Depth of factors of square free monomial ideals

Let $I$ be an ideal of a polynomial algebra over a field, generated by $r$ square free monomials of degree $d$. If $r$ is bigger (or equal, if $I$ is not principal) than the number of square free monomials of $I$ of degree $d+1$, then $\depth_SI= d$. Let $J\subsetneq I$, $J\not =0$ be generated by square free monomials of degree $\geq d+1$. If $r$ is bigger than the number of square free monomials of $I\setminus J$ of degree $d+1$, or more generally the Stanley depth of $I/J$ is $d$, then $\depth_SI/J= d$. In particular, Stanley's Conjecture holds in theses cases.


Introduction
Let S = K[x 1 , . . . , x n ] be the polynomial algebra in n variables over a field K, d a positive integer and I J, be two square free monomial ideals of S such that I is generated in degrees ≥ d, respectively J in degrees d + 1. Let ρ d (I) be the number of all square free monomials of degree d of I. It is easy to note (see Lemma 1.1) that depth S I/J ≥ d. Our Theorem 2.2 gives a sufficient condition which implies depth S I/J = d, namely this happens when ρ d (I) > ρ d+1 (I) − ρ d+1 (J).
Suppose that this condition holds. Then the Stanley depth of I/J (see [11], [2], or here Remark 2.6) is d and if Stanley's Conjecture holds then depth S I/J ≤ d, that is the missing inequality. Thus to test Stanley's Conjecture means to test the equality depth S I/J = d, which is much easier since there exist very good algorithms to compute depth S I/J but not so good to compute the Stanley depth of I/J. After a lot of examples computed with the computer algebra system SINGULAR we understood that a result as Theorem 2.2 is believable. The above condition is not necessary as Example 2.4 shows. Necessary and sufficient conditions could be possible found classifying some posets (see Remark 2.5) but this is not the subject of this paper.
The proof of Theorem 2.2 uses the Koszul homology and can be read without other preparation. Our first section gives easy proofs in special cases, but they are mainly an introduction in the subject. Remarks 1.7, 1.9 show that the Koszul homology seems to be the best tool in our problems. Section 2 starts with one example, where we give the idea of the proof of Theorem 2.2.
The support from grant ID-PCE-2011-1023 of Romanian Ministry of Education, Research and Innovation is gratefully acknowledged.
If I is generated by more (or equal, if I is not principal) square free monomials of degree d than n d+1 , or more general than ρ d+1 (I), then depth S I = d as shows our Corollary 3.4. This extends [9,Corollary 3], which was the starting point of our research, the proof there being easier. Remark 3.5 says that the condition of Corollary 3.4 is tight.
The conditions above are consequences of the fact that sdepth I/J = d, as we explained in Remark 2.6, and we saw that they imply depth I/J = d. But what happens if we just suppose that sdepth I/J = d? Then there exists a monomial square free ideal I ′ ⊂ I such that ρ d (I ′ ) > ρ d+1 (I ′ ) − ρ d+1 (I ′ ∩ J) using our Theorem 4.1 (somehow an extension of [10,Lemma 3.3]) and it follows also depth S I/J = d by our Theorem 4.3.
We owe thanks to a Referee who found gaps in a preliminary version of our paper.

Factors of square free monomial ideals
Let J I, be two nonzero square free monomial ideals of S and d a positive integer. Let ρ d (I) be the number of all square free monomials of degree d of I. Suppose that I is generated by square free monomials f 1 , . . . , f r , r > 0, of degrees ≥ d and J is generated by square free monomials of degree ≥ d + 1. Set s := ρ d+1 (I) − ρ d+1 (J) and let b 1 , . . . , b s be the square free monomials of I \ J of degree d + 1. Proof. We may suppose that in E there exist no monomial generator of F . In the exact sequence 0 → J/E → I/E → I/J → 0 we see that the first end is isomorphic with F/(F ∩ E) and has depth ≥ d + 2 by Lemma 1.1. Apply the Depth Lemma and we are done.
Before trying to extend the above lemma it is useful to see the next example.  Proof. By induction on the number of the generators of H it is enough to consider the case H = (u) for some square free monomial u ∈ I of degrees d + 1. In the exact sequence 0 → I/J → (I + (u))/J → (I + (u))/I → 0 we see that the last term is isomorphic with (u)/I ∩ (u) and has depth ≥ d + 1 by Lemma 1.1, since I ∩ (u) has only monomials of degrees > d + 1. Using the Depth Lemma the first term has depth d if and only if the middle has depth d, which is enough.
Using Lemmas 1.2, 1.4 we may suppose always in our consideration that I, J are generated in degree d, respectively d + 1, in particular f i have degree d.  Proof. We may suppose i = 1. By our hypothesis J : f 1 is generated by (n − d) variables. If r = 1 then the depth of I/J ∼ = S/(J : f 1 ) is d. If r > 1 apply the above lemma for e = 1.
Remark 1.7. Suppose in the proof of the above lemma that f 1 = x 1 · · · x d . Then the hypothesis says that (x d+1 , . . . , x n )f 1 ⊂ J. It follows that z = f 1 e σ 1 , e σ 1 = e d+1 ∧. . .∧e n induces a nonzero element in the Koszul homology module H n−d (x; I/J) of I/J (some details from Koszul homology theory are given in Example 2.1). Thus depth S I/J ≤ d by [1, Theorem 1.6.17], the other inequality follows from Lemma 1.1. This gives a different proof of the above lemma using stronger tools, which will be very useful in the next section. We also remind that Proof. Apply induction on r ≥ 2. Suppose that r = 2. By hypothesis the greatest common divisor u = (f 1 , f 2 ) have degree d − 1 and after renumbering the variables we may suppose that f i = x i u for i = 1, 2. By hypothesis the square free multiples of f 1 , f 2 by variables x i , i > 2 belongs to J. Thus we see that I/J is a finite module over a polynomial ring in (d + 1)-variables and we get depth S I/J ≤ d since I/J it is not free. Now it is enough to apply Lemma 1.1. If r > 2 then apply Lemma 1.5 for e = 2. Remark 1.9. We see in the proof of the above lemma (similarly as in Remark Proof. If there exists i ∈ [r] such that f i has in J all square free multiples of degree d + 1, then we apply Lemma 1.6. Otherwise, each f i has a square free multiple of degree d + 1 which is not in J. By hypothesis, there exist i, j ∈ [r], i = j such that f i , f j have the same multiple b of degree d + 1 in I \ J. Now apply the above lemma.
Corollary 1.11. Suppose that r > s ≤ 1. Then depth S I/J = d.
Proof. Using Lemma 1.5 for e = 3 we reduce to the case r = 3. By Lemma 1.6 we may suppose that each f i divides b 1 , or b 2 . By Proposition 1.10 we may suppose that Similarly, if f 3 |b 1 then f 3 |b 2 and we may apply Lemma 1.8 to f 2 , f 3 . Thus we reduce to the case when f 3 |b 2 and f 3 |b 1 . We may suppose that Case i = j Then we may suppose i = j = 3 and f 1 = x 3 u for a square free monomial u of degree d − 1. It follows that f 2 = x 2 u, f 3 = x 1 u. Let S ′ be the polynomial subring of S in the variables x 1 , x 2 , x 3 and those dividing u. Then for each variable x k ∈ S ′ we have f i x k ∈ J and so I/J ∼ = I ′ /J ′ , where I ′ = I ∩ S ′ , J ′ = J ∩ S ′ . Changing from I, J, S to I ′ , J ′ , S ′ we may suppose that n = d + 2 and u = Π n i>3 x i . Then I/J ∼ = (I : u)/(J : Case i = j Then we may suppose i = 3, j = 4 and f 1 = x 3 x 4 v for a square free monomial v of degree d − 2. It follows that  Proof. By Lemma 1.13 we may suppose that I = (x 1 , . . . , x r ) with r < n. Using Lemma 1.6 we may suppose that each x i , i ∈ [r] divides a certain b k . Apply induction on s, the case s ≤ 2 being done in Proposition 1.12. Assume that s > 2. We may suppose that each b k is a product of two different x i , i ∈ [r] because if let us say b s is just a multiple of one x i , i ∈ [r], for example x r , then we may take I ′ = (x 1 , . . . , x r−1 ), J ′ = J ∩ I ′ and we get depth S I ′ /J ′ = 1 by induction hypothesis on s since r − 1 > s − 1, that is depth S I/J = 1 by Lemma 1.5. But if each b k is a product of two different x i , i ∈ [r] we see that b j ∈ S ′ = K[x 1 , . . . , x r ] for all j ∈ [s] and we may apply again Lemma 1.13.

Main result
We want to extend Proposition 1.14 for the case d > 1. Next example is an illustration of our method.

5
Then the element are the only monomials of degree 3 which are not in J. Note that in a term ue σ of an element from Im ∂ 5 we have u of degree ≥ 3 because I is generated in degree 2. Thus z ∈ Im ∂ 5 induces a nonzero element in H 4 (x; I/J). By [1, Theorem 1.6.17] we get depth S I/J ≤ 2, which is enough.  . This is a system of s homogeneous linear equations in r variables Y , which must have a nonzero solution in K because r > s. As in the above example z ∈ Im ∂ n−d+1 if I is generated in degree d (this may be supposed by Lemmas 1.2, 1.4).
The condition given in Theorem 2.2 is tight as shows the following two examples.
We have r = s = 6 and b 1 = Remark 2.5. The above example 2.4 shows that one could find a nice class of factors of square free monomial ideals with r = s but depth S I/J = d similarly as in [9,Lemma 6]. An important tool seems to be a classification of the possible posets given on f 1 , . . . , f r , b 1 , . . . , b s by the divisibility. Remark 2.6. Given J I two square free monomial ideals of S as above one can consider the poset P I\J of all square free monomials of I \ J (a finite set) with the order given by the divisibility. Let P be a partition of P P I\J in intervals [u, v] = {w ∈ P I\J : u|w, w|v}, let us say P I\J = ∪ i [u i , v i ], the union being disjoint. Define sdepth P = min i deg v i and sdepth S I/J = max P sdepth P, where P runs in the set of all partitions of P I\J . This is the Stanley depth of I/J, in fact this is an equivalent definition (see [11], [2]). If r > s then it is obvious that sdepth S I/J = d and so Theorem 2.2 says that Stanley's Conjecture holds, that is sdepth S I/J ≥ depth S I/J. In general the Stanley depth of a monomial ideal I is greater than or equal with the Lyubeznik' size of I increased by one (see [3]). Stanley's Conjecture holds for intersections of four monomial prime ideals of S by [5] and [7] and for square free monomial ideals of K[x 1 , . . . , x 5 ] by [6] (a short exposition on this subject is given in [8]). Also Stanley's Conjecture holds for intersections of three monomial primary ideals by [13]. In the case of a non square free monomial ideal I a useful inequality is sdepth I ≤ sdepth √ I (see [4,Theorem 2.1]).

Around Theorem 2.2
Let S ′ = K[x 1 , . . . , x n−1 ] be a polynomial ring in n − 1 variables over a field K, S = S ′ [x n ] and U, V ⊂ S ′ , V ⊂ U be two square free monomial ideals. Set W = (V + x n U)S. Actually, every monomial square free ideal T of S has this form because then (T : x n ) is generated by an ideal U ⊂ S ′ and T = (V + x n U)S for V = T ∩ S ′ .    We have I = (V + x n (U + V ))S as above. Renumbering the variables we may suppose that U, V = 0. Note that µ(I) = r + ρ d (V ) and ρ d+1 ( . By hypothesis, µ(I) ≥ ρ d+1 (I) and so r > ρ d (U) − ρ d (U ∩ V ). Applying Theorem 3.3 we get depth S S/I = d − 1, which is enough. , that is the hypothesis of the above corollary are not fulfilled. This is the reason that depth S ′ L = 4 by Lemma 3.2 since depth S I/J = 3. Thus the condition of the above corollary is tight.

Minimal Stanley depth
Let S = K[x 1 , . . . , x n ] be the polynomial algebra in n-variables over a field K, d a positive integer and J I, be two square free monomial ideals of S. Let ρ d (I) be the number of all square free monomials of degree d of I. Suppose that ρ d (I) > 0 and I is generated in degree ≥ d. It follows that sdepth S I/J ≥ d. (1) sdepth S I/J = d (2) there exist some square free monomials of degree d in I, which generate an ideal I ′ such that ρ d (I ′ ) > ρ d+1 (I ′ ) − ρ d+1 (I ′ ∩ J).
Proof. If J = 0 then J is generated in degree ≥ d, even we may suppose that J is generated in degree ≥ d + 1 using an easy isomorphism. Let