3 Minimal Generating Set of (Q:X)
We want to extend Proposition 2.8 to the ring of polynomials in d indeter- minates. By Theorem 2.6, (Q : X) is generated by the generators of Q and the Q-corner-elements. Let us begin with some observations on the relationship between the generators and corner-elements of Q in order to get the range of v(Q : X).
Lemma 3.1. Assume Q is a monomial ideal in R = k[X1, X2, . . . , Xd] with minimal monomial basis {g1, g2, ..., gn}. Let w1, w2, ..., wt be the Q-corner- elements. Then
1. wi ∈ (g/ j) for all 1 ≤ i ≤ t, 1 ≤ j ≤ n.
2. wi ∈ (w/ j) for all 1 ≤ i 6= j ≤ t.
3. gi ∈ (g/ j) for all 1 ≤ i 6= j ≤ n.
4. If gi ∈ (wj), then gi = wjXs for some 1 ≤ s ≤ d.
Proof. 1. and 3. are immediate from the definition of a corner-element and that of a minimal basis.
For 2., if wi ∈ (wj), then wi = wjm for some monomial m. Since wi and wj are Q-corner-element and since wjm = wi ∈ Q, we have m = 1. That is/ wi = wj, which contradicts our assumption.
For 4., suppose wj = X1a1X2a2· · · Xdadand gi = X1a1+k1X2a2+k2· · · Xdad+kd, where k1, k2, ..., kd ≥ 0. Since wj ∈ Q, k/ i 6= 0 for some 1 ≤ i ≤ d. With- out loss of generality, assume k1 6= 0. Then X1a1+1X2a2· · · Xdad divides gi. But X1a1X2a2· · · Xdad is a Q-corner element, gs divides X1a1+1X2a2· · · Xdad for some 1 ≤ s ≤ n. In other words, gs divides gi, hence gi = gs. Therefore, X1a1+1X2a2· · · Xdad divides gi = gs, which divides X1a1+1X2a2· · · Xdad. Then we have gi = X1a1+1X2a2· · · Xdad = wjX1.
Definition 3.2. Given any real number x, the floor of x, denoted bxc and the ceiling of x, denoted dxe, are defined as follows:
bxc= the unique integer n such that n ≤ x < n + 1, and
dxe= the unique integer n + 1 such that n < x ≤ n + 1.
Proposition 3.3. Let R = k[X1, X2, . . . , Xd] and X = (X1, X2, . . . , Xd). If Q is a monomial ideal in R with v(Q) = n, then v(Q : X) ≥ dnde.
Proof. Suppose w1, w2, . . . , wtare the Q-corner-elements and assume {g1, g2, . . . , gn} is the minimal monomial basis of Q. By Theorem 2.6,
(Q : X) = (g1, g2, . . . , gn, w1, w2, . . . , wt).
By Lemma 3.1, if v((g1, g2, . . . , gn, w1, w2, . . . , wt)) < n + t, there must exist 1 ≤ i ≤ n and 1 ≤ j ≤ t such that gi ∈ (wj). By Lemma 3.1 again, if gi ∈ (wj), gi = wjXs for some 1 ≤ s ≤ d. Hence wj has at most d multiples wjX1, wjX2, . . . , wjXd in {g1, g2, . . . , gn}. Therefore dnde corner-elements are needed to replace all the n generators g1, g2, . . . , gn, so v(Q : X) ≥ dnde.
We will now turn our attention to the upper bound of v(Q : X). A study has been made on the number of corner-elements as follow.
Theorem 3.4. [A, Lemma 17]. The maximal number of corner-elements of any monomial ideal minimally generated by n monomials in three variables is 2n − 5, for all n ≥ 3.
Hence we have the following result on the range of v(Q : X) in k[x, y, z].
Proposition 3.5. Let R = k[x, y, z] and let X = (x, y, z). If Q is a monomial ideal in R with v(Q) = n ≥ 3, then dn3e ≤ v(Q : X) ≤ 3n − 5.
Proof. By Proposition 3.2, we have v(Q : X) ≥ dn3e. Since (Q : X) is generated by the generators and corner-elements of Q, together with Theorem 3.4 we have v(Q : X) ≤ n + (2n − 5) = 3n − 5.
Conversely, given an integer n ≥ 3, we can show that for each intermediate integer t such that dn3e ≤ t ≤ 3n − 5, there exists a monomial ideal Q such that v(Q) = n and v(Q : X) = t. We separate the construction into 9 cases:
Case 1 dn3e ≤ t ≤ bn2c, 3|n.
Case 2 dn3e ≤ t ≤ bn2c, 3|n − 1.
Case 3 dn3e ≤ t ≤ bn2c, 3|n − 2.
Case 4 bn2c + 1 ≤ t ≤ n − 2, n is odd.
Case 5 bn2c + 1 ≤ t ≤ n − 2, n is even.
Case 6 t = n − 1.
Case 7 t = n.
Case 8 n + 1 ≤ t ≤ 2n − 3.
Case 9 2n − 2 ≤ t ≤ 3n − 5.
The appearances of the constructions for the first three cases are approx- imately the same. From Lemma 3.1, we find that we can achieve the lower bound dn3e if every minimal monomial generator of the monomial ideal is of the form w · Xi, where w is a corner element of the ideal. After omitting the difference between these three cases, such an ideal can be constructed as the ideal
w1x , w2x , w3x , . . . , wlx, w1y , w2y , w3y , . . . , wly, w1z , w2z , w3z , . . . , wlz
with corner elements w1, w2, w3, . . . , wl. If we control the corner elements such that wky = wk+1x for some wk’s, then (Q : X) can be increased step by step.
Lemma 3.6. Let n be an integer with n ≥ 3 and 3|n. For each intermediate integer t such that dn3e ≤ t ≤ bn2c, there exists a monomial ideal Q such that v(Q) = n and v(Q : X) = t.
Proof. Let s be an integer with 1 ≤ s < n3 and set fi = x2n3+1−2iy2(i−1), for i = 1, . . . ,n3 − s hi =
( x2n3 −2iy2i−1, for i = 1, . . . ,n3 − s
xn3+s−iyn3−s−1+i, for i = n3 − s + 1, . . . ,n3 + s gi =
( x2n3 −2iy2(i−1)z, for i = 1, . . . ,n3 − s xn3+s−iyn3−s−2+iz, for i = n3 − s + 1, . . . ,n3 wi =
( x2n3 −2iy2(i−1), for i = 1, . . . ,n3 − s xn3+s−iyn3−s−2+i, for i = n3 − s + 1, . . . ,n3
Take
Q =
f1, . . . , fn
3−s, h1, . . . , hn
3−s, hn
3−s+1, . . . , hn
3, hn
3+1, . . . , hn
3+s, g1, . . . , gn
3−s, gn
3−s+1, . . . , gn
3
.
We first claim that v(Q) = n. Note that every generator of Q listed above has degree 2n3 − 1, hence we only need to show that they are all distinct. First of all, z divides gi for all i, but z divides none of f1, . . . , fn
3−s, h1, . . . , hn
3+s, hence we can discuss these two parts, g1, . . . , gn
3 and f1, . . . , fn
3−s, h1, . . . , hn
3+s, separately. After rearranging fi’s and hi’s as
f1, h1, f2, h2, . . . , fn
3−s, hn
3−s, hn
3−s+1, hn
3−s+1, . . . , hn
3+s,
we can observe that the power of y is increasing. The same statement holds for g1, g2, . . . , gn
3. Therefore, v(Q) = (n3 − s) + (n3 + s) + (n3) = n.
Next, we claim that w1, w2, . . . , wn
3 are precisely the Q-corner-elements. For any Q-corner-element w, we have wz ∈ Q. Then there exists some generator g of Q such that g divides wz, that is to say, wz = gh for some monomial h. If z does not divides g, then z divides h, i.e., h = xaybzc for some a, b ∈ N ∪ {0}, c ∈ N, so wz = g · xaybzc. Take one z away from both side of the equality, then we get w = g · xaybzc−1. Since c − 1 ≥ 0, w = g · xaybzc−1 ∈ Q, which is a contradiction. Thus we have that z divides g. Recall that g is a generator of Q, hence g = gi for some i = 1, 2, . . . ,n3. Notice that gi = wiz for all i. Together with wz = gh, we have wz = gh = gih = wizh. Thus w = wih.
Since
wix =
( fi, if i = 1, . . . ,n3 − s hi−1, if i = n3 − s + 1, . . . ,n3 wiy = hi
wiz = gi
(∗)
and since w is a Q-corner-element, h = 1 and so w ∈ {w1, w2, . . . , wn
3}.
Conversely, since wi has degree 2n3 − 2 for all i and since every generator of Q has degree 2n3 − 1, wi ∈ Q. Moreover, by (∗), w/ ix, wiy, wiz are in Q, so wi is a Q-corner-element. Therefore, w1, . . . , wn
3 are the Q-corner-elements.
Moreover, we have
(Q : X) =
f1, . . . , fn
3−s, h1, . . . , hn
3−s, hn
3−s+1, . . . , hn
3, hn
3+1, . . . , hn
3+s, g1, . . . , gn
3−s, gn
3−s+1, . . . , gn
3, w1, . . . , wn
3−s, wn
3−s+1, . . . , wn
3
= (w1, w2, . . . , wn
3, hn
3+1, hn
3+2, . . . , hn
3+s).
Observing the monomials in the above generating set of (Q : X), the power of y is increasing when that of x is decreasing. That is to say, this is the minimal monomial generating set of (Q : X). Therefore v(Q : X) = n3 + s.
Note that 1 ≤ s ≤ n3 − 1, thus n3 + 1 ≤ n3 + s ≤ 2n3 − 1. In other words, we construct a monomial ideal Q such that v(Q) = n and v(Q : X) = t for all
n
3 + 1 ≤ t ≤ 2n3 − 1. Furthermore, we have (2n3 − 1) − bn2c =
( 2n
3 − n−12 − 1, if n is odd,
2n
3 − n2 − 1, if n is even.
=
( n−3
6 , if n is odd,
n−6
6 , if n is even.
Recall that n ≥ 3 and 3|n. If n is odd, then n ≥ 3; if n is even, then n ≥ 6.
Therefore, (2n3 − 1) − bn2c ≥ 0, i.e., 2n3 − 1 ≥ bn2c. Therefore, we construct a monomial ideal Q such that v(Q) = n and v(Q : X) = t for all n3+ 1 ≤ t ≤ bn2c.
Hence, it remains to construct a monomial ideal Q such that v(Q) = n and v(Q : X) = dn3e = n3. Set
fi = x2n3 +1−2iy2(i−1), for i = 1, . . . ,n3 hi = x2n3 −2iy2i−1, for i = 1, . . . ,n3 gi = x2n3 −2iy2(i−1)z, for i = 1, . . . ,n3 wi = x2n3 −2iy2(i−1), for i = 1, . . . ,n3 and set
Q =
f1, . . . , fn
3, h1, . . . , hn
3, g1, . . . , gn
3
.
Note that
wix = fi, wiy = hi, wiz = gi.
With a similar proof as in the previous case, we have v(Q) = n and the Q- corner-elements are w1, . . . , wn
3. Then
(Q : X) =
f1, . . . , fn
3, h1, . . . , hn
3, g1, . . . , gn
3, w1, . . . , wn
3
= (w1, w2, . . . , wn
3).
Therefore, v(Q : X) = n3. This completes the proof.
We use the following example to demonstrate the construction.
Example 3.7. Suppose n = 12 in Lemma 3.6. We now want to construct ideals Q such that v(Q) = 12 and v(Q : X) = 4, 5 and 6, respectively.
1. v(Q) = 12 and v(Q : X) = 4.
Let
Q =
x7, x5y2, x3y4, xy6, x6y, x4y3, x2y5, y7, x6z, x4y2z, x2y4z, y6z
.
Then the Q-corner-elements are x6, x4y2, x2y4, y6. Hence
(Q : X) =
x7, x5y2, x3y4, xy6, x6y, x4y3, x2y5, y7, x6z, x4y2z, x2y4z, y6z, x6, x4y2, x2y4, y6
= (x6, x4y2, x2y4, y6).
2. v(Q) = 12 and v(Q : X) = 5.
Let
Q =
x7, x5y2, x3y4,
x6y, x4y3, x2y5, xy6, y7, x6z, x4y2z, x2y4z, xy5z
.
Then the Q-corner-elements are x6, x4y2, x2y4, xy5. Hence
(Q : X) =
x7, x5y2, x3y4,
x6y, x4y3, x2y5, xy6, y7, x6z, x4y2z, x2y4z, xy5z, x6, x4y2, x2y4, xy5
= (x6, x4y2, x2y4, xy5, y7).
3. v(Q) = 12 and v(Q : X) = 6.
Let
Q =
x7, x5y2,
x6y, x4y3, x3y4, x2y5, xy6, y7, x6z, x4y2z, x3y3z, x2y4z
.
Then the Q-corner-elements are x6, x4y2, x3y3, x2y4. Hence
(Q : X) =
x7, x5y2,
x6y, x4y3, x3y4, x2y5, xy6, y7, x6z, x4y2z, x3y3z, x2y4z,
x6, x4y2, x3y3, x2y4
= (x6, x4y2, x3y3, x2y4, xy6, y7).
Lemma 3.8. Let n be an integer with n ≥ 3 and 3|n−1. For each intermediate integer t such that dn3e ≤ t ≤ bn2c, there exists a monomial ideal Q such that v(Q) = n and v(Q : X) = t.
Proof. Let s be an integer with 1 ≤ s < n−13 and set
fi =
( x2n−23 , for i = 1
x2n+13 −2(i−1)y2i−3, for i = 2, . . . ,n−13 − s + 1 hi =
( x2n−23 −2iy2i, for i = 1, . . . ,n−13 − s
xn−13 +s−iyn−13 −s+i, for i = n−13 − s + 1, . . . ,n−13 + s gi =
( x2n−23 −2iy2i−1z, for i = 1, . . . ,n−13 − s xn−13 +s−iyn−43 −s−iz, for i = n−13 − s + 1, . . . ,n−13 wi =
( x2n−23 −2iy2i−1, for i = 1, . . . ,n−13 − s xn−13 +s−iyn−43 −s−i, for i = n−13 − s + 1, . . . ,n−13 Take
Q =
f1, f2, . . . , fn−1
3 −s+1, h1, . . . , hn−1
3 −s, hn−1
3 −s+1, . . . , hn−1
3 , hn−1
3 +1, . . . , hn−1
3 +s, g1, . . . , gn−1
3 −s, gn−1
3 −s+1, . . . , gn−1
3
.
Since every generators of Q listed above has degree 2n−23 , v(Q) = n if they are all distinct. The detail can be checked by the same method as in the proof of Lemma 3.6.
Note that
wix =
( fi+1, if i = 1, . . . ,n−13 − s hi−1, if i = n−13 − s + 1, . . . ,n−13 wiy = hi
wiz = gi
With the same argument as in the proof of Lemma 3.6, we can show that w1, w2, . . . , wn−1
3 are the Q-corner-elements. Moreover, we have (Q : X)
=
f1, f2, . . . , fn−1
3 −s+1, h1, . . . , hn−1
3 −s, hn−1
3 −s+1, . . . , hn−1
3 , hn−1
3 +1, . . . , hn−1
3 +s, g1, . . . , gn−1
3 −s, gn−1
3 −s+1, . . . , gn−1
3 , w1, . . . , wn−1
3 −s, wn−1
3 −s+1, . . . , wn−1
3
= (f1, w1, w2, . . . , wn−1
3 , hn−1
3 +1, hn−1
3 +2, . . . , hn−1
3 +s).
Observing the monomials in the above generating set of (Q : X), the power of y is increasing when that of x is decreasing. That is to say, this is the minimal monomial generating set of (Q : X). Therefore v(Q : X) = 1+n−13 +s = n+23 +s.
Note that 1 ≤ s ≤ n−43 , thus n+23 + 1 ≤ n+23 + s ≤ n+23 + n−43 . In other words, we construct a monomial ideal Q such that v(Q) = n and v(Q : X) = t for all n+53 ≤ t ≤ 2n−23 . Furthermore, we have
2n−2
3 − bn2c =
( 2n−2
3 − n−12 , if n is odd,
2n−2
3 − n2, if n is even.
=
( n−1
6 , if n is odd,
n−4
6 , if n is even.
Recall that n ≥ 3 and 3|n − 1. If n is odd, then n ≥ 7; if n is even, then n ≥ 4. Therefore, 2n−23 − bn2c ≥ 0, i.e., 2n−23 ≥ bn2c. Therefore, we construct a monomial ideal Q such that v(Q) = n and v(Q : X) = t for all n+53 ≤ t ≤ bn2c.
Hence, it remains to construct a monomial ideal Q such that v(Q) = n and
v(Q : X) = n+23 = dn3e. Set f1 = x2n−23
fi = x2n+13 −2(i−1)y2i−3, for i = 2, . . . ,n−13 + 1 hi = x2n−23 −2iy2i, for i = 1, . . . ,n−13
gi = x2n−23 −2iy2i−1z, for i = 1, . . . ,n−13 wi = x2n−23 −2iy2i−1, for i = 1, . . . ,n−13 and set
Q =
f1, f2, . . . , fn−1
3 +1, h1, . . . , hn−1
3 , g1, . . . , gn−1
3
.
Note that
wix = fi+1 wiy = hi wiz = gi
With a similar proof as in the previous case, we have v(Q) = n and the Q- corner-elements are w1, . . . , wn−1
3 . Then
(Q : X) =
f1, f2, . . . , fn−1
3 +1, h1, . . . , hn−1
3 , g1, . . . , gn−1
3 , w1, . . . , wn−1
3
= (f1, w1, w2, . . . , wn−1
3 ).
and v(Q : X) = 1 +n−13 = n+23 = dn3e. This completes the proof.
We use the following example to demonstrate the construction.
Example 3.9. Suppose n = 13 in Lemma 3.8. We now want to construct ideals such that v(Q) = 13 and v(Q : X) = 5 and 6, respectively.
1. v(Q) = 13 and v(Q : X) = 5.
Let Q be the ideal
x8, x7y, x5y3, x3y5, xy7, x6y2, x4y4, x2y6, y8, x6yz, x4y3z, x2y5z, y7z
.
Then the Q-corner-elements are x6y2, x4y4, x2y6, y8. Hence
(Q : X) =
x8, x7y, x5y3, x3y5, xy7, x6y2, x4y4, x2y6, y8, x6yz, x4y3z, x2y5z, y7z, x6y, x4y3, x2y5, y7
= (x8, x6y, x4y3, x2y5, y7).
2. v(Q) = 13 and v(Q : X) = 6.
Let Q be the ideal
x8, x7y, x5y3, x3y5,
x6y2, x4y4, x2y6, xy7, y8, x6yz, x4y3z, x2y5z, xy6z
.
Then the Q-corner-elements are x6y, x4y3, x2y5, xy7. Hence
(Q : X) =
x8, x7y, x5y3, x3y5,
x6y2, x4y4, x2y6, xy7, y8, x6yz, x4y3z, x2y5z, xy6z, x6y, x4y3, x2y5, xy6
= (x8, x6y, x4y3, x2y5, xy6, y8).
Lemma 3.10. Let n be an integer with n ≥ 3 and 3|n − 2. For each interme- diate integer t such that dn3e ≤ t ≤ bn2c, there exists a monomial ideal Q such that v(Q) = n and v(Q : X) = t.
Proof. Let s be an integer with 1 ≤ s < n−23 and set
fi =
( x2n−43 , for i = 1
x2n+53 −2iy2i−3, for i = 2, . . . ,n−23 − s + 1 hi =
( x2n−43 −2iy2i, for i = 1, . . . ,n−23 − s
xn−23 +s−iyn−23 −s+i, for i = n−23 − s + 1, . . . ,n−23 + s
gi =
x2n−73 z, for i = 1
x2n+23 −2iy2i−3z, for i = 2, . . . ,n−23 − s + 1 xn+13 +s−iyn−83 −s+iz, for i = n−23 − s + 2, . . . ,n−23 + 1
wi =
x2n−73 , for i = 1
x2n+23 −2iy2i−3, for i = 2, . . . ,n−23 − s + 1 xn+13 +s−iyn−83 −s+i, for i = n−23 − s + 2, . . . ,n−23 + 1
