Ifu1x1+u2x2+· · ·+uqxq = 0, thenhu1,u2,. . .,uqi=u∈S(M); so there arev1,v2,. . .,vp∈S such that
u=v1s1+v2s2+· · ·+vpsp. This is equivalent to the following:
u1 = v1s1,1+v2s2,1+· · ·+vpsp,1 u2 = v1s1,2+v2s2,2+· · ·+vpsp,2 .. . uq = v1s1,q+v2s2,q+· · ·+vpsp,q.
2.5
S-Polynomials
Definition 2.5.1 (S-Polynomials) Let S be a ring; R =S[x1, . . ., xn]
be a ring of polynomials overS;G⊆Rbe a finite subset; and F ={f1, . . . , fq} ⊆G. Set T =nt1= Hcoef(f1), . . . , tq= Hcoef(fq) o ⊆S and J = (t1, . . . , tq),
the ideal inS generated byT. From the previous discussion, we can write the syzygy ofJ,S(J)⊆Sq, by its system of generators,{s
1,. . .,sp}, where si∈Sq,i= 1, . . .,p. That is, S(J) = hs1,1, s1,2, . . . , s1,qi, .. . hsp,1, sp,2, . . . , sp,qi .
We define the set ofS-polynomials ofF, which we denote bySP(F), to be the set{h1,F,. . .,hp,F}, where, fori= 1,. . ., p,
hi,F =si,1· m
Hterm(f1)·
f1+· · ·+si,q· m Hterm(fq)·
fq, wherem=LCM(Hterm(f1), . . . ,Hterm(fq)).
Note that if we simplify the expression for anS-polynomial, say hi,F, we get hi,F = si,1t1·m+si,1· m Hterm(f1)· Tail(f1) +· · · +si,qtq·m+si,q· m Hterm(fq)·Tail(fq) = (si,1t1+· · ·+si,qtq)m +si,1· m Hterm(f1)· Tail(f1) +· · ·+si,q· m Hterm(fq)· Tail(fq) = si,1· m Hterm(f1)· Tail(f1) +· · ·+si,q· m Hterm(fq)· Tail(fq).
Thus, Hterm(hi,F)<
Am= LCM
Hterm(fi) :fi∈F
.
Theorem 2.5.1 Let F = {f1, . . ., fq} be as in the previous definition. Assume that for someu1,. . .,uq∈S,p1,. . .,pq ∈PP(x1,. . .,xn),
h= q
X
i=1 uipifi, and that p1Hterm(f1) =· · ·=pqHterm(fq) =M >
A Hterm(h). Thus, q X i=1 uipiHmono(fi) = 0.
Then we can expresshin terms of the S-polynomialsof the setF as follows: h=
p
X
j=1
vj,F rF hj,F,
where vj,F ∈ S, rF ∈PP(x1, . . ., xn) and hj,F’s are S-polynomialsof F, SP(F). Furthermore, rFHterm hj,F < A M =piHterm(fi). proof. Let
t1= Hcoef(f1), . . . , tq = Hcoef(fq) and J = (t1, . . . , tq). Let a system of generators for the syzygy ofJ,S(J), be given as follows:
S(J) = hs1,1, s1,2, . . . , s1,qi, .. . hsp,1, sp,2, . . . , sp,qi .
Section 2.5 S-Polynomials 57 Since q X i=1 uipiHmono(fi) =M · q X i=1 uiti = 0, there existv1, . . .,vp∈S such that
u1= p X j=1 vjsj,1, . . . , uq = p X j=1 vjsj,q.
Now, if we letm= LCMHterm(f1), . . . ,Hterm(fq)
,then it is clear that m|piHterm(fi), for alli,
i.e., there is some power productr∈PP(x1, . . .,xn) such that m·r=M =pi·Hterm(fi), for alli. Thus, we can rewritehas follows:
h = p X j=1 vjsj,1 r· m Hterm(f1)· f1+· · · + p X j=1 vjsj,q r·Hterm(m f q)· fq = v1r·h1,F +· · ·+vpr·hp,F. Also, for alli, Hterm(r) Hterm(hi,F) <
A r·m=piHterm(fi).
Definition 2.5.2 (Syzygy Condition for a Finite Set G) LetS be a ring, and let
G={g1, g2, . . . , gm} ⊆S[x1, . . . , xn],
be a finite set of polynomials. We say G satisfies thesyzygy condition if for every subset F ⊆ G and every S-polynomial h ∈ SP(F), h can be expressed as h= m X i=1 figi, wherefi∈S[x1, . . . , xn] and Hterm(h)≥
A
Hterm(fi) Hterm(gi).
Now we are ready to give a new characterization of a Gr¨obner basis in terms of the syzygy condition:
Theorem 2.5.2 LetI be an ideal inR=S[x1,. . .,xn], and G={g1, . . . , gm} ⊆I,
a finite subset ofI. Then the following three statements are equivalent: 1. Head(G) = Head(I). 2. ∀f ∈I hf =Pgi∈Gfigi i , where fi∈S[x1,. . .,xn] andHterm(f)≥ A
Hterm(fi) Hterm(gi), for alli. 3. (G) =I andGsatisfies the syzygy condition.
Section 2.5 S-Polynomials 59
proof.
(1⇔2):
The equivalence of (1) and (2) is simply Theorem 2.3.3. (2⇒3):
Note first that (2) establishesGas a Gr¨obner basis (by Theorem 2.3.3) and that (G) =I (by Theorem 2.3.1). Furthermore, ifF ⊆Gandh∈SP(F), thenh∈I; thus by condition (2) itself,
h= X
gi∈G
figi,
where fi∈S[x1, . . . , xn] and Hterm(h)≥
A
Hterm(fi) Hterm(gi). But this is precisely the syzygy condition.
(3⇒2):
We assume to the contrary, i.e., condition (3) holds, but not (2); we shall derive a contradiction.
We first defineheight of a sequence
H =Dhi: 1≤i≤m, andhi∈R
E
(with respect toGand the admissible ordering <
A) as Height(H) = max > A n Hterm(hi) Hterm(gi) : 1≤i≤mo.
Since by (3), (G) =I, everyf ∈I can be expressed as m
X
i=1
figi, fi∈S[x1, . . . , xn].
Furthermore, since (2) is assumed to be false, we can choose anf ∈Isuch that
HeightDf1, . . . , fm
E
>
AHterm(f).
We assume that the representation off is so chosen that it is of aminimal height,M. Let
F =ngi∈G: Hterm(fi) Hterm(gi) =M
o
.
Without loss of generality, we may assume that F consists of the first k elements ofG. Thus f = k X i=1 Hmono(fi)gi+ k X i=1 Tail(fi)gi+ m X i=k+1 figi,
where
Hterm(fi) Hterm(gi) = M, 1≤i≤k, Hterm(Tail(fi)) Hterm(gi) <
A M, 1≤i≤kand Hterm(fi) Hterm(gi) < A M, k+ 1≤i≤m. Sincef < AM, we see that
Hterm(f1) Hterm(g1) = · · · = Hterm(fk) Hterm(gk) = M > A Hterm Xk i=1 Hmono(fi)gi .
As noted before (Theorem 2.5.1), we see that the expressionPki=1Hmono(fi)gi can be expressed in terms ofSP(F), theS-polynomials ofF. Thus we may writef as f =X j vj,F rF hj,F + k X i=1 Tail(fi)gi+ m X i=k+1 figi.
But sincerFHterm(hj,F) <
A Hterm(fi) Hterm(gi) =M, 1≤i≤k,
and since, by the syzygy condition, each hj,F = X i fi,j,F gi with Hterm(hj,F)≥ A
Hterm(fi,j,F) Hterm(gi),we get
f =X j X i vj,F rF fi,j,F gi+ k X i=1 Tail(fi)gi+ m X i=k+1 figi. But, by the previous arguments,
Htermvj,F rF fi,j,F
Hterm(gi) ≤
A
Hterm(rF) Hterm(fi,j,F) Hterm(gi)
≤
A
Hterm(rF) Hterm(hj,F) <
A M,
and hence, we can expressf differently as f = m X i=1 fi′gi, with Height(hf′ 1, . . . , fm′ i)<
AM, and thus contradicting our choice of the
Problems 61
Problems
Problem 2.1Show the following:
(i)An elementr∈Rof the ringRis a unit if and only if (r) = (1). (ii)Ifris a unit andxis nilpotent in the ringR, thenr+xis again a unit.
Problem 2.2
Prove that ifI1 and I2, as well as I1 and I3 are coprime, thenI1 and I2I3 are coprime. Hence, ifI1,I2,. . .,In are pairwise coprime, then
I1I2· · ·In =I1∩I2∩ · · · ∩In.
Problem 2.3
LetI1,I2⊆Rbe two ideals with basesB1andB2, respectively. Which
of the following statements are true? Justify your answers. (i)B1∪B2 is a basis forI1+I2.
(ii){b1·b2 : b1∈B1 andb2∈B2} is a basis forI1I2.
(iii){bi1bi2· · ·bin : bi1, bi2, . . . , bin ∈B1}is a basis forI
n
1.
(iv)Let f ∈R, and C1, a basis for I1∩(f). Then, every c1 ∈ C1 is
divisible byf, and {c1/f : c1∈C1}is a basis for (I1: (f)).
Problem 2.4
LetI,J,K be ideals in a ringR. Prove the following:
(i)Modular law: IfI⊇J orI⊇K, thenI∩(J+K) = (I∩J)+(I∩K). (ii)I⊆I:J, and (I:J)J⊆I. (iii)(TiIi:J) =Ti(Ii:J), and (I:PiJi) = T i(I:Ji). (iv)√I⊇I, and p√I=√I. Problem 2.5
For eachk= 1, . . . , n, we define a functionUk as follows:
Uk : PP(x1, x2, . . . , xn)→R
: p=xa1
1 xa22· · ·xann7→uk,1a1+uk,2a2+· · ·+uk,nan,
where each ofuk,l(k= 1, . . . , n,l= 1, . . . , n) is a nonnegative real number. Assume thatUk’s are so chosen that hU1(p), . . . , Un(p)i=h0, . . . ,0iif and
only ifp= 1. We define an ordering >
U on the power products as follows:
given two power products p, q ∈ PP(x1, x2, . . . , xn), we say p >
U
q, if the first nonzero entry of
is positive.
Show that the ordering >
U
is an admissible ordering. Characterize the admissible orderings >
LEX, TLEX> and TRLEX> in terms of appropriately chosen
functionsUk.
Problem 2.6
Let (x2+y2), (xy)
⊆Q[x,y] be two ideals with Gr¨obner bases{x2+ y2
}and {xy}, respectively, with respect to the >
TRLEX (withx >TRLEXy). Is
{x2+y2, xy
}a Gr¨obner basis for (x2+y2) + (xy) under >
TRLEX?
Problem 2.7 Let <
A be a fixed but arbitrary admissible ordering on PP(x1,. . .,xn).
Consider the following procedure (possibly, nonterminating) to compute a basis for an ideal in the polynomial ringR=S[x1, . . . , xn]:
G:={0};
while(G)6=I loop
Choosef∈I\(G), such that Hmono(f) is the smallest among all such elements with respect to <
A
;
G:=G∪ {f}; end{loop }
Which of the following statements are true? Justify your answers. (i) The procedure terminates ifS is Noetherian.
(ii)The procedure is an effective algorithm.
(iii)Letfi be the element ofIthat is added to Gin theith iteration. Then Hterm(f1)≤ A Hterm(f2)≤ A · · · ≤A Hterm(fn)≤ A · · ·
(iv) The setGat the termination is a Gr¨obner basis forI.
Problem 2.8
Consider the polynomial ring R = S[x1, . . ., xn], with total reverse
lexicographic admissible ordering >
TRLEX such thatx1 > TRLEX · · · > TRLEX xn. The homogeneous part of a polynomialf ∈R of degreed(denoted fd) is simply the sum of all the monomials of degree din f. An idealI ⊆R is said to be homogeneous if the following condition holds: f ∈Iimplies that for alld≥0,fd∈I
Prove that: If G is a Gr¨obner basis for a homogeneous ideal I with respect to >
TRLEX inR, thenG∪ {xi, . . . , xn} is a Gr¨obner basis for (I, xi,
. . ., xn) (1≤i≤n) with respect to >
TRLEX in R.
Solutions to Selected Problems 63
Problem 2.9
LetKbe a field andIan ideal inK[x1,x2,. . .,xn]. LetGbe a maximal subset ofI satisfying the following two conditions:
1. Allf ∈Gare monic, i.e., Hcoef(f) = 1.
2. For allf ∈Gandg∈I, Hterm(f)6= Hterm(g) implies that Hterm(g) does not divide Hterm(f).
(i)Show thatGis finite.
(ii)Prove thatGis a Gr¨obner basis forI.
(iii)A Gr¨obner basis Gmin for an ideal I is a minimal Gr¨obner basis
forI, if no proper subset ofGminis a Gr¨obner basis forI.
Show thatG(defined earlier) is a minimal Gr¨obner basis forI. (iv)LetGbe a Gr¨obner basis (not necessarily finite) for an idealI of K[x1, x2, . . . , xn]. Then there is a finite minimal Gr¨obner basisG′⊆Gfor I.
Problem 2.10
(i) Given as input a0, a1, . . ., an andb0, b1, . . .,bn integers, write an algorithm that computes all the coefficients of
(a0+a1x+· · ·+anxn)(b0+b1x+· · ·+bnxn). Your algorithm should work inO(n2) operations.
(ii) Show that givenM andv, two positive integers, there is a unique polynomial
p(x) =p0+p1x+· · ·+pnxn satisfying the following two conditions:
1. p0,p1,. . .,pn are all integers in the range [0, M−1], and 2. p(M) =v
(iii) Devise an algorithm that on input a0, a1, . . ., an; M integers
evaluates
a0+a1M+· · ·+anMn. Your algorithm should work inO(n) operations.
(iv) Use (ii) and (iii) to develop an algorithm for polynomial multi- plication [as in (i)] that uses O(n) arithmetic operations (i.e., additions, multiplications, divisions, etc.).
(v) Comment on your algorithm for (iv). What is the bit-complexity of your algorithm?
Solutions to Selected Problems
Problem 2.3All but the last statement are true. Since the proofs for the first three assertions are fairly simple, we shall concentrate on the last statement.
(iv)Here is a counterexample (due to Giovanni Gallo of University of Catania):
LetR=Q[x,y]/(y5) be a ring,f =y2∈R andI
1 = (xy), an ideal
in R. Since xy2∈(y2) andxy2 ∈(xy), clearly,xy2∈I∩(f). Conversely, ifa ∈I∩(f), then amust be r·xy2 for some r ∈R. Hence, a∈(xy2), and (xy2) =I∩(f).
On the other hand, as, for all k≥3,
yk·f ≡y2+k ≡0 mod (y5), yk
∈I: (f). But, for anyk >0,yk
6∈(x). Therefore,{x}is not a basis for I: (f). In fact, for this example,{x, y3
} is a basis forI: (f) = (xy) : (y2),
for the following reasons: Ifa∈(xy) : (y2), then the following statements
are all equivalent.
ay2∈(xy) ⇔ ay2∈(xy)∩(y2) = (xy2)
⇔ ay2≡rxy2mod (y5), r∈R
⇔ (a−rx)y2=sy5, s∈R
⇔ a=rx+sy3, r, s∈R.
However, iff ∈ R is not a nonzero zero divisor, or if R is an integral domain, then the statement holds. Leta6= 0∈I: (f). Thenaf ∈I, and af 6= 0. Thus,af = Pci∈Criciand a = Pci∈Crici/f exists. Hence,a∈
({c/f :c∈C}).As, for allci∈C,ci/f∈I: (f), we see that{c/f :c∈C} is a basis forI: (f).
In general, the following holds:
Letf ∈R,Cbe a basis forI∩(f), andDbe a basis forann (f). Then, every c∈C is divisible byf, and {c/f :c∈C} ∪D is a basis for I: (f). af ∈I ⇔ af ∈I∩(f) = (C) ⇔ af− X ci∈C rici= 0, ri∈R ⇔ a− X ci∈C ri ci f ! f = 0 ⇔ a− X ci∈C ri ci f ! ∈ann (f)
Solutions to Selected Problems 65 ⇔ a= X ci∈C rici f + X di∈D sidi, ri, si∈R ⇔ a∈ c f |c∈C ∪D . Thus{c/f :c∈C} ∪D is a basis forI: (f). Problem 2.7
LetGi denote{f1,. . .,fi}.
(i) True. The termination of the procedure directly follows from the ascending chain condition (ACC) property of the Noetherian ring as
(G1) (G2) (G3) · · ·.
(ii) False. This algorithm is not effective because it does not say how to findf ∈I\(G).
(iii)True. The condition Hterm(f1)≤
A
Hterm(f2)≤
A · · ·
follows immedi- ately from the fact that, at each step, we choose anfisuch that Hmono(fi) is the smallest among all elements ofI\(Gi−1). Indeed, if it were not true,
then there would beiandj (j < i) such that Hterm(fi)<
AHterm(fj).
Assume that among all such elementsfi is the first element which violates the property. But, since (Gj−1)⊂(Gi−1), and since fi ∈I\(Gi−1), we
havefi ∈I\(Gj−1). Since, Hterm(fi)<
AHterm(fj), the algorithm would
have chosenfiin thejth step, instead offj, contradicting the hypothesis. (iv) False. Consider the ideal I = (xy, x2+y2) ⊆Q[x,y]. Let the
first polynomial chosen bef1=xy∈I\(0) as Hmono(f1) is the smallest
among all such elments (with respect to <
TRLEX
). Similarly, let the second polynomial chosen bef2=x2+y2∈I\(xy) as Hmono(f2) is the smallest
among all such elments (with respect to <
TRLEX). As I = (f1, f2), the
algorithm terminates withG={f1,f2}. Thus Head(G) = (xy,x2). Since y3 =y(x2+y2)−x(xy)∈I, we have y3 ∈Head(I)\Head(G), and Gis
not a Gr¨obner basis ofI. Problem 2.8
SinceG⊆I, we have (G,xi,. . .,xn)⊆(I,xi,. . .,xn) and Head(G,xi, . . .,xn)⊆Head(I,xi,. . .,xn). Hence, we only need to show that Head(G, xi, . . .,xn)⊇Head(I,xi, . . .,xn).
Following the hint, we proceed as follows: Letfd∈(I,xi,. . .,xn) be a homogeneous polynomial of degreed. Then, if Hmono(fd)∈(xi, . . ., xn),
then, plainly, Hmono(fd)∈(Head(I), xi, . . ., xn). Otherwise, Hmono(fd) is not divisible by any of thexj’s andfd can be expressed as follows:
fd=gd+hixi+· · ·+hnxn, gd∈I, andhi, . . . , hn∈S[x1, . . . , xn].
Since Hmono(fd) ∈ PP(x1, . . ., xn), by the choice of our admissible
ordering, Hmono(fd) >
TRLEXHmono(hjxj), and thus
Hmono(fd) = Hmono(gd)∈Head(I)⊆(Head(I), xi, . . . , xn). This proves that
(Head(I), xi, . . . , xn) ⊇ Head(I, xi, . . . , xn)
⊇ (Head(I), xi, . . . , xn).
Since Head(G, xi, . . .,xn)⊇(Head(G),xi, . . .,xn), using the hint, we have
Head(G, xi, . . . , xn)
⊇ (Head(G), xi, . . . , xn)
= (Head(I), xi, . . . , xn) (Gis a Gr¨obner basis forI)
⊇ Head(I, xi, . . . , xn), as claimed.
Problem 2.9
SinceK is a field, for alla∈K,a−1∈K. Hence,
∀f ∈I ∃f′∈I hHterm(f) = Hterm(f′), but Hcoef(f′) = 1i. Moreover, for alla∈K,af′ ∈I.
(i) Let the set of power products,Gb, be the set of head terms ofG:
b
G={Hterm(f) :f ∈G}.
Let Hterm(f) and Hterm(g) be two distinct power products in Gb. Then f ∈ G ⊆ I and g ∈ G ⊆ I. Thus, by condition (2), Hterm(f) is not a multiple of Hterm(g), nor the converse. But, then by Dickson’s lemma,Gb is finite, and so isG, since there is a bijective map betweenGandGb.
(ii)We claim that
∀h∈I ∃f ∈G hHterm(h) =p·Hterm(f)i, wherep∈PP(x1, . . . , xn).
Solutions to Selected Problems 67
Indeed if it were not true, then we could choose a “smallest”h∈I[i.e., the Hterm(h) is minimal under the chosen admissible ordering ≤
A
] violating the above condition. Without loss of generality, we assume thathis monic. If g ∈ I is a polynomial with a distinct Hterm from h such that Hterm(g) divides Hterm(h), then Hterm(g)<
AHterm(h), and by the choice
ofh, Hterm(g) is a multiple of Hterm(f), for somef ∈G. But, this con- tradicts the assumption that Hterm(h) is not divisible by Hterm(f), as f ∈G.
Thus, for all g∈I, Hterm(h)6= Hterm(g) implies that Hterm(g) does not divide Hterm(h). But, if this is the case, then G∪ {h} also satisfies conditions (1) and (2), which contradicts the maximality ofG.
Now, we see that if h∈I, then
Hmono(h) = Hcoef(h)·Hterm(h) = Hcoef(h)·pHterm(f), where f ∈G, p∈PP(x1, . . . , xn)
⇒ Hmono(h)∈Head(G).
Therefore Head(I)⊆Head(G); henceGis a Gr¨obner basis forI.
(iii) SupposeGis not a minimal Gr¨obner basis forI. Let G′ Gbe a minimal Gr¨obner basis. Let f ∈G\G′. By the construction ofG, for all g ∈ G′ ⊂ I, Hterm(g) does not divide Hterm(f). Thus Hterm(f) ∈ Head(G)\Head(G′). This leads us to the conclusion that
Head(G′)6= Head(G) = Head(I). ThusG′ is not a Gr¨obner basis, as assumed.
(iv) Let G′ ⊆ G be a minimal Gr¨obner basis for I. Without loss of generality, assume that all the polynomials ofG′ are monic. Letf,g∈G′ with distinct Hterm’s. We claim that Hterm(f) does not divide Hterm(g). Since, otherwise,
Head(I) = Head(G′) = Head (G′\ {g}),
and, G′ \ {g} G′ would be a Gr¨obner basis for I, contradicting the minimality ofG′. Thus, by Dickson’s lemma the set of power products,
c
G′ ={Hterm(g)|g∈G′},