2.3.3
Finite Gr¨obner Bases
Theorem 2.3.9 Let S be a Noetherian ring. Then every ideal of R = S[x1,x2,. . .,xn]has a finite Gr¨obner basis.
proof.
Since S is Noetherian, by Hilbert’s basis theorem, so is R = S[x1, x2, . . ., xn]. Let <
A be an arbitrary but fixed admissible ordering on
P P(x1, x2, . . . , xn).
LetI be an ideal inR, and Head(I), the monomial ideal generated by the head monomials of the polynomials in I. Let us choose a polynomial g1 ∈ I; if G1 = {g1} ⊆ I is not a Gr¨obner basis of I, then Head(G1)
Head(I), and there is a polynomialg2∈Isuch that Hmono(g2)∈Head(I)\
Head(G1). Clearly,G2={g1, g2} ⊆I and Head(G1) Head(G2).
In the (k+1)thstep, assume that we have chosen a setG
k={g1,g2,. . ., gk} ⊆I. Now, ifGk is not a Gr¨obner basis forI, then there is agk+1∈I
such that
Hmono(gk+1)∈Head(I)\Head(Gk),
andGk+1 =Gk∪ {gk+1} ⊆Iand Head(Gk) Head(Gk+1). But, sinceR
is Noetherian, it cannot have a nonstationary ascending chain of ideals Head(G1) Head(G2) · · · Head(Gk) · · ·,
and there is somen≥1 such that Head(Gn) = Head(I). But sinceGn⊆I, we see thatGn={g1, g2, . . . , gn}is a finite Gr¨obner basis forIwith respect to the admissible ordering <
A.
2.4
Modules and Syzygies
Definition 2.4.1 (Modules) Given a ringS, an Abelian groupM, and a mapping
µ : S×M →M
: hs, xi 7→sx,
we sayM is anS-module if, for all s, t∈ S and x, y ∈ M, the following axioms are satisfied:
s(x+y) = sx+sy, (s+t)x = sx+tx,
(st)x = s(tx), 1x = x.
Thus, an S-module is an additive Abelian group M on which the ringS acts linearly.
IfS is a fieldK, then aK-module is said to be a K-vector space. Note that if S is any ring, then any ideal I ⊆ S is an S-module. In particular, S itself is an S-module. Also, every Abelian group (G, +, 0) is a Z-module: here, the mapping hn, xi 7→ nx (n ∈ Z, x ∈ G) has the following outcome: nx= x+x+· · ·+x | {z } n ifn >0; 0 ifn= 0; (−x) + (−x) +· · ·+ (−x) | {z } n ifn <0.
LetS 6={0}be a ring,T ⊆S, a multiplicatively closed subset andM, anS-module. Consider the following equivalence relation “∼” onM ×T:
∀ hx1, a1i,hx2, a2i ∈M×T h hx1, a1i ∼ hx2, a2i iff (∃a3∈T) [a3(a2x1−x2a1) = 0] i . Let MT = M ×T / ∼ be the set of equivalence classes on M ×T with respect to the equivalence relation ∼. The equivalence class containing
hx, ai is denoted by x/a. MT can be made into an ST-module with the obvious definitions of addition and scalar multiplication. MT is called the module of fractions of M with denominator setT.
Definition 2.4.2 (Module Homomorphisms) LetS be a ring and let M andN beS-modules. Then a mapping
φ:M →N
is said to be anS-module homomorphism if, for alls∈S andx,y∈M, φ(x+y) =φ(x) +φ(y) and φ(sx) =sφ(x),
i.e.,S acts linearly with respect toφ.
Letφbe anS-module homomorphism as before. We define the kernel ofφto be
kerφ={x∈M : φ(x) = 0} and theimage ofφto be
imφ={φ(x)∈N : x∈M}. It can be verified that kerφand imφare bothS-modules.
Section 2.4 Modules and Syzygies 51
Definition 2.4.3 (Submodule) Let S be a ring and M an S-module. ThenM′ is a said to be asubmodule of M ifM′ is a subgroup of M and M′ is anS-module, i.e.,M′ is closed under multiplication by the elements ofS.
Definition 2.4.4 (Quotient Submodule) GivenS,M andM′as in the previous definition (Definition 2.4.3), we make the quotient Abelian group M/M′ an S-module by allowing it to inherit an S-module structure in a natural manner. In particular, we make the natural definition for multipli- cation inM/M′: fors∈S and x∈M,
s(x+M′) =sx+M′.
This definition is consistent, since, if x+M′ =y+M′, i.e.,x−y ∈ M′ thens(x−y) =sx−sy∈M′ andsx+M′=sy+M′. The axioms for an S-module (as in Definition 2.4.1) follow quite easily.
The S-moduleM/M′ thus defined is called the quotient submodule of M by M′, and the mapping
φ : Monto→ M/M′ : x7→x+M′, is a surjectiveS-module homomorphism.
Definition 2.4.5 (Module Operations) LetS be a ring,I be an ideal ofS,M be anS-module and
M= (Mi)i∈I, a family of submodules ofM; then:
1. Sum: X i∈I Mi= ( X i∈I
xi: xmany ofi∈Mi,xand all but finitelyi’s are zero
)
is a submodule of M. Thus PMi consists of all sums formed by taking exactly one element from each submodule in a finite subfamily ofM.
2. Intersection: \
i∈I Mi, is a submodule ofM.
3. Product: IM =n n X i=1 aixi:ai∈I, xi∈M andn∈Z o , is a submodule ofM.
4. Quotient: LetN andP be two submodules ofM: N :P =na∈S:aP ⊆No,
is an ideal of S. The quotient 0 : M is an ideal and is called the annihilator ofM (denoted annM).
AnS-moduleM isfaithful if annM = 0.
Definition 2.4.6 (Generators) Let S be a ring and let M be an S- module. Note that, for anyx∈M,
Sx={sx:s∈S} is a submodule ofM. Let
X ={xi}i∈I
be a (possibly, infinite) subset ofM. X is said to be asystem of generators ofM, if
M =X
i∈I Sxi.
Equivalently,X is a system of generators ofM if every elementx∈M can be expressed in the form
X
i∈J sixi,
whereJ ⊆Iis a finite subset of the index set andsi∈S andxi∈ X. IfS is a ring andM is anS-module,M is said to befinitely generatedif M has a finite set of generators, andcyclic(ormonogenic) if it is generated by only one element. A system of generatorsu1,. . .,unof anS-moduleM is a basis ofM, if X aiui= 0 ⇒ ∀i hai= 0 i ,
i.e., M has a linearly independent system of generators. M is called free (of rankn) if it has abasis of size n.
Section 2.4 Modules and Syzygies 53
IfS is a ring, then it is natural to make Sn into anS-module,M, by defining, for anyhs1,. . .,sni,ht1,. . .,tni ∈Sn and any s∈S,
0 = h0, . . . ,0i,
hs1, . . . , sni+ht1, . . . , tni = hs1+t1, . . . , sn+tni and sht1, . . . , tni = hst1, . . . , stni.
It is easy to see thatM =Sn is a freeS-module of rankn.
Definition 2.4.7 (Noetherian Modules) AnS-moduleMis calledNoe- therian if every submoduleN ofM is finitely generated.
Proposition 2.4.1 If S is a Noetherian ring, then Sn (n <
∞) is a Noetherian S-module.
proof.
LetNbe a submodule ofSn. We proceed by induction onn: Ifn= 1, then there is nothing to prove, since, in this case, N ⊆S1=S is a submodule
and hence an ideal inS, thus possessing a finite set of generators. Ifn >1, then let
I=ns∈S: (∃s2, . . . , sn∈S) [hs, s2, . . . , sni ∈N]
o
.
Iis clearly an ideal inS, soI has a finite set of generators{s1,1, . . . , s1,k}. Picks1,. . ., sk ∈N such that for i= 1, . . ., k, si has as first component s1,i. For an arbitrary element s=hs1,s2,. . ., sni ∈N, we can express s1
as s1= k X i=1 ris1,i, for some r1, . . . , rk ∈S. Now, note that
s′=s− k X i=1 risi is of the formh0,s∗
2,. . .,s∗ni ∈N′, whereN′ is the following submodule of Sn:
N′ ={s=h0, s2, . . . , sni:s∈N}. But the mapping
φ : N′→Sn−1
: h0, s2, . . . , sni 7→ hs2, . . . , sni
is a homomorphism with the kernel (0, 0,. . ., 0). Thusφis an isomorphism ofN′ into its image imφ⊆Sn−1. Hence, by induction, imφa submodule
ofSn−1has a finite system of generators, and so doesN′. Let{t
1,. . .,tl} be such a system of generators. Then, since
s′ =s− k X i=1 risi= l X i=1 qiti, whereqi, ri ∈S and s= k X i=1 risi+ l X i=1 qiti,
{s1,. . .,sk,t1,. . .,tl} is a finite system of generators ofN.
Definition 2.4.8 (Syzygies) LetS be a ring and letM = (x1, . . ., xq)
be a finitely generatedS-module. Note that φ : Sq →M
: hs1, . . . , sqi 7→s1x1+· · ·+sqxq is anS-module homomorphism. Thus
K= kerφ={hs1, . . . , sqi ∈Sq|s1x1+. . .+sqxq = 0},
is a submodule ofSq;K is said to be the (first module of)syzygies ofM [with respect to the system of generators{x1,. . .,xq}ofM] and is denoted S(M).
Proposition 2.4.2 IfS is a Noetherian ring andM is a finitely generated S-module, thenS(M), the syzygy ofM, is finitely generated.
proof.
If M = (x1, x2, . . ., xq), then S(M) is a submodule of a NoetherianS- module, Sq. Thus, by Proposition 2.4.1, S(M), the syzygy of M is also finitely generated.
GivenS, a Noetherian ring, andM = (x1,. . .,xq), a finitely generated S-module, we haveS(M) = (s1,. . .,sp) where
s1 = hs1,1, s1,2, . . . , s1,qi s2 = hs2,1, s2,2, . . . , s2,qi .. . sp = hsp,1, sp,2, . . . , sp,qi and, fori= 1,. . .,p si,1x1+· · ·+si,qxq = 0.