Normality and Related Properties of Forcing Algebras
aa r X i v : . [ m a t h . A C ] J u l NORMALITY AND RELATED PROPERTIES OF FORCINGALGEBRAS
DANNY DE JES ´US G ´OMEZ-RAM´IREZ AND HOLGER BRENNER
Abstract.
We present a sufficient condition for irreducibility of forcingalgebras and study the (non)-reducedness phenomenon. Furthermore,we prove a criterion for normality for forcing algebras over a polynomialbase ring with coefficients in a perfect field. This gives a geometricalnormality criterion for algebraic (forcing) varieties over algebraicallyclosed fields. Besides, we examine in detail an specific (enlightening)example with several forcing equations. Finally, we compute explicitlythe normalization of a particular forcing algebra by means of findingexplicitly the generators of the ideal defining it as an affine ring.
Mathematical Subject Classification (2010): 13B22, 14A15, 14R25, 54D05Keywords: Forcing algebra, normality, normalization, reduceness, irreducibil-ity Introduction
Let R be a commutative ring, I = ( f , . . . , f n ) a finitely generated idealand 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 (forexample to the radical, the integral closure, the plus closure, the solid closure,the tight closure, among others). To answer this question the concept of aforcing algebra introduced by Mel Hochster in the context of solid closure [13]is important (for more information on forcing algebras see [3], [4]): Definition 1.
Let R be a commutative ring, I = ( f , . . . , f n ) an ideal and f ∈ R another element. Then the forcing algebra of these (forcing) data is A = R [ T , . . . , T n ] / ( f T + . . . + f n T n + f ) . Intuitively, when we divide by the forcing equation f T + . . . + 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 , thereexists a (non-unique) homomorphism of R -algebras θ : A → S . This paper should be cited as follows D. A. J. G´omez-Ram´ırez and H. Brenner. Nor-mality and Related Properties of Forcing Algebras. Communications in Algebra. Volume44, Issue 11, pp. 4769-4793, 2016.
Furthermore, the formation of forcing algebras commutes with arbitrarychange of base. Formally, if α : R → S is a homomorphism of rings, then S ⊗ R A ∼ = S [ T , ..., T n ] / ( α ( f ) T + · · · + α n ( f n ) T n + α ( f ))is the forcing algebra for the forcing data α ( f ) , . . . , α ( f n ) , α ( f ) . In par-ticular, if p ∈ X = Spec R , then the fiber of (the forcing morphism) ϕ : Y := Spec A → X := Spec R over p , ϕ − ( p ), is the scheme theoretical fiberSpec( κ ( p ) ⊗ R A ), where κ ( p ) = R p / p R p is its residue field. In this case, thefiber ring κ ( p ) ⊗ R A is the forcing algebra over κ ( p ) corresponding to theforcing data f ( p ) , . . . , f n ( p ) , f ( p ), where we denote by g ( p ) ∈ κ ( p ), the im-age (the evaluation) of g ∈ R inside the residue field κ ( p ) = R p / p R p . Also,note that for any f i A f i ∼ = R f i [ T , ..., ˇ T i , ..., T n ], via the R f i − homomorphismsending T i
7→ − P j = i ( f j /f i ) T j − ( f /f i ) and T r T r for r = i .An extreme case occurs when the forcing data consists only of f . Then, wedefine I as the zero ideal. Therefore A = R/ ( f ).Besides, if n = 1, then intuitively the forcing algebra A = R [ T ] / ( f T − f )can be consider as the graphic of the “rational” function f /f . We will explorethis example in more detail in chapter two.By means of forcing algebras and forcing morphisms one can rewrite thefact that the element f belongs to a particular closure operations of I . Weshall illustrate this now.Firstly, the fact that f ∈ I is equivalent to the existence of a homomorphismof R − algebras α : A → R , which is equivalent at the same time to theexistence of a section s : X → Y , i.e. ϕ ◦ s = Id X .Secondly, f belongs to the radical of I if and only if ϕ is surjective. Infact, suppose that ϕ is surjective and let us fix a prime ideal p ∈ X con-taining I . Then, ϕ − ( x ) = Spec κ ( p ) ⊗ A = ∅ , that means, κ ( p ) ⊗ A = κ ( p )[ T , ..., T n ] / ( f ( p ) T + · · · + f n ( p ) T n + f ( p )) = 0. But, each f i ( p ) = 0,since f i ∈ p , therefore f ( p ) is also zero, thus f ∈ p . In conclusion, f ∈∩ p ∈ V ( I ) p = rad I . Conversely, suppose that f ∈ rad I and take an arbitraryprime p ∈ X . Then, if I is not contained in p , then some f j ( p ) = 0 and so κ ( p ) ⊗ A = 0, that means ϕ − ( p ) = ∅ . Lastly, if I ⊆ p then f ∈ p , andtherefore κ ( p ) ⊗ A = κ ( p )[ T , ..., T n ] = 0 and thus ϕ − ( p ) = A n − κ ( p ) = ∅ . Inconclusion ϕ is surjective.Thirdly, let us review the definition of the tight closure of an ideal I ofa commutative ring R of characteristic p >
0. We say that u ∈ R belongsto the tight closure of I , denoted by I ∗ , if there exists a c ∈ R not in anyminimal prime, such that for all q = p e ≫ cu q ∈ I [ q ] , where I [ q ] denotesthe expansion of I under the e − th iterated composition of the Frobenius ho-momorphism F : R → R , sending x → x p . Tight Closure is one of themost important closure operations in commutative algebra and was intro-duced in the 80s by M. Hochster and C. Huneke as an attempt to provethe “Homological Conjectures” (for more information [14]). Let ( R, m ) be
ORMALITY AND RELATED PROPERTIES OF FORCING ALGEBRAS 3 normal local domain of dimension two. Suppose that I = ( f , ..., f n ) is an m − primary ideal and f is an arbitrary element of R . Then, f ∈ I ∗ if andonly if D ( IA ) = Spec A r V ( IA ) is not an affine scheme, i.e. is not of theform Spec D for any commutative ring D (see [4, corollary 5.4.]).Forth, the origin of the forcing algebras comes from the definition of thesolid closure, as an effort to defining a closure operation for any commutativering, independent the characteristic (see [13]). Explicitly, let R be a Noether-ian ring, let I ⊆ R an ideal and f ∈ R . Then, f belongs to the solid closure of I if for any maximal ideal m of R and any minimal ideal q of its comple-tion b R m , for the complete local domain ( R ′ = b R m / q , m ′ ) holds that the d − thlocal cohomology of the forcing algebra A ′ , obtained after the change of base R ֒ → R ′ , H dm ( A ′ ) = 0, where d = dim R ′ (see [2, Definition 2.4., p. 15]).Fifth, let us consider an integral domain R and an ideal I ⊆ R . Then, u belongs to the plus closure of I , denoted by I + , if there exists a finiteextension of domains R ֒ → S , such that f ∈ IS . If R is a Noetherian domainand I = ( f , ...f n ) ⊆ R is an ideal and f ∈ R , then f ∈ I + if and onlyif there exists an irreducible closed subscheme e Y ⊆ Y = Spec A such thatdim e Y = dim X , ϕ ( e Y ) = X and for each x ∈ X , ϕ − ( x ) ∩ e Y is finite (for anprojective version of this criterion see [2, Proposition 3.12]).Finally, if R denotes an arbitrary commutative ring and I ⊆ R is an ideal,then we say that u belongs to the integral closure of I , denote by I , if thereexist n ∈ N , and a i ∈ I i , for i = 1 , ..., n , with u n + a u n − + · · · + a n = 0 . We proved in [5, Chapter 2] that f ∈ I , where I = ( f , ..., f n ) ⊆ R , if andonly if the corresponding forcing morphism ϕ is universally connected, i.e.Spec( S ⊗ R A ) is a connected space for any Noetherian change of base R → S ,such that Spec S is connected.From this we derive a criterion of integrity for fractions r/s ∈ K ( R ), where R denotes a Noetherian domain, in terms of the universal connectedness ofthe natural forcing algebra A := R [ T ] / ( sT + r ).In view of this results, it seems very natural to study in commutativealgebra the question of finding a closure operation with “good” properties (see[8]), in terms of finding suitables algebraic-geometrical as well as topologicalor homological properties of the forcing morphism. This approach goes closerto the philosophy of Grothendieck’s EGA of defining and studying the objectsin a relative context (see [11] and [10]). A simple and deep example of thisapproach is the counterexample to one of the most basic and important openquestions on tight closure: the Localization Problem i.e., the question whethertight closure commutes with localization. This was done by H. Brenner and P.Monsky using vector bundles techniques and geometric deformations of tightclosure (see [2]). DANNY DE JES ´US G ´OMEZ-RAM´IREZ AND HOLGER BRENNER
Besides, another good example going in this direction is a general defini-tion of forcing morphism for arbitrary schemes. Specifically, let X and Y be arbitrary schemes. Suppose that i : Z → X is a closed subscheme and f ∈ Γ( X, O X ) is a global section. Then, a morphism ϕ : Y → X is a forcingmorphism for f and Z , if i) the pull-back of the restriction of f to Z , f | Z = i ♯Z ( f ) is zero, i.e. ϕ ♯ | ϕ − ( Z ) ( f | Z ) = 0; ii) for any morphism of schemes ψ : W → X with the same property, i.e. ψ ♯ | ψ − ( Z ) ( f | Z ) = 0, there exists a (non-unique) morphism e ψ : W → Y suchthat ψ = ϕ ◦ e ψ . It is a natural generalization of the universal property of aforcing algebra but in the relative context and in a category including that ofcommutative rings with unity.In general, for a integral base ring there are two kinds of irreducible compo-nents for the prime spectrum of a forcing algebra. In fact, let R be a noether-ian domain, I = ( f , . . . , f n ) an ideal, f ∈ R and A = R [ T , . . . , T n ] / ( f T + . . . + f n T n + f ) the forcing algebra for these data. For I = 0 there existsa unique irreducible component H ⊆ Spec A (“horizontal component”) withthe property of dominating the base Spec R (i.e. the image of H is dense).This component is given (inside R [ T , . . . , T n ]) by p = R [ T , . . . , T n ] ∩ ( f T + . . . + f n T n + f ) Q ( R )[ T , . . . , T n ] , where Q ( R ) denotes the quotient field of R .All other irreducible components of Spec A (“vertical components”) are ofthe form V ( q R [ T , . . . , T n ])for some prime ideal q ⊆ R which is minimal over ( f , . . . , f n , f ) (for a com-plete proof of this fact and more information see [5, Lemma 2.1.]).Finally, let us describe briefly the content of the following sections of thispaper.We study the case corresponding to a sub-module N of a finitely generatedmodule M and an arbitrary element s ∈ M . This case corresponds to forcingalgebras with several forcing equations A = R [ T , . . . , T n ] / * f . . . f n ... . . . ... f m . . . f mn · T ... T n + f ... f m + . Even very basic properties of forcing algebras are not yet understood, andthese paper deal in some extent with these questions.For example, we describe how to perform elementary row and column oper-ations on the forcing algebra by means of considering elementary affine linear
ORMALITY AND RELATED PROPERTIES OF FORCING ALGEBRAS 5 isomorphisms and a specific relation between the regular sequences of forcingelements and the fitting ideals of the corresponding forcing matrix ( § I is bigger or equal that2 ( § § I is the whole base R , then we get a complete characterization of theintegrity of the forcing algebra over UFDs ( § I and the ideal I + D , where D is generated bythe partial derivatives of the data. In the case that we are working over analgebraic closed field and our base is the ring of coordinates of an irreduciblevariety X , the normality of the (forcing) hyperplane defined by the forcingequation can be characterized by imposing the condition that the codimensionof the singular locus of X in the whole affine space is a least three ( § § § Forcing Algebras with several Forcing Equations
Now, we study just a few elementary properties of forcing algebras whichare defined by several forcing equations and which leads us in a natural wayto the understanding of the linear algebra over the base ring R . This section DANNY DE JES ´US G ´OMEZ-RAM´IREZ AND HOLGER BRENNER could be understood as a simple invitation to this barely explored field ofmathematics. Here we recommend for further reading [3]. In this case we canwrite the forcing algebra in a matrix form: A = R [ T , . . . , T n ] / * f . . . f n ... . . . ... f m . . . f mn · T ... T n + f ... f m + . This corresponds to a submodule N ⊆ M of finitely generated R -modulesand an element f ∈ M via a free representation of these data (see [3, p. 3]).Now, we study how the forcing algebra behaves when we make elementaryrow or column operations in the associated matrix M . Remember that thematrix notation in the forcing algebra just means that we are consideringthe ideal generated by the rows of the resulting matrix, after performing thematrix multiplications and additions.First, if l , ..., l m denote the rows of M , and c ∈ R denote an arbitraryconstant, making a row operation, l j cl j + l i , ( i = j ; that is changing the jth row by c times the ith row plus the jth row) just means changing thegenerators h , ..., h m to the new generators h , ..., h j − , ch i + h j , h j +1 , ..., h m .The ideal generated by these two groups of forcing elements coincides andtherefore the associated forcing algebra are the same. Similarly, if we makeoperations of the form l i l j and l i cl i , where c is an invertible element of R , which correspond to change two rows and to multiply a row by an elementin R , then the forcing algebra does not change.For the column operation, the problem is a little bit more subtle. Let { C , ..., C n } be the columns of the matrix A . Consider the column operation dC i + C j C j , where d ∈ R . Now, define the following automorphism ϕ of thering of polynomials R [ T , ..., T n ] sending T s T s , for s = i , and T i cT j + T i .Now, ϕ ( h r ) = f r T + ... + f ri ( cT j + T i ) + ... + f rn T n = f r T + ... + ( cf ri + f rj ) T j + ... + f rn T n . and then ϕ induces an isomorphism between the forcing algebra with matrix M and the forcing algebra with matrix obtained from M performing theprevious column operation. Similarly, for operations of the form C i C j and C i dC i , where d ∈ R is an invertible element, the resulting forcingalgebras coincide. Now, if R is a field and the rank of the associated matrix M is r , where r ≤ min( m, n ), then performing row and column operations onthe associated matrix we can obtained a matrix form by the r × r identitymatrix in the upper-left side and with zeros elsewhere.Therefore, the elements h i have just the following simple form: h i = T i + g i ,for i = 1 , ..., r and h i = g i , for i > r , and some g i ∈ R (this g i could appearjust in the nonhomogeneous case, corresponding to the changes made on theindependent vector form by the f j ). Thus the forcing algebra A is isomorphic ORMALITY AND RELATED PROPERTIES OF FORCING ALGEBRAS 7 either to zero (in the case that there exists g i = 0, for some i > r ) or to k [ T r +1 , ..., T n ]. This allow us to present the following lemma describing thefibers 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 algebracorresponding 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 orisomorphic to the affine space A n − rκ ( p ) .Proof. We know by a previous comment of the introduction that the fiberring over p is κ ( p ) ⊗ R A which is just κ ( p )[ T , . . . , T n ] / * f ( p ) . . . f n ( p )... . . . ... f m ( p ) . . . f mn ( p ) · T ... T n + f ( p )... f m ( p ) + . Now, making elementary row and column operations on the matrix ( f ij ( p )),as indicated before, we can obtain a matrix with zero entries except for thefirst r entries of the principal diagonal which are ones, plus an independentvector.In conclusion, after performing all the necessarily elementary operations,we obtain an isomorphism from A to a very simple forcing algebra B = κ ( p )[ T , ..., T n ] / ( T + g , ..., T r + g r , g r +1 , ..., g n ) , corresponding to the matrix with zero entries except for the first r entries ofthe principal diagonal, which are ones. But then B is clearly isomorph to theaffine ring κ ( p )[ T r +1 , ..., T n ], if g r +1 = · · · = g n = 0, and A = 0 otherwise,proving our lemma. (cid:3) If κ ( p ) is algebraically closed, then the fiber over a point p ∈ Spec R ofthis forcing algebra is just the solution set of the corresponding system ofinhomogeneous linear equations over κ ( p ). If the vector ( f , . . . , f m ) is zero,then we are dealing with a “homogeneous” forcing algebra. In this case thereis a (zero- or “horizontal”) section s : X = Spec R → Y = Spec A comingfrom 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 , . . . , T n )+ p ∈ Y . Remark 1.2.
If all f k are zero, and m = n , then the ideal a is defined bythe linear forms h i = f i T + · · · + f in T n , and in this case we can “translate”the fact of multiplying by the adjoint matrix of M , denoted by adj M , just tosaying that the elements det M T i ∈ a . In fact, det M T ...det M T n = det M · I nn · T ... T n = adj M · M · T ... T n = adj M · h ... h n DANNY DE JES ´US G ´OMEZ-RAM´IREZ AND HOLGER BRENNER where the entries of the last vector belong to a . From this fact we deducethat, when the determinant of M is a unit in R , then a = ( T , ..., T n ) andthe forcing algebra is isomorphic to the base ring R . Note that the previousargument works also in the nonhomogeneous case.Now, we study the homogeneous case when the elements { h , ..., h m } forma regular sequence. First, we need the following general fact about the purecodimension of regular sequences in Noetherian rings: Let S be a Noetherianring and { r , ..., r m } ⊆ S a regular sequence and I the ideal generated bythese elements. Then the pure codimension of I is m . For a proof see [6].Besides, if j ∈ { , ..., min( m, n ) } , then we define the Fitting ideals I j asthe ideals generated by the minors of size j of the matrix M . This definitioncorresponds to the standard definition of Fitting ideals regarding M as a R -homomorphism of free modules (see [7, p. 497]). Proposition 1.3.
Let R be a Noetherian integral domain and A the homo-geneous forcing algebra corresponding to the data { f ij } , with i = 1 , ..., n and j = 1 , ..., m . Suppose that the forcing equations { h , ..., h m } form a regularsequence in B := R [ T , ..., T n ] . Then m ≥ n and I min( m,n ) = (0) . Proof.
First, note that the ideal I generated by the forcing elements is con-tained in the homogeneous ideal P = ( T , ..., T n ), therefore we see that thedimension of A is smaller or equal to the dimension of B/P , which is exactlythe dimension of R . On the other hand, if we consider a saturated chain ofprimes in A , P * P * ... * P dim A , where P is a minimal prime over I , then the former comments ht( P ) = m andthus completing the former chain with a saturated chain for P in B of length m , we see that dim B ≥ m + dim A. Now, noting that dim B = dim A + n ,since A is Noetherian, we get n + dim A ≥ m + dim A , which implies that n ≥ m .For the second part, let’s consider the matrix M in the field of fractions K of R . It is an elementary fact that the rank of M is ≤ s if and only if everyminor of size s + 1 of M is zero. This follows from the fact that performingrow operations on a matrix change the values of fixed minor of size r of theoriginal matrix just by a nonzero constant term of another minor of size r of the changed matrix (this is a general way of saying that performing a rowoperation is just multiplying by an invertible matrix and therefore the factthat the determinant of the matrix is zero or not is independent of the rowoperation).Now, suppose by contradiction that I min( m,n ) = 0, then the rank of M in K is strictly smaller than min( m, n ) and thus the rows of M are lin-early dependent in K . Without loss of generality, assume that there is j ∈ { , ..., min( m, n ) } , such that the jth row of M , l j , is a linear combination ORMALITY AND RELATED PROPERTIES OF FORCING ALGEBRAS 9 of the former ones, that is, there exist some α i ∈ K , such that l j = P j − i =1 α i l i .Now, after multiplying by a nonzero common multiple β ∈ R of the denomi-nators, we get an equation of the form βl j = P j − i =1 γ i l i , for some γ i ∈ R . But,seeing this equality in A , (which means just multiplying this equality by the n × T i ) we see that βh j = P j − i =1 γ i h i , which implies that h j is a zero divisor in B/ ( h , ..., h j − ), because β / ∈ ( T , ..., T n ) ( β is a nonzeroconstant polynomial in B ) and therefore β / ∈ ( h , ..., h j − ). This contradictsthe fact that I is generated by a regular sequence. (cid:3) The converse of the previous proposition is false as the following exampleshows.
Example 1.4.
Consider R = k [ x ]; B = R [ T , T ]; h = xT − xT and h = xT + xT , where k is a field of char k = 2. Then m = n = 2 and thedeterminant of the associated matrix is 2 x , but the sequence { h , h } is notregular, in fact, the ideal I generated by its elements has height just one,because it is contained in the principal ideal ( x ), and therefore by formercomments, I cannot be generated by a regular sequence. Geometrically, thevariety defined by I is the union of a line V ( T , T ) and a plane V ( x ).Intuitively, this example comes from the following observation. Supposethat we have the forcing algebra with equations h ′ = T + T and h = T − T .If we consider the line V ( T − T , T + T ) = V ( T , T ) (whose associateddeterminant is 2 = 0) and multiplying these equations by x , we obtain avariety that is automatically the union of this line with the plane V ( x ) ⊆ k ,which has bigger dimension, but the associated determinant of the new variety(our former example) is just x times the former determinant. This processgives us a new variety with nonzero determinant but with an ideal with smallercodimension. 2. Irreducibility
Here we shall see that if A is a forcing algebra over a Noetherian integraldomain such that ht( f, f , ..., f n ) ≥
2, where { f , ..., f n , f } is the forcing data,then A is an irreducible ring (i.e. A has just one minimal prime). Theorem 2.1.
Let R be a Noetherian integral domain; A = R [ T , ..., T n ] / ( f T + · · · + f n T n + f ); h = f T + · · · + f n T n + f , where f , ..., f n , f ∈ R and J = ( f, f , ..., f n ) .Assume that ht J ≥ , then A is an irreducible ring.Proof. By Lemma [5, Lemma 2.1.(2)], it is enough to see that for any minimalprime q ∈ R of J , q B is not minimal over ( h ), because on that case, A hasjust the horizontal component, and therefore is irreducible.Let q ∈ R be minimal over J . Then,ht q B ≥ ht q ≥ ht J ≥ . Therefore q B is not minimal over ( h ), since by Krull’s Principal Ideal Theoremthe minimal primes over a principal ideal have height smaller or equal thanone. (cid:3) (Non)Reduceness In this section we study the (non)reducedness of forcing algebras over areduced base ring R . First, for a base field k [5, Lemma 3.1] shows thatany forcing algebra is isomorphic to a ring of polynomials over k or the zeroalgebra, therefore it is reduced.Now, if R is a local ring, let us first stay an elementary remark concerning ageneralization of the Monomial Conjecture (MC) (see [12]) in dimension one.In dimension one (CM) just says that if x ∈ m does not belong to anyminimal prime ideal of R then x n / ∈ ( x n +1 ) for all nonnegative integer n . Inthe next remark we will prove a generalization of this fact for a quasi-localring, that is, not necessarily Noetherian. Remark 3.1.
Let (
R, m ) be a quasi-local ring and x ∈ m . Then, there existsa positive integer n such that x n / ∈ ( x n +1 ), if and only if x is a nilpotent or aunit.In fact, one direction is trivial, for the other one, assume that x is neithernilpotent nor a unit and that there exists n ∈ N and y ∈ R such that x n = yx n +1 , thus x n (1 − yx ) = 0, but 1 − yx / ∈ m , therefore it is a unit, then x n = 0,which is a contradiction. Example 3.2.
Let (
R, m ) be a quasi-local reduced ring, which is not a field,and f ∈ m r { } . Then, the trivial forcing algebra A := R/ ( f ) is non-reduced because clearly f ∈ nil A and by the previous Remark f = A . So,there are always non-reduced forcing algebras over quasi-local reduced baserings, which are not a field.Now, we want to study in which generality we can guarantee the existenceof an element f ∈ R such that f / ∈ ( f ). The following Proposition gives acompact characterization of the fact that any element f ∈ R belongs to ( f ). Proposition 3.3.
A commutative ring with unity R is reduced of dimensionzero if and only if for any element f ∈ R , holds that f ∈ ( f ) . In particular,if R is noetherian then it is equivalent to the fact that R is a finite directproduct of fields.Proof. Assume that R is a reduced zero-dimensional ring. Then it is enoughto check the desired property locally. But, on that case R is a ring with aunique prime ideal, which is at the same time reduced, therefore it is a fieldand in particular f ∈ ( f ), for all f ∈ R .For the other direction, let P be a prime ideal of R . Clearly, the sameproperty holds for R/P , thus, for any g ∈ R/P there exists c ∈ R/P , such
ORMALITY AND RELATED PROPERTIES OF FORCING ALGEBRAS 11 that g = cg . Therefore, g (1 − cg ) = 0, implying that either g = 0 or 1 − cg = 0,because R is an integral domain. So R/P is a field and then R has dimensionzero. Finally, from the hypothesis follows that for any f ∈ R , f ∈ ( f m ), forevery natural number m . In particular, if f is nilpotent and f m = 0, for some m ∈ N , then f ∈ ( f m = (0)). In conclusion, f is reduced. The second part isa direct consequence of the Chinesse Remainder Theorem. (cid:3) Remark 3.4.
The previous proposition guarantees the existence of non-reduced forcing algebras over any noetherian ring which is not a finite directproduct of fields. Specifically, as before, we choose an element f ∈ R , suchthat f / ∈ ( f ) and define A := R/ ( f ).Finally, we present a more interesting example of an irreducible but notreduced forcing algebra over an affine domain base ring R such that thecodim(( f , ..., f n ) , R ) is arbitrary large. Example 3.5.
Consider R = k [ x , ..., x n − , z ] / ( x z, ..., x n +1 z ), h = x T + · · · + x n +1 T n +1 + z and A = R [ T , ..., T n +1 ) / ( h ). Then,codim(( x , ..., x n +1 ) , R ) = n, because the ring of polynomials is catenary. Besides, it is straightforward toverify that z / ∈ ( h ), and z = z h ∈ ( h ). Therefore A is non-reduced.4. Integrity over UFD
Now, we prove an integrity criterion for forcing algebras over UFD as basering involving just the height of the forcing elements f , ..., f n . Lemma 4.1.
Let R be a Noetherian UFD which is not a field, J = ( f , . . . , f n , f ) ,where some f i = 0 , and let A be the forcing algebra corresponding to this dataand B = R [ T , ..., T n ] . Then A is an integral domain if and only if J = R , or ht J ≥ .Proof. Along the proof we will use the basic fact that in a UFD the notionsof prime and irreducible element coincide. We will prove the negation of theequivalence (( h ) ∈ Spec B ) ⇔ ( I = R ∨ ht J ≥ h ) / ∈ Spec B ) ⇔ ( I = R ∧ ht J ≤ J ≤
1, assuming implicitly that ht I is well defined, i.e., I = R . So we will see that A is not an integral domain if and only if ht J ≤ J = 0 and therefore ht J = 1. Choose a prime ideal P of R such that P contains J and ht P = 1. Choose a = 0 ∈ P . Now, some of theprime factors of a , say p , belongs to P and therefore P = ( p ), due to the factthat both prime ideals have height one. Thus, there exist g i , g ∈ R such that f i = pg i and f = pg , hence h = f T + . . . + f n T n + f = p ( g T + . . . + g n T n + g )is the product of p and an element which is not a unit since some of the f i is different from zero. Therefore h is not irreducible, which is equivalent ofbeing a non prime element. In conclusion, A is not an integral domain. Conversely, assume that A is not an integral domain, or equivalently that h = f T + . . . + f n T n + f is not irreducible. Hence, there exists polynomials Q , Q ∈ R [ T , . . . , T n ], not units, such that h = Q Q . Now, the degree of h is the sum of the degrees of Q and Q , because R is an integral domain. Thenone of the two factors has degree zero, say Q . Comparing the coefficients weget that each f i = Q g i and f = Q g , and Q = g T + . . . + g n T n + g . Inconclusion, J ⊆ ( Q ) R and therefore by Krull’s Theorem ht( J ) ≤ (cid:3) A Normality Criterion for Polynomials over a Perfect Field
Now we will try to understand under what conditions on the elements f , . . . , f n , f ∈ R the associated forcing algebra is a normal domain in thecase that R is the ring of polynomials over a perfect field. For some exam-ples, results and intuition we assume a very basic and modest knowledge ofalgebraic geometry, mainly relating affine varieties (see, for example [9] and[11, Chapter I]).We state explicitly the statement of a corollary of the Jacobian Criterion,which we use later. For proofs see [7, Theorem 16.19, Corollary 16.20]. Corollary 5.1.
Let R [ x , . . . , x r ] /I be an affine ring over a perfect field k and suppose that I has pure codimension c , i.e., the height of any minimalprime over I is exactly c . Suppose that I = ( f , . . . , f n ) . If J is the idealof R generated by the c × c minors of the Jacobian matrix ( ∂f i /∂x j ) , then J defines the singular locus of R in the sense that a prime P of R contains J ifand only if R P is not a regular local ring. Besides, let us recall Serre’s Criterion for normality for any Noetherian ring(see [7, Theorem 11.2.]). Remember that a ring is normal if it is the directproduct of normal domains:
Theorem 5.2.
A Noetherian ring S is normal if and only if the followingtwo conditions holds:(1) (S2) For any prime ideal P of S holds depth P ( S P ) ≥ min(2 , dim( S P )) . (2) (R1) Every localization of S on primes of codimension at most one isa regular ring. Remark 5.3. If R = k [ x , . . . , x r ] and h = f T + . . . + f n T n + f ∈ B := R [ T , . . . , T n ], h = 0, then the forcing algebra A = R [ T , . . . , T n ] / ( h ) isequidimensional of dimension dim A = r + n − ht(( h )) = r + n −
1, since= R [ T , . . . , T n ] is catenary and h has pure codimension one, because everyminimal prime over ( h ) has height one by Krull’s principal ideal theorem.Therefore in the case that k is a perfect field we deduce from the corollary of ORMALITY AND RELATED PROPERTIES OF FORCING ALGEBRAS 13 the Jacobian criterion that the singular locus of the forcing algebra is exactlythe prime spectrum of the following ring A S = A/ (( ∂h/∂x j ) , ( ∂h/∂T i )) = R [ T , . . . , T n ] / ( h, ( ∂h/∂x j ) , ( ∂h/∂T i )) . Now, ( ∂h/∂x j ) = P ni =1 ( ∂f i /∂x j ) T i + ( ∂f /∂x j ) and ∂h/∂T i = f i . Thus weget J := ( h, ( ∂h/∂x j ) , ( ∂h/∂T i )) = ( h, n X i =1 ( ∂f i /∂x j ) T i + ( ∂f /∂x j ) , f i )= ( f, f i , n X i =1 ( ∂f i /∂x j ) T i + ( ∂f /∂x j )) , where i, j ∈ { , ..., n } . We can write the last set of generators in a compactway using matrices: ∂f /∂x . . . ∂f n /∂x ... ... ∂f /∂x r . . . ∂f n /∂x r · T ... T n + ∂f /∂x ... ∂f /∂x r . We will denote by J the class of J in A .Now we rewrite the normality condition for the forcing algebra A in termsof the codimension of its singular locus V ( J ) ∈ Spec A , or in terms of thecodimension of the corresponding closed subset V ( J ) ⊆ Spec( R [ T , . . . , T n ]),which are isomorphic as affine schemes. On this section we set I = ( f, f , ..., f n ) ∈ R and D = ( ∂f /∂x i , ∂f j /∂x i ) for i, j ∈ { , ..., n } . Note that J ⊆ ( I + D ) B . Inparticular. V ( IB ) ∩ V ( DB ) ⊆ V ( J ) ⊆ Spec B .First, let’s consider the trivial case R = k . By previous comments we knowthat if A = 0 then A = k [ T , ..., ˇ T i , ..., T n ], so A is regular and thus a normaldomain. In conclusion, A is a normal domain if and only if all f i and f arezero, or there exists some f i = 0. Lemma 5.4. let R = k [ x , . . . , x r ] be the ring of polynomials over a perfectfield k and h = f T + . . . + f n T n + f ∈ B := R [ T , . . . , T n ] , with h = 0 , and A = R [ T , . . . , T n ] / ( h ) . Then, the following conditions are equivalent:(1) A is a normal ring.(2) codim( J , A ) ≥ , or J = A .(3) codim( J, B ) ≥ , or J = B .Proof. (1) ⇒ (2) Assume that A is a normal ring, then the Serre’s Criteriontells us that for any prime ideal q of A with ht q ≤ A q is a regular ring(remember that in dimension zero regularity is equivalent to being a field). Now, suppose that J ( A . Then, we know that for any prime P of A thatcontains J , A P is not regular, therefore ht P ≥
2, thus codim(
J, A ) ≥ ⇒ (1) We know that A is C-M because it is the quotient of C-M R [ T , . . . , T n ] by an ideal ( h ) of height one generated by exactly one element,(see [7, Theorem 18.13]). Therefore, for any prime ideal P of A the local ring A P is C-M. Then,depth( A P ) = dim( A P ) ≥ min(2 , dim( A P )) . Thus A satisfies the condition (S2) of the Serre’s Criterion. Besides, A satisfiescondition (R1). In fact, any prime ideal P of A of height at most 1 does notcontain J , because codim( J , A ) = ht A ( J ) ≥
2, or J = A , hence P is not inthe singular locus of A , that means the regularity of the local ring A P .Since, J = A if and only if J = B then, for the equivalence between (2)and (3) we can assume that J ( A (respectively J ( B ).(2) ⇒ (3) . Let P be a prime ideal of B that contains J , then by hypothesisht A ( P ) ≥
2. Let P $ P $ P = P be a chain of primes in A , then one cansee the corresponding chain of prime ideals in B adding the zero ideal, whichis a prime ideal, Q = (0) $ Q = P $ Q = P $ Q = P = P , that meanscodim( J, B ) ≥ ⇒ (2) Let P be a prime ideal of A that contains J , and let Q be theprime ideal of B that correspond to P . Clearly, J ⊆ Q as subsets of B . Weknow that ht( Q ) ≥ h ) ⊆ Q , therefore Q contains a minimal prime idealof ( h ), say Q , which has height one by Krull’s Principal Ideal Theorem. Invirtue of this, we know that there exists a saturated chain of primes ideals of B , Q = (0) $ Q $ Q $ Q ⊆ Q, since B is a catenary ring and ht Q ≥
3, and therefore any saturated chainof prime ideals from (0) to Q has the same length, that is, ht( Q ), which is aleast three. Therefore, looking at the corresponding chain in A , and denotingby P i − the prime ideal of A corresponding to Q i , we get, starting with theclass of Q , the following saturated chain: P $ P $ P ⊆ P , then ht P ≥ J , A ) ≥ (cid:3) Remark 5.5.
An important fact is that for R = k [ x , . . . , x r ], I an ideal of R and B = R [ T , ..., T n ] we know that the codim( I, R ) = codim(
IB, B ), becauseby previous results we get n + r − codim( IB, B ) = dim(
B/IB ) = dim((
R/I )[ T , . . . , T n ]) =dim( R/I ) + n = dim R − codim( I, R ) + n = n + r − codim( I, R ) . We want to find necessary and sufficient conditions for the forcing data f , ..., f n and f on the base ring of polynomials R = k [ x , ..., x n ], such thatthe associated forcing algebra turns out to be a normal domain. The previ-ous lemma gives a condition over A and the Jacobian ideal J of the partial ORMALITY AND RELATED PROPERTIES OF FORCING ALGEBRAS 15 derivatives of the forcing equation, which involves, as seen before, again theforcing ideal and new forcing equations defined by the partial derivatives ofthe original forcing data. This suggests that a suitable condition for normal-ity over the base R should involve the forcing data and its partial derivatives.The following collection of examples start to give us a good first intuition ofthe phenomenon. Example 5.6.
Let k be a perfect field and let’s define R = k [ x, y ]; B = k [ x, y, T , T ]; A = B/ ( h ) and h = x a T + y b T + x c y d , where a, b, c and d are nonnegative integers. After computations we have thatthe Jacobian ideal J = ( x a , y b , x c y d , ax a − T + cx c − y d , by b − T + dx c y d − ) . Let D ⊆ R be the ideal generated by all the partial derivatives of the genera-tors of the forcing ideal I = ( f , f , f ) = ( x a , y b , x c y d ), i.e., D = ( ax a − , by b − , cx c − y d , dx c y d − ) . By Lemma 4.1, A is a domain for any nonnegative values of the exponents.After elementary considerations we see that codim( J, B ) ≥ J = B ifand only if some of the following seven cases occur:i) a = 0.ii) a = 1.iii) b = 0.iv) b = 1.v) d = c = 0.vi) c = 1 and d = 0.vii) c = 0 and d = 1.In fact, in any other case J ⊆ ( x, y ) B , and therefore codim( J, B ) ≤ I + D is equal to R .In conclusion, in virtue of the previous Lemma, A is a normal domain ifand only if I + D = R . Remark 5.7.
Suppose that k is an algebraically closed field. Continu-ing with the notation of the former example, let’s write V = V ( I ) ⊆ k , W = V ( D ) ⊆ k , Y = V ( h ) ⊆ k and S = V ( J ) ⊆ k denote the corre-sponding affine varieties and π : S → V the natural projection to the firsttwo coordinates. Geometrically, Example 5.6 suggests that the normality ofthe variety X (which is equivalent to the normality of the forcing algebra,see [11, Exercise I.3.17]), is related to the intersection of V and W , because V ∩ W = ∅ , if and only if I + D = R . In fact, this is true for arbitrarypolynomial data f , f and f ∈ R as we will see. First, by Lemma 4.1, A is an integral domain if and only if ht I ≥ I = R . So, let’s assume that A is a domain and I ( R , otherwise V = ∅ and J = B , being A normal, by Lemma 5.4. Thus, ht I ≥
2, which means thatthe minimal prime ideals over I are just finitely many maximal ideals, sincedim R = 2. But, by the Nullstellensatz (see [1, Exercise 7.14]) this pointscorrespond exactly to the points of V . Therefore, let’s write V = { v , ..., v r } .Moreover, let S be the singular locus of Y in the sense that, if we consider S as a subvariety of Y . By previous comments S is the finite union of its(singular) fiber varieties S v i = π − ( v i ). Now, by Lemma 5.4, Y is a normalvariety if and only if codim( S, K ) ≥ S, Y ) ≥ V ∩ W = ∅ , i.e., I + D ( R , and let’s prove that Y is notnormal. In fact, we know that J ⊆ ( I + D ) B . Therefore, by Remark 5.5codim( S, k ) = codim( J, B ) ≤ codim(( I + D ) B, B ) = codim( I + D, R ) ≤ , implying that Y is not normal.Conversely, assume that V ∩ W = ∅ . Then, for any point v ∈ V , thereexists some ∂f i ( v ) /∂x j = 0, because if not all the partial derivatives of theforcing data would be zero at v (the elements ∂f ( v ) /∂x j are also zero, becausewe can write them as a linear combinations of the ∂f i ( v ) /∂x j , see Remark5.3), implying that v ∈ W , but that is impossible.Clearly, S v = V ( G ), where v = ( a, b ) ∈ k and G = ( x − a, y − b, ∂f ( v ) /∂xT + ∂f ( v ) /∂xT + ∂f ( v ) /∂x,∂f ( v ) /∂yT + ∂f ( v ) /∂yT + ∂f ( v ) /∂y ) . But, under the condition that some ∂f i ( v ) /∂x j = 0, it is elementary to seethat codim( G, B ) ≥
3. In conclusion, codim( S v , k ) ≥
3, implying thatcodim(
S, k ), being the minimum of the codimension of its singular fibers,is bigger or equal than three, which means the normality of Y .Besides, if we move to the next dimension, i.e., R = k [ x , x , x ] and B = R [ T , T , T ], then, it is possible to see in a natural way that a necessarycondition for the normality of Y is that (dim V ∩ W ) < − V ∩ W ≥
1. For any point v ∈ V ∩ W , by Remark 5.3 and Lemma 1.1the fiber S v ∼ = A k . Therefore, ( V ∩ W ) × A k ⊆ S . But, dim( V ∩ W ) × A k ≥ S ≥
4, thus, codim(
S, k ) ≤
2, implying that Y isnot normal. Note that this argument works independent from the number ofvariables. However, this case was very suitable to obtain the right intuitionabout the desired condition i.e., dim( V ∩ W ) < r − S by knowing the generalbehavior of the dimension of the fibers S v and the dimension of the basespace V . Now, by Lemma 1.1 the fibers S v have maximal dimension exactly ORMALITY AND RELATED PROPERTIES OF FORCING ALGEBRAS 17 when the rank of the forcing matrix is minimal, i.e., when the point v belongsto W ∩ V . Therefore, to guarantee that the dimension of Y is not so big(in order to maintain the codimension big enough), we need to bound thedimension of the subvariety of V with maximal dimensional singular fibers,i.e., the dimension of V ∩ W . In fact, assuming that Y is irreducible, the rightnecessary and sufficient condition for Y being an (irreducible) normal varietyis that (dim V ≤ r − V ∩ W ≤ r −
3, where
V, W ⊆ k r .First, in order to get a better intuition about the fibers, the followingproposition tells us that the points of Spec R with fibers completely singularare exactly the points of V ( I ) ∩ V ( D ). Proposition 5.8.
Let R = k [ x , . . . , x r ] be the ring of polynomials over aperfect field k ; B = R [ T , . . . , T n ] ; h = f T + · · · + f n T n + f ; f, f , . . . , f n ∈ R ; A = B/ ( h ) ; I = ( f, f , . . . , f n ) ; D = ( ∂f /∂x j , ∂f i /∂x j ) and J := ( h, ( ∂h/∂x j ) , ( ∂h/∂T i )) . Let ϕ : Y = Spec A → X = Spec R be the forcing morphism. Choose a point x ∈ Y with nonempty fiber ϕ − ( x ) . Then x ∈ X has fiber completely singulari.e., ϕ − ( x ) ⊆ V ( J ) ∈ Y if and only if x ∈ V ( I + D ) ⊆ X .Proof. We know from the Corollary of the Jacobian Criterion that for anyprime ideal y ∈ Y , A y is not regular if and only if y ∈ V ( J ). Let x ∈ V ( I + D )and Q ∈ ϕ − ( x ). Then, ( I + D ) B ∈ Q and so J ∈ Q , meaning that Q ∈ V ( J ).Conversely, let’s consider a point x ∈ X , such that ϕ − ( x ) ⊆ V ( J ). Now,it is elementary to see that the last condition means that ϕ − ( x ) = V ( J x ),where ϕ − ( x ) = A = k ( x )[ T , . . . , T n ] / ( f ( x ) T + . . . + f n ( x ) T n + f ( x )) , and J x = ( P ni =1 ( ∂f i ( x ) /∂x j ) T i + ( ∂f ( x ) /∂x j ) , for i, j ∈ { , ..., n } .Firstly, if f i / ∈ x , for some i , then the fiber ϕ − ( x ) is completely regular,because, by previous comments (Ch. 1 § ϕ − ( x ) ∼ = A n − k ( x ) .Secondly, if f / ∈ x , then f ( x ) = 0. But, we know that f ( x ) = · · · = f n ( x ) = 0, therefore the fiber is empty, since h = f ( x ) = 0 ∈ k ( x ). But, itcontradicts our hypothesis. Note that, until now, we know that h = f ( x ) T + . . . + f n ( x ) T n + f ( x ) = 0.Thirdly, suppose that ∂f i /∂f j / ∈ x , for some i, j ∈ { , ..., n } , that means, ∂f i ( x ) /∂f j = 0. We consider two cases: Suppose that ∂f ( x ) /∂f j = 0. Then,since h = 0, the ideal Q = ( T , ..., T n ) ∈ ϕ − ( x ), but n X i =1 ( ∂f i ( x ) /∂x j ) T i + ( ∂f ( x ) /∂x j ) / ∈ Q. Therefore
Q / ∈ V ( J x ), a contradiction. In the second case, i.e., ∂f ( x ) /∂f j =0 , , the prime ideal Q ′ = ( T , ..., T i − , ..., T n ) ∈ ϕ − ( x ), but n X i =1 ( ∂f i ( x ) /∂x j ) T i + ( ∂f ( x ) /∂x j ) = n X i =1 ( ∂f i ( x ) /∂x j ) T i / ∈ Q ′ . So, again, Q ′ / ∈ V ( J x ), a contradiction.Lastly, if ∂f ( x ) /∂x j = 0, for some j , then, due to the last results n X i =1 ( ∂f i ( x ) /∂x j ) T i + ( ∂f ( x ) /∂x j ) = ∂f ( x ) /∂x j ∈ J x , thus ϕ − ( x ) = V ( J x ) = ∅ . But, this is not possible, because the fiber is notempty.In conclusion, ϕ − ( x ) ⊆ V ( J ) ∈ Y , as desired. (cid:3) Now, we present the statement of the normality criterion for forcing alge-bras over the ring of polynomials with coefficients in a perfect field.
Theorem 5.9.
Let R = k [ x , . . . , x r ] be the ring of polynomials over a per-fect field k ; B = R [ T , . . . , T n ] ; f, f , . . . , f n ∈ R ; I = ( f, f , . . . , f n ) ; D =( ∂f /∂x j , ∂f i /∂x j ) , for i, j ∈ { , ..., n } . Then, the forcing algebra for this data A is a normal domain if and only if the following two conditions hold:(a) codim( I, R ) ≥ , or I = R .(b) codim( I + D, R ) > , or I + D = R .Moreover, in the case that all f i = 0 , then (b) is a necessary and sufficientcondition for A being a normal ring.Proof. We have already proved in Lemma 4.1 that (a) is a necessary andsufficient condition for A being an integral domain. Let’s prove that (b) isequivalent to normality. Effectively, following Lemma 5.4 we just need to seethe condition (b) is equivalent to codim( J, B ) >
2, or J = B . Let’s denote thelast condition by (b’). By Remark 5.3 we know that J ⊆ ( I + D ) B . Supposethat (b’) holds. First, if J = B , then ( I + D ) B = B , implying I + D = R .Second, if codim( J, B ) >
2, then by Remark 5.5 we getcodim( I + D, R ) = codim(( I + D ) B, B ≥ codim( J, B ) > . Conversely, assume that (b) holds and J = B . We prove that codim =( J, B ) >
2. Let Q be a prime ideal of B that contains J . First, assume that( I + D ) B ⊆ Q , then I + D = R , therefore codim( I + D, R ) >
2, so, again byRemark 5.5 codim(( I + D ) B, B ) >
2, which implies that codim(
Q, B ) > I + D * Q , then necessarily one of the partial derivatives ∂f /∂x j or ∂f i /∂x j is not contained in Q , because IB ⊆ J ⊆ Q . In fact, there ORMALITY AND RELATED PROPERTIES OF FORCING ALGEBRAS 19 exits some b ∈ , . . . , n and some c ∈ , . . . , r with ∂f b /∂x d / ∈ Q , cause if not,all ∂f i /∂x j would be contained in Q and also the elements P ni =1 ( ∂f i /∂x j ) T i + ∂f /∂x j and therefore ∂f /∂x j for any j , thus D would be also contained in J , which is not the case. For simplicity suppose that Q not contained theelement α := ∂f /∂x and let’s write l := P ni =1 ( ∂f i /∂x ) T i + ∂f /∂x . Let ψ be the following homomorphism of R ( α ) algebras ψ : B ( α ) ≅ R ( α ) [ T , . . . , T n ] −→ R ( α ) [ T , . . . , T n ] , that sends T to g := − α − ( P ni =2 ( ∂f i /∂x ) + ∂f /∂x ) and T j to T j , for j ≥
2. Clearly, ψ is surjective. Moreover, ker ( ψ ) = ( T − g ). To see this let S ∈ ker ( ψ ). Then using the binomial expansion we can write it in the form: S = S ( x , . . . , x r , ( T − g ) + g, . . . , T n ) = S ( x , . . . , x r , ( T − g ) , . . . , T n )+ S ( x , . . . , x r , g, . . . , T n ) , = S ( x , . . . , x r , ( T − g ) , . . . , T n ) + ψ ( S )= S ( x , . . . , x r , ( T − g ) , . . . , T n ) , with S being divisible by T , which implies that the former expression isdivisible by T − g . Thus S ∈ ( T − g ).On the other hand, in the ring R ( α ) [ T , . . . , T n ] we know that ( T − g ) =( l ), therefore ψ induces an isomorphism between R ( α ) [ T , . . . , T n ] / ( l ) and R ( α ) [ T , . . . , T n ]. Denote by Q the image under ψ of QR ( α ) [ T , . . . , T n ], andassume for the sake of contradiction that codim( Q, B ) ≤ d := dim( B/Q ) = dim B − codim( Q, B ) = n + r − codim( Q, B ) ≥ n + r − . Besides, B is a Jacobson ring, hence there exists a maximal ideal m con-taining Q such α / ∈ m , otherwise α would be contained in the intersection ofall the maximal ideals containing Q , which is Q , absurd. Now, let’s consider asaturated chain of primes ideals from Q to m , which exits in virtue of Zorn’slemma. Besides, this chain has length exactly d because B/Q is an affinedomain and therefore, d is the length of any saturated chain of primes on it(see fundamental results on Chapter 1). Then, Q = Q $ Q $ , . . . , $ Q d − $ Q d = m. Now, we can consider this chain in R ( α ) [ T , . . . , T n ], because no Q i contains α . This shows that dim( R ( α ) [ T , . . . , T n ]) /Q e ≥ d and, in fact, the equalityholds because we are localizing and thus the dimension cannot be biggerthat the dimension of the original ring. Besides, ψ induces an isomorphismbetween R ( α ) [ T , . . . , T n ] /Q e and R ( α ) [ T , . . . , T n ] /Q , then finally, recallingthat codim( I, R ) ≥ l ∈ Q we get d = dim( R ( α ) [ T , . . . , T n ]) /Q e ) = dim( R ( α ) [ T , . . . , T n ]) /Q ) ≤ dim( R ( α ) [ T , . . . , T n ]) /I e ) ≤ dim( R [ T , . . . , T n ]) /I e ) = dim(( R/I )[ T , . . . , T n ]) = dim( R/I ) + n − R − codim( I, R ) + n − ≤ r + n − − < n + r − . Which is a contradiction with the former estimate of d . Finally, if all f i = 0then J = I + D and then from the fact that codim(( I + D ) , R ) = codim(( I + D ) , B ) we deduce from Lemma 5.4 that condition (b) is equivalent to thenormality of A . (cid:3) Now, we state a direct application of the previous Theorem to normal affinevarieties. As said before, our convention is that dim ∅ = − Corollary 5.10.
Let R = k [ x , . . . , x r ] be the ring of polynomials over analgebraically closed field k ; B = R [ T , . . . , T n ] ; f, f , . . . , f n ∈ R ; I = ( f, f , . . . , f n ) and D = ( ∂f /∂x j , ∂f i /∂x j ) . Assume that ( h ) is a radical ideal, where h = f T + · · · + f n T n + f. Let’s denote by V = V ( I ) ⊆ k r and W = V ( D ) ⊆ k r theaffine varieties defined by I and D , respectively. Then, X = V ( H ) ⊆ k n + r isa normal (irreducible) variety if and only if the following two conditions holdssimultaneously(1) dim V ≤ r − .(2) dim( V ∩ W ) < r − .Moreover, in the case that all f i = 0 , then (2) is a necessary andsufficient condition for X being a normal (irreducible) variety.Proof. Recall that a variety is normal if for any point x ∈ X , the stalk O X,x is a normal domain (see [11, Exercise I.3.17]). Since ( h ) is a radical ideal, weknow that the forcing algebra A = B/ ( h ) is exactly the ring of coordinatesof X . Since X is affine and normality is a local property we have that X isa normal (irreducible) variety if and only if A is a normal domain. Besides,from Hilbert’s Nullstellensatz we getdim V = dim( R/I ( V )) = dim( R/ rad( I )) = dim( R/I )= dim R − codim( I, R ) = r − codim( I, R ) , and analogously dim( V ∩ W ) = r − codim( I + D, R ) . From this and the fact that V = ∅ (or V ∩ W = ∅ ), if and only if I = R (or I + D = B ), we rewrite the conditions (a) and (b) of the former theorem as(1) and (2). (cid:3) As a comment, we say that the discussion beginning at Example 5.6 isessentially the way in which the above criterion of normality was discovered.Lastly, in order to support the former intuition we dedicate the next pair ofsections to study two interesting and enlightening examples.
ORMALITY AND RELATED PROPERTIES OF FORCING ALGEBRAS 21 An Enlightening Example
In this section we study an specific example of a forcing algebra with sev-eral forcing equations and we explore the some interesting properties. Thisexample shows how rich and interesting is the formal study of forcing algebrason its own.Let R = k [ x, y ] be the ring of polynomials over a (perfect) field k , B = R [ T , T ], A = B/H , where H = ( h , h ) = ( xT + yT , yT + xT ) = (cid:18)(cid:18) x yy x (cid:19) · (cid:18) T T (cid:19)(cid:19) . The determinant of the associated matrix M is x − y = ( x + y )( x − y ).It is easy to check that h is irreducible and that h does not belong to theideal generated by h . Therefore h , h ⊆ B is a regular sequence and hence,by former comments H has pure codimension 2.Let P be a minimal prime of H . Then, by a previous remark, P containsthe elements det M T i = ( x − y )( x + y ) T i for i = 1 ,
2. If det
M / ∈ P , then T i ∈ P , and therefore P = ( T , T ). Now, assume that det M ∈ P , then x − y ∈ P or x + y ∈ P . In the first case, h − T ( x − y ) = y ( T + T )should be in P . But, if y ∈ P then x = ( x − y ) + y ∈ P , which implies that P = ( x, y ). If T + T ∈ P then it is easy to check that P = ( x − y, T + T ),since this is a prime ideal containing H . On the other hand, if x + y ∈ P , then,similarly we see that P = ( x, y ), or P = ( x + y, T − T ). In conclusion, theminimal primes of H (which are, in fact, the associated primes of H , because A is a Cohen-Macaulay ring) are the four ideals P = ( T , T ), P = ( x, y ), P = ( x − y, T + T ) and P = ( x + y, T − T ).This example shows that Theorem 2.1 is false for several forcing equations,since Spec A is not an irreducible space but the ideal generated by the forcingdata ( x, y ) has height two.Let V i = V ( P i ) ⊆ k be the affine variety define by P i , which correspondto the irreducible components of V = V ( H ). Now, the intersections of anycouple of this components correspond to singular points of V (we assume fora while that k is algebraically closed, and we replace H by rad H in order towork with the corresponding variety V ), because the ring of coordinates of V localized at the maximal ideal corresponding to such a points has at leasttwo irreducible components and therefore it is not an integral domain, inparticular, it is not a regular local ring, since local regular rings are domains.This is a way to see geometrically the non-normality of V , because the nor-mality is a local property and the localization at these intersection points, say p ∈ Spm( A ), is not a normal ring. In fact, a local ring has clearly a connectedspectrum, therefore A p cannot be a direct product of normal domains ([5] § A p cannot be neither a normal domain. Returning to our computations, we see that the intersection of these irre-ducible components are, in general, defined by lines and, in two cases, definedby just one point. In fact, V ∩ V = V ( x, y, T , T ); V ∩ V = V ( T , T , x − y ); V ∩ V = V ( T , T , x + y ); V ∩ V = V ( x, y, T + T ); V ∩ V = V ( x, y, T − T )and V ∩ V = V ( x, y, T , T ).Furthermore, It is easy to see that Spec A is connected (see [5, Proposition1.2.]), since we are in the homogeneous case. Moreover, V ( P ) is an hori-zontal component, V ( P ) a vertical component and V ( P ) and V ( P ) behavelike “mixed” components i.e., they do not dominate the base nor are theythe preimage of a subset of the base. Besides, Spec A is also locally (over thebase) connected because every pair of minimal components have non-emptyintersection and the elementary fact that the minimal primes of a localiza-tion are exactly the minimal primes of the original ring not intersecting themultiplicative system.In the case that k is a perfect field, we can use also the Jacobian Criterionin order to prove again that A is not a normal ring. In fact, as seen before,the pure codimension of H is two, since { h , h } is a regular sequence. So, thesingular locus in Spec A is given by the 2 × h i , that is, J = (cid:0) T − T , x − y , yT − xT , xT − yT (cid:1) . Thus in order to test normality we should find the codimension of J in A anddetermine if it is bigger or equal than two. Since the pure codimension of H is two we can translate our problem to the ring of polynomial in four variables B = k [ x, y, T , T ] and to test if the corresponding Jacobian ideal J = (cid:0) T − T , x − y , yT − xT , xT − yT , h , h (cid:1) has codimension bigger or equal to four (in general, the codimension of aprime ideal decreases in n , if we mod out by ideals of pure codimension n ,mainly because an affine domain is catenary and its dimension is the lengthof any maximal chain of prime ideals [7, Corollary 13.6]. But, after somecomputations we can show that the prime ideals that contain J are exactlythe ideals defining the varieties corresponding to the intersections of pairsof the irreducibles components of V . That is, ( x, y, T , T ), ( T , T , x − y ),( T , T , x + y ), ( x, y, T + T ) and ( x, y, T − T ). Therefore codim( J , B ) = 3,and then, codim( J , A ) = 1 <
2, implying that A does not satisties Serre’scondition (R1). Hence, by Serre’s Normality Criterion A is not a normalring. Moreover, by the same reason B/ rad H is not a normal ring, and thisis equivalent to the non-normality of the variety V ( H ) ⊆ k .Geometrically, if k is an algebraic closed field, it means just that the sin-gular points of V , which correspond to the maximal ideals containing J , areexactly the points in the intersections of the different irreducible componentsof the variety, which correspond to the geometrical intuition of singularities. ORMALITY AND RELATED PROPERTIES OF FORCING ALGEBRAS 23
This example suggests the following conjecture.
Conjecture 6.1.
In the homogeneous case, assume that R = k [ x , ..., x r ],and suppose H = ( h , ..., h m ) = P ∩ ... ∩ P s , where P i are the minimalprimes, for i = 1 , ..., s . Then V ( P i ) ∩ V ( T , ..., T n ) = ∅ .7. An Example of Normalization
On this section we will compute explicitly the normalization of a forcingalgebra by elementary methods illustrating how good examples lead us in anatural way to the study of general basic properties of normal domains.Let k be a perfect field. Our example is a particular case of the Example 5.6.Let R = k [ x, y ], B = R [ t, s ], A = B/ ( h ), where h = x t + y s + xy . Now, withthe notation of section 3, I = ( x , y , xy ), D = ( x, y ), and so, I + D = ( x, y ).By Theorem 5.9 A is a non-normal domain, because codim( I, B ) ≥
2, butcodim( I + D, B ) = 2. Besides, the integral closure, or normalization of A , A is a module-finite extension of A (in general, this is true for finitely generatedalgebras over complete local rings, see [15, Exercise 9.8]).Now, we will give an explicit description of A as an affine domain.First, let K = K ( A ) be the field of fractions of A and let u = tx/y ∈ K .Then, if we consider the forcing equation h in K [ t, s ], we get the followingintegral equation for u , after multiplication by t/y :( tx/y ) + ( tx/y ) + st = 0 . Let A ′ = A [ u ] be the A − subalgebra of K generated by u . So, we rewrite h considered in A ′ , by means of yu = xt , to obtain the equation 0 = h = y ( xu + ys + x ). But, y = 0, therefore xu + ys + x = 0.Let C = k [ X, Y, T, S, U ] be the ring of polynomials. Define φ : C → A ′ the homomorphism of k − algebras sending each capital variable into itscorresponding small variable. Note that from the previous considerations theideal P = ( Y U − XT, XU + Y S + X, U + U + T S ) ⊆ ker φ . We will see that P = ker φ . Effectively, let’s write E = k [ X, Y, U, T ] / ( Y U − XT ). Then, E isa forcing algebra and by Theorem 5.9 is a normal domain.First, we prove that P is a prime ideal. Define Q = K ( E ), then, informallyif we consider the equations XU + Y S + X = U + U + T S = 0in the variable S and solve them, it lead us to obtain the equality S = − ( U + U ) /T = − ( XU + X ) /Y in a “suitable” field of fractions. But, in fact, it hodsthat − ( U + U ) /T = − ( XU + X ) /Y ∈ Q, because − Y ( U + U ) = − T XU − XT = − T ( XU + X ) ∈ D , due to the fact that Y U = XT ∈ E . Write S ′ = − ( U + U ) /T = − ( XU + X ) /Y ∈ Q and consider the natural homomorphism ψ : E [ S ] → E [ S ′ ] ⊆ Q ,where E [ S ] denote the ring of polynomials in the variable S . We will provethat ker ψ = ( XU + Y S + X, U + U + T S ). For that we need the followingbasic lemma about normal domains:
Lemma 7.1.
Let R be a normal domain, q ∈ K ( R ) , I = ( bx − a ∈ R [ x ] : q = a/b ; a, b ∈ R } , and ( R : q ) = { b ∈ R : bq ∈ R } be the denominator ideal. Consider thehomomorphism of R − algebras ϕ : R [ x ] → R [ q ] ⊆ K ( R ) , sending x to q . Then the following holds:(1) If q / ∈ R , then codim(( R : q ) , R ) = 1 .(2) Suppose that ( R : q ) = ( b , ..., b m ) ∈ R , such that q = a i /b i , for some a i ∈ R . Then, I = ( b x − a , ..., b m x − a m ) .(3) ker ϕ = I .Proof. (1) It is a well known fact that any normal Noetherian domain isthe intersection of its localizations on primes of height one (see [7, Corollary11.4]). We argue by contradiction. If codim(( R : q ) , R ) ≥
2, then ( R : q ) isnot contained in any prime ideal P ⊆ R of height one. In particular, thereexists for every such prime ideal P an element b P / ∈ P , but b P ∈ ( R : q ),meaning that there is a P ∈ R , with q = a P /b P ∈ R P . In conclusion, q ∈∩ ht P =1 R P = R .(2) Let bx − a ∈ I . That means, in particular, that b ∈ ( R : q ). So, we canwrite b = c b + · · · + c r b r ∈ R , for some c i ∈ R , i = 1 , ..., r . Now, let a i ∈ R be elements such that q = a i /b i . Since, a = bq = n X i =1 c i b i q = n X i =1 c i a i , it is straightforward to verify bx − a = P ri =1 c i ( b i x − a i ), as desired.(3) Clearly I ⊆ ker ϕ . For the other containment, let f ∈ ker ϕ we argueby induction on the degree of f . Write f = v n x n + · · · + v . The case n = 1is clear. So, assume n ≥
2. First, we know that v n q n + · · · + v = 0 ∈ K ( R ) , then after multiplying by v n − n , we get the integrity equation for v n q ,( v n q ) n + v n − v n ( v n q ) n − + · · · + v v n − n = 0 . So, v n q ∈ R , because R is a normal domain. Therefore, there exists d ∈ R such that q = d/v n . Now, f − x n − ( v n x − d ) ∈ ker ϕ , and it has lower degree.Thus, by the induction hypothesis f − x n − ( v n x − d ) ∈ I , and then f ∈ I ,because v n x − d ∈ I . (cid:3) ORMALITY AND RELATED PROPERTIES OF FORCING ALGEBRAS 25
We continue with our discussion, by abuse of notation we write with thesame capital letters its classes in E . Now, we know that Y, T ∈ ( E : S ′ ).Besides, ( X, T ) ∈ E in a prime ideal of codimension one in E , therefore invirtue of Lemma 7.1(1), ( Y, T ) = ( E : S ′ ). Hence, applying again Lemma7.1(2)-(3) we see thatker ψ = (( Y ) S + ( XU + X ) , ( T ) S + ( U + U )) , as desired. In conclusion, E [ S ] / ( XU + Y S + X, U + U + T S ) ∼ = E [ S ′ ]is an integral domain, therefore C/P ∼ = E [ S ] / ( XU + Y S + X, U + U + T S )so is.On the other hand, since the extension A → A ′ is integral, both rings havethe same dimension (it is a direct consequence from the Going Up, see [7,Proposition 4.15]). But, dim A = dim B − ht( h ) = 3, and then3 = dim A ′ = dim C/ ker φ = 5 − ht(ker φ ) , implying ht(ker φ ) = 2. Besides, it is easy to check that P ⊆ ker φ is a (prime)ideal of height strictly bigger that one, therefore both ideals coincide. Finally,we can apply Corollary 5.1 to the affine domain C/P . After computations weverify that ( U + 1)(2 U + 1) , U (2 U + 1) , U ( U + 1) + ST, ST ∈ J, where J denotes the Jacobian ideal, defining the singular locus of C/P . But,easily we check that C = (( U + 1)(2 U + 1) , U (2 U + 1) , U ( U + 1) + ST, ST ) , therefore the singular locus is empty and then C/P is regular, and in partic-ular, normal. In conclusion, an explicit description of the normalization of A as an affine ring is A ∼ = k [ X, Y, T, S, U ] / ( Y U − XT, XU + Y S + X, U + U + T S ) . Remark 7.2.
One can go forward in a natural way by computing the normal-ization for forcing algebras with forcing equations of the form h = x n t + y n + xy ,for n ≥
2. However, just for the case n = 3, new methods seem to be needed.In particular, we get an ideal P = ( Y U − X T, XU + X + Y S, U + U + XY ST ) . But, in order to apply Lemma 7.1, the most challenging part appears to befinding an explicit description of the generators of the corresponding denom-inator ideal, cause S ′ = − ( X + U X ) /Y = − ( U + U ) /XY T, and therefore we just know that Y , XY T ∈ ( D : S ′ ), where E = k [ X, Y, U, T ] / ( Y U − X T ) . But, on this case the ideal ( Y , XY T ) is not prime as in the argument beforewhere we get the prime ideal ( X, Y ) as denominator ideal.This section suggests on its own a way for forthcoming research on com-puting the normalization of forcing algebras.
Acknowledgement
Danny de Jes´us G´omez-Ram´ırez would like to thank to his parents Jos´eOmar G´omez Torres and Luz Stella Ram´ırez Correa for all the assistance, loveand inspiration. Besides, he wishes to thank the German Academic ExchangeService (DAAD) for the financial and academic support.
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Vienna University of Technology, Institute of Discrete Mathematics and Ge-ometry, wiedner Hauptstae 8-10, 1040, Vienna, Austria.
ORMALITY AND RELATED PROPERTIES OF FORCING ALGEBRAS 27
Universit¨at Osnabr¨uck, Fachbereich 6: Mathematik/Informatik, Albrechtstr.28a, 49076 Osnabr¨uck, Germany
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