1. science
2. physics
3. electromagnetism

TRANSITIVITY FOR WEAK AND STRONG GRO BNER BASES
W. W. ADAMS, A. BOYLE, AND P. LOUSTAUNAU
Abstract. Let be a Noetherian integral domain which is graded by an
ordered group ? and let x be a set of variables with a term order. It is
shown that a nite subset of [x] is a weak (respectively strong) Grobner
basis in [x] graded by ? Zn if and only if is a weak Grobner basis in
[x] graded by f0g Zn and certain subsets of the set of leading coecients
of the elements of form weak (respectively strong) Grobner bases in . It
is further shown that any ?-graded ring for which every ideal has a strong
Grobner basis is isomorphic to [ 1
n ], where is a PID.
R
n
F
R
R
F
R
F
R
R
k x ;::: ;x
k
1. Introduction
Let y and x be sets of variables, each with a term order and an elimination order
between them. Let k denote a Noetherian commutative ring. In several places in
the literature (e.g. Adams and Boyle (1992), Bayer and Stillman (1987), Gianni,
Trager, and Zacharias (1988), Spear (1977), Shtokhamer (1988)), the problem of
lifting Grobner bases from the ring k[y] to the ring k[y; x] has been examined.
This entails understanding the dierence between a Grobner basis in (k[y])[x] and
a Grobner basis in k[y; x]. We refer to this as the transitivity question. This
problem was examined in Adams and Boyle (1992) mainly for the case when x = x
consisted of a single variable. There it was shown that certain subsets of the leading
coecients with respect to x must form Grobner bases in k[y] in order to go from
a Grobner basis in (k[y])[x] to a Grobner basis in k[y; x].
In this paper we are able to answer the transitivity question when x is more than
one variable by considering so-called saturated sets of polynomials as introduced by
Moller (1988) (where they were called maximal sets). We found that graded rings
are a natural setting for this question. So, let R be a Noetherian integral domain
and assume that R is graded by an ordered group ?. The concept of Grobner basis
can be extended to such graded rings R (see, for example Miola and Mora (1988)).
If x denotes a set of n variables, the ring R[x] can be graded both by ? Zn and
f0g Zn . Then the transitivity question in this setting becomes: when is a Grobner
basis in R[x], graded by f0g Zn , also a Grobner basis in R[x], graded by ? Zn .
The solution is stated in terms of certain sets of leading coecients corresponding
to the saturated subsets being Grobner bases in R.
Grobner bases for k[x1; : : : ; xn], where k is a eld, were originally dened by
Buchberger, cf. Buchberger (1985). There, a nite set F is called a Grobner basis
for an ideal I, if given any element a of I, there is an f 2 F such that the leading
term of f divides the leading term of a. When the concept of Grobner bases was
extended to a commutative ring k, the leading term of a was only required to be
a linear combination of leading terms of elements of F. Following Moller (1988),
we refer to the rst concept as a strong Grobner basis and the latter as a weak
Grobner basis. The transitivity question is answered in the current paper for both
1
2
W. ADAMS, A. BOYLE, P. LOUSTAUNAU
strong Grobner bases and weak Grobner bases. Using this we can take a Grobner
basis in R[x] graded with respect to f0g Zn and explicitly show how to construct
a Grobner basis in R[x] graded with respect to ? Zn .
The nal section of this paper is concerned with graded rings in which every ideal has a strong Grobner basis. Moller (1988) remarks that if every ideal
in k[x1; : : : ; xn], where k is a unique factorization domain (UFD), has a strong
Grobner basis, then k is necessarily a principal ideal domain (PID). We show that
every ?-graded ring R with this property must look like R = k[x1; : : : ; xn], where
k is a PID.
We now give a more detailed account of the sections in this paper. In Section
2 we give the basic denitions of weak and strong Grobner bases with respect to
graded rings; derive some of their elementary properties; and, in Theorem 2.7, give
some of the usual characterizations of Grobner bases. In Section 3 we consider the
transitivity question for weak Grobner bases. The main characterization is given in
Theorem 3.2 where we show that a nite subset F of R[x] is a weak Grobner basis
with respect to ? Zn if and only if F is a weak Grobner basis in R[x] with respect
to f0g Zn and for all saturated subsets S of f1; :::; sg, the set of leading coecients
of the polynomials indexed by S is a weak Grobner basis in R with respect to ?.
We use this result to show in Corollary 3.3, that if we can compute weak Grobner
bases in R[x] with respect to f0g Zn , and in R with respect to ?, then we can
compute weak Grobner bases in R[x] with respect to ? Zn . We go on to give
alternate characterizations of the conditions in Theorem 3.2. In Proposition 3.7
we determine certain subsets of a weak Grobner basis that are also weak Grobner
bases. Examples are given illustrating some of these results. In Section 4 we give
the analogous results for strong Grobner bases. Theorem 4.1 gives the main result
and Corollaries 4.2 and 4.3 show that if we can compute weak Grobner bases in R[x]
with respect to f0g Zn , and strong Grobner bases in R with respect to ?, then
we can compute strong Grobner bases in R[x] with respect to ? Zn. In Corollary
4.6 we show that for a eld k and a single variable y, a strong Grobner basis in
(k[y])[x] is automatically a (strong) Grobner basis in k[y; x]. Finally, in Section 5,
we give some structure theorems for graded rings with strong Grobner bases. We
rst show in Theorem 5.1, that if R is a ?-graded UFD with only nitely many
homogeneous irreducibles of positive value, then R is isomorphic to a polynomial
ring over a UFD. We then show that this hypothesis is satised if the so-called
irrelevant ideal of R has a strong Grobner basis. This enables us to prove that
if R is a ?-graded UFD and every ideal has a strong Grobner basis then R is a
polynomial ring over a PID.
2. Graded Rings
Let ? be an additive abelian group which is totally ordered with respect to an
order, denoted by \<", and where we assume that the order respects the group
law. The latter means that for all ; ; 2 ?, we have that < implies that
+ < + . Two examples are most important in this paper. One is the trivial
group ? = f0g. The other is ? = Zn (where Z denotes the integers and n > 0 is an
integer) and ? has a term order (c.f Buchberger (1985) and Robbiano (1989)).
that R is a Noetherian integral domain graded by ?. Thus R =
L We Rassume
,
where
each R is an additive abelian group and R R R + for all
2?
; 2 ?. Let ?0 = f 2 ?jR 6= 0g. Since 0 2 ?0 (1 = 12 2 R0 ), we see that ?0
TRANSITIVITY FOR GRO BNER BASES
3
is, in fact, a submonoid of ?. We will assume that ?0 generates ?. We will also
assume that ?0 is well ordered.
In the rst example mentioned above we have for ? = f0g that R = R0 for any
integral domain R0. In other words an important consideration for us is when the
ring has a trivial grading. In the second example where ? = Zn with a term order,
we consider R = k[x1; :::; xn] for indeterminates x1; :::; xn and an integral domain
k. In this example ?0 = Nn , where N denotes the natural numbers.
We rst observe
Lemma 2.1. ?0 is well ordered if and only if 0 for all 2 ?0.
Proof. Assume that ?0 is well ordered and for some 2 ?0 we have < 0. Choose
a 2 R , a 6= 0. Then an 2 Rn and an 6= 0 so that n 2 ?0 , for all n 2 N.
Also, 0 > implies that > 2, etc. So we have an innite descending chain
> 2 > 3 > in ?0 , which is a contradiction. Conversely, let 1 ; 2 ; ::: be in
?0 such that 1 > 2 > . Choose ai 6= 0 in Ri for each i. Consider the chain of
ideals ha1i ha1 ; a2i P . We see that this chain is strict since a +1 2 ha1 ; :::; ai
implies that a +1 = i=1 riai where, since the a's are homogeneous, we may
assume each ri is homogeneous, say of degree i . Then for all i such that ri 6= 0,
we have i + i = +1 . Also for all such i we have i 2 ?0 and so i 0.
Thus +1 = i + i i , which is a contradiction. Thus the chain of ideals
above is strictly increasing and this violates the assumption that the ring R is
Noetherian.
From Lemma 2.1 and our assumption that ?0 is well ordered, we are justied in
using (complete) induction to prove results involvingP?0.
Now for each a 2 R (a 6= 0) we may write a = 0 a , with a 2 R and
a0 6= 0. We dene lt(a) = a0 and v(a) = 0 . Set lt(0) = 0 and v(0) = 0. We call
lt(a) the leading term of a and v(a) the value of a. Since R is an integral domain,
we have, for all a; b 2 R, lt(ab) = lt(a)lt(b), and if ab 6= 0 then v(ab) = v(a) + v(b).
For a subset F R, we set Lt(F) = hlt(a)ja 2 F i. (Here the symbol h i denotes
the ideal generated by .) Clearly Lt(F) is a homogeneous ideal.
In this paper we will study two dierent denitions of Grobner bases (cf Moller
(1988)). These denitions are given now.
Denition 2.2. Let I be an ideal in R and let F be a subset of I . We call F a
weak Grobner basis for I provided that F is nite and Lt(F) = Lt(I).
Denition 2.3. Assume that R is a unique factorization domain (UFD). Let I be
an ideal in R and let F be a subset of I . We call F a strong Grobner basis for I
provided that F is nite and for all a 2 I there exists b 2 F such that lt(b)jlt(a).
We will say that a subset F of R is a weak (respectively strong) Grobner basis
provided that F is a weak (respectively strong) Grobner basis of the ideal, hF i,
that it generates.
As an example let R = k[x1; :::; xn], where k is a UFD and ? = Zn with a
term order (so ?0 = Nn ). If k is a eld, then it is well known that these two
concepts coincide, c.f. Buchberger (1985) and Robbiano (1989). However, if k is
not a eld, then these concepts are dierent (c.f. the example after Proposition 2.6
and Corollary 4.3).
As a second example let ? = f0g. Then one easily sees that any subset of R is a
weak Grobner basis. On the other hand, in Moller (1988) it is noted that for a UFD
4
W. ADAMS, A. BOYLE, P. LOUSTAUNAU
R, every ideal can have a strong Grobner basis if and only if R is a principal ideal
domain (PID). In this case F is a strong Grobner basis if and only if gcd(F) 2 F.
We will give a generalization of this result in Theorem 5.4 below.
We will now make a few observations concerning these denitions.
Proposition 2.4. Let R be a UFD, I be an ideal in R and F be a subset of I . If
F is a strong Grobner basis for I , then it is also a weak Grobner basis for I .
Proof. If a 2 I, then we can nd b 2 F such that lt(b)jlt(a) and so lt(a) 2 Lt(F).
Proposition 2.5. If F is a subset of the ideal I in R and F is a weak Grobner
basis of I, then I = hF i.
Proof. Let F = ff1 ; : : : ; fsg. Let aP2 I. We use induction on v(a) to show that
a 2 hF i. By assumption lt(a) = si=1 bilt(fi ) for some bi 2 R. Since lt(a) is
homogeneous, we may assume that each bi is homogeneous and v(fi ) v(a) for all
i such that bi 6= 0. If v(a) = 0, then
v(lt(f )) = v(f ) = 0 for all i with b 6= 0, and
P
s b f .i On the iother hand, if v(a) > i0 we have
so lt(f
)
=
f
,
lt(a)
=
a
and
a
=
Pi i
P i=1 i i
a ? si=1 bifi 2 I and v(a ? si=1 bi fi ) < v(a) and so we are done by induction.
Proposition 2.6. Every ideal I in R has a weak Grobner basis.
Proof. Since R is Noetherian, there are b1;P
: : :bt 2 Lt(I) such that Lt(I) = hb1 ; : : : ; bti.
Now each of the bi can be written as bi = cij lt(aij ) with aij 2 I. Then F = faij g
is a weak Grobner basis for I.
As noted above the existence of strong Grobner bases is more problematical. For
example, let R = k[x; y; z] graded with the integers Z on the z variable (i.e. the z
degree of a polynomial in R; so R0 = k[x; y]). Consider the ideal I = hx; y; z i in
R. This ideal does not have a strong Grobner basis. This is because there are an
innite number of homogeneous non-associate irreducible polynomials in the two
variables x; y in I and each of these would have to have leading term (which is the
irreducible polynomial itself) divisible by the leading term of an element of a strong
Grobner basis. This forces an innite number of elements in the strong Grobner
basis which violates the denition of strong Grobner basis. The results in Section
5 bring out this point more clearly.
Finally the following Theorem provides in our context a couple of the usual
equivalent conditions for a set to be a weak Grobner basis. The third condition
is the usual statement about the module of syzygies of the leading terms of the
elements of the weak Grobner basis F. As usual it would have suced to replace
statement 3 in the Theorem below by making the same statement about a nite
homogeneous basis of the syzygy module.
Theorem 2.7. Let F = ff1 ; : : : ; fs g be a nite subset of R. Then the following
three statements are equivalent:
1. F is a weak Grobner basis
2. for every a 2 hF i we can write
(1)
a=
and ri 2 R
s
X
i=1
rifi where v(a) = 1max
(v(r f ))
is i i
TRANSITIVITY FOR GRO BNER BASES
5
3. for all sequences (b1 ; : : : ; bs) where bi 2 R is homogeneous, P
and where for
some 2 ?, v(bi lt(fi )) = for all i such that bi 6= 0, and where si=1 bilt(fi ) =
0; we have that
s
X
a = bi fi
i=1
has a representation as in Equation 1 above.
Proof. We rst assume that Lt(F) = Lt(hF i), that is, F is a weak Grobner basis
and
statement 2 is valid. Let a 2 hF i. Then lt(a) 2 Lt(F) and so lt(a) =
Ps show
c
lt(f
i=1 i i ) for some ci 2 R. Since lt(a) is homogeneous, we mayPassume ci is
homogeneous and is either zero or lies in Rv(a)?v(fi) . Setting b = a ? si=1 cifi , we
have that v(b) < v(a) and we may proceed by induction. (The case where v(a) = 0
is trivial.) That statement 2 implies statement 3 is clear since it is a special case of
statement 2. Finally
P we show that statement 3 implies statement 1. So, let a 2 hF i
and write a = si=1 rifi where ri 2 R. Set = max1is v(ri fi ). If v(a) = , we
are done, since then
X
lt(ri )lt(fi ):
lt(a) =
v(ri fi )=
If not, then v(a) < and so
X
v(ri fi )=
lt(ri )lt(fi ) = 0:
Then by hypothesis we can write
X
v(ri fi )=
lt(ri )fi =
such that for all i, v(ti fi ) < . Then we have
a =
=
s
X
i=1
ti fi
s
s
X
X
X
lt(ri )fi + ti fi
rifi ?
i=1X
i=1X
v(ri fi )=
v(ri fi )=
(ri ? lt(ri ) + ti )fi +
v(ri fi )<
(ri + ti )fi :
Every term in the second sum has value < as does every term in the rst, and so
we may proceed by induction.
3. Transitivity for Weak Grobner Bases.
Let x = fx1; :::; xng be variables. Assume that we have a term order \<" on
Zn. (We note that this is just an order on the group Zn in the sense given in the
previous section.) Given any integral domain R there is the natural Zn grading on
R[x] whose non-zero homogeneous summands are indexed precisely by Nn . Now
assume that R is, in fact, a ?-graded ring as above. Set = ? Zn. Then we can
dene a -grading on R[x] where, for 2 ? and 2 Zn we have
R[x](; ) = R x
provided that 2 Nn and is f0g otherwise. (By x we mean x11 xnn , where
= (1 ; :::; n).) We see that 0 = ?0 Nn . We will dene an order on as
follows.
6
W. ADAMS, A. BOYLE, P. LOUSTAUNAU
Denition 3.1. The elimination order on is dened as (1; 1) < (2; 2) if and
only if 1 < 2 or 1 = 2 and 1 < 2 .
This generalizes the concept of elimination order that occurs in the literature,
for example, in the computer algebra system Macaulay, and also in Adams and
Boyle (1992). It is easily seen that Denition 3.1 makes into an ordered group
satisfying the conditions assumed above.
We note that we are using the symbol \<" in (at least) three contexts, but it
should always be clear from what is being compared which context is the correct
one. We will also, as is usual in the theory of Grobner bases, use the term order
\<" on Zn on the variables themselves. That is, we will say that x1 < x2 if and
only if 1 < 2 .
Now given a 2 R we use the notation lt? (a) and v? (a) to specify the leading
term and value of a as dened in the previous section. If f 2 R[x] we denote the
same concepts with respect to by lt (f) and v (f) respectively. Write f = ax +
lower terms in x, where a 2 R and a 6= 0. Then set ltx (f) = ax , lpx (f) = x ,
lcx (f) = a and vx (f) = . Of course, ltx and vx are the leading term and value
concepts in R[x] with respect to the group f0gZn . Also, lpx (f) is called the leading
power product of f and lcx (f) is called the leading coecient of f. We note that
lt (f) = lt (ltx (f)) = lt? (lcx (f))lpx (f) and v (f) = (v? (lcx (f)); vx (f)). We also
dene Lt? , Ltx and Lt as in the last section. The former will give homogeneous
ideals in R and the latter will give ideals in R[x], homogeneous with respect to
f0g Zn and respectively.
We are interested in the following transitivity question: let F be a nite subset
of R[x]; is there a relation between the following two statements:
Is F a weak Grobner basis in R[x] with respect to = ? Zn ?
Is F a weak Grobner basis in R[x] with respect f0g Zn ?
Relationships such as these have been examined in Adams and Boyle (1992), Bayer
and Stillman(1987), Gianni, Trager, and Zacharias (1988), Spear (1977), Shtokhamer
(1988). (We will consider similar questions for strong Grobner bases in the next
section.)
The relationship between them involves certain subsets of R as well, which we
will now dene. Let F = ff1 ; :::; fsg be a set of polynomials in R[x]. We adopt the
notation that
fi = ai Xi + lower terms in the x variables,
where Xi is a power product in the x variables, and ai 2 R. That is, lcx (fi ) = ai
and lpx (fi ) = Xi . We also dene G = fa1 ; ::; asg. We will continue using this
notation throughout the paper.
For each subset S of f1; :::; sg, we dene
DS = lcmi2S Xi = lcmi2S flpx (fi )g,
FS = ffi jXi divides DS g, and
GS = faijfi 2 FS g.
Also let S = fijfi 2 FS g. We say that S is saturated in f1; : : : ; ng if S = S.
We are now in a position to state our main result on transitivity of weak Grobner
basis. This result is a generalization of Theorem 3.2 in Adams and Boyle (1992).
The important dierence here is that in the previous paper we had to assume that
there was only one x variable, that is, n = 1. Another dierence was that here we
have the more general situation of the graded ring R. The proof of the rst part of
TRANSITIVITY FOR GRO BNER BASES
7
the theorem below was given in Adams and Boyle (1992) for arbitrary n but for a
polynomial ring instead of the graded ring R and the proof given here parallels it
fairly closely, but for the sake of completeness we will include it.
Theorem 3.2. F is a weak Grobner basis in R[x] with respect to = ? Zn if
and only if
1. F is a weak Grobner basis in R[x] with respect to f0g Zn and
2. For all saturated subsets S of f1; :::; sg, GS is a weak Grobner basis in R with
respect to ?.
Proof. Let us rst assume that F is a weak Grobner basis in R[x] with respect to .
We need to prove that Ltx (F) = Ltx (hF i). One inclusion is trivial, so let f 2 hF i.
Since F is a weak Grobner basis, we may apply Theorem 2.7 to , and write
f=
Xt
j =1
rj Tj fij ;
where rj 2 R is homogeneous, Tj is a power product in x, the v (rj Tj fij )'s are
decreasing (not necessarily strictly), and v (f) = v (r1 T1fi1 ). Let j0 be least such
that v (rj0 Tj0 fij0 ) > v (rj0 +1 Tj0 +1 fij0 +1 ). Then, since v (f) = v (r1T1 fi1 ), we
see that
j0
X
j0
X
rj lt? (aij ) 6= 0:
j =1
Choose j1 j0 least such that Tj1 lpx (fij1 ) = Tj1 Xij1 > Tj1 +1 Xij1 +1 = Tj1 +1 lpx (fij1 +1 ).
From the denition of the order > on , we have that Tj1 Xij1 > Tj Xij for all j > j1 .
lt (f) =
j =1
rj Tj lt (fij ) = T1Xi1
It then follows that
ltx (f) =
j1
X
j =1
rj Tj aij Xij = (
P
provided that h = jj1=1 rj aij 6= 0. However
j1
X
j =1
rj aij )T1 Xi1
j0
X
rj lt? (aij ) 6= 0;
j =1
as was noted above. Since ltx (fi ) = ai Xi , we see that ltx (f) 2 Ltx (F), as desired.
To prove statement 2, we need to show that for every saturated set S, Lt?(GS ) =
Lt? (P
hGS i). Let S be a saturated subset of f1; :::; sg, and let a 2 hGS i. Then
a = i2S bi ai, where bi 2 R. Set
X
X
f = bi DXS fi = bi DXS (ai Xi + lower x terms )
i
i
i2S
i2S
lt? (h) =
= aDS + lower x terms :
Note that f is in hF i. Furthermore, by denition of our elimination ordering
lt (f) = lt (aDS ) = lt? (a)DS :
By assumption Lt (F) = Lt (hF i), and hence we can write
lt (f) =
s
X
i=1
kilt (fi );
8
W. ADAMS, A. BOYLE, P. LOUSTAUNAU
where ki 2 R[x]. As usual we may assume that each ki is homogeneous, say
ki = ciXi0 , where ci 2 R, is homogeneous, and Xi0 is a power product in x. We can
also assume that DS = lpx (f) = Xi0 Xi , for all i such that ci 6= 0. Then
X
lt?(a)DS = lt (f) = ci lt?(ai )Xi0 Xi :
ci 6=0
Now for all i such thatPci 6= 0, Xi divides DS , and so ai is in GS , by the denition
of GS . Thus lt? (a) = i2S ci lt? (ai ) as desired.
To prove the converse we need to show that Lt (F) = Lt (hF i). Let f 2 hF i.
By hypothesis 1 and Theorem 2.7,
f=
s
X
i=1
hi fi ; where lpx (f) = 1max
v (h ) + vx (fi ):
is x i
Dene X = lpx (f) and S0 = fi 2 f1; :::; sg j lpx (hi )lpx (fi ) = X g. Set S = fi 2
f1; :::; sg j Xi divides X g. Then S is saturated and S S0 . Now
X
lcx (hi )ai X + lower x terms:
i2S0
P
Hence lcx (f) = i2S0 lcx (hi )ai is in hGS i. By hypothesis 2, GS
basis and hence Lt? (GS ) = Lt? (hGS i). Thus we can write
f=
lt?(lcx (f)) =
X
i2S
is a weak Grobner
ci lt? (ai );
where for all i such that ci 6= 0,
v?(ci lt? (ai )) = v?(lcx (f)):
P
Finally lt (f) = lt? (lcx (f))lpx (f) = i2S cilt? (ai )X. By denition of X, Xi
divides X for all i 2 S. So
X
X
lt (f) = ci XX lt? (ai )Xi = ci XX lt (fi ):
i2S
i
i2S
i
Thus lt (f) is in Lt (F) as desired.
Note. Let S be any subset of f1; :::; sg and let S be its saturation. Since
GS = GS , statement 2 in Theorem 3.2 can be replaced by \For all subsets S of
f1; :::; sg;GS is a weak Grobner basis in R with respect to ?."
Moller (1988) has shown how to construct weak Grobner bases in R[x] with
respect to f0g Zn for a certain class of rings, the so-called computable rings (see
Moller (1988) and Zacharias (1978)). Our next result shows that if we can compute
weak Grobner bases in R[x] with respect to f0g Zn , and in R with respect to ?,
then we can compute weak Grobner bases in R[x] with respect to ? Zn .
Corollary 3.3. Let R be a ?-graded ring, and F = ff1 ; : : : ; fsg be a weak Grobner
basis in R[x] with respect to f0g Zn. For each saturated subset S of f1; : : : ; sg
let faS;1 ; : : : ; aS;tS g be a weak Grobner basis of GS in R with respect to ?. Write
aS;i =
X
j 2S
bS;i;j aj
TRANSITIVITY FOR GRO BNER BASES
and dene
=
fS;i
for all i = 1; : : : ; tS . Then
F =
X
j 2S
[
S saturated
9
S
bS;i;j D
X fj
j
g[F
ffS; 1 ; : : : ; fS;t
S
is a weak Grobner basis with respect to .
Proof. We need to show statements 1 and 2 of Theorem 3.2. Note rst that hF i =
hF i, and since F is a weak Grobner basis with respect to f0g Zn , so is F . To
prove statement 2, let T be a saturated subset of
[
f(S; 1); : : : ; (S; tS )g [ f1; : : : ; sg:
S saturated
Note that if (S; i) is in T, then S T, and (S; j) 2 T, for j = 1; : : : ; tS , since T
is saturated. Therefore DT = DS for the saturated subset S = T \ f1; : : : ; sg of
f1; : : : ; sg. But then
[
GT =
faS 0 ;1 ; : : : ; aS 0 ;tS0 g [ fai j i 2 S g;
S0 S
S 0 saturated
and hence GT is a weak Grobner basis in R with respect to ?, since hGT i = hGS i and
faS;1; : : : ; aS;tS g was dened to be a weak Grobner basis of GS = fai j i 2 S g.
If we specialize to the case R = k[y], where k is an integral domain, y =
fy1 ; :::; ymg are variables with a term order, and we have an elimination order
between the y variables and the x variables with the x variables bigger than the y
variables, then we have the following immediate consequence of Theorem 3.2.
Corollary 3.4. A subset F of k[y; x] is a weak Grobner basis in k[y; x] over k if
and only if
1. F is a weak Grobner basis in (k[y])[x] over k[y] and
2. For all saturated subsets S of f1; :::; sg, GS is a weak Grobner basis in k[y]
over k.
We note that the statement F is a weak Grobner basis in R[x] with respect to
f0g Zn in Theorem 3.2 is just the usual concept in the literature of a (weak)
Grobner basis over a commutative ring. We will now give an equivalent condition
for a set to satisfy this condition involving the concept of saturated sets above.
Although the result does not involve the grading on the domain R, in fact does not
require that we have a domain, but just a commutative ring, we will state it in the
terminology above to have notational consistency throughout the paper.
Proposition 3.5. Let F = ff1; :::; fsg be a subset of R[x]. Then the following
statements are equivalent:
1. F is a weak Grobner basis in R[x] with respect to f0g Zn ;
2. For all f 2 hF i, there exists a saturated subset S of f1; :::; sg such that DS
divides lpx (f) and lcx (f) 2 hGS i.
10
W. ADAMS, A. BOYLE, P. LOUSTAUNAU
Proof. Assume 1, and let f 2 hF i. Then ltx (f) =
Ps h lt (f ), where we may
i=1 i x i
assume that hi is a monomial in R[x]. Now dene S to be the saturation of
fi 2 f1; :::; sg j hi 6= 0g. Since each Xi such that hi 6= 0 divides lpx (f), their least
common multiple, DS , divides lpx (f). Moreover after cancelling lpx (f) from both
sides of the above equation, we have lcx (f) expressed as a linear combination of
the ai , with i 2 S. Conversely, assume 2. Let f 2 hF i. Then, by hypothesis there
exists a saturated subset S of f1;P:::; sg such that lpx (f) = XDS , where X is a
power product in x, and lcx (f) = i2S bi ai, where bi 2 R, for all i 2 S. Now
X
X
ltx (f) = lcx (f)lpx (f) = bi ai XDS = bi X DXS ltx (fi ):
i
i2S
i2S
Therefore ltx (f) is in Ltx (F).
The next proposition provides an equivalent condition for statement 2 of theorem
3.2 in terms of a dierent transitivity property. This extends Proposition 2.4 of
Adams and Boyle (1992) where it is assumed that x is a single variable. Moreover
the proof that the second statement implies the rst is contained in Lemma 2.2 in
Adams and Boyle (1992) in the slightly less general situation where the domain R
is not a graded ring but is a polynomial ring over a commutative ring.
Proposition 3.6. The following statements are equivalent:
1. For all saturated subsets S of f1; :::; sg, GS is a weak Grobner basis in R;
2. For all saturated subsets S of f1; :::; sg, Lt (Ltx (FS )) = Lt (FS ).
Proof. We rst assume statement 1 and let S be a saturated subset of f1; :::; sg.
Clearly Lt (FS ) Lt (Ltx (FS )). Conversely, let h be in Ltx (FS ), say
h=
X
j 2S
hj ltx (fj ) =
P
X
j 2S
hj aj Xj
where hj belongs to R[x]. Set P
hj = Pdkj=0 hj;k Xk0 , where hj;k 2 R, and Xk0 is a
power productPin x. Then h = j 2S dkj=0 hj;k aj Xk0 Xj . Let X = lpx (h), and let
a = lcx (h) = Xk0 Xj =X hj;k aj . Note that lt (h) = lt?(a)X. Let T0 = fj 2 S j
Xj Xk0 = X for some k, 0 k dj g. Let T be the saturation of T0 . Since S is
saturated we see that T S. Since Xj divides X for all j 2 T, we have that DT
divides X. We have
P that a 2 hGT i and by hypothesis, GT is a weak Grobner basis.
Hence lt? (a) = j 2T rj lt? (aj ), where the rj 's are in R. Finally,
X
X
lt (h) = rj lt? (aj )X = rj XX lt (fj ):
j
j 2T
j 2T
Thus lt (h) is in Lt (FT ) Lt (FS ).
Conversely assume statementP2. Let S be a saturated subset of f1; :::; sg. Let
a 2 hGS i, a 6= 0, and write a = i2S bi ai for bi 2 R. Set f = aDS . Then
X
X
f = bi DXS ai Xi = bi DXS ltx (fi ) 2 Ltx (FS ):
i
i
i2S
i2S
So
x (FS )) = Lt (FS ), by hypothesis. Hence we can write lt (f) =
P lth(f)lt 2(fLt).(LtSince
lt (f) is homogeneous with respect to we may assume
i2S i i
that hi = ci Xi0 where ci is homogeneous in R and Xi0 is a power
P product in x. Of
course, lt (f) = lt?(a)DS , and so we obtain that lt? (a)DS = i2S ciXi0 lt?(ai )Xi .
TRANSITIVITY FOR GRO BNER BASES
11
Since thenP DS = Xi0 Xi for all i such that ci 6= 0 we cancel these terms to obtain
lt? (a) = i2S ci lt? (ai ), as desired.
In Theorem 3.1 in Adams and Boyle (1992) the sets FS dened right above
Theorem 3.2 also turned out to be weak Grobner bases. It is unfortunately not
true that for saturated sets S the sets FS are weak Grobner bases as the following
example shows.
Example. Let k be the rational numbers and let R = k[y] with the usual
Z grading. Let n = 2 and consider R[x1; x2] with the lexicographical ordering
on x1 ; x2 with x1 > x2 . Thus k[y; x1; x2] has the lexicographical ordering with
x1 > x2 > y. Let f1 = y2 x42 +x32 ; f2 = yx1 +x2 ; f3 = x1x22 ?yx42 ; f4 = x21x2 +x42 ; f5 =
x31 + x52y and let F = ff1 ; f2; f3 ; f4; f5g. Then F is a weak Grobner basis in
k[y; x1; x2]. S = f2; 5g is a saturated set, but ff2 ; f5g is not a weak Grobner basis.
However, FS is a weak Grobner basis for certain saturated sets S. Let F =
ff1 ; :::; fsg be a subset of R[x]. A subset S of f1; :::; sg is called full if for all j not
in S we have DS < lpx (fj ).
Proposition 3.7. If F is a weak Grobner basis in R[x] with respect to , then
FS is also a weak Grobner basis in R[x] with respect to for all full subsets S of
f1; :::; sg.
Proof. We will use the third criterion in Theorem 2.7. So let S be a full subset of
f1; :::; sg, and suppose that Pi2S ci Xi0 lt (fi ) = 0, where ci 2 R, homogeneous, Xi0
is a power
product in x, and v (ci Xi0 lt(fi )) is constant for all i such that ci 6= 0. Set
P
f = i2S ci Xi0 fi . We need a representation as in Equation 1 of Theorem 2.7 for f
with respect to . It is clear that we may assume that gcd(Xi0 j ci 6= 0) = 1. It then
follows that for all i such that ci 6= 0, we get Xi0 Xi j DS , and so lpx (f)P
DS . Now F
is a weak Grobner basis in R[x] with respect to implies that f = si=1 hifi where
v (f) = max1is v (hi fi ) (from Theorem 2.7). Then if there is an i 2= S such
that hi 6= 0, we have Xi > DS and thus lpx (hi)lpx (fi ) lpx (fi ) > DS lpx (f).
This contradiction shows that the representation for f above is of the correct type
for FS and we are done.
In the example above, the smallest full set containing f2; 5g is in fact f1; 2; 3; 4;5g.
Also, in the same example, the set f1; 2; 3g is full, and hence FS = ff1; f2 ; f3g is a
weak Grobner basis.
It is now natural to ask whether Theorem 3.2 holds if in statement 2, saturated
is replaced by full. It is not true in general as the following example shows.
Example. We use the same setup for R and the variables that we used in the
previous example. Let f1 = x32; f2 = y(1+y)x22 x1 , and f3 = y(2+y)x2 x21. Note that
F = ff1; f2 ; f3g is a weak Grobner basis in (k[y])[x1; x2], since the three polynomials
are monomials in (k[y])[x1; x2]. So statement 1 of Theorem 3.2 is satised. The full
subsets of f1; 2; 3g are f1g and f1; 2; 3g. For S equal to either of these sets, GS is a
weak Grobner basis in k[y]. However F is not a weak Grobner basis in k[y; x1; x2],
since the S-polynomial of f2 and f3 is yx22 x21 which cannot be reduced by F. Note
that S = f2; 3g is saturated and GS is not a weak Grobner basis.
We give another example to illustrate some of the ideas of this section. It is a
continuation of the example at the end of Moller (1988).
Example. We consider the polynomials in Z[x; y; z], f1 = ?6x3 + 51x, f2 =
(2x2y ? 17y)z, f3 = (?85y2 + 6x2)z 2 , and f4 = (5xy2 ? 3x)z 2. We will use the
lexicographical ordering with x < y < z. We consider, in the notation above, R =
12
W. ADAMS, A. BOYLE, P. LOUSTAUNAU
Z[x; y], ? = Z2 , = Z3, and a1 = ?6x3 + 51x; a2 = 2x2y ? 17y; a3 = ?85y2 + 6x2;
and a4 = 5xy2 ? 3x. Let F = ff1 ; f2; f3 ; f4g. We will use Theorem 3.2 to prove that
F is a weak Grobner basis in Z[x; y; z]. First note that since the polynomials in F
are homogeneous in the variable z, condition 1 of Theorem 3.2 is trivially satised.
Also, the saturated subsets of f1; 2; 3; 4g are f1g; f1; 2g, and f1; 2; 3; 4g, and hence
we need to show that the sets fa1g; fa1; a2 g, and fa1; a2; a3; a4g are weak Grobner
bases in Z[x; y]. Moller (1988) showed that fa1; a2; a3; a4g is a weak Grobner basis
. Also, the full subsets of f1; 2; 3; 4g with respect to ? are exactly f1g; f1; 2g, and
f1; 2; 3; 4g. So, by Proposition 3.7, the sets fa1g and fa1 ; a2g are also weak Grobner
bases.
4. Transitivity for Strong Grobner Bases.
In this section we will assume that R is a unique factorization domain. Otherwise we will maintain exactly the same setup as we had in the previous section.
Corresponding to Theorem 3.2 we have the following
Theorem 4.1. F is a strong Grobner basis in R[x] with respect to = ? Zn if
and only if
1. F is a weak Grobner basis in R[x] with respect to f0g Zn and
2. For all saturated subsets S of f1; :::; sg, GS is a strong Grobner basis in R
with respect to ?.
Proof. We rst assume that F is a strong Grobner basis in R[x] with respect to
= ? Zn. Since, by Proposition 2.4 F is a weak Grobner basis in R[x] with
respect to = ? Zn as well, statement 1 followsPfrom Theorem 3.2. It remains
to prove statement 2. Let a 2 hGS i and write a = i2S bi ai where bi 2 R. Set
X
f = bi DXS fi = aDS + lower x terms.
i
i2S
Since f 2 hF i, there is a j such that lt (fj ) j lt (f). That is
lt? (lcx (fj ))lpx (fj ) j lt? (a)DS
and so lt? (lcx (fj )) j lt? (a). Moreover, since lpx (fj ) j DS and S is saturated then
lcx (fj ) 2 GS and so we are done.
Conversely assume statements 1 and 2. Let f 2 hF i and
P write, using the assumption of statement 1 and Theorem 2.7 statement 2, f = si=1 hi fi where hi 2 R[x]
and lpx (f) = max1isflpx(hi )lpx (fi )g. Set X P
= lpx (f). Let S be the saturation of S0 = P
fijlpx(hi )lpx (fi ) = X g. Then f = ( i2S0 lcx (hi )lcx (fi ))X+ lower x
terms. Also i2S0 lcx (hPi )lcx (fi ) 2 hGS i and so by statement 2 there is a j 2 S
such that lt? (aj ) j lt?( i2S0 lcx (hi )lcx (fi )). Then we see that lt (fj ) j lt (f),
since lpx(fj ) j DS j lpx (f).
We now give the analog to Corollary 3.3 for strong Grobner bases. The proof is
similar to the one for the weak case, and we do not include it.
Corollary 4.2. Let R be a ?-graded UFD, and F = ff1 ; : : : ; fsg be a weak Grobner
basis in R[x] with respect to f0g Zn. For each saturated subset S of f1; : : : ; sg
let faS;1; : : : ; aS;tS g be a strong Grobner basis of GS in R with respect to ?. Write
X
aS;i = bS;i;j aj
j 2S
TRANSITIVITY FOR GRO BNER BASES
and dene
=
fS;i
for all i = 1; : : : ; tS . Then
F =
X
j 2S
[
S saturated
13
S
bS;i;j D
X fj
j
g[F
ffS; 1 ; : : : ; fS;t
S
is a strong Grobner basis with respect to .
The above Corollary is a generalization of a result of Moller (1988). There
? = 0, and R is a PID and the polynomials fS;i 's are called T-polynomials. The
next corollary essentially states that result.
Corollary 4.3. Let R be a PID with a trivial grading and let F = ff1; : : : ; fsg be
a weak Grobner basis in R[x]. For each saturated subset S of f1; :::; sg, let
P
aS = gcd(ai j i 2 S) and write aS = i2S li ai .
Dene
fS =
Then
X DS
li X fi :
i
i2S
F = ffS j S is a saturated subset of f1; :::; sgg [ F
is a strong Grobner basis in R[x].
Corollary 4.4. If R is a PID with the trivial grading, then every ideal in R[x] has
a strong Grobner basis.
Proof. By Proposition 2.6 R[x] has a weak Grobner basis so by the last corollary
R[x] has a strong Grobner basis.
As we did in Corollary 3.4 for Theorem 3.2, we restate Theorem 4.1 in the case
R = k[y], where k is an integral domain, y = fy1 ; :::; ymg are variables with a term
order, and we have an elimination order between the y variables and the x variables
with the x variables bigger than the y variables.
Corollary 4.5. A subset F of k[y; x] is a strong Grobner basis in k[y; x] over k if
and only if
1. F is a weak Grobner basis in (k[y])[x] over k[y] and
2. For all saturated subsets S of f1; :::; sg, GS is a strong Grobner basis in k[y]
over k.
In the next corollary we specialize to the case where the variables y are a single
variable y. We also assume that k is a eld.
Corollary 4.6. Let F be a nite subset of k[y; x]. Then F is a strong Grobner
basis in (k[y])[x] if and only if F is a weak (and hence strong) Grobner basis in
k[y; x].
Proof. Applying Theorem 4.1 we have that F is a strong Grobner basis in (k[y])[x]
if and only if F is a weak Grobner basis in (k[y])[x] and GS is a strong Grobner
basis in k[y] with respect to the trivial grading. The last statement is equivalent
to GS containing the gcd of its elements, which is equivalent to GS being a weak
14
W. ADAMS, A. BOYLE, P. LOUSTAUNAU
Grobner basis with respect to the canonical Z grading. So nally using Corollary
3.4 we get the result.
As mentioned in Section 3, Moller (1988) gives an algorithm for computing a
weak Grobner basis of an ideal I in (k[y])[x] with respect to the grading f0g Zn .
Corollaries 4.3 and 4.6 then indicate a method for computing a Grobner basis of
an ideal I in k[y; x]: rst compute a weak Grobner basis of I with respect to the
grading f0g Zn using Moller's algorithm, then use Corollary 4.3 to obtain a strong
Grobner basis of I with respect to f0g Zn ; by Corollary 4.6, this is a weak Grobner
basis in k[y; x].
We now go back to the general situation. We observed in the Example directly
above Proposition 3.7, that even though F is a weak Grobner basis, it need not be
true that each of the sets FS , for saturated sets S, is a weak Grobner basis. The
same situation occurs for strong Grobner bases. Indeed, in the same example we
note that the grading on R[x1; x2] = k[y; x1; x2] is just a term order over a eld
and so weak Grobner basis and strong Grobner basis are equivalent. Thus F being
a strong Grobner basis does not even imply that FS is a weak Grobner basis, let
alone a strong Grobner basis. Again for full subsets this problem does not occur,
i.e. analogously to Proposition 3.7 we have
Proposition 4.7. If F is a strong Grobner basis in R[x] with respect to , then
FS is also a strong Grobner basis in R[x] with respect to for all full subsets S of
f1; : : : ; sg.
Proof. Let f 2 hFS i and write f = aX+ lower x terms, where a 2 R and X is a
power product in x. We need to nd a j 2 S such that lt (fj ) j lt (f), that is,
lt? (aj ) j lt? (a) and Xj j X. We rst note that by Proposition 2.4, F is a weak
Grobner basis with respect to , and so by Proposition 3.7, FS is a weak Grobner
basis with respect to . Thus by Theorem 3.2, FS is a weak Grobner basis with
respect to f0g Zn. Hence, using Proposition 3.5, there is a saturated subset T of
S such that DT j X and a 2 hGT i. Now T is also a saturated subset of f1; : : : ; sg
since if i 2 f1; : : : ; sg, then Xi j DT implies Xi j DS which in turn implies that
i 2 S, since S is saturated in f1; : : : ; sg, and so nally i 2 T since T is saturated in
S. Thus by Theorem 4.1, GT is a strong Grobner basis with respect to ?. Hence
there is a j 2 T such that lt? (aj ) j lt? (a). Since j 2 T we have Xj j DT and so
Xj j X.
Example. We will continue the Example given at the end of the last section. We
add to the weak Grobner basis F the polynomial f5 = (x2y2 ? 51y2 + 3x2)z 2 . One
can show that this is the polynomial that we need to add to F using Corollary 4.2
to see that ff1 ; : : : ; f5 g is a strong Grobner basis in Z[x; y; z] (see the corresponding
example in Moller (1988)). Alternatively, since in Moller (1988) it is shown that
fa1; : : : ; a5g is a strong Grobner basis we can apply Proposition 4.7 as we did in
the previous example to show that ff1; : : : ; f5g is a strong Grobner basis.
5. Structure Theorem for Graded Rings with Strong Grobner Bases
Moller (1988) states that all ideals in R = k[x1; : : : ; xn] for a UFD k can have
a strong Grobner basis only when k is a PID. Here, in our terminology, R has
the grading associated to a term order on x and k has the trivial grading. In this
section we show that any ?-graded ring R with strong Grobner bases is of the form
R = k[x1; : : : ; xn], where k is a PID and ? is Zn with a term order (see Theorem
TRANSITIVITY FOR GRO BNER BASES
15
5.4). The statement that the gradings are all basically Zn gradings was already
noted by Robbiano in Robbiano (1986). We will continue our assumptions made
in section 2 concerning the group ?. Namely, ? is generated by ?0 and ?0 is well
ordered.
Theorem 5.1. Assume that R is a ?-graded UFD and that there are only nitely
many homogeneous irreducibles of positive value. Then R is isomorphic to R0[x1; : : : ; xn]
where R0 is a UFD, fx1; : : : ; xng are indeterminates. Furthermore ? is isomorphic
to Zn, the ordering on ? is a term ordering and the grading on R is the usual one
associated to a term ordering.
Proof. Since R is a Noetherian UFD, we know that R0 is also a Noetherian UFD.
Let p1 ; : : : ; ps be all the homogeneous irreducibles of positive value. Then since every homogeneous element must factor as a product of homogeneous irreducibles we
see that R = R0[p1; : : : ; ps]. Also we see that ?0 is generated by v? (p1 ); : : : ; v?(ps ).
Moreover ? is torsion free since if 2 ? and > 0 then < 2 < 3 < (and
similarly if < 0). Hence ? is isomorphic to Zn for some positive integer n. From
now on we will assume that ? = Zn . Let v? (pi ) = i 2 Zn .
Let A = (i ) denote the matrix of these vectors. We now show that s = rank(A).
It suces to show that s rank(A) so assume to the contrary that s > rank(A).
Then there existsPa non empty subset K of f1; : : : ; sg and non-zero integers ai for
i 2 K satisfying
ii = 0. Let K1 = fi 2 KQj ai > 0g and let K2Q= K ? K1 .
i2K aP
P
Set = i2K1 ai i = i2K2 (?ai )i . Let b1 = i2K1 pai i and b2 = i2K2 pi?ai .
Then b3 = b1 + b2 is in R and none of its irreducible homogeneous divisors can
be a pi with i Q
2 K1 [ K2. Set K3 = fi j pi divides b3g and note that K3 6= ;.
Write b3 = r3 i2K3 pai i where r3 2 R0, and ai > 0 for all i 2 K3. Now let
b4 = b1b2 +b1 b3 +b2 b3 2 R2 . Again b4 can have none of its irreducible homogeneous
divisors among the pi with i 2 K1 [ K2 [ K3 . So continue by dening K4 for b4
analogously to the above and set b5 = b1 b2b3 +b1 b2 b4 +b1 b3b4 +b2 b3 b4. We continue
the argument until we have an element of an R for some > 0 with no irreducible
homogeneous divisors.
P This is impossible and so s = rank(A), as desired.
Now set L = f si=1 ai i j ai 2 Ng. We show that for all 2 Zn we have
80
if 2= L
>< s
s
R = > Y
ai R0 if = X ai i 2 L:
p
: i=1 i
i=1
Since
= s, we have for all 2 L there is a unique decomposition
Q =
Ps arank(A)
Rs be non-zero. Then c = r0 si=1 pbi i ,
i=1 i i . First assume 2= L and let c 2 P
where r0 2 RP
2 L. This is a contradiction.
0, and bi 2 N. But then = i=1 bii Q
Now let = si=1 aii 2PLs and c 2 RP s. Then c = r0 si=1 pbi i , where r0 2 R0, and
bi 2 N. Therefore = i=1 bi i =Qs i=1 ai i which implies that ai = bi for each
i, since rank(A) = s. Hence c is in i=1 pai i R0 .
Finally, s = n, since L = ?0 , and ?0 generatesPZsn . Also L is isomorphic to
n
N since every element of ? is written uniquely as i=1 ai i . Also p1; : : : ; pn are
algebraically independent because of the direct sum decomposition of R.
Corollary 5.2. P
Assume that R is a ?-graded UFD and let I be the so-called irrel-
evant ideal, I =
>0 R . Assume that
I has a strong Grobner basis. Then R is
16
W. ADAMS, A. BOYLE, P. LOUSTAUNAU
isomorphic to R0[x1; : : : ; xn] where R0 is a UFD and fx1; : : : ; xng are indeterminates. Furthermore ? is isomorphic to Zn, the ordering on ? is a term ordering
and the grading on R is the usual one associated to a term ordering.
Proof. By the previous Theorem, it suces to show that there are only nitely many
homogeneous irreducibles of positive value in R. If I = 0 then R = R0 and we are
done. Otherwise I 6= 0. Let fb1; : : : ; btg be a strong Grobner basis for I. Since I
is a homogeneous ideal we may assume that each of the bi is homogeneous. Then
if c is any homogeneous irreducible, there is an i such that bi = lt? (bi ) j c = lt?(c).
Since c is irreducible and bi is not a unit (one readily sees that I \ R0 = f0g) we
see that c is an associate of bi and so fb1; : : : ; btg contains all the homogeneous
irreducibles of positive value.
Proposition 5.3. Assume that R is a ?-graded UFD and every ideal in R generated by elements of R0 has a strong Grobner basis. Then R0 is a PID.
Proof. It suces to show that for any c1 ; c2 2 R0 with gcd(c1 ; c2) = 1, the ideal
I = hc1; c2 i in R0 is the unit ideal R0. Let J = IR. Since any divisor of an
element of R0 is in R0 (Lemma 2.1) and by hypothesis, J has a strong Grobner
basis, we see that there exist b1; : : : ; bt in R0 such that for all r 2 I there is an i
such that bi j r. Thus
there exist integers < such that for some i, bi divides
both c1 + c2 and c1 + c2 (this follows since there are innitely many such c1 + c2
and only nitely many bi ). Then bi divides the dierence c1 ? c1 = c1 (c1? ? 1).
Note that d = gcd(bi ; c1) = 1, for otherwise d divides both c1 and c2 which is a
contradiction to gcd(c1 ; c2) = 1. Thus bi divides c1? ? 1, say c1? ? 1 = abi and
hence 1 2 I.
Theorem 5.4. Assume that R is a ?-graded UFD and every ideal in R has a
strong Grobner basis. Then R is isomorphic to k[x1; : : : ; xn] where k is a PID and
fx1; : : : ; xng are indeterminates. Furthermore ? is isomorphic to Zn , the ordering
on ? is a term ordering and the grading on R is the usual one associated to a term
ordering.
Proof. This is immediate from the previous three results.
Corollary 5.5. Let R be a ?-graded PID and suppose that every ideal in R has a
strong Grobner basis with respect to ?. If ? 6= 0, then ? is isomorphic to Z and R
is isomorphic to k[x] where k is a eld, x is a single variable and the grading is the
canonical grading.
We note that Corollary 4.4 may be viewed as a converse to Theorem 5.4.
[1]
[2]
[3]
[4]
[5]
References
Adams, W.W., Boyle, A.K. (1992). Some Results on Grobner basis over Commutative Rings.
J. Symb. Comp. 13, 473{484.
Bayer, D., Stillman, M. (1987). A Theorem on Rening Division Orders by the Reverse
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