Table Of ContentAdmissible Orderings and Bounds for Gro(cid:127)bner
Basis Normal Form Algorithms
1
T. Dub(cid:19)e B. Mishra Chee-Keng Yap
Courant Institute of Mathematical Sciences
New York University
251 Mercer Street
New York, NY 10012.
July 6, 1995
1
Supported in part by NSF grants #DCR-84-01898 and #DCR-84-01633.
1
Abstract
The concepts of admissible orderings and normal form algorithm
are basic in Buchberger's Gr(cid:127)obner basis algorithm. We present a con-
structive and elementaryproof ofRobbiano'scharacterization theorem
for admissibleorderings. Using this characterization, we give a bound
on the complexityof the normalformalgorithmfor arbitrary admissi-
ble orderings. Using a simplere(cid:12)nement ofthe normalformalgorithm
(ordered reductions), we obtain signi(cid:12)cantly improvedbounds.
1 INTRODUCTION 2
1 Introduction
Gr(cid:127)obner basis has become an important algorithmic tool in computational
algebraic geometry [Buchberger1985]. Much of the pioneering work is due
to Buchberger. In particular, Buchbergergave an algorithm for constructing
a Gro(cid:127)bner basis. We refer to [Mishra and Yap 1986] for a self-contained
introduction to the subject.
InGro(cid:127)bnerbasis,weareinterestedinthepolynomialringR= K[x1;:::;xn]
for some (cid:12)eld K. The fundamentalconcept here is the `reduction'of polyno-
mials. In order to introduce this, we (cid:12)rst let PP = PP(x1;:::;xn) be the
set of all power products
Yn
ei
x
i
i=1
where ei (cid:21) 0 are natural numbers. A total ordering > on the set PP is said
A
to be admissible if the following two axioms are satis(cid:12)ed.
1. xi>1 for 1 < i< n
A
2. p>q =) rp>rq for all p;q;r2 PP
A A
Therearetwonaturalexamplesofadmissibleorderings,thelexicographic
and the total degree orderings (see next section). Power products are also
called terms, and admissible orderings are also called termorderings or mul-
tiplicative orderings. Relative to such an ordering >, we may de(cid:12)ne the head
A
monomial, Hmono(f), of any polynomial f 2 R to be that monomial in f
whose power product is the greatest under >.
A
Now we are ready to de(cid:12)ne reduction. Given two polynomials f;g 2 R,
we say f is reducibleby g if Hmono(g)divides somemonomialm in f. Then
m =c(cid:1)Hmono(g)for some monomialc. We say the polynomial h =f (cid:0)c(cid:1)g
is the reduct of f by g and denote the relationship by
g
f (cid:0)!h:
We say that the monomial m (or the corresponding power product p) is
eliminated by application of g in this case. If G is a set of polynomials,
G g
we write f(cid:0)!h if f(cid:0)!h holds for some g 2 G. We denote the re(cid:13)exive
G G
transitive closure of (cid:0)! by (cid:0)!. If f is not reducible by any g 2 G, we
(cid:3)
indicate this by writing
G
f (cid:0)!f:
2 ADMISSIBLE ORDERINGS 3
G G
We say h is a G-normal form of f if f (cid:0)!h(cid:0)!h, and we write NFG(f) for
(cid:3)
the set of all G-normalformsof f. It is importantto realize thata G-normal
form of f is not unique in general, and the central idea in Gro(cid:127)bner basis is
to enlarge G so that it becomes unique: A (cid:12)nite set G (cid:18) R is said to be a
Gr(cid:127)obner basis (for the ideal generated by G) if the G-normal form of every
polynomial f 2 R is unique, i.e., jNFG(f)j=1.
Given a (cid:12)nite set F (cid:18) R of polynomials, we de(cid:12)ne a (trivial) non-
deterministic algorithm that, for any input polynomial f, repeatedly apply
F
thereductionstep(cid:0)!tof anditsreductsuntilanormalformoff isreached.
This simple algorithm will be called the normal form algorithm. Let nfF(f)
denote a (cid:12)nal normal form so obtained, if the process halts at all. It can
be shown that this process must halt regardless of the choice of reduction {
see [Mishra and Yap 1986] fora proof. The normalformalgorithmis a basic
step in Buchberger'salgorithm for constructingGro(cid:127)bner bases.
Inthispaperweareinterestedinaboundonthenumberofreductionsteps
in the normalformalgorithm. Previously, the only bounds knownare forthe
simplecasewhere>isthetotaldegreeordering. In[Mishra and Yap 1986],a
A
boundforthelexicographicorderingwasgiven. We nowextendthisboundto
the generalcase. Along theway,wewill develop anelementaryandconstruc-
tive proof of a characterization theorem for all admissible orderings. The
characterization was (cid:12)rst given by [Robbiano 1985] but his proof is highly
non-constructive.
2 Admissible Orderings
Example 1: Lexicographic Ordering ( > )
LEX
Let A = xa11x2a2:::xnan and B = xb11xb22:::xbnn. Then A > B if ai 6= bi
LEX
for some i, and we have ai > bi for the minimum such i. To illustrate
this, consider PP(x;y;z). Then, assuming x > y > z we have:
LEX LEX
3 2 2 2
x > y z , xy > xz, and y z > yz
LEX LEX LEX
Example 2: Total-Degree Ordering (>)
T
Let A = x1a1xa22:::xann and B = xb11xb22:::xbnn. Then A>B if deg(A)>
T
deg(B) where deg(A)= a1+a2+:::+an.
2 ADMISSIBLE ORDERINGS 4
Considering again the example PP(x;y;z),we have,
3 2 2 2
y >xy, xyz>y , and y z >xyz
T T T
Noticehowever,thattotal-degreealonedoesnotprovideatotalordering
sinceitdoesnotallowcomparisonoftwopowerproductswiththesame
degree. Among the many ways in which total-degree can be extended
into a total (admissible) ordering are:
Total-Degree(Lex): We say A>B if:
TL
either deg(A)> deg(B)
or deg(A)=deg(B) & A > B
LEX
Total-Degree(Recursive): We say A>B if:
TT
either deg(A)> deg(B)
a1 a2 ak(cid:0)1 b1 b2 bk(cid:0)1
or deg(A)= deg(B) & x1 x2 :::xk(cid:0)1 T>Tx1 x2 :::xk(cid:0)1
where k = maxfijai 6=0 or bi 6= 0g
It is easily veri(cid:12)ed that both are admissible. To see that these two
orderings are in fact di(cid:11)erent, notice that on PP(x;y;z)we have
x>y>z and x>y>z
TL TL TT TT
3 3
xyz>y but y >xyz
TL TT
Remark: Usually,theordering > issimplycalled the`total-degreeordering'.
TL
Although PP is de(cid:12)ned with the natural numbers as exponents, many
of our proofs simplify if we extend the exponents to the integers and also
to the rational numbers. We write PP(x;Z) and PP(x;Q) to indicate these
extensions. We also write PP(x;N) for PP. Admissible orderings for these
extensions are de(cid:12)ned in the same way:
1. xi>1 for 1 < i< n
A
2. p>q =) rp>rq for all p;q;r2 PP(x;Q) (resp:PP(x;Z))
A A
2 ADMISSIBLE ORDERINGS 5
Lemma 1 Let > be an arbitrary admissible ordering on PP(x;Q),
A
For any M;N 2 PP(x;Q); r 2 Q,
r r
if r > 0 then M>N () M >N
A A
r r
if r < 0 then M>N () N >M
A A
Proof.
(r 2 N) For r = 1, the result is trivial, so assume the result holds for r(cid:0)1,
r r(cid:0)1 r r r
then M>N implies M > M N > N . Conversely,if M >N , then
A A A A
r r
M>N since M 6 >N leads to the contradictionM 6>N .
A A A
s
(r > 0) Letr beapositiverationaloftheform ,thentheresultfollowsfrom
t
s s
Mt>Nt () Ms>Ns () M>N (by two applications of the case
A A A
r 2 N).
(cid:0)1 (cid:0)1
(r < 0) ItiseasytocheckthatM>N () N >M . Thenanapplication
A A
(cid:0)1 (cid:0)1 r r
of the case r > 0 shows N >M () N >M
A A
Q.E.D.
If> is anadmissibleorderingonPP(x;Q), thenit isclear that> induces
Q Q
an admissible ordering > on PP(x;N) , namely > is the restriction of > to
N N Q
PP. Theconverserelationisalsotrue. Everyadmissibleorder>onPP(x;N)
N
induces a relation > on the rationalsde(cid:12)ned by:
Q
c c
M>N () M U>N U
Q N
c c
where c 2 N is chosen such that M ,N 2 PP(x;Z) and U 2 PP(x;Z) is
c c
chosen such that M U;N U 2 PP(x;N). The reader can verify that the
induced relation > is an admissible ordering. It is seen that, for every pair
Q
of monomials M;N 2 PP(x;N),
M>N () M>N
Q N
so the induced admissible orderings on PP(x;Q) are simply extensions of
the admissible orderings on PP(x;N). These are in fact the only admis-
sible orderings on PP(x;Q). For any admissible ordering > on PP(x;Q),
A
c c
M>N () M U>N U , so the ordering is completely speci(cid:12)ed by the
A A
ordering of power products with natural number exponents. This proves:
3 CHARACTERIZINGADMISSIBLE ORDERINGS 6
Lemma 2 Thereisanaturalbijectionbetweenthesetofadmissibleorderings
on PP(x;N) and the set of admissible orderings on PP(x;Q).
A useful characterizationof admissible orderings is:
Lemma 3 If > and > are admissible orderings on PP(x;Q), then > and >
A B A B
are identical if and only if the following sets are equal:
(cid:26) (cid:27)
SA = M>1 : M 2 PP(x;Q)
A
(cid:26) (cid:27)
SB = M>1 : M 2 PP(x;Q)
B
(cid:0)1 (cid:0)1
Proof. If SA =SB, then M>N () MN 2 SA () MN 2 SB ()
A
M>N. So > and > are identical.
B A B
Otherwise, if SA 6= SB, then without loss of generality assume that there
exists an N 2 SA(cid:0)SB. Then, > and >are di(cid:11)erent since N>1, but N 6>1.
A B A B
Q.E.D.
3 Characterizing Admissible Orderings
The main result of this sectionis a new and constructiveproofof Robbiano's
theorem [Robbiano 1985]:
Theorem 4 Any admissible ordering > on PP(x;Q) can be characterized
A
by a set of linear `weight functions' W1;W2;:::;Wn given by
Xn
Wk(x(cid:11)11x(cid:11)22:::x(cid:11)nn) = wk;i(cid:11)i for 1(cid:20) k (cid:20) n
i=1
where the wk;i's are real, such that if M is a power product M =
(cid:11)1 (cid:11)2 (cid:11)n
x1 x2 :::xn then,
M>1() (9k)(Wk(M)> 0 & (8j<k)Wj(M)= 0) (1)
A
Thus,
(cid:0) (cid:1)
(cid:0)1
Wk(M) = (cid:0)Wk M and Wk(MN) = Wk(M)Wk(N)
Note that (1) is a characterizationbecause of lemma 3.
3 CHARACTERIZINGADMISSIBLE ORDERINGS 7
Essentially, we are reduced to a lexicographic ordering on the tuples
(W1(M), :::, Wn(M)). The existence of the wk;i's will be demonstrated
th
by an explicit construction. We will do the constructionin stages. In the k
stage,we will de(cid:12)ne the function Wk, and also choose a new variablezk from
fx1;:::;xng which was not chosen in a previous stage.
Stage 1:
De(cid:12)ne z1 to be xj where for all xi, xj(cid:21)xi.
(cid:26) A (cid:27)
s
Then choose w1;i to be sup s 2 Qj xi(cid:21)z1 .
A
(
X1 = fx1;:::;xng(cid:0)fz1g
Set .
Z1 = fz1g
Stage k (k = 2 to n):
Assume that Wj has been de(cid:12)ned for j = 1;:::;k(cid:0)1.
Case 1:
There exists xi 2 Xk(cid:0)1 such that for all N 2 PP(Zk(cid:0)1;Q)
(9j<k)Wj(Nxi)6=0 :
Choose zk to be any such xi (for example the least).
For every j, set wk;j = 0.
Case 2:
For every xi 2 Xk(cid:0)1, there exists N 2 PP(Zk(cid:0)1;Q) such that for
all j < k, Wj(Nxi)=0.
De(cid:12)ne Mk(M), (cid:13)k;i, and x^i as follows:
For all M 2 PP(x;Q), and xi 2 Xk(cid:0)1, Mk(M) = NM where
(8j<k)Wj(NM)= 0 and N 2 PP(Zk(cid:0)1;Q).
We will show later that Mk(M) is well de(cid:12)ned.
(
1 if Mk(xi)(cid:21)1
(cid:13)k;i = A
(cid:0)1 otherwise
(cid:13)k;i
x^i = xi
Choose zk to be the xj 2 Xk(cid:0)1 such that for all xl 2 Xk(cid:0)1,
Mk(x^j)(cid:21)Mk(x^l),
A
and let z^k = x^j where zk = xj.
3 CHARACTERIZINGADMISSIBLE ORDERINGS 8
Then for all l = 1;:::;n choose,
8
< 0 (cid:26) if xl 62 Xk(cid:0)1 (cid:27)
wk;l = s :
: (cid:13)k;lsup s 2 QjMk(x^l)(cid:21)Mk(z^k)
A
(Either case:)
(
Xk = Xk(cid:0)1 (cid:0)fzkg
Set .
Zk = Zk(cid:0)1 [fzkg
To verify the correctness of this construction, we must show that the
resulting weight functions satisfy (1). To show that (1) holds in the (=
direction, we will prove at each stage:
Lemma Ak :
Wk(M)> 0 & 8j<kWj(M) =0 =) M>1
A
To show that (1) also holds in the =) direction, we will prove at each
stage:
Lemma Bk :
For any M 2 PP(Zk); (8j(cid:20)k)((Wk(M)= 0)=) (M = 1))
Now, at stage n, this includes all power products M 2 PP(x;Q). Let M
(cid:0)1
be any power product with M>1 ,and so M <1. Lemma Bn requires at
A A
(cid:0)1 (cid:0)1
least one of the Wi(M ) be non-zero,for otherwiseM would be 1. Let k
(cid:0)1
be the smallest i for which Wi(M ) is non-zero, then
(cid:0) (cid:1)
(cid:0)1 (cid:0)1
Wk(M ) 6=0 & 8j<kWj M =0:
(cid:0)1 (cid:0)1 (cid:0)1 (cid:0)1
If Wk(M )> 0,then by lemmaAk, M >1. But, M <1,so Wk(M )<
A A
0,andthereforeWk(M)> 0. Therefore,thetwolemmasimplythecorrectness
of the wk;i construction.
Now we are ready to prove lemmas Ak and Bk.
Stage 1:
t qt
Lemma 5 for 1< i < n, 8q;t2Q (qt< w1;it) =) (xi (cid:21) z1 )
A
qt t
Proof. Otherwise,there exists i;q, and t such that qt < w1;it, and z1 > xi
A
3 CHARACTERIZINGADMISSIBLE ORDERINGS 9
Case I: t > 0 . Then
qt t q
z1 >xi =) z1>xi
A A
qt < w1;it =) q < w1;i
From the de(cid:12)nition of w1;i, for each (cid:15) > 0 it is possible to (cid:12)nd an s
such that
w1;i(cid:0)s < (cid:15)
and,
s
xi (cid:21) z1
A
Choose (cid:15) < w1;i(cid:0)q, then s(cid:0)q > 0, which leads to the contradiction
q s q (cid:0)q s (cid:0)q
z1>xi(cid:21)z1 =) z1z1 >z1z1
A A A
s(cid:0)q
=) 1>z1
A
=) 1>z1
A
Case II: t = 0 .
This leads to the immediate contradiction1>1.
A
Case III: t < 0 .
qt t q
z >x =) x>z
A A
qt < w1;it =) w1;i < q
(cid:26) (cid:27)
s
But, by de(cid:12)nition, w1;i =sup sjx>z so, w1;i (cid:21) q.
A
Q.E.D.
Using this result we prove:
Lemma A1 : W1(M)> 0 =) M>1
A
(cid:11)1 (cid:11)2 (cid:11)n
Proof. LetM beanypowerproductM = x1 x2 :::xn suchthatW1(M)>
0. Since W1(M) > 0 it is possible to (cid:12)nd a c 2 Q suct that, c > 0 and
W1(M)>c. Then, for each i, choose qi satisfying
c
w1;i(cid:11)i (cid:0) < qi(cid:11)i < w1;i(cid:11)i
n
