Definition 2.3.1 (Head Monomial Ideal) Let G ⊆ R be a subset of R. Thehead monomial ideal ofG[denoted Head(G)] is the ideal generated by the head monomials of the elements ofG, i.e.,
Head(G) ={Hmono(f) :f ∈G}.
Definition 2.3.2 (Gr¨obner Basis) A subset Gof an ideal I ⊆R in R is called aGr¨obner basis of the idealI if
Head(G) = Head(I),
i.e., if the set of monomials {Hmono(f) : f ∈ G} is a basis of Head(I).
If, in the definitions of head monomial, head monomial ideal and Gr¨obner basis, the underlying admissible ordering >
A is not readily decipherable,
then we explicitly state which ordering is involved by a suitable subscript. Notice that sinceG= I satisfies the above condition, every ideal has a Gr¨obner basis. Also, an ideal can have many distinct Gr¨obner bases. For instance, ifGis a Gr¨obner basis for I, then so is every G′,G⊆G′ ⊆ I. However, since a Gr¨obner basis need not be finite, as such, it is not computationally very beneficial. Also, in general (for arbitrary ringS), we do not know how to compute a Gr¨obner basis effectively.
Further, notice that if G ⊆ I, then Head(G) ⊆ Head(I). Hence, to demonstrate that G is a Gr¨obner basis of I, it suffices to show that Head(G)⊇Head(I).
The following theorem justifies the term “basis” in a Gr¨obner basis: Theorem 2.3.1 LetI ⊆R be an ideal ofR, andGa subset of I. Then
Head(G) = Head(I) ⇒ (G) =I. That is, every Gr¨obner basis of an ideal generates the ideal.
proof.
SinceG⊆I, we have (G)⊆I. If (G)6=Ithen we may choose anf ∈I\(G) such that Hmono(f) is minimal with respect to the underlying admissible well-ordering, say ≤
A
, among all such polynomials. Thus, Hmono(f) ∈ Head(I) = Head(G):
Hmono(f) = X gi∈G
Section 2.3 Gr¨obner Bases 45
and
f′= Tail(f)−X tiTail(gi) =f−X tigi∈I.
Clearly,f′∈I\(G), since, otherwise,f =f′+Ptigiwould be in (G). But, Hmono(f′)<
AHmono(f), since every monomial in Tail(f) as well as every
monomial in each oftiTail(gi) is smaller than Hmono(f); consequently, we have a contradiction in our choice off. Hence, (G) =I, i.e., every Gr¨obner basis of an ideal generates the ideal.
Corollary 2.3.2
1. Two ideals I and J with the same Gr¨obner basis G are the same: I= (G) =J.
2. IfJ ⊆I are ideal ofR, andHead(J) = Head(I), thenJ =I.
proof.
(1) is simply a restatement of the previous theorem. (2) follows from (1) sinceJ is a Gr¨obner basis for both I andJ.
Theorem 2.3.3 The subset G⊆I is a Gr¨obner basis of an ideal I of R with respect to an admissible ordering >
A if and only if each polynomial
h∈I can be expressed as
h= X
gi∈G
figi, fi∈R,
such that Hterm(fi) Hterm(gi)≤
A
Hterm(h).
proof.
(⇒) Leth∈I. Inductively, assume that the theorem holds for allh′<
Ah.
Since Hmono(h)∈Head(I) = Head(G), it is possible to write Hmono(h) = X
gi∈G
aipiHmono(gi), ai∈S, pi∈PP(x1, . . . , xn)
such thatpiHterm(gi) = Hterm(h). Let
h′= Tail(h)−XaipiTail(gi) =h−
X
aipigi. Since Hmono(h′)<
A Hmono(h), by the inductive hypothesis, we can writeh
as h=h′+Xa ipigi= X f′ igi′+ X aipigi, such that Hterm(f′
i) Hterm(g′i)≤
A
Hterm(h′)<
AHterm(h) and piHterm(gi)
≤
A
(⇐) Without loss of generality, we assume thath∈I is expressed as h=a1p1g1+· · ·+akpkgk, ai∈S, pi∈PP(x1, . . . , xn), gi∈G, such thatpiHterm(gi)≤
A
Hterm(h). LetLbe the set of indices such that
L=ni∈ {1, . . . , k}: Hterm(h) =piHterm(gi)o. Since Hterm(h)≥
A
piHterm(gi),L6=∅.
Equating terms of equal degree in the previous expression forh, we get Hcoef(h) =Pi∈LaiHcoef(gi).Hence
Hmono(h) = Hcoef(h) Hterm(h)
= X i∈L aiHcoef(gi)piHterm(gi) = X i∈L aipiHmono(gi), i.e., Head(G)⊇Head(I), andGis a Gr¨obner basis ofI.
2.3.1
Gr¨obner Bases in
K[x
1, x
2, . . . , x
n]
LetKbe any arbitrary field, andK[x1, x2, . . . , xn] be the polynomial ring
over the fieldK in variablesx1,x2,. . ., andxn.
Theorem 2.3.4 Every ideal I of K[x1,x2, . . ., xn]has a finite Gr¨obner basis.
proof.
Let <
A be an arbitrary but fixed admissible ordering on PP(x1,. . .,xn).
LetX ={Hterm(f) :f ∈I} be a subset of PP(x1, x2, . . . , xn). Then by Dickson’s lemma, there is a finite subsetY ⊆X such that every power product ofX is divisible by a power product ofY. Define an injective map Φ:Y →I, as follows: for each power product p∈Y choose a polynomial g= Φ(p)∈Isuch that Hterm(g) =p. This map is well defined, since every p∈Y is a head term of some polynomial inI; it is injective, sincep,q∈Y andp6=qimplies that Hterm(Φ(p))= Hterm(Φ(6 q)), and Φ(p)6= Φ(q).
LetG= Φ(Y)⊆I. From the finiteness ofY, it trivially follows thatG is finite. But, by proceeding as in the proof of Theorem 2.2.2, we see that
Head(G) = (Y) = (X) = Head(I), andGis a Gr¨obner basis forI.
Corollary 2.3.5 For any field K,
1. Every ideal ofK[x1, x2, . . . , xn]has a finite system of generators.
Section 2.3 Gr¨obner Bases 47
2.3.2
Hilbert’s Basis Theorem
Proposition 2.3.6 Let R be a ring. Then the following three statements are equivalent:
1. R is Noetherian.
2. The ascending chain condition (ACC) for ideals holds: Any ascending chain of ideals of R
I1⊆I2⊆ · · · ⊆In⊆ · · ·
becomes stationary. That is, there exists an n0 (1 ≤ n0) such that
for all n > n0,In0 =In.
3. The maximal condition for ideals holds:
Any nonempty set of ideals of R contains a maximal element (with respect to inclusion).
proof.
(1⇒2):
For a chain of ideals ofR
I1⊆I2⊆ · · · ⊆In ⊆ · · ·
I=S∞n=1In is also an ideal ofR. (f,g∈I implies that for large enough n0,f,g∈In0; hencef−g∈In0 ⊆I. f ∈I implies that for large enough
n0,f ∈In0; hence for allh∈R,h·f ∈In0⊆I.)
By hypothesis,I is finitely generated: I= (f1, f2, . . . , fm),fi∈R. For sufficiently largen0 we havefi∈In0 (i= 1, . . . , m). Thus
I= (f1, f2, . . . , fm)⊆In0 ⊆In0+1⊆ · · · ⊆I,
and for alln > n0,In0 =In =I.
(2⇒1):
Assume to the contrary. Then there is an idealIofR, which is not finitely generated. Iff1,f2, . . ., fm∈I, then (f1, f2, . . . , fm) I. Hence there is
anfm+1∈I,fm+16∈(f1, f2, . . . , fm). Thus
(f1, f2, . . . , fm) (f1, f2, . . . , fm, fm+1).
Thus we can construct an infinite (nonstationary) ascending chain of ideals (f1) (f1, f2) (f1, f2, f3) · · ·,
in direct contradiction to our hypothesis. (2⇒3):
For each I1 ∈ I there is an I2 ∈ I with I1 I2. In this way one can
construct a nonstationary ascending chain of ideals: I1 I2 · · · In · · ·, contradicting the hypothesis.
(3⇒2):
Apply the maximal condition to the set of ideals in a chain of ideals to obtain an In0, maximal among the ideals (under inclusion). Thus for all
n > n0,In06⊂In, i.e.,In0 =In.
Theorem 2.3.7 (Hilbert’s Basis Theorem) IfRis a Noetherian ring, so isR[x].
proof.
Assume thatR is Noetherian, butR[x] is not. We shall derive a contra- diction.
ThenR[x] must contain an idealI, which is not finitely generated. Let f1 ∈ I be a polynomial of least degree. If fk (k ≥ 1) has already been chosen, choose fk+1, the polynomial of least degree in I\(f1, f2, . . . , fk).
SinceI is not finitely generated such a sequence of choices can be carried on.
Letnk= deg(fk) andak ∈R, the leading coefficient offk(k= 1,2, . . .). Observe that
• n1≤n2≤ · · ·, simply by the choice offk’s;
• (a1)⊆(a1, a2)⊆ · · ·(a1, a2, . . . , ak)⊆(a1, a2, . . . , ak, ak+1)⊆ · · ·is a
chain of ideals that must become stationary, asRis Noetherian. That is, for some k, (a1, a2, . . . , ak) = (a1, a2, . . . , ak, ak+1), and ak+1 = b1a1+b2a2+· · ·bkak,bi∈R.
Now consider the polynomial
g=fk+1−b1xnk+1−n1f1− · · · −bkxnk+1−nkfk.
Notice that (1) degg < degfk+1, (2) g ∈ I and (3)g 6∈ (f1, f2, . . . , fk).
[Otherwise, it would imply that fk+1 ∈(f1, f2, . . . , fk).] But this contra- dicts our choice offk+1 as a least-degree polynomial inI\(f1, f2, . . . , fk).
Corollary 2.3.8
1. IfRis a Noetherian ring, so is every polynomial ringR[x1, x2, . . . , xn].
2. Let R be a Noetherian ring and S an extension ring of R that is finitely generated over R, in the ring sense. (S is a homomorphic image of a polynomial ring R[x1, . . . , xn].)Then S is Noetherian.