Connectivity of hyperplane sections of domains
Matteo Varbaro




















 
Connectivity of hyperplane sections of domains
    
Matteo Varbaro Supported by the program for abroad research activity of the University of Genoa, Pr. N, 100021-2016-MV-ALTROUNIGE_001, and by the GNSAGA-INdAM 

Dip. di Matematica
 Università  di Genova


<varbaro@dima.unige.it> 
    
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 To Gennady Lyubeznik on his 60th birthday 




During the conference held in 2017 in Minneapolis for his 60th birthday, Gennady Lyubeznik proposed the following problem:
Find a complete local domain R and an element x∈ R having three minimal primes _1,_2 and _3 such that _i+_j
has height 2 for all i≠ j and _1+_2+_3 has height 4. In this note this beautiful problem will be discussed,
and will be shown that the principle leading to the fact that such a ring cannot exist is false. The specific problem, though, remains open.




§ INTRODUCTION


Given an element x in a domain R, at what extent does the ring R/xR remember that R was a domain? 
More precisely, if R is a complete local domain, and x∈ R∖, does the punctured spectrum ^o (R/xR)
have any special feature?
The wish to know the answer to this question
led Gennady Lyubeznik to pose the problem mentioned in the abstract. 

The purpose of this note is to convince the reader that ^o (R/xR) does not have special features apart from the obvious ones and from the fact that, if R has dimension at least 3,
it must be connected by the Grothendieck's connectedness theorem <cit.>. What we mean by “has not special features” will be made precise in Problem <ref>: we are not able to solve it, as well as we are not 
able to solve Lyubeznik's problem, however we will exhibit an example of a 4-dimensional -graded domain R and a linear form x∈ R such that 
^o (R/xR) has an unexpected property (Example <ref>), contrary to the prediction of Lyubeznik. On the other hand, we show that his prediction holds in many cases (Theorem <ref>).



§ THE LYUBEZNIK COMPLEX OF A -GRADED -ALGEBRA


Let  be a separably closed field and R=⊕_i∈R_i be a -graded -algebra 
(i.e. a -graded Noetherian ring such that R_0=). Denote by =⊕_i>0R_i.
Notice that the minimal prime ideals of R are homogeneous, so we have

    ∈(R) ⊆.

Therefore, in this situation we can define the simplicial complex (R) as follows:


  * The vertex set of (R) is {1,… ,s}=[s], where (R)={_1, …, _s}.

  * {i_1, …, i_k} is a face of (R) if and only if √(_i_1 + … + _i_k)≠.

We will refer to the simplicial complex (R) as the Lyubeznik complex of R. 


The Lyubeznik complex can be defined in the same way for any local ring. If R is a -graded -algebra  with homogeneous maximal ideal ,
then

    (R)=(R^).

(here R^ denotes the -adic completion of R). Furthermore, R is a domain if and only if R^ is a domain: both these facts follow, for example, from <cit.>. This justifies why we consider -graded -algebras: to provide examples, which is our purpose, that is enough.



In what follows, we will state some results proved in the standard graded case for -graded -algebras. This is because,
for the problems we will deal with, it is often possible to reduce to the standard graded case: If R is a -graded -algebra,
then there exists a high enough natural number m such that the Veronese ring R^(m)=⊕_i∈R_im is standard graded.
Furthermore, 

    (R^(m))=(R).

Notice also that R^(m)≥ R for all m≥ 1 by <cit.>.



Let S=[X_1,… ,X_n] be the polynomial ring in n variables over a field . 
Given a finite simplicial complex Δ on [n], let

    I_Δ=(∏_i∈σX_i:σ∈ 2^[n]∖Δ)⊆ S

the Stanley-Reisner ideal of Δ, and [Δ]=S/I_Δ the Stanley-Reisner ring of Δ.
Denoting by () the set of facets of  and by x_i=X_i∈[Δ], we have

    ([])={(x_i:i∈[n]∖σ):σ∈()}.

In this case, so, ([]) is the simplicial complex with () as vertex set and such that

    𝒜⊆() ⋂_σ∈𝒜σ≠∅.

So, ([]) is the nerve of  with respect to the covering given by its facets. The Borsuk's nerve lemma <cit.> therefore implies that 
([]) and  are homotopically equivalent. In particular,

    H_i(([]);)≅H_i(;)    ∀ i∈.

Hence, by Hochster's formula <cit.>, we infer that, if R is a Stanley-Reisner ring over  such that R≥ k, then

    H_0((R);)=H_1((R);)=… = H_k-2((R);)=0.




It is not known whether the same vanishing of the above example holds for any -graded -algebra.  What is known so far is:


If R is a -graded -algebra such that R≥ 2, then H_0((R);)=0.



If R is a -graded -algebra such that R≥ 3, then H_0((R);)=H_1((R);)=0.


Another interesting problem concerns the behavior of the Lyubeznik complex under hyperplane sections. For example, in the case in which R is a domain, then (R) is just a point, but there can be some graded element x∈ R such that (R/xR) is quite complicated. Not arbitrarily complicated however, indeed:


If R is a -graded domain of dimension at least 3 and x∈ R_i for some i>0, then H_0((R/xR);)=0.


(In the previous theorem, and in the rest of the paper, by -graded domain we mean -graded -algebra which is a domain). In view of the above discussion about the depth,
is hard to resist from asking:


Let R be a -graded domain of dimension at least 4 and let x∈ R be a graded element of positive degree. Is it true that H_0((R/xR);)=H_1((R/xR);)=0?


With the purpose of inquiring on the above question Lyubeznik, during the conference held in 2017 in Minneapolis for his 60th birthday, proposed the following problem:


Find a -graded 4-dimensional domain R and a graded element x∈ R of positive degree such that 

    Min(xR)={_1,_2,_3},   (_i+_j)=2  ∀ i≠ j,   (_1+_2+_3)=4.




If R and x∈ R were like in <ref>, then (R/xR) would be the (empty) triangle ⟨{1,2},{2,3},{1,3}⟩. In particular H_1((R/xR);)≅, so R and x∈ R would provide a negative answer to Question <ref>. In this note we will give, in Example <ref>, a negative answer to <ref>, but without solving <ref>. Before moving to the proof of <ref>, however, I would like to briefly discuss on the existence of rings like in  <ref>. The same proof of <cit.> gives the following:


Let d,δ be two natural numbers such that d> δ+1, and assume char()=0. For all δ-dimensional simplicial complex Δ, there exists a d-dimensional domain R standard graded over , and a homogeneous ideal I⊆ R of height 1 such that:


  (i) (R/I)=Δ;

  (ii) If (I)={_1, …, _s} and {i_1, …, i_k}∈Δ, then (_i_1+… +_i_k)=k.

  (iii) If H_0(Δ;)=0, everything can be chosen so that R/I≥ 2;

  (iv) If H_0(Δ;)=H_1(Δ;)=0, everything can be chosen so that R/I≥ 3;



In particular, by choosing d=4, δ=1 and Δ the (empty) triangle, we can find a 4-dimensional domain standard graded over  and a height 1 ideal I⊆ R such that 

    Min(I)={_1,_2,_3},   (_i+_j)=2  ∀ i≠ j,   (_1+_2+_3)=4.

If I were principal up to radical we would have solved Problem <ref>. Let us see how the construction works  in this particular case:


Consider J=(X,Y,Z),  H=(XYZ) ⊆ S=[X,Y,Z,W], and 

    R=[J_3]⊆ S.
 
The ideal I=H∩ R has three minimal primes, 

    _1=(X)∩ R,  _2=(Y)∩ R,  _3=(Z)∩ R,
 
satisfying the properties 

    (_i+_j)=2  ∀ i≠ j,   (_1+_2+_3)=4.

However this example does not solve Problem <ref>, since I is not principal: in fact, I=(XYZ,  XYZW^3,  XYZW^6). (Up to radical I is generated by XYZ and XYZW^6, but one can argue, and this will also follow by the next theorem, that I is not principal even up to radical).



Let R be a -graded -algebra, and assume that char()=0.
If X= R is connected, satisfies (S_3), is Du Bois in codimension 2, and H^1(X,Ø_X)= 0, then H_1((R/xR);)=0 ∀ x∈ R_i with i>0. In particular, if X is a smooth variety and H^1(X,Ø_X)= 0, then H_1((R/xR);)=0  ∀ x∈ R_i with i>0.


First we notice that we can assume that R is standard graded: in fact, take m∈ large enough so that R^(m) is standard graded. We have that R≅ R^(m) and ( R/xR)_ red≅( R^(m)/x^mR^(m))_ red. In particular

    (R/xR)=(R^(m)/x^mR^(m)).

So, if we prove the statement in the standard graded case we are done.

Let S=[X_1,… ,X_n] be the standard graded polynomial ring, I⊆ S the homogeneous ideal such that R=S/I and J=I+(f), where f∈ S is a representative of x∈ R.  We will show that (S;J)≤ n-3, and therefore conclude by <cit.>.

Consider the short graded exact sequence 0→ R(-i) R → R/xR→ 0. By taking the long exact sequence in local cohomology

    …→ H^2_(R)_-i→ H^2_(R)_0→ H^2_(R/xR)_0→ H^3_(R)_-i→…

By <cit.>, H^2_(R)_-i=H^3_(R)_-i=0. So H^2_(R/xR)_0≅ H^2_(R)_0=0, that is H^1(Y,Ø_Y)=0, where Y =  R/xR. By the exponential sequence we have an injection 

    H^1(Y;)↪ H^1(Y,Ø_Y),
 
therefore H^1(Y;)=0. It is now enough to show that H^n-2_J(S) is supported at the irrelevant ideal (X_1,… ,X_n)⊆ S, and we will conclude by <cit.>, <cit.> and the universal coefficient theorem. Let P∈ S containing J, and denote by =P/I. If (P)=n-1, then R_ has dimension at least 3, and satisfies (S_2) (indeed (S_3)). So the punctured spectrum of S_P/JS_P≅R_/xR_ is connected by <cit.>, thus H^n-2_J(S)_P=H^n-2_JS_P(S_P)=0 by <cit.>. If P∈ S is a prime ideal containing J of height n-2, in the same way the Hartshorne-Lichtembaum vanishing theorem will show that H^n-2_J(S)_P vanishes, and if (P)<n-2 the Grothendieck vanishing theorem will do the job.



In Example <ref>, R=X is the blow-up of ^3 at a point; in particular it is a smooth variety with H^1(X,Ø_X)=0, so it cannot serve to solve Problem <ref>. In general, the construction of Theorem <ref> produces a smooth rational variety R, so it cannot be used to answer negatively <ref>. 


Theorem <ref> gives hope that Question <ref> has a positive answer.



§ QUESTION <REF> HAS A NEGATIVE ANSWER


Throughout this section, let us fix a polynomial ring S=[X_1,… , X_n] and a monomial order < on S.


Assume that the answer to Question <ref> is positive. Then, if ⊆ S is a prime ideal such that S/≥ 3 and _<()=I_ is square-free, H_1(;)=0.


Take a weight w=(w_1,… ,w_n)∈^n such that _<()=_w() (see, for example, <cit.>). Let Z be a new indeterminate over S, and P=S[Z] the polynomial ring with grading
(X_i)=w_i and (Z)=1. The homogenization _w()⊆ P of ⊆ S is still a prime ideal. So, R=P/_w() is a -graded domain of dimension at least 4. If the answer to Question <ref> were positive, then 

    H_1((R/ZR);)=0.

But R/ZR≅ S/_<()=[], and we saw in Example <ref> that the singular homologies of ([]) and  agree.


The conclusion of Proposition <ref> looks suspicious. It is related to the following:


Is there a prime ideal ⊆ S such that _<() is square-free but S/ is not Cohen-Macaulay? 


At a first thought, probably your answer to the above question would be: “Of course!”. And in fact, as we will briefly see, <ref> has an affirmative answer; however, before we wish to outline that most 
known examples of prime ideals with square-free initial ideal are Cohen-Macaulay:



  * Determinantal-type ideals (<cit.>);

  * Orlik-Terao ideals (<cit.>;

  * Multigraded multiplicity-free ideals (<cit.>);

  * Prime ideals ⊆ S such that S/≤ 2 (<cit.>).
 

We will soon provide an example of a prime ideal as in <ref> and contradicting the conclusion of <ref> (and so its assumption). We need to know that “Segre products and Gröbner deformations commute", which is a special case of the results achieved by Shibuta in <cit.>. We will give a simple proof of the case we need for the convenience of the reader.

Let A=[Y_0,… ,Y_a] and B=[Z_0,… ,Z_b] be two standard graded polynomial rings over . Given two homogeneous ideals I⊆ A and J⊆ B, their Segre product I♯ J 
is the ideal of P=[X_ij:i=0,… ,a, j=0,… ,b] defined as the kernel of the -algebra homomorphoism from P to the Segre product (A/I)♯ (B/J)=⊕_d∈(A/I)_d⊗_ (B/J)_d mapping X_ij to Y_i·Z_j. 









Let >_a the lexicographical monomial order on A extending the linear order
Y_0>… >Y_a and >_b the lexicographical monomial order on B extending Z_0>… >Z_b. Finally, let > be the lexicographical monomial order on P extending the linear order
X_00>X_01>… >X_0b>X_10>… >X_1b>… >X_a0>… >X_ab.


With the above notations, if _<_a(I) and _<_b(J) are square-free, then _<(I♯ J) is square-free. More precisely, if u_1,… ,u_r∈ A (resp. v_1,… ,v_s∈ B) are square-free monomial generators of _<_a(I) (resp. of _<_b(J)), then the following are
square-free monomial generators of _<(I♯ J):






  * u_i(X_0i_0,… ,X_ai_a) such that i=1,… ,r and b≥ i_0≥…≥ i_a≥ 0;

  * v_j(X_j_00,… ,X_j_bb) such that j=1,… ,s and a≥ j_0≥…≥ j_b≥ 0;

  * X_ijX_hk such that 0≤ i<h≤ a and 0≤ j<k≤ b.



Since (A/I)♯ (B/J) and (A/_<_a(I))♯ (B/_<_b(J)) have the same Hilbert function, _<(I♯ J) and _<(_<_a(I)♯_<_b(J)) must coincide; so we can assume that I=(u_1,… ,u_r) and J=(v_1,… ,v_s). First of all we claim that I♯ J is generated by:


  (i) u_i(X_0i_0,… ,X_ai_a) such that i=1,… ,r and b≥ i_0≥…≥ i_a≥ 0;

  (ii) v_j(X_j_00,… ,X_j_bb) such that j=1,… ,s and a≥ j_0≥…≥ j_b≥ 0;

  (iii) X_ijX_hk-X_ikX_hj such that 0≤ i<h≤ a and 0≤ j<k≤ b.

To see this consider the homomorphisms of -algebras PA♯ B and A♯ B(A/I)♯ (B/J) defined by ϕ(X_ij)=Y_iZ_j and ϕ(Y_iZ_j)=Y_iZ_j. Then, 

    I♯ J=(π∘ϕ)=(ϕ)+ϕ^-1((π)).

It is well known that (ϕ) is the ideal of 2-minors of the matrix X=(X_ij), that is generated by the polynomials in (iii). Moreover, it is easy to see that a system of generators of (π) is given by
u_i· m such that i=1,… ,r and m∈ B is a monomial of degree (u_i) and m· v_j such that j=1,… ,s and m∈ A is a monomial of degree (v_j). It is now clear that ϕ^-1((π)) is generated by the monomials in (i) and (ii). 

It remains to prove that the polynomials in (i),(ii) and (iii) form a Gröbner basis with respect to <. It is well known that the polynomials in (iii) alone are a Gröbner basis.
Therefore, it  just remains to prove that the S-polynomials of the polynomials in (iii) against the monomials in (i) and (ii) 
reduce to zero. This is simple to check.


Finally, we are ready to negatively answer Question <ref>:


Let A=[Y_0,Y_1] and B=[Z_0,Z_1,Z_2]. Furthermore, take 

    g=Z_0Z_1Z_2+Z_1^3+Z_2^3∈ B,

I={0}⊆ A and J=(g)⊆ B. Notice that E= B/J is an elliptic curve of ^2
(not smooth, but irreducible), and that P/(I♯ J)≅^1× E. So =I♯ J is a prime ideal.
Since _<_b(g)=Z_0Z_1Z_2, Proposition <ref> implies that _<() is generated by: 

    X_10X_11X_12,  X_10X_11X_02,  X_10X_01X_02,  X_00X_01X_02,      X_00X_11,  X_00X_12,  X_01X_12.


So _<()=I_Δ⊆ P=[X_ij:(i,j)∈{0,1}×{0,1,2}] where Δ is the following simplicial complex on the vertices {0,1}×{0,1,2}:






    < g r a p h i c s >




Since H_1(Δ;)≅, this example contradicts the conclusion of Proposition <ref>, so it provides a negative answer to Question <ref>. Notice that it also provides a positive answer to Question <ref>: in fact P/ is a 3-dimensional domain and _<() is square-free, but P/ is not Cohen-Macaulay by <cit.>.

Using the notations of Proposition <ref>, R=P[Z]/_w()  is a -graded 4-dimensional domain and H_1((R/ZR);)≠ 0.


I would like to conclude the paper by proposing two problems:


Given a connected simplicial complex Δ and d> Δ+1, find a d-dimensional standard graded domain R and a linear form x∈ R such that:


  (i) (R/xR)=Δ;

  (ii) If (xR)={_1, …, _s} and {i_1, …, i_k}∈Δ, then (_i_1+… +_i_k)=k.



Notice that a positive answer to the above problem would supply a positive answer to Lyubeznik's problem <ref>. In order to solve it, one might try to deform the examples coming from Theorem <ref>.
The last problem I want to propose, that is related to Question <ref>, came out from discussions with Alexandru Constantinescu and Emanuela De Negri, who I thank:


Find a prime ideal ⊆ S such that _<() is square-free, S/ is smooth, but S/ is not Cohen-Macaulay. 


If one found an example of a smooth Calabi-Yau projective variety with a reduced Gröbner degeneration, then the same construction of Example <ref> would provide an answer to the above problem. Unfortunately I am not aware of such Calabi-Yau varieties: for sure there is no elliptic curve in ^2 like this.



















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