The strong Lefschetz property of monomial complete intersections in two variables
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Abstract
In this paper we classify the monomial complete intersections, in two variables, and of positive characteristic, which has the strong Lefschetz property. Together with known results, this gives a complete classification of the monomial complete intersections with the strong Lefschetz property.
1 Background
A graded algebra \(A=\bigoplus _{i \ge 0} A_i\) is said to have the strong Lefschetz property (SLP) if there is a linear form such that multiplication by any power of this linear form has maximal rank in every degree. Let A be a monomial complete intersection, that is \(A=K[x_1, \ldots , x_n]/(x_1^{d_1}, \ldots , x_n^{d_n})\), where K is a field and \(d_1, \ldots , d_n\) some positive integers. In characteristic zero, A always has the SLP, which was first proved by Stanley in [12]. When the characteristic is positive, the algebra does not always have the SLP. A first result is that A has the SLP when \(p>\sum (d_i1)\), where p is the characteristic. This was proved in the case \(n=2\) by Lindsey in [8], and later in the general case by Cook II in [5].
A classification of all monomial complete intersections in three or more variables with the SLP is provided in [9]. Notice that the problem is trivial when \(n=1\), so the remaining case is \(n=2\), which will be treated in this paper. The sufficient conditions in [9] hold also in two variables, but it turns out that there is an additional class of algebras \(K[x,y]/(x^a,y^b)\) with the SLP. This is indicated by Cook II in [5], where the two special cases, when \(a=b\), and when the characteristic is two, is studied. Cook II solves these cases, under the assumption that the residue field K is infinite.
The main result of this paper is Theorem 3.2, which is a classification of the algebras \(K[x,y]/(x^a,y^b)\) with the SLP, where K is a field of characteristic \(p \ge 3\). The classification is given in terms of the base p digits of the integers a and b. Together with the mentioned earlier results, this gives a complete classification of the monomial complete intersections with the SLP, see Theorem 3.4.
The technique used both in [5] and in this paper, is the theory of the syzygy gap function, introduced by Monsky in [10]. The syzygy gap function deals with the degrees of the relations on \(x^a, y^b\) and \((x+y)^c\). This can then be connected to the SLP using results of Brenner and Kaid in [1] and [2]. In [1, 2], and [10] the residue field is required to be algebraically closed. We will see in Sect. 4 that this assumption can be dropped. We will also give a new proof of the connection to the SLP, when working with monomial complete intersections.
2 The strong Lefschetz property
Let \(A=\bigoplus _{i \ge 0} A_i\) be a graded algebra. A linear map \(A_i \rightarrow A_{j}\) is said to have maximal rank if it is injective or surjective. Each homogeneous element \(f \in A_d\) induces a family of linear maps \(A_i \rightarrow A_{i+d}\) by \(a \mapsto f \cdot a\). Let such maps be denoted by \(\cdot f: A_i \rightarrow A_{i+d}\). For short, we say that multiplication by f has maximal rank in every degree, if all the maps induced by f have maximal rank.
Definition 2.1
A graded algebra A is said to have the strong Lefschetz property (SLP) if there exists an \(\ell \in A_1\) such that the maps \(\cdot \ell ^m: A_i \rightarrow A_{i+m}\) have maximal rank for all \(i\ge 0\) and all \(m\ge 1\). In this case, \(\ell \) is said to be a strong Lefschetz element.
We say that A has the weak Lefschetz property (WLP) if there exists an \(\ell \in A_1\) such that the maps \(\cdot \ell : A_i \rightarrow A_{i+1},\) have maximal rank for all \(i\ge 0\). In this case, \(\ell \) is said to be a weak Lefschetz element.
Let now K be a field, and \(A=K[x_1, \ldots , x_n]/I\), where I is a monomial ideal. In [9, Proposition 4.3] it is proved that A has the WLP if and only if \(x_1+ \dots + x_n\) is a weak Lefschetz element. The corresponding is also true for the strong Lefschetz property.
Theorem 2.2
Let \(R=K[x_1, \ldots , x_n]\), where K is a field, and let \(I \subset R\) be a monomial ideal. Then R / I has the SLP (WLP) if and only if \(x_1 + \ldots + x_n\) is a strong (weak) Lefschetz element.
Proof
The Hilbert function of a graded algebra \(A=\bigoplus _{i \ge 0}A_i\) with residue field K is a function \({{\mathrm{HF}}}_A:{\mathbb {Z}}_{\ge 0} \rightarrow {\mathbb {Z}}_{\ge 0}\) defined by \({{\mathrm{HF}}}_A(i)={{\mathrm{vdim}}}_K A_i\), i. e. the vector space dimension of \(A_i\) over K. The Hilbert series of A, denoted \({{\mathrm{HS}}}_A\), is the generating function of the sequence \({{\mathrm{HF}}}(i)\), that is \({{\mathrm{HS}}}_A(t)=\sum _{i \ge 0} {{\mathrm{HF}}}(i)t^i\).
Let now A be a monomial complete intersection, \(A=K[x_1, \ldots , x_n]/(x_1^{d_1}, \ldots , x_n^{d_n})\), for some positive integers \(d_1, \dots , d_n\). Let \(t=\sum _{i=1}^n(d_i1)\). This is the highest possible degree of a monomial in A, and hence \({{\mathrm{HF}}}_A(i)=0\) when \(i>t\). It can also be seen that the Hilbert function is symmetric about t / 2, and that \({{\mathrm{HF}}}_A(i) \le {{\mathrm{HF}}}_A(i+d)\) when \(i \le (td)/2\). For a multiplication map to have maximal rank in every degree in A, it shall then be injective up to some degree i, and surjective for larger i. It can be proved that the injectiveness in this case implies the surjectiveness.
Proposition 2.3
Let \(A=K[x_1, \ldots , x_n]/(x_1^{d_1}, \ldots , x_n^{d_n})\) and \(t=\sum _{i=1}^n(d_i1)\), and let \(f\in A\) be a form of degree d. The maps \(\cdot f:A_i \rightarrow A_{i+d}\) all have maximal rank if and only if the maps with \(i \le (td)/2\) are injective.
Proof
See e. g. [9, Proposition 2.6]. \(\square \)
In other words, multiplication by a form f has maximal rank in every degree if all homogeneous zero divisors of f are of degree greater than \((td)/2\). Another interesting fact is that if we consider forms of the type \(\ell ^d\), and \(td\) is even, then multiplication by \(\ell ^{d+1}\) has maximal rank in every degree if multiplication by \(\ell ^d\) does. This result will be important for the classification of algebras with the SLP when \(n=2\).
Proposition 2.4
Let \(A=K[x_1, \ldots , x_n]/(x_1^{d_1}, \ldots , x_n^{d_n})\) and \(t=\sum _{i=1}^n(d_i1)\). Let \(\ell \in A\) be a linear form, and d a positive integer such that \(td\) is even. If the maps \(\cdot \ell ^d: A_{i} \rightarrow A_{i+d}\) have maximal rank for all \(i\ge 0\), so does the maps \(\cdot \ell ^{d+1}: A_{i} \rightarrow A_{i+d+1}.\)
Proof
Assume that \(\cdot \ell ^d: A_{i} \rightarrow A_{i+d}\) have maximal rank for all \(i\ge 0\). By Proposition 2.3 all zero divisors of \(\ell ^d\) are of degree at least \((td)/2\). Suppose that there is a homogeneous element f such that \(\ell ^{d+1}f=0\). By Proposition 2.3, we are done if we can prove that \(\deg (f)>(t(d+1))/2=(td)/21/2\). Since \(td\) is even, the right hand side is not an integer, and it is enough to prove \(\deg (f)>(td)/21\). Consider first the case when \(\ell ^df=0\). That is, f is a zero divisor of \(\ell ^d\), and it follows that \(\deg (f)>(td)/2\). Consider instead the case when \(\ell ^d f \ne 0\). We know that \(\ell ^{d+1}f=0\), that is \(\ell f\) is a homogeneous zero divisor of \(\ell ^d\). Then \(\deg (\ell f)>(td)/2\), and \(\deg (f)>(td)/21\), which finishes the proof. \(\square \)
Proposition 2.5
Proof
The “only if”part follows from Theorem 2.2.
3 Classifying the monomial complete intersections with the strong Lefschetz property
A classification of the monomial complete intersections with the SLP, in three or more variables, is given in [9, Theorem 3.8]. Here we give a slightly reformulated version of the theorem, to make the notation similar to that used later in the case of two variables. We will prove that the formulation here is equivalent to that in [9].
Theorem 3.1
 1.
\(t <p\),
 2.
\(d_1 \ge p\), \(d_i < p\) for \(i =2, \ldots , n\) and \(\sum _{i=2}^n(d_i1) \le \min (r_1,pr_1)\).
Proof
The two conditions in Theorem 3.1 above can be generalized to the case \(n=2\). Next we will prove that in two variables, and characteristic \(p>2\), the algebra A has the SLP in these two cases, but also in an additional one.
Theorem 3.2
 1.
When \(a,b<p\), A has the SLP if and only if \(a+b \le p+1\).
 2.
When \(a<p\) and \(b \ge p\), A has the SLP if and only if \(a \le \min (b_0,pb_0) +1\).
 3.When \(a,b \ge p\), A has the SLP if and only if the following three conditions are satisfied.
 (a)
\(a_0= \frac{p\pm 1}{2}\), and \(b_0=\frac{p\pm 1}{2}\),
 (b)
\(a_i=b_i=\frac{p1}{2}\) for \(i=1, 2, \ldots , k1\),
 (c)
\(a_k+b_k \le p1\), and \(b_k \ge a_k\) when \(\ell >k\).
 (a)
Notice that there are no restrictions on \(b_i\) for \(i>k\), in the case \(\ell >k\). The theorem will be proved later in this section.
In [5, Theorem 4.9] Cook II proves the special case \(a=b\) of Theorem 3.2. Cook II also proves the characteristic two case.
Theorem 3.3
 1.
\(a=2\) and b is odd,
 2.
\(a=3\) and \(b \equiv 2 \mod 4\).
Theorems 3.1, 3.2, and 3.3 can now be combined into a complete classification of the monomial complete intersections with the SLP.
Theorem 3.4
 1.
\(n=1\),
 2.
\(n=2\), \(p=2\), and one of the following holds, for \(d_1 \le d_2\)

\(d_1=2\) and \(c_{20}=1\),

\(d_1=3, c_{21}=1\), and \(c_{20}=0\),
 3.
\(n=2\), \(p>2\) and all the following conditions are satisfied, for \(k_1 \le k_2\)

\(c_{10}=\frac{p \pm 1}{2}, c_{20}=\frac{p \pm 1}{2}\),

\(c_{1j}=c_{2j}=\frac{p1}{2}\), for \(j=1, \ldots , k_11\),

\(c_{1k_1}+c_{2k_1} < p\), and \(c_{2k_1} \ge c_{1k_1}\) if \(k_1<k_2\),
 4.
\(n \ge 2\), and \(\sum _{i=1}^n (d_i1) <p\),
 5.
\(n \ge 2\), and there is a j such that \({d_j \ge p}\), \({d_i<p}\) for all \({i \ne j}\), and \(\sum _{i \ne j}(d_i1) \le \min (c_{j0},pc_{j0})\).
Proof
The case \(n=1\) is trivial. Condition 3 is condition 3 of Theorem 3.2, and Condition 4 is Theorem 3.2 with \(b=d_2\) written in base 2. The conditions 4 and 5 are the conditions 1 and 2 from Theorems 3.1 and 3.2 combined. Notice that 4 and 5 are not satisfied when \(p=2\). \(\square \)
Both proofs of [5, Corollary 4.8] and [5, Theorem 4.9] use Theorem 3.5 below. This will also be the key to the proof of Theorem 3.2.
Theorem 3.5
Theorem 3.5 is proved in Sect. 4.
We will now prove that Theorem 3.5 can be reformulated as the following proposition.
Proposition 3.6
 1.
If \(m_i>0\), then \(r_i \ge s_i1\),
 2.
If \(n_i>0\), then \(s_i \ge r_i1\),
 3.
If \(m_i >0\) and \(n_i>0\), then \(r_i+s_i \ge p^i1\),
 4.
\(r_i+s_i \le p^i+1\).
Proof
 I.
 \(u=m_i\) and \(v=n_i\) HereTo obtain \(aup^i + bvp^i + a+b2cwp^i\le p^i1\) it is necessary that \(r_i+s_i \le p^i1\).$$\begin{aligned} aup^i + bvp^i = r_i+s_i. \end{aligned}$$Suppose first that \(r_i+s_i=p^i1\). Since \(u+v+w=m_i+n_i+w\) is supposed to be odd, we must have \(w=m_i+n_i 2d+1\), for some integer d. Thenwhich is an odd number, and thus \(a+b2cwp^i\ge 1\). We get$$\begin{aligned} a+b2cwp^i&= n_ip^i + r_i + m_ip^i+s_i 2c (m_i+n_i 2d+1)p^i\\&= r_i+s_i2c + (2d1)p^i = 2dp^i2c1, \end{aligned}$$and we can conclude that \(aup^i + bvp^i + a+b2cwp^i\ge p^i\) for all w and c, when \(r_i+s_i=p^i1\). Now suppose that \(r_i+s_i \le p^i2\). We want to find out what the smallest possible value of \(a+b2cwp^i\) is. For this purpose we choose the largest w such that \(u+v+w\) is odd, and \(a+bwp^i>0\). After that we choose the value for c that makes \(a+b2cwp^i\) as small as possible. Since \(r_i+s_i \le p^i2\), the largest w with the required properties is \(w=m_i+n_i1\). Then$$\begin{aligned} aup^i + bvp^i + a+b2cwp^i \ge p^i1+1=p^i, \end{aligned}$$If \(m_i=0\), then \(\min (a,b) = \min (r_i,b) \le r_i\) and \(c \le r_i1\). Then$$\begin{aligned} a+bwp^i = p^i+r_i+s_i. \end{aligned}$$In a similar way we see that \(aup^i + bvp^i + a+b2cwp^i>p^i\) if \(n_i=0\). Suppose now that \(m_i>0\) and \(n_i>0\). Then we choose \(c=[(p^i+r_i+s_i)/2]\), where \([\ldots ]\) denotes the integer part. This gives$$\begin{aligned} a+b2cwp^i \ge p^i+r_i+s_i 2(r_i1) = p^ir_i+s_i+2, ~~\text {and} \\ aup^i + bvp^i + a+b2cwp^i \ge r_i+s_i + p^ir_i+s_i+2>p^i. \end{aligned}$$The conclusion, in this case, is that \(aup^i + bvp^i + a+b2cwp^i<p^i\), exactly when \(m_i,n_i>0\) and \(r_i+s_i \le p^i2\). This corresponds to condition 3 in the proposition.$$\begin{aligned}&a+b2cwp^i = 0 ~ \text {or} ~ 1, ~~\text {and} \\&aup^i + bvp^i + a+b2cwp^i \le r_i + s_i +1 \le p^i 1. \end{aligned}$$
 II.
 \(u=m_i\) and \(v=n_i+1\) HereTo obtain \(aup^i + bvp^i + a+b2cwp^i\le p^i1\) it is necessary that \(r_i+p^is_i \le p^i1\), that is \(r_i \le s_i1\). Let us first consider the case when \(r_i=s_i1\). Since \(u+v+w\) is supposed to be odd we must have \(w=n_i+m_i 2d\), for some integer d. This gives$$\begin{aligned} aup^i + bvp^i = r_i+p^is_i. \end{aligned}$$which is odd. Then \(a+b2cwp^i\ge 1\), and$$\begin{aligned} a+bwp^i = r_i+s_i+2dp^i = 2r_i+1 + 2dp^i, \end{aligned}$$Suppose instead that \(r_i \le s_i2\). We use that same idea as in case 1, and choose first w, and then c, such that \(a+b2cwp^i\) has the smallest possible value. The best option for w is \(w=n_i+m_i\). This gives$$\begin{aligned} aup^i + bvp^i + a+b2cwp^i \ge r_i+p^is_i+1 =p^i. \end{aligned}$$If \(m_i=0\), then \(\min (a,b)=\min (r_i,b)=r_i\), thus \(c=r_i1\) is the largest allowed value of c. Then$$\begin{aligned} a+bwp^i=r_i+s_i. \end{aligned}$$If \(m_i>0\) on the other hand, we are allowed tho choose \(c=s_i1\). Then we get$$\begin{aligned}&a+b2cwp^i = r_i+s_i2(r_i1) = s_ir_i+2, ~~\text {and} \\&aup^i + bvp^i + a+b2cwp^i = r_i+p^is_i + s_ir_i+2=p^i+2. \end{aligned}$$instead. Note that this is a nonpositive number. This gives$$\begin{aligned} a+b2cwp^i = r_is_i+2 \end{aligned}$$The conclusion, in this case, is that \(aup^i + bvp^i + a+b2cwp^i<p^i\), exactly when \(m_i>0\) and \(r_i \le s_i2\). This corresponds to condition 1 in the proposition.$$\begin{aligned} aup^i + bvp^i + a+b2cwp^i = r_i+p^is_i + s_ir_i2=p^i2. \end{aligned}$$
 III.

\(u=m_i+1\) and \(v=n_i\)
In the same way as above, we see that this corresponds to condition 2.
 IV.

\(u=m_i+1\) and \(v=n_i+1\)
Hereso for this to be smaller than \(p^i\) we must have \(2p^ir_is_i \le p^i1\), which is \(r_i+s_i \ge p^i+1\). Consider first the case when \(r_i+s_i=p^i+1\). Then we must choose \(w=m_i+n_i2d+1\), for some integer d. Then$$\begin{aligned} aup^i + bvp^i = 2p^ir_is_i, \end{aligned}$$and \(a+b2cwp^i\ge 1\). Then we get$$\begin{aligned} a+bwp^i = r_i+s_i +(2d1)p^i = 2dp^i+1, \end{aligned}$$Suppose now that \(r_i + s_i \ge p^i +2\). We choose \(w=m_i+n_i+1\) and \(c=[(r_i+s_ip^i)/2]\), because this gives$$\begin{aligned} aup^i + bvp^i + a+b2cwp^i \ge 2p^ir_is_i+1 = p^i. \end{aligned}$$This shows that \(aup^i + bvp^i + a+b2cwp^i<p^i\) when \(r_i + s_i \ge p^i +2\), which is condition 4. \(\square \)$$\begin{aligned}&a+b2cwp^i = r_i+s_ip^i 2c = 0 ~ \text {or} ~ 1,~~ \text {and} \\&aup^i + bvp^i + a+b2cwp^i \le 2p^ir_is_i +1 \le p^i 1. \end{aligned}$$
Proposition 3.6 will be used later in this section to prove Proposition 3.7, which says something about the structure of an algebra that does not have the SLP. Now we shall use Proposition 3.6, with \(p>2\), to prove Theorem 3.2.
Proof of Theorem 3.2
Let \(A=K[x,y]/(x^a,y^b)\), and suppose throughout this proof that the characteristic of K is greater than 2. Write a and b in base p as \(a=a_kp^k + \dots + a_1p+a_0\) and \(b=b_\ell p^\ell + \dots + b_1p+b_0\), where \(0 \le a_i,b_i <p\). We assume that \( \ell \le k\). With the notation \(a=m_ip^i+r_i\) from Proposition 3.6 we have \(r_i=a_{i1}{p^{i1}} + \dots + a_1p+a_0\), and \(m_i=a_kp^{ki} + a_{k1}p^{ki1}+ \dots + a_i\), and similar for b.
If \(a,b < p\) then \(n_i=m_i=0\) in Proposition 3.6, for all i, and the conditions 1, 2 and 3 are trivially satisfied. Since \(a+b<2p\) condition 4 is satisfied for \(i>1\). The only restriction we get comes from condition 4 when \(i=1\), and states that A has the SLP if and only if \(a+b \le p+1\).
If \(a < p\) and \(b \ge p\) we get \(b_0 \ge a_01\) and \(a_0+b_0 \le p+1\) from the conditions 2 and 4 with \(i=1\). These two inequalities can be written as \(a_0 \le \min (b_0, pb_0)+1\). In condition 1 and 3 there is nothing to check, and for \(i>1\) all conditions are satisfied. We get that A has the SLP if and only if \(a_0 \le \min (b_0, pb_0)+1\).
Assume now that \(a, b \ge p\). The idea now is to translate the four conditions of Proposition 3.6 into the base p digits of a and b.
The proof in [9] of when an algebra in three or more variables does not have the SLP, is carried out by finding a monomial zerodivisor of \((x_1+ \dots + x_n)^m\), for some m. We will now see that this can also be done in two variables. This gives an alternative proof of the ”only if”part of Theorem 3.2.
Proposition 3.7
Let \(A=K[x_1, \ldots , x_n]/(x_1^{d_1}, \ldots , x_n^{d_n})=\bigoplus _{i \ge 0} A_i\) be an algebra of characteristic \(p>0\) which does not possess the SLP. Let \(\ell \) be a linear form in A. Then there are integers d and m such that \({{\mathrm{HF}}}_A(d) \le {{\mathrm{HF}}}_A(d+m)\), and the kernel of the multiplication map \(\cdot \ell ^m:A_d \rightarrow A_{d+m}\) contains a nonzero monomial.
Proof
For the case \(n \ge 3\), see [9].
4 The syzygy gap
The main purpose of this section is to prove Theorem 3.5. If we require the residue field to be algebraically closed, the theorem follows from combining a theorem by Han [6] and results by Brenner and Kaid in [1] and [2]. Han’s result is also proved in a different way by Monsky in [10]. Monsky deals with the syzygy module of three pairwise relatively prime polynomials in two variables, and the so called ”syzygy gap”, while Brenner and Kaid connects this to the Lefschetz properties. We will go through the results from [10], and give a new proof of the connection to the SLP in the case of monomial complete intersections. The reason to go though the results of [10] is to prove that the residue field does not need to be algebraically closed, but also to give a deeper understanding of Theorem 3.5 and the theory behind it.
4.1 Mason–Stothers’ Theorem
First we need a review of Mason–Stothers’ Theorem. Suppose f is a polynomial in \(K[x_1, \ldots , x_n]\), where K is some field. The polynomial f can be factorized as \(f=\prod _{i=1}^sp_i^{e_i}\), where the \(p_i\)’s are distinct irreducible factors. Define \({{\mathrm{r}}}(f)=\deg (\prod _{i=1}^sp_i)\). Note that \({{\mathrm{r}}}(fg) \le {{\mathrm{r}}}(f)+{{\mathrm{r}}}(g)\), with equality when f and g are relatively prime. Let \(f'_{x_j}\) denote the formal derivative of f w. r. t. the variable \(x_j\). When in a polynomial ring with just one variable, we write \(f'\) for the derivative. Mason–Stothers’ theorem is usually formulated over one variable, as follows.
Theorem 4.1

f, g and h are pairwise relatively prime,

\(f', g'\) and \(h'\) are not all zero,

\(f+g+h=0\).
An elementary proof can be found in [11]. There is also a version of this theorem for homogeneous polynomials in two variables. For clarity we will prove how it can be deduced from Theorem 4.1.
Theorem 4.2

f, g and h are pairwise relatively prime,

\(f'_x, f'_y, g'_x, g'_y, h'_x\) and \(h'_y\) are not all zero,

\(f+g+h=0\).
Proof
4.2 The syzygy gap
Lemma 4.3
Let \(f_1,f_2\) and \(f_3\) be nonzero, pairwise relatively prime homogeneous polynomials in K[x, y], with \(d_i=\deg (f_i)\). Then \(\Delta (f_1,f_2,f_3) \equiv d_1+d_2+d_3 \mod 2\).
We shall also see some other properties of the function \(\Delta \).
Lemma 4.4
Proof
Let us now investigate what happens with \(\Delta (f_1,f_2,f_3)\) when, for example, \(f_1\) is replaced by \(\ell f_1\), for some linear form \(\ell \). By Lemma 4.3, \(\Delta (f_1,f_2,f_3)\) and \(\Delta (\ell f_1,f_2,f_3)\) has different parity, so they can not be equal. If we have a relation \(A_1f_1+A_2f_2+A_3f_3=0\), we also get a relation on \(\ell f_1, f_2, f_3\) by multiplying the expression by \(\ell \). This means that the two elements that generates \({{\mathrm{Syz}}}(\ell f_1,f_2,f_3)\) can have degrees at most \(\alpha +1\) and \(\beta +1\). On the other hand, a relation \(A_1\ell f_1+A_2f_2+A_3f_3=0\) on \(\ell f_1, f_2, f_3\) can also be considered a syzygy \((A_1\ell , A_2, A_3)\) on \(f_1,f_2,f_3\). Hence, the two generators of \({{\mathrm{Syz}}}(\ell f_1,f_2,f_3)\) have degrees at least \(\alpha \) and \(\beta \). This shows that \(\Delta \) must either increase of decrease by 1 when \(f_1\) is replaced by \(\ell f_1\). We summarize this in a lemma.
Lemma 4.5
We shall look more carefully into two special cases where Lemma 4.5 applies. Let \((A_1, A_2, A_3)\) be the element in \({{\mathrm{Syz}}}(f_1,f_2,f_3)\) of the lowest degree \(\alpha \). If \(\ell  A_1\) then \((\ell ^{1}A_1 ,A_2,A_3)\) is a syzygy of \(\ell f_1,f_2,f_3\) of degree \(\alpha \). The other generating syzygy can have degree \(\beta \) or \(\beta +1\), as we saw above. But since \(\Delta (\ell f_1,f_2,f_3)\ne \Delta (f_1,f_2,f_3)\) it must have degree \(\beta +1\). Hence, \(\Delta (\ell f_1,f_2,f_3)=\Delta (f_1,f_2,f_3) + 1\) in this case.
It follows also from Lemma 4.5 that \(\Delta (\ell ^{1} f_1,f_2,f_3)=\Delta (f_1,f_2,f_3) \pm 1\), if \(\ell  f_1\). If, in addition, \(\ell  A_2\), it follows from the equality \(A_1f_1+A_2f_2+A_3f_3=0\) that \(\ell \) also divides \(A_3\). Then we can divide the whole expression by \(\ell \), and get a syzygy \((A_1,\ell ^{1}A_2, \ell ^{1}A_3)\) on \(\ell ^{1}f_1, f_2, f_3\), of degree \(\alpha 1\). We see that we must have \(\Delta (\ell ^{1} f_1,f_2,f_3)=\Delta (f_1,f_2,f_3) + 1\), in this case.
This, together with Theorem 4.2, can now be used to prove the following proposition.
Proposition 4.6
( [10, Theorem 8]) Let K be a field of characteristic \(p>0\). Let \(f_1, f_2,\) and \(f_3\) be homogeneous relatively prime polynomials in K[x, y]. Assume there is a linear form \(\ell \) such that \(f_1=\ell ^m h\), where \(\ell \not \mid h\) and \(p \not \mid m\). Assume also that \(\Delta (f_1,f_2,f_3)\) decreases when \(f_1\) is replaced by \(\ell f_1\) or \(\ell ^{1}f_1\). Then \(\Delta (f_1,f_2,f_3) \le {{\mathrm{r}}}(f_1f_2f_3)2\).
Proof
4.3 Application of the syzygy gap function to monomial complete intersections
We will now specialize to the case \(f_1=x^{d_1}\), \(f_2=y^{d_2}\), and \(f_3=(x+y)^{d_3}\). This is allowed, since these polynomials are pairwise relatively prime. For an easier notation we introduce a new function \(\delta :{\mathbb {Z}}_+^3 \rightarrow {\mathbb {Z}}_{\ge 0}\) defined by \(\delta (d_1,d_2,d_3)=\Delta (x^{d_1}, y^{d_2}, (x+y)^{d_3})\). We will now see how the theory of the syzygy gap connects to the SLP.
Proposition 4.7
Let \(S=K[x,y]/(x^{d_1}, y^{d_2})\). The maps \(\cdot (x+y)^{d_3}: S_i \rightarrow S_{i+d_3}\), with \(d_3 <d_1+d_2\), have maximal rank for all i if and only if \(\delta (d_1,d_2,d_3) \le 1\).
This result can be proved for general \(f_1, f_2\) and \(f_3\) using [1, Theorem 2.2] and [2, Corollary 3.2]. Below follows an easier proof for this special case.
Proof
It remains to prove that \(A_3 \ne 0\). If \(A_3=0\) we would have a relation \(A_1f_1+A_2f_2=0\). Since \(f_1\) and \(f_2\) are relatively prime, this gives \(A_1=cf_2\) and \(A_2=cf_1\), for some \(c \in K\). Then \(\alpha =d_1+d_2\), and since \(\alpha +\beta =d_1+d_2+d_3\), we get \(\beta =d_3\). But \(\beta \ge \alpha \) and \(d_3 < d_1+d_2\) yields a contradiction. \(\square \)
This result combined with Proposition 2.5 now gives the following.
Theorem 4.8
Proof
It follows directly from Propositions 4.7 and 2.5 that \(K[x,y]/(x^{d_1}, y^{d_2})\) has the SLP if and only if \(\delta (d_1,d_2,d_1+d_22c)\le 1\). By Lemma 4.3 \(\delta (d_1,d_2,d_1+d_22c)\) is even, so it must be 0 in this case. \(\square \)
Lemma 4.9
Let \((d_1,d_2,d_3) \in L_=\). Then \(\delta (d_1,d_2,d_3)=0\).
Proof
Suppose \(d_1 \le d_2 <d_3=d_1+d_2\). We are in the situation when \(x^{d_1}\), \(y^{d_2}\), \((x+y)^{d_3}\) is not a minimal generating set; there are polynomials g and h such that \( (x+y)^{d_1+d_2}=gx^{d_1}+hy^{d_2}\) As we saw in the beginning of Sect. 4.2, the module \({{\mathrm{Syz}}}(x^{d_1},y^{d_2},(x+y)^{d_1+d_2})\) is, in this case, generated by \((g,h,1)\) and \((y^{d_2}, x^{d_1}, 0)\). Both these relations have degree \(d_1+d_2\), which gives \(\delta (d_1,d_2,d_3)=0\).
The case when \(d_1\) or \(d_2\) is the largest among \(d_1,d_2,d_3\) follows from the above after a linear change of the variables x and y. \(\square \)
Lemma 4.10
Proof
Recall from Lemma 4.5 that \(\delta (d_1,d_2,d_3)\) increases or decreases by 1 when we ”take a step” in \({\mathbb {Z}}_+^3\), that is when one \(d_i\) is replaced by \(d_i \pm 1\). This proves (3).
Theorem 4.11
Proof
Proof of Theorem 3.5
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