On the connectedness of the spectrum of forcing algebras

We study the connectedness property of the spectrum of forcing algebras over a noetherian ring. In particular we present for an integral base ring a geometric criterion for connectedness in terms of horizontal and vertical components of the forcing algebra. This criterion allows further simplifications when the base ring is local, or one-dimensional, or factorial. Besides, we discuss whether the connectedness is a local property. Finally, we present a characterization of the integral closure of an ideal by means of the universal connectedness of the forcing algebra


Introduction
Let R be a commutative ring, I = (f 1 , . . . , f n ) a finitely generated ideal and f an arbitrary element of R. A very natural and important question, not only from the theoretical but also from the computational point of view, is to determine if f belongs to the ideal I or to some ideal closure of it (for example to the radical, the integral closure, the solid closure, the tight closure, among others). To answer this question the concept of a forcing algebra introduced by Mel Hochster in the context of solid closure [8] is important (for more information on forcing algebras see [2], [3] and [4]): Definition 0.1. Let R be a commutative ring, I = (f 1 , . . . , f n ) an ideal and f ∈ R another element. Then the forcing algebra of these (forcing) data is A = R[T 1 , . . . , T n ] (f 1 T 1 + · · · + f n T n + f ).
Intuitively, when we divide by the forcing equation f 1 T 1 + · · · + f n T n + f we are "forcing" the element f to belong to the expansion of I in A. Besides, it has the universal property that for any R-algebra S such that f ∈ IS, there exists a (non-unique) homomorphism of R-algebras θ : A → S. Furthermore, the formation of forcing algebras commutes with arbitrary change of base. Formally, if α : R → S is a homomorphism of rings, then S ⊗ R A is the forcing algebra for the forcing data α(f 1 ), . . . , α(f n ), α(f ) [1,Chapter 2]. In particular, if p ∈ X = Spec R, then the fiber of (the forcing morphism) ϕ : Spec A → Spec R over p, ϕ −1 (p), is the scheme theoretical fiber Spec κ(p) ⊗ R A , where κ(p) = R p /pR p is its residue field. In this case, the fiber ring κ(p) ⊗ R A is the forcing algebra over κ(p) corresponding to the forcing data f 1 (p), . . . , f n (p), f (p), where we denote by g(p) ∈ κ(p), the image (the evaluation) of g ∈ R inside the residue field κ(p).
One can also define forcing algebras by several forcing equations and write them in a matrix form: This corresponds to a submodule N ⊆ M of finitely generated R-modules and an element f ∈ M via a free representation of these data (see [2]). The fiber over a point p ∈ Spec R of this forcing algebra is just the solution set of the corresponding system of inhomogeneous linear equations over κ(p). If the vector (f 1 , . . . , f m ) is zero, then we are dealing with a "homogeneous" forcing algebra. In this case there is a (zero-or "horizontal") Section s : X = Spec R → Y = Spec A coming from the homomorphism of R-algebras from A to R sending each T i to zero. This Section sends a prime ideal p ∈ X to the prime ideal (T 1 , . . . , T n ) + p ∈ Y .
In this paper we discuss when (the spectrum of) a forcing algebra is connected. The fibers of a forcing algebra are affine spaces hence connected unless they are empty (Lemma 1.1). In the homogeneous case it is easy to show that if X = Spec R is a connected topological space, so is Y = Spec A (Proposition 1.2 (4)). However, in the non-homogeneous case this question is more subtle. We establish in the noetherian case (in the setting of an ideal, i.e. m = 1) a condition equivalent to the connectedness of Y in terms of the horizontal and the vertical (to be defined) irreducible components of Spec A (Theorem 3.1). In the following we specialize then to the cases where R is one-dimensional (Corollary 3.2), local (Corollary 3.3), a factorial domain (Corollary 4.3) and a principal ideal domain (Corollary 4.5). In the fifth Section we study the problem whether the connectedness of a forcing algebra is a local property over the base. We give a quite general result (Theorem 5.1) that local connectedness (over the base) implies connectedness. In the one-dimensonal domain case the converse is true for forcing algebras (Corollary 5.2), but this can neither be extended to higher dimension nor to one-dimensional rings which are not domains. In the final Section we relate the integral closure of an ideal to the universal connectedness of the forcing algebra. This paper arose during the research stay of Danny de Jesús Gómez-Ramírez at the University of Osnabrück in 2011. He thanks all the department there for the help and support, in particular, to Daniel Brinkmann for inspiring discussions on the subject. He also thanks the Universidad Nacional de Colombia and the German Academic Exchange Service (DAAD) for financial support. Finally, we want to thank the referee for careful reading and many suggestions.

Connectedness
Recall that a topological space X is connected if there exists exactly two subsets (namely ∅ and X = ∅) which are open as well as closed. A connected component of X is a maximal connected subspace, i.e., not strictly contained in any connected subspace of X. Every connected component is necessarily closed because its closure is a closed connected set containing it. Moreover, the connected components form a partition of the space X.
For a commutative ring A, the spectrum Spec A is connected if and only if A = 0 and if it is not possible to write A = A 1 × A 2 with A 1 , A 2 = 0. Equivalently, there exist exactly two idempotent elements, namely 0 and 1 (see for example [7, Exercise 2.19, Chapter II] or [5, Exercise 2.25]). Hence domains and local rings are connected. If A is a C-algebra of finite type, then the connectedness of Spec A (in the Zariski topology) is equivalent to the connectedness of the complex space (Spec A) C in the complex topology, because irreducible varieties are connected over C (the real spectrum might be disconnected though). Now, we present a lemma describing the fibers of a forcing algebra as affine spaces over the base residue field. Lemma 1.1. Let R be a commutative ring and let A be the forcing algebra corresponding to the data {f ij , f i }. Let p ∈ X be an arbitrary prime ideal of R and r the rank of the matrix f ij (p) . Then the fiber over p is empty or isomorphic to the affine space A n−r κ(p) .
Proof. We know by a previous remark that the fiber ring over P is κ(p) ⊗ R A which is just Hence we are dealing with a forcing algebra over a field. With the usual techniques of linear algebra (elementary row operations and exchange of columns) we can bring the matrix to triangular form without changing (up to isomorphism) the algebra. The linear system has no solution if and only if the forcing algebra is 0. Otherwise we can use r rows to eliminate r variables showing that the algebra is isomorphic to a polynomial algebra in n − r variables.
Since an affine space is irreducible and hence connected, the preceding lemma tells us that the fibers of a forcing algebra are connected unless they are empty. The easiest example of an empty forcing algebra is K[T ]/(0T − 1), where K is a field. A forcing algebra may be connected though some fibers may be empty, an example is given by In the following we will mainly deal with the case where all fibers are nonempty. This is equivalent to say that f belongs to the radical of the ideal I (or, by definition, to the radical of the submodule N , see [2]). Proposition 1.2. Let A be a forcing algebra over R with the corresponding morphism ϕ : Y = Spec A → X = Spec R. Then the following hold.
(1) The connected components of Y are of the form ϕ −1 (Z) for suitable subsets Z ⊆ X.
(2) If ϕ : Y → X is surjective, then these Z are uniquely determined.
(3) If ϕ : Y → X is surjective and Y is connected, then X is connected.
(4) If the forcing data are homogeneous, then there is a bijection between the connected components of X and Y . In particular, X is connected if and only if Y is connected.
(5) Suppose that ϕ : Y → X is (surjective and) a submersion. Then there is a bijection between the connected components of X and Y . In particular, X is connected if and only if Y is connected. Proof.
(1). By Lemma 1.1 the fibers are connected. Hence a connected component of Y which contains a point of a fiber contains already the complete fiber. (2) and (3) are clear. (4). First note that in the homogeneous case there exists an R-algebrahomomorphism A → R given by T i → 0, which induces a section s : X → Y given by p → (p, T 1 , . . . , T n ). Therefore ϕ : Y → X is surjective. Suppose first that X is connected. Then the image s(X) of the section is connected. Hence two points Q 1 , Q 2 ∈ Y lie in the connected subset ϕ −1 ϕ(Q 1 ) ∪ s(X) ∪ ϕ −1 ϕ(Q 2 ) and so Y is connected. This argument holds for every subset Z ⊆ X, so the statement about the components follows.  i.e. a disjoint union of a line over a point and a hyperbola dominating the base (its image is the pointed affine line, hence dense). But, Spec R is the affine line which is connected. Note that the element X belongs to the radical of (X 2 ), but does not belong to (X 2 ) nor to its integral closure. Hence ϕ : Spec A → Spec R is surjective, but not submersive (see also Section 6).

Horizontal and Vertical Components
We describe now the irreducible components of the spectrum of a forcing algebra over an integral base in the ideal case. We will identify prime ideals inside the forcing algebra for these data. Then the following hold.

This component is given (inside
where Q(R) denotes the quotient field of R.
(2) All other irreducible components of Spec A are of the form for some prime ideal q ⊆ R which is minimal over (f 1 , . . . , f n , f ). Proof.
(1). For I = 0 the polynomial , hence the intersection of this principal ideal with R[T 1 , . . . , T n ] gives a prime ideal in this polynomial ring and therefore in The minimality is clear, since it holds in a localization. Because of R ∩ p = 0, this component dominates the base. On the other hand, let p be a minimal prime ideal that is minimal over Then there exists r, s ∈ R, r = 0, and a polynomial G ∈ R[T 1 , . . . , T n ] such that rh = sG(f 1 T 1 + · · · + f n T n + f ). This element belongs to p and since r / ∈ p we deduce h ∈ p . Hence p = p.
(2). Let (f 1 T 1 + · · · + f n T n + f ) ⊆ q be a minimal prime ideal different from p. In every localization A fi there is only one minimal prime ideal, namely the horizontal component (as in (1)), since f 1 T 1 + · · · + f n T n + f is a prime element whenever some f i is a unit. Therefore q / ∈ Spec A fi and hence f i ∈ q. Because q contains the forcing equation we also deduce f ∈ q. Hence (f 1 , . . . , f n , f ) ⊆ q and by the minimality condition q is minimal over the extended ideal (f 1 , . . . , f n , f )R[T 1 , . . . , T n ]. Therefore q must be the extended ideal of a minimal prime ideal of (f 1 , . . . , f n , f ) in R (the minimal prime ideals above (f 1 , . . . , f n , f )R, above (f 1 , . . . , f n , f )R[T 1 , . . . , T n ] and above (f 1 , . . . , f n , f )A are in bijection). . . , f n , f ) of height one, then the extended ideal in the polynomial ring has also height one and can therefore not contain the horizontal prime ideal. Therefore such prime ideals yield vertical components.
It is possible that all the V (q j ), where q j ⊇ (f 1 , . . . , f n , f ) is a minimal prime ideal, lie on the horizontal component. In this case there exists no vertical component.  So let q be a minimal prime ideal over (d). Then, by Krull's Theorem, the height of q is 1 and hence it is also minimal over J ⊆ (d). Moreover, the height of qR[T 1 , . . . , T n ] is also 1 Conversely, let q denote a minimal prime ideal of J such that h does not belong to qR[T 1 , . . . , T n ]. Assume that d / ∈ q. Then, because of f 1 d, .
. . , f n d, f d ∈ q, we get immediately f 1 , . . . , f n , f ∈ q and hence the contradiction h ∈ qR[T 1 , . . . , T n ]. So we must have d ∈ q and q must also be minimal over (d).
If R is a (not necessarily noetherian) factorial domain (a unique factorization domain), then there exists always a factorization h = dh with h a prime element in B. The minimal prime ideals over (d) are given by the prime factors p of d, and pB has also height 1. Hence the argument of the lemma goes through, also in this case. Example 4.7 below shows that a forcing equation does not always have a prime decomposition as in the lemma. Then it is more complicated to determine the vertical components.
The following example shows that the irreducible components in the module case are more complicated. In particular, Lemma 2.1 (2) is not true.
The horizontal component of Spec A is given by the prime ideal (T 1 −Y, T 2 − X). The algebra is connected, since this component is a section. The other minimal prime ideals are (X, T 2 ), (Y, T 1 ) and (X, Y ). Only the last one is the extension of a prime ideal of the base.

Connectedness Results
The following is our main general connectedness result on forcing algebras. Proof. This is clear from Lemma 2.1 since the connected components of a noetherian scheme are just the unions of its irreducible components which intersect chainwise. Note also that the Z i can be determined in Spec R alone. Proof. This follows from Theorem 3.1 since the minimal prime ideals of I = 0 in a one-dimensional domain are maximal ideals. These maximal ideals form the connected components of V (I).
Note that this corollary is not true in higher dimension, see Example 4.6 in the next section. We specialize now to the local case. Proof. Let q j , j ∈ J, be the minimal prime ideals of (I, f ), disregarding whether the V j = V (q j R[T 1 , . . . , T n ]) give rise to vertical components of Spec A or not. Note that V j ∩ V i = ∅ for all i, j, because we are over a local ring. Suppose first that for at least one j we have V j = V (q j ) ⊆ H. Then V i ∩H = ∅ and hence the forcing algebra is connected by Theorem 3.1. But this assumption also means that p ⊆ q j R[T 1 , . . . , T n ] ⊆ mR[T 1 , . . . , T n ] and therefore p+(I, f ) = (1), because m does not extend to the unit ideal. So, under this assumption, the two properties are equivalent.
Suppose next that V j H for all j, meaning that all V j are vertical components of Spec A. The subsets V (q j ) inside the local ring Spec R are connected, hence there is exactly one Z in the notation of Theorem 3.

Connectedness over Factorial Domains
We deal now with the case where R is a factorial domain. Note that if R is factorial, then B = R[T 1 , . . . , T n ] is factorial as well. So, if h = f 1 T 1 + · · · + f n T n + f ∈ B is a forcing equation, then one can factor out a greatest common divisor of all the coefficients f 1 , . . . , f n and f , say d, and obtain a representation of h as a product of an element d in R and an irreducible polynomial h = (f 1 /d)T 1 +· · ·+(f n /d)T n +(f /d) in B (for n ≥ 1), which generates a prime ideal because B is a factorial domain. This hypothesis appeared already in Lemma 2.4 and is also crucial in the following sufficient condition for connectedness. h := f 1 T 1 + · · · + f n T n + f = d(f 1 T 1 + · · · + f n T n + f ) be a forcing equation such that h := f 1 T 1 +· · ·+f n T n +f is a prime polynomial. Assume (f 1 , . . . , f n ) = 0. Suppose that (f 1 , . . . , f n ) is not contained in any minimal prime ideal of (d). Then Spec A is connected, where A = B/(h).
Proof. By Lemma 2.4, the horizontal component of Spec A is V (h ) and the vertical components correspond to the minimal prime ideals q over (d). We will show that V (q) intersects the horizontal component.
This can be established after the base change R → κ(q). Now, at least one of the f i becomes a unit in κ(q) and therefore h is not a unit over κ(q). So, κ(q)[T 1 , . . . , T n ]/(h ) = 0.
Note also that if d is a unit, then h = h is a prime polynomial by assumption and then the forcing algebra is integral, hence connected anyway. Now, we shall deduce a Corollary in the case that R is a factorial domain. In this kind of rings we can define a greatest common divisor of a finite set of elements a 1 , . . . , a m , denoted by gcd(a 1 , . . . , a m ), using the prime factorization, and it is well defined up to a unit in R and defined as the unity in R in the case that the elements have no irreducible common factor.
be the forcing equation and let d = p 1 · · · p k be a prime factorization of d. Suppose that {1, . . . , k} = i∈I J i such that the j∈Ji V (p j ) are the connected components of V (d). Then Spec B/(h) is connected if and only if for every i there exists some j ∈ J i such that (f 1 T 1 + · · · + f n T n + f , p j ) = (1).
Proof. By Lemma 2.4 and Lemma 4.2, p = (f 1 T 1 + · · · + f n T n + f ) describes the horizontal component and every vertical component corresponds to a prime factor of d. Hence the statement follows directly from Theorem 3.1.
The condition that (f 1 T 1 + · · · + f n T n + f , p) = (1) is true if p does not divide all f 1 , . . . , f n or if f is not a unit modulo p.
Because Y is not a unit modulo X, the forcing algebra is connected. For h = X(XT + X + 1) we also have d = X = p. Now X + 1 is a unit modulo X and the forcing algebra is not connected. For h = XY (XT 1 + Y T 2 + f ) we have d = XY , but f 1 = X, f 2 = Y do not have a common prime factor and hence the forcing algebra is connected.
Let p be a prime factor of d. Then p does not divide some f i . But then f 1 T 1 + · · · + f n T n + f is not a unit modulo p, and the condition of Corollary 4.3 holds even for every p.
Finally, we assume that R is a principal ideal domain and that where e ∈ R is not a unit. Let p ∈ R be a prime element dividing e. We still work with the factorization h = dh , where h is irreducible and describes the horizontal component. The elements f 1 /d, . . . , f n /d, f /d do not have a common prime factor, hence p does not divide f /d. Therefore in the field R/(p) the element f /d becomes a unit u and the polynomial h becomes 0T 1 +· · ·+0T n +u. Therefore the horizontal component V (h ) and the vertical component V (p) are disjoint and the forcing algebra is not connected by Corollary 3.2.
Example 4.6. The condition of being a principal ideal domain for R in the last part of Corollary 4.5 is necessary, as the following example shows (see also Figure 1, for the case K = R). With the notation from above we consider the following setting:

Local Properties
An interesting question is whether the connectedness of Y = Spec A is a local property over the base X = Spec R. Specifically, is it true that Y is a connected space if and only if X is connected and for every p ∈ X, Spec A p is connected, where A p denotes the localization of A at the multiplicative system R p (considered in A). The next theorem gives a positive answer to the "if" part of this question in general. For a forcing algebra, the converse holds for a onedimensional domain, but neither over a reducible curve nor over the affine plane.
Theorem 5.1. Let ψ : R → A be a ring homomorphism. Set X := Spec R and Y := Spec A. Suppose that X is a connected space and that for all p ∈ X, Spec A p is connected, where A p := A R p , that is, Y is locally (over the base) connected. Then Y is connected.
Proof. We first show that we can assume that ψ is injective: first, note that for any minimal prime p ∈ X, the space Spec A p is not empty, because it is connected (our convention is that the empty set is not connected). Let Q ∈ Spec A p be a prime ideal, then ψ −1 (Q) is a prime ideal of R contained in p, because ψ −1 (Q)∩(R p) = ∅, moreover it is equal to p in view of the minimality of p. Therefore, for any minimal prime in R, there exists a prime ideal Q in A lying over it. Therefore for any a ∈ ker ψ, we know that ψ(a) = 0 ∈ ∩ Q∈Y Q and then a ∈ ∩ Q∈Y ψ −1 (Q) ⊆ ∩ p∈min R p = nil R, that is ker ψ ⊆ nil R.
In consideration of this it is enough to reduce to the case of R being reduced. For this reduction consider the natural homomorphism ψ red := R red → A red induced by ψ, killing the nilpotent elements. Now, our hypothesis of locally (over the base X) connected and the conclusion holds for Y if and only if holds over the base X red = Spec(A red ) for Y red := Spec(A red ). In fact, clearly X X red and Y Y red as topological spaces; besides, for any p ∈ X red , (A red ) p A p /(nil A)A p and (nil A)A p ⊆ nil(A p ), hence Spec (A red ) p Spec(A p / nil A p ) Spec A p . In conclusion, it is enough to prove the theorem in the reduced case for injective ψ. Now, we assume that Y is not connected, which is equivalent to say that there exists nontrivial idempotents e 1 , e 2 ∈ A with e 1 + e 2 = 1, e 1 e 2 = 0 and e 1 , e 2 = 0, 1. Set J i = Ann R (e i ) for i = 1, 2. We claim that J 1 + J 2 R. Otherwise, there exists y i ∈ J i such that y 1 +y 2 = 1, and then y 1 y 2 = y 1 y 2 (e 1 + e 2 ) = y 2 (y 1 e 1 ) + y 1 (y 2 e 2 ) = 0 + 0 = 0. Therefore X = V (y 1 ) V (y 2 ), that is, we can write X as a disjoint union of two closed subsets, which implies, in view of the connectedness of X, that one of these closed subsets is empty, or what is the same, one of the y i is a unit. Hence, e i = y −1 i (y i e i ) = y −1 i 0 = 0, a contradiction.
So let J 1 + J 2 ⊆ P be a prime ideal. By assumption, A R p is connected, hence either e 1 or e 2 become 0 in this ring. This means (in the first case) that there exists s ∈ R p such that se 1 = 0 in A. But then we get the contradiction s ∈ J 1 . In conclusion, Y is a connected space.
We deal next with the one-dimensional case.
Corollary 5.2. Suppose that R is a noetherian domain of dimension 1. Let I = (f 1 , . . . , f n ) = 0 be an ideal, f ∈ R an element and A = R[T 1 , . . . , T n ]/(f 1 T 1 + · · · + f n T n + f ) the forcing algebra for these data. Then Spec A is connected if and only if Spec A is locally connected, i.e., for every prime ideal p ∈ Spec R is A R p connected.
Proof. The global property follows from the local property by Theorem 5.1. So suppose that Spec A is connected. By the assumption I = 0 we know that a horizontal component exists. Hence the fiber over the generic point (0) is nonempty, thus connected by Lemma 1.1. The connectedness of Spec A means, by Corollary 3.2, that the horizontal component meets every vertical component. The vertical components of Spec A p over Spec R p for a maximal ideal p in Spec R are empty or V (pA) and, in the second case, they are a vertical component of Spec A. By the intersection condition the horizontal component and this vertical component (if it exists) intersect, so Spec A p is connected.
The following example shows that for a non-integral one-dimensional base ring, connectedness is not a local property.  . . , f n ) be an ideal, f ∈ R an element inside the radical of I and A = R[T 1 , . . . , T n ]/(f 1 T 1 + · · · + f n T n + f ) be the forcing algebra for these data. Then the following are equivalent: (1) Spec A is connected.
(2) Spec A is locally connected, i.e. for every prime ideal P ∈ Spec R is A R P connected.
Proof. The equivalence between (1) and (2) follows from Corollary 5.2. For the equivalence with (3) we may assume that R is local, i.e., a discrete valuation domain. Let p be a generator of its maximal ideal. We may assume at once that I = 0, because else f = 0 due to the radical assumption, and also that all f i (If f i = 0 we can omit this without changing any property) and f are not 0. We write f i = u i p ki and f = up k with units u i , u. Assume that f / ∈ I. Then k < min(k 1 , . . . , k n ). We write the forcing equation as where the exponents k i − k are all positive. Because of the radical assumption we have k ≥ 1. But then the forcing algebra has the two components V (p) and V (u 1 p k1−k T 1 + · · · + u n p kn−k T n + u) which are disjoint. The other direction follows from Proposition 1.2 (4).
For a non-normal one-dimensional domain this equivalence can never be true because of Corollary 6.3 below. The next trivial example shows that this statement is also not true without the radical assumption.
Example 5.5. For R = K[X], K a field, the forcing algebra K[X, T ]/(0T − X) ∼ = K[T ] is connected, but the fiber over the generic point is empty, hence not connected.
In higher dimension, even for a factorial domain, the converse of Theorem 5.1 is also not true.
Example 5.6. We continue with Example 4.6 (see Figure 2 for K = R) i.e., R := K[X, Y ] is the ring of polynomials in two variables, B := R[T ], h := X 2 Y T − XY and A := B/(h). The morphism Spec A → Spec R is surjective, since XY ∈ rad(X 2 Y ). We know already that Spec A has the three irreducible components V (X), V (Y ) and V (XT − 1) that it is connected.
However, if we localize A in p := (X) ∈ Spec R, that is, if we consider the ring A p = A R p , then Spec A p has just two irreducible components, namely, V (X)A p (plane) and V (XT + 1)A p (hyperbolic surface), because the minimal primes of A p are just (X)A p and (XT + 1)A p , since the remaining minimal prime ideal (Y ) meets R p, since Y ∈ R p. Moreover, these two irreducible components are disjoint, because In conclusion, Spec A p is not connected. For getting a better intuition compare Figure 2 with Figure 1.

Integral Closure and Connectedness
In this final section we relate the integral closure of an ideal to the universal connectedness of the forcing algebra. Recall that the integral closure, written I, of an ideal I ⊆ R is defined as I := f ∈ R : f n + a 1 f n−1 + · · · + a n−1 f 1 + a n = 0, where a j ∈ I j . Definition 6.1. Let ϕ : Y → X be a morphism between affine schemes. We say that ϕ is a universally connected if W × X Y is connected for any affine noetherian change of base W → X, with W connected. Now, we prove a criterion for belonging to the integral closure in terms of the universal connectedness of the corresponding forcing morphism. Theorem 6.2. Let A be a forcing algebra over a noetherian ring R and ϕ : Y := Spec A → X := Spec R the corresponding forcing morphism. Then the following conditions are equivalent: (1) f belongs to the integral closure of I, i.e. f ∈ I.
where D is a discrete valuation domain. Proof.
(1) ⇒ (2). Recall that a submersion is universal if it remains a submersion under noetherian change of base. Due to the valuative criterion for the integral closure (see [9,Theorem 6.8.3]) and the fact that (2) can be checked after change of base to a discrete valuation domain D (see [6, Remarque 2.6]), we can assume that R = D and that f ∈ I and we have to prove that ϕ : Spec A → Spec D is a submersion. But f ∈ I if and only if there exists a section s : Spec D → Spec A, i.e. ϕ • s = Id Spec D . By an elementary topological argument the existence of a section implies that ϕ is a submersion.
(2) ⇒ (3). Let W → X be an affine noetherian connected change of base, then, since ϕ W : W × X Y → W is a submersion, by Proposition 1.2 (5) W × X Y is connected.
(4) ⇒ (1). Let W = Spec D → X be a change of base, where D is a discrete valuation domain. By the valuative criterion for integral closure it is enough to show that f ∈ ID. First, note that f ∈ rad(ID), which is equivalent to say that ϕ W : W × X Y → W is surjective (see [2]). If the fiber over p ∈ W were empty, then the morphism Spec κ(p) x −→ Spec κ(p) −→ W −→ X would yield a contradiction, since then the pull-back (Spec κ(p) x )× X Y would be empty (hence not connected) and κ(p) x is a discrete valuation domain. Thus f ∈ rad(ID) and, by Corollary 5.4, the connectedness of W × X Y is equivalent to f ∈ ID. Now, we prove a corollary of this theorem charaterizing the property that a fraction belongs to the integral closure (or normalization) of an integral domain. Proof. It is an elementary fact that r/s is integral over R if and only if r ∈ (s). Then by the former theorem r ∈ (s) if and only if ϕ is universally connected.
In our final example we show that a forcing algebra over a non-normal curve might be connected but not universally connected. In fact the pull-back to the normalization is already not connected. Example 6.4. Let K be a field and consider the ring-homomorphism . Since x 2 − x − u = 0, the extension R → K[X] is integer, but K[X] is integrally closed, therefore the integral closure (or normalization) of R is K[X]. We consider the forcing algebra A = R[T ]/(vT − u). It consists of a horizontal component given by V (vT + u, vT 3 − T + 1) (check that this is a prime ideal) (twisted hyperbola) and the vertical component V (u, v) (middle vertical line). They intersect in V (u, v, T −1), hence the forcing algebra is connected (see Figure 3). When we pull-back this situation to the normalization we get Now we have one horizontal component and two vertical components (lines), and the horizontal hyperbola meets exactly one of them, hence this forcing algebra is not connected by Corollary 3.2 (see Figure 4). Heuristically we get