   \(\pi _{k}\) is biregulous.

4. (4)

   The ringed spaces \((Y(k),\mathcal K_{Y(k)}^0)\) and \((X(k),\mathcal K_{X(k)}^0)\) are isomorphic.

5. (5)

   \(\pi _{k}\) is a homeomorphism for the R-topology.

6. (6)

   \(\pi _{k}\) is a homeomorphism for the Zariski topology.

7. (7)

   \(\pi \) is a homeomorphism.

8. (8)

   \(\pi \) is radicial.

Proof

The equivalence between (6) and (7) is given by the Nullstellensatz. Using the Nullstellensatz, by Proposition 1.1 and proceeding similarly to Bernard’s proof of [6, Thm. 3.1], we get the equivalence between (1), (2) and (7). It is clear that (5) implies (2). The proof that (2) implies (5) is a direct consequence of Lemma 4.4. Indeed assuming (2), the map \(\pi _k\) admits an inverse, and this inverse is continuous with respect to the R-topology since \(\pi _k\) is a closed map by Lemma 4.4. The equivalence between (1) and (8) is given by Proposition 3.3.

Assuming \(\pi \) to be subintegral, it follows from Proposition 1.11 that X and Y have the same seminormalization. So \(\mathcal K_{X(k)}^0\rightarrow (\pi _k)_*\mathcal K_{Y(k)}^0\) is an isomorphism by Theorem 4.10. It shows (1) implies (4).

We show (4) implies (2) by contradiction. Assume \(\pi _k\) is not bijective. We can suppose X and Y are affine. We may separate two different points \(y,y'\) in the fibre \(\pi ^{-1}_k(x)\) of some \(x\in X(k)\) by a regular function f on Y(k). But such a function is continuous with respect to the R-topology, and does not belong to the image of \(\mathcal K^0(X(k))\rightarrow \mathcal K^0(Y(k))\) (given by the composition by \(\pi _k\)) since it is not constant on the fibres of \(\pi _k\).

A biregulous map is bijective so (3) implies (2), and conversely by (4) we know that the inverse of \(\pi _k\) is continuous rational, so regulous by Corollary 4.11.

\(\square \)

Remark also that a morphism satisfying the conditions of Theorem 4.13 is (bijective thus) automatically birational.

We prove now that the seminormalization determines a variety up to biregulous equivalence. This result goes in the direction of the problems considered by Kollár & al. [9, 23]. Before that, we need to prove that a biregulous map between algebraic varieties over k is a homeomorphism for the Zariski topology.

Proposition 4.14

Let X and Y be algebraic varieties over k and let \(h:Y(k)\rightarrow X(k)\) be a biregulous map. Then h is a homeomorphism for the Zariski topology.

Proof

Let Z be an irreducible closed subset of Y(k). By restriction we get a biregulous map \(Z\rightarrow h(Z)\). We claim that h(Z) is a closed subset of X(k). Since h is biregulous then h(Z) is closed in X(k) for the R-topology. To prove the claim, it is sufficient to assume that \(\overline{h(Z)}=X(k)\), X is irreducible and thus we have to prove that \(h:Z\rightarrow X(k)\) is surjective. However h is biregulous hence biregular between Zariski open subsets of Z and X(k). In particular the image h(Z) contains a Zariski open subset U of X(k). We have \(\overline{U}^R\subset \overline{h(Z)}^R=h(Z)\) and thus \(h:Z\rightarrow X(k)\) is surjective by Lemma 4.2.

It shows that \(h^{-1}\) is continuous for the Zariski topology. Similarly, h is continuous for the Zariski topology. \(\square \)

The following theorem implies Theorem D stated in the introduction.

Theorem 4.15

Let X and Y be algebraic varieties over k. Then, X(k) and Y(k) are biregulously equivalent if and only if \(X^+\) and \(Y^+\) are isomorphic.

Proof

Assume \(h:Y(k)\rightarrow X(k)\) is a biregulous map. From Proposition 4.14h is a homeomorphism for the Zariski topology and thus the ringed spaces \((Y(k),\mathcal K_{Y(k)}^0)\) and \((X(k),\mathcal K_{X(k)}^0)\) are isomorphic. By Theorem 4.10 it follows that \(X^+\) and \(Y^+\) are isomorphic.

Since \(X^+(k)\rightarrow X(k)\) and \(Y^+(k)\rightarrow Y(k)\) are biregulous morphisms then we easily get the converse implication. \(\square \)

Note that in the statement of the previous theorem, we do not require that there is a morphism from X to Y nor from Y to X. Consider a smooth irreducible curve Y over k without non-trivial automorphisms (e.g \({{\,\mathrm{\mathbb P}\,}}^1_{k}\) minus sufficiently many points). Let \(P_1,P_2\) be two distincts points of Y. Let \(X_1\) (resp. \(X_2\)) be the curve obtained from Y by creating a cusp at \(P_1\) (resp. \(P_2\)) i.e associated to the module \(2P_1\) (resp. \(2P_2\)). We have \(Y=X_1^+=X_2^+\) and thus \(X_1(k)\) is biregulously equivalent to \(X_2(k)\). However, there is no morphism from \(X_1\) to \(X_2\) (nor from \(X_2\) to \(X_1\)) because otherwise it would lift to a finite and birational morphism \(Y\rightarrow Y\) sending \(P_1\) to \(P_2\) (see [14, Ex. 5.5]) and since Y is normal then this morphism is an isomorphism, a contradiction.

Recall that k is an algebraically closed field of characteristic zero. We fix a real closed field R such that \(R[\sqrt{-1}]=k\). In this section we prove Theorems B and C stated in the introduction.

Given a morphism \(\pi :Y\rightarrow X\) between algebraic varieties over k, we are looking for algebraic conditions on \(\pi \) which are respectively equivalent to the topological property that \(\pi _{k}\) is a homeomorphism for the R-topology and \(\pi \) is a homeomorphism for the Zariski topology.

In case \(\pi \) is finite then we already know the answer. Indeed, from Theorem 4.13 the two topological properties stated above are equivalent to each other, and moreover equivalent to the two algebraic conditions that \(\pi \) is subintegral or \(\pi \) is radicial.

In the sequel, we compare the four properties:

1. (1)

   \(\pi _{k}\) is a homeomorphism for the R-topology.

2. (2)

   \(\pi \) is a homeomorphism.

3. (3)

   \(\pi \) is subintegral.

4. (4)

   \(\pi \) is radicial

when we do not assume the finiteness of \(\pi :Y\rightarrow X\).

After some generalities on the relation between homeomorphism with respect to Zariski topology, isomorphism and normality, we provide a complete solution to this problem for R-homeomorphisms. As consequences, we give a partial answer to the problem for Zariski homeomorphisms and we provide statements that explain exactly when R-homeomorphisms and Zariski homeomorphisms are isomorphisms similarly to Vitulli’s result [32, Thm. 2.4].

5.1 Bijection, birationality, homeomorphism

We aim to compare the notions of bijection, birational morphism, homeomorphism with respect to Zariski topology at the spectrum level and homeomorphism with respect to Zariski topology at the level of closed points.

To begin with, recall that it follows from the Nullstellensatz that the property for a morphism to be a homeomorphism with respect to the Zariski topology is already decided at the level of closed points.

It is clear that a homeomorphism induces a bijection at the level of k-rational points, the converse being false as illustrated by Example 5.2 below. Restricting our attention to curves, note that the converse holds true in the case of a morphism between irreducible algebraic curves. The irreducibility of the source space is crucial here, consider for instance the disjoint union of a point with a line minus a point, sent to a line. However even for morphisms between irreducible curves, a birational homeomorphism need not be an isomorphism as illustrated by the normalization of the cuspidal curve with equation \(y^2=x^3\).

A contribution to these questions is the classical fact that the bijectivity at the level of closed points induces the birationality for irreducible varieties, by Zariski Main Theorem.

Proposition 5.1

Let X and Y be irreducible varieties over k. A morphism from Y to X inducing a bijection at the level of k-rational points is quasi-finite and birational. If in addition X is normal, it is an isomorphism.

The following example shows that a morphism which gives a bijection at the level of k-rational points needs not be a homeomorphism.

Example 5.2

1. (1)

   Consider the varieties of Example 3.6, for which we have an open immersion \(\psi \) and a finite morphism \(\phi \) as follows:

   $$\begin{aligned} \pi : Y\times {{\,\mathrm{\mathbb A}\,}}_k^n \xrightarrow {\psi } X'\times {{\,\mathrm{\mathbb A}\,}}_k^n \xrightarrow {\phi } X\times {{\,\mathrm{\mathbb A}\,}}_k^n. \end{aligned}$$

   Even if \(\pi _k\) is still bijective, the morphism \(\pi \) is no longer a homeomorphism when \(n>0\). To see it, it suffices to consider the case \(n=1\). Denote by \(O\in (X\times {{\,\mathrm{\mathbb A}\,}}_k^1)(k)\) the origin, and by \(P=(0,1,0)\) and \(Q=(0,-1,0)\) the two points in the fiber \(\phi ^{-1}(O)\), where the coordinates are \((x,z,t_1)\) in the notation of Example 3.6. Let C be the curve in \(X'\times {{\,\mathrm{\mathbb A}\,}}_k^1\) given by intersection with the plane \(x-z+t_1+1=0\) in \({{\,\mathrm{\mathbb A}\,}}_k^2\times {{\,\mathrm{\mathbb A}\,}}_k^1\). Note that \(P\in C\), \(Q\notin C\), \(C\setminus \{P\}\) is not a closed subset of \(X'\times {{\,\mathrm{\mathbb A}\,}}_k^1\) but it is a closed subset of \(Y\times {{\,\mathrm{\mathbb A}\,}}_k^1\) since \(Y=X'\setminus \{P\}\). If \(\pi \) was a homeomorphism then also \(\pi _k\) and thus \(\pi _{k}(C(k)\setminus \{P\})\) should be Zariski closed in \((X\times {{\,\mathrm{\mathbb A}\,}}_k^1)(k)\). However O is in the closure of \(\pi _{k}(C(k)\setminus \{P\})\).

2. (2)

   This example is also interesting to consider relatively to Grothendieck’s notion of universal homeomorphism [17, Defn. 3.8.1]. Recall that a morphism \(Y\rightarrow X\) is a universal homeomorphism if \(Y\times _X Z\rightarrow Z\) is a homeomorphism for any morphism \(Z\rightarrow X\). The morphism \(Y\rightarrow X\) of Example 3.5 is a homeomorphism but not a universal homeomorphism. Indeed, let \(Z=X\times {{\,\mathrm{\mathbb A}\,}}_k^1\) and consider the base change \(Z\rightarrow X\) given by the first projection. We have already checked that \(Y\times _X Z=Y\times {{\,\mathrm{\mathbb A}\,}}_k^1\rightarrow Z=X\times {{\,\mathrm{\mathbb A}\,}}_k^1\) is not closed.

5.2 Main results

We focus now on the four properties (1), (2), (3) and (4) stated at the beginning of the section. By Theorem 4.13, these four properties are equivalent when \(\pi \) is finite. Considering the open immersion \({{\,\mathrm{\mathbb A}\,}}_{k}^1\rightarrow {{\,\mathrm{\mathbb P}\,}}_{k}^1\), it is radicial but not bijective. Since finite morphisms are surjective by Proposition 1.1 then if we do not assume \(\pi \) to be finite then we have to replace the last property above by:

(4) \(\pi \) is radicial and surjective.

The equivalence between the four above properties is no longer true without the finiteness hypothesis. In particular, a homeomorphism with respect to the Zariski topology need not be a homeomorphism with respect to the R-topology, even for irreducible affine curves.

Example 5.3

1. (1)

   Consider the morphism \(\pi :Y\rightarrow X\) from Example 3.5. We have already shown that \(\pi \) is a homeomorphism, \(\pi \) is radicial but \(\pi \) is not subintegral since it is not an integral morphism. The morphism \(\pi _k\) is not a homeomorphism with respect to the R-topology. Indeed, consider a small open ball B of Y(k) containing the point that is sent to the singular point of X(k) by \(\pi _k\). The image of \(Y(k)\setminus B\) is not closed.

2. (2)

   Let \(n>0\) and consider the morphism \(\pi _n: Y\times {{\,\mathrm{\mathbb A}\,}}_k^n\rightarrow X\times {{\,\mathrm{\mathbb A}\,}}_k^n\) from Example 3.6. We have already proved that \(\pi \) is radicial and surjective, \( \pi \) is not subintegral nor a homeomorphism. Note that \((\pi _n)_k\) is not a homeomorphism with respect to the R-topology either, following the same proof as in (1).

In view of the foregoing explanation, the only equivalence which remains possible to obtain is between the properties for a morphism to be subintegral and a homeomorphism for the R-topology. It will be the subject of our main result.

The following two results measure the rigidity of the R-topology.

Proposition 5.4

Let \(\pi :Y\rightarrow X\) be a morphism between algebraic varieties over k. If \(\pi _{k}\) is a homeomorphism with respect to the R-topology, then \(\pi \) is finite.

Proof

It is sufficient to assume that Y and X are irreducible. By Proposition 5.1, \(\pi \) is quasi-finite. By Grothendieck’s form of Zariski Main Theorem, \(\pi \) factorizes into an open immersion \(g:Y\rightarrow Z\) and a finite morphism \(h:Z\rightarrow X\). We consider Y(k) embedded as an open subset of Z(k) for the R-topology. We also assume Y to be Zariski dense in Z, and thus Y(k) is dense in Z(k) for the R-topology by Lemma 4.2.

Since \(\pi _{k}\) is bijective, the finite morphism h is birational by Proposition 5.1. Moreover \(h_{k}\) is surjective by Proposition 1.1. Let us prove that \(h_{k}\) is also injective. If not, there exist \(y\in Y(k)\) and \(z\in Z(k){\setminus } Y(k)\) with \(h_{k}(y)=h_{k}(z)\). Denote this point by \(x\in X(k)\). Let \(V_y\) be a closed neighborhood of y in Y(k) for the R-topology, and \(V_z\) a closed neighborhood of z in Z(k) for the R-topology disjoint from \(V_y\). Then \(V_x=h_{k}(V_y)\) is a closed neighborhood of x in X(k) for the R-topology by assumption on \(\pi _{k}\).

By the Curve Selection Lemma [7, Thm. 2.5.5], there is a continuous semialgebraic curve \(\gamma :[0,1)\rightarrow Z(k)\) with \(\gamma (0)=z\) and \(\gamma (0,1)\subset V_z\cap Y(k)\). Then \(h_{k}\circ \gamma :[0,1)\rightarrow X(k)\) is a continuous semialgebraic curve with \( h_{k} \circ \gamma (0)=x\), so it meets \(V_x\setminus \{x\}=h_{k}(V_y\setminus \{y\})\). As a consequence \(V_y\) and \(V_z\) cannot be disjoint because \(\pi _{k}\) is bijective.

Therefore \(h_{k}\) is bijective and thus \(g_{k}\) is also bijective. The Nullstellensatz forces g to be a bijective open immersion, thus an isomorphism. As a consequence \(\pi \) is finite like h. \(\square \)

Corollary 5.5

Let \(\pi :Y\rightarrow X\) be a morphism between algebraic varieties over k. If \(\pi _{k}\) is a homeomorphism with respect to the R-topology, then \(\pi \) is a homeomorphism.

Proof

The finiteness follows from Proposition 5.4. Being a bijection on the closed points, it is a homeomorphism by Theorem 4.13. \(\square \)

Corollary 5.5 admits a converse, for varieties of dimension at least two.

Proposition 5.6

Let \(\pi :Y\rightarrow X\) be a morphism between irreducible algebraic varieties over k of dimension at least two. If \(\pi \) is a homeomorphism, then \(\pi _{k}\) is a homeomorphism with respect to the R-topology.

Proof

By [32, Theorem 2.2], the morphism \(\pi \) is finite (it is also birational by Proposition 5.1), so we conclude using Theorem 4.13. \(\square \)

Proposition 5.6 is however not true for curves as illustrated by Example 5.3 (1).

We are now able to get the main result of the paper: the first statement of Theorem B stated in the introduction. We prove that, for a given morphism between algebraic varieties over k, the topological property to be a homeomorphism for the R-topology does not depend of the chosen real closed field since it is equivalent to the algebraic property to be subintegral.

Theorem 5.7

Let \(\pi :Y\rightarrow X\) be a morphism between algebraic varieties over k. The following properties are equivalent:

1. (1)

   \(\pi _{k}\) is a homeomorphism with respect to the R-topology.

2. (2)

   \(\pi \) is subintegral.

3. (3)

   \(X_Y^+=Y\).

Proof

From Proposition 1.11 we know that (2) and (3) are equivalent. Assume (3). By Proposition 1.8 then \(\pi \) is finite and we get (1) using Theorem 4.13.

Assume \(\pi _{k}\) is a homeomorphism with respect to the R-topology. The morphism \(\pi \) is finite by Proposition 5.4 and thus \(\pi \) is subintegral again by Theorem 4.13. \(\square \)

Note that we cannot replace in Theorem 5.7 the topological assumption (1) on \(\pi _{k}\) by \(\pi _{k}\) is bijective or even \(\pi \) or \(\pi _k\) is a homeomorphism, as illustrated by Examples 3.5 and 5.3.

As illustrated by Proposition 5.1, the normality of the target space plays a role to upgrade a bijection into an isomorphism. The next result, which is a direct consequence of Theorem 5.7 is the first statement of Theorem C stated in the introduction. It shows that relative seminormality is the correct notion to associate to a homeomorphism with respect to the R-topology in order to obtain an isomorphism.

Corollary 5.8

Let \(\pi :Y\rightarrow X\) be a morphism between algebraic varieties over k such that \(\pi _{k}\) is a homeomorphism with respect to the R-topology. Then \(\pi \) is an isomorphism if and only if X is seminormal in Y.

We obtain an alternative version of [32, Thm. 2.4], where we replace the Zariski topology by the R-topology. In particular our statement is valid without any restriction on dimension.

Corollary 5.9

Let \(\pi :Y\rightarrow X\) be a morphism between algebraic varieties over k. If \(\pi _{k}\) is a homeomorphism with respect to the R-topology and X is seminormal, then \(\pi \) is an isomorphism.

Proof

From Proposition 3.4 then X is seminormal in Y and the result follows from Corollary 5.8. \(\square \)

We end the section by some results with a slightly different flavour. Forgetting about the R-topology, we compare now homeomorphisms with radicial and surjective morphisms. So we compare the two properties quoted at the beginning of the section, properties that do not appear in the equivalence of Theorem 5.7. We may also wonder what is the correct assumption to add to a homeomorphism in order to obtain an isomorphism in the spirit of Corollary 5.8. We start by proving that a homeomorphism between varieties over k is radicial. It gives the second part of Theorem B stated in the introduction.

Proposition 5.10

Let \(\pi :Y\rightarrow X\) be a morphism between algebraic varieties over k. If \(\pi \) is a homeomorphism then \(\pi \) is radicial.

Proof

We may assume X and Y irreducible, since the irreducible components of X and Y are homeomorphic one-by-one. Assume first that the dimension of X and Y is at least two. Then \(\pi _{k}\) is a homeomorphism with respect to the R-topology by Proposition 5.6, so \(\pi \) is finite by Proposition 5.4. By Theorem 5.7 we get that \(\pi \) is subintegral and thus radicial (Proposition 3.3). Assume X and Y are curves. Then X and Y are birational by Proposition 5.1. Since moreover \(\pi _{k}\) is bijective then \(\pi \) is bijective and equiresidual. By definition \(\pi \) is radicial. \(\square \)

Contrary to Theorem 5.7, the converse implication in Proposition 5.10 is not valid as Examples 5.2 (2) and 5.3 (2) show. Anyway, we get the analogue of Corollary 5.8 with respect to the Zariski topology. It gives the second part of Theorem C stated in the introduction. It is a generalization of [32, Thm. 2.4] in any dimension for varieties over k.

Corollary 5.11

Let \(\pi :Y\rightarrow X\) be a morphism between algebraic varieties over k such that \(\pi \) is a homeomorphism. Then \(\pi \) is an isomorphism if and only if X is saturated in Y.

Proof

Assume X is saturated in Y, the converse implication being trivial. By Proposition 5.10 then \(\pi \) is radicial. Then \(Y{\mathop {\rightarrow }\limits ^{Id}} Y{\mathop {\rightarrow }\limits ^{\pi }} X\) is a radicial sequence of morphisms. We conclude that \(\pi \) is an isomorphism by Proposition 2.11. \(\square \)

We get the analogue of Corollary 5.9 with respect to the Zariski topology only in dimension \(\ge 2\) (cf. Example 3.5 for the case of curves). It is a kind of reformulation of [32, Theorem 2.4] in characteristic zero.

Corollary 5.12

Let \(\pi :Y\rightarrow X\) be a morphism between irreducible algebraic varieties over k of dimension at least 2. If \(\pi \) is a homeomorphism and X is saturated, then \(\pi \) is an isomorphism.

Proof

From Propositions 5.4, 5.6 and 5.10 then \(\pi \) is radicial and finite. It follows from Proposition 3.3 that \(\pi \) is subintegral and X is seminormal in Y. We conclude \(\pi \) is an isomorphism from Corollary 5.8. \(\square \)

Note that, even though the statements of Proposition 5.10 and Corollaries 5.11, 5.12 does not mention the R-topology, it plays a crucial role in our proofs.