on projective modules and computation of dimension of...
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International Scholarly Research NetworkISRN AlgebraVolume 2011, Article ID 926165, 12 pagesdoi:10.5402/2011/926165
Research ArticleOn Projective Modules andComputation of Dimension of a Module overLaurent Polynomial Ring
Ratnesh Kumar Mishra, Shiv Datt Kumar, and Srinivas Behara
Department of Mathematics, Motilal Nehru National Institute of Technology, Allahabad 211004, India
Correspondence should be addressed to Ratnesh Kumar Mishra, [email protected]
Received 18 April 2011; Accepted 9 May 2011
Academic Editors: A. Fialowski and A.-G. Wu
Copyright q 2011 Ratnesh Kumar Mishra et al. This is an open access article distributed underthe Creative Commons Attribution License, which permits unrestricted use, distribution, andreproduction in any medium, provided the original work is properly cited.
We give a procedure and describe an algorithm to compute the dimension of a module overLaurent polynomial ring. We prove the cancellation theorems for projective modules and alsoprove the qualitative version of Laurent polynomial analogue of Horrocks’ Theorem.
1. Introduction
Hilbert [1] introduced free resolution on iterated syzygies of a finitely generated gradedmodule M over polynomial ring R = K[x1, . . . , xn]. A choice of a0 = dim (M⊗RK)homogeneous generators ofM defines a surjective homomorphism Ra0 → M, and its kernelis the first syzygy module of M. Hilbert proved that the syzygies are also finitely generated(Hilbert basis theorem). Hilbert’s syzygy theorem states that “every finitely generatedmoduleM over K[x1, . . . , xn] has finite free resolution, that is, the lth syzygy is Ral for someal.” The number of generators ai of the syzygies is chosen minimally, they are independent ofthe choice of generators. These ai are called the Betti numbers ofM and the minimal lengthof free resolution (i.e., i) is dimension of a moduleM. WhenM is projective, the dimensionof M is called projective dimension of M (denoted by pd(M)). A projective dimension is ameasure of how far the module is being projective.
Kaplansky [2] described how homological dimension changes when one passes froma ring R to a quotient ring R/(X), where (X) is the ideal generated by a nonzero divisor,without using Ext or Tor. Shanuel noticed that there is an elegant relation between differentprojective resolutions of the same module. Using the ideas of Shanuel, Kaplansky defined theprojective dimension of a module.
2 ISRN Algebra
The global dimension (or homological dimension) of a ring R denoted by gldim(R),is a non-negative integer or infinity, and is a homological invariant of the ring. It is definedto be the supremum of the set of projective dimensions of all (finite or cyclic) R-modules.Global dimension is an important technical notion in the dimension theory of Noetherianrings. Eisenbud [3] proved that a commutative Noetherian local ring R is regular if and onlyif it has finite global dimension, in which case the global dimension coincides with the Krulldimension of R. This theorem opened the door for application of homological methods tocommutative algebra. Now we state a theorem of Eisenbud [3] which implies that everymodule has a finite free resolution of length at most l <∞.
Theorem 1.1 (see [3]). The following conditions on a ring R are equivalent:
(1) gldim(R) ≤ l, that is, pd(M) ≤ l for every R-moduleM,
(2) pd(M) ≤ l for every finitely generated moduleM.
Serre [4] taught a course on multiplicities at the college de France. Part of that coursefocussed upon the simple inequality pdR(M) ≤ pdR(S) + pdS(M) for module M over anR-algebra S. Auslander and Buchsbaum [5] realized that Serre’s method could be used tostudy the close connection between the dimension and multiplicity over a local ring. This ledthem to the Auslander-Buchsbaum equality [6]: let M be a finitely generated module overa Noetherian local ring (R,m). If pd(M) < ∞, then depth(R) = depth(M) + pd(M). Gago-Vargas [7] computed the projective dimension of a module over Weyl algebra. Projectivedimension is a main ingredient to compute the tilted algebra which nowadays plays a veryimportant role in the representation theory of algebras. In this paper, we give a procedureand describe an algorithm to compute the projective dimension of a module which is validover the Laurent polynomial ring.
We also prove the cancellation theorems for projective modules over the Laurentpolynomial ring [8, Section 5] motivated by Swan’s work. Horrocks proved the followingtheorem.
Theorem 1.2 (see [9]). Let R be a local ring, A = R[x] and S be the set of monic polynomials of A.If P is a finitely generated projectiveA-module such that PS is free overAS, then P is a freeA-module.
This theorem draw the attention of several mathematicians, as a result severalanalogues of Horrocks’ theorem can be found in the literature [10, 11]. Mandal proved theLaurent polynomial version of Horrocks’ theorem in [12] whereas Nashier [13] proved thequalitative version of the same theorem over polynomial ring. But it is observed that thequalitative version of Horrocks’ theorem over Laurent polynomial ring is not found anywherein the literature. One of the objectives of this paper is to fill this gap. In this paper, we provethe qualitative version of Horrocks’ theorem over the Laurent polynomial ring.
2. Preliminary Notes
In this section we define some terms used in this paper and state certain standard resultswithout proof. We hope that this will improve the readability and understanding of the proofof the paper.
A polynomial f in the Laurent polynomial ring R[X,X−1] is said to be doubly monicpolynomial if coefficient of the highest degree term and the lowest degree term are unit.
ISRN Algebra 3
We need the following result of Kunz [14] to compute the projective resolution inSection 3.
Definition 2.1 (see [14]). An exact sequence
Fi+1αi−−−→ FiF1
α0−−−→ F0ε−−−−→M → 0 (2.1)
with only free (resp., projective) modules Fi(i = 0, 1, 2, . . .) is called a free (resp., projective)resolution ofM.
Projective modules over the Laurent polynomial ring R[X,X−1] are stably free. In [15]it is proved that a projectivemodule is stably free provided it possesses a finite free resolution.The minimum length of free resolution is called the projective dimension of a moduleM.
Let R be a ring and M be an R-module. M[X,X−1] means the analogously definedLaurent polynomial module.We identifyM⊗RR[X,X−1]withM[X,X−1], whereM[X,X−1] ={∑∞
n=−∞mnXn | mn ∈ M and mn = 0 for all but finite n}. If f : M → N is a homomorphism
of R-modules, then this induces a homomorphism
ψ :M[X,X−1
]−→N
[X,X−1
],
defined by ψ(∑
mnXn)=∑
f(mn)Xn,
(2.2)
where ψ can be identified with f⊗RR[X,X−1]. Given any R-moduleM and f ∈ EndR(M), wecan make M as R[X,X−1]-module whose scalar multiplication is defined as m(
∑anX
n) =an
∑fn(m). We denote this R[X,X−1]-module by fM. There is a canonical R[X,X−1]-
epimorphism ϕf :M[X,X−1] → fM defined by ϕf(∑mnX
n) =∑fn(mn). The characteristic
sequence of f is an exact sequence which proved in Proposition 4.3.The following is the Laurent polynomial version of a Horrocks Theorem which we
state as follows.
Theorem 2.2 (see [12]). Suppose R[X,X−1] is a Laurent polynomial ring over a local Noetheriancommutative ring R, and P is a projective R[X,X−1]-module. If Pf is free for some doubly monicLaurent polynomial f , then P is free.
In the proof of the Theorem 5.1, we need the following theorem.
Theorem 2.3 (see [12]). Suppose that R[X,X−1] is a Laurent polynomial ring over a localNoetherian commutative ring R. Also suppose that P and P ′ are two projective R[X,X−1]-moduleswith RankP ′ < RankP . If P ′
f is a direct summand of Pf for some doubly monic polynomial f , thenP ′ is also a direct summand of P .
3. Projective Dimension of Module
Lemma 3.1. LetM1,M2,M3 be R-modules and
0 −→M1α1−−→M2
α2−−→M3 −→ 0 (3.1)
4 ISRN Algebra
be a split short exact sequence, and β1 and β2 are the splittings corresponding to α1 and α2, respectively.Then the following sequence
0 −→M3β2−−−→M2
β1−−−→M1 −→ 0 (3.2)
is an exact sequence.
Proof. Since β1 and β2 are the splittings corresponding to α1 and α2, respectively, then β1oα1 =IM1 and α2oβ2 = IM3 . Let x ∈ M2. Then α2(x − β2(α2(x))) = α2(x) − α2(β2(α2(x))) = α2(x) −α2(x) = 0. Hence x − β2(α2(x)) ∈ Ker(α2) = Im(α1). Since α1 is injective, there exists a uniqueβ1(x) ∈ M1 such that α1(β1(x)) = x − β2(α2(x)). We need to show that Ker(β1) ⊂ Im(β2).Let x ∈ Ker(β1). Then β1(x) = 0 and α1(β1(x)) = α1(0) = 0, by injectivity of α1. Hencex − β2(α2(x)) = 0 ⇒ x = β2(α2(x)). Thus there exists α2(x) ∈ M3 such that β2(α2(x)) = x.Hence x ∈ Im(β2). Therefore Ker β1 ⊂ Im(β2). Conversely, let x ∈ Im(β2). Then there existsy ∈ M3 such that β2(y) = x. Applying α2 on both sides, we get y = α2(x). Now applying β2on both sides, we get β2α2(x) = x and therefore x − β2α2(x) = 0. Using the defining propertyof β1 and injectivity of α1, α1(β1(x)) = 0 = α1(0). Therefore β1(x) = 0. Hence x ∈ Ker(β1). ThusIm(β2) ⊂ Ker(β1). Therefore Im(β2) = Ker(β1)
By applying the technique discussed in [7] for Weyl algebra, we give a procedure tocalculate the projective dimension of a module over the Laurent polynomial ring. We alsodescribe an algorithm to compute the projective dimension of a module.
Procedure. The first step is to define the Laurent polynomial ring R[x±11 , x
±12 , . . . , x
±1n ] in terms
of quotient ring as
R[x±11 , x
±12 , . . . , x
±1n
] ∼= R[x1, y1, x2, y2, . . . , xn, yn
]
(x1y1 − 1, x2y2 − 1, . . . , xnyn − 1
) . (3.3)
The second step is to test whether the R[x±11 , . . . , x
±1n ]-module M is projective or not. IfM is
projective, then again we test whether M is stably free or not, that is, we find a matrix thatdefines an isomorphismM ⊕ R[x±1
1 , . . . , x±1n ]r ∼= R[x±1
1 , . . . , x±1n ]s, for some positive integers r
and s. If M is stably free, then the next step is to find finite free resolution of the projectivemoduleM. We denote the homomorphisms with their matrices to simplify the notation.
Given an R[x±11 , . . . , x
±1n ]-module M defined by a system of generators in some
R[x±11 , . . . , x
±1n ]t, choose a free R[x±1
1 , . . . , x±1n ]-module P0 and a surjection σ : P0 → M with
kernel C0, then we get an exact sequence
0 −→ C0ψ0−−−→ P0
σ−−→M −→ 0. (3.4)
Now, choose a free R[x±11 , . . . , x
±1n ]-module P1 together with a surjection map φ1 : P1 → C0
with kernel C1. Again we choose a free R[x±11 , . . . , x
±1n ]-module P2 together with a surjection
φ2 : P2 → C1 with kernel C2, then we get an exact sequence
0 −→ C2ψ2−−−→ P2
φ2−−−→ C1ψ1−−−→ P1
φ1−−−→ C0ψ0−−−→ P0
σ=α0−−−−→M −→ 0. (3.5)
ISRN Algebra 5
Continuing this process, we have an exact sequence
℘ : 0Plαl−−−→ Pl−1
αl−1−−−→ P1α1−−−→ P0
σ=α0−−−−→M −→ 0, (3.6)
which is a free resolution of the moduleM, with rank(Pi) = ti, where αl = ψl−1φl. For l = 1
0 −→ P1α1−−−→ P0
α0−−−→M −→ 0 (3.7)
is a free resolution of moduleM. SinceM is a projective module, this sequence splits, so thereexists β1 : P0 → P1 such that β1α1 = It1 . We can compute the matrix β1 from the rows of thematrix α1. We express each vector of the canonical basis of P1 as a linear combination of therows of α1, and with these coefficients we construct the matrix β1. Using Lemma 3.1, the exactsequence in (3.7) splits, giving another exact sequence
0 −→Mβ0−−−→ P0
β1−−−→ P1 −→ 0. (3.8)
Then P1 ⊕M ∼= P0 ∼= Ker(β1) ⊕ P1. SinceM is a projective module, the short exact sequence
0 −→ Ker(α0) −→ P0α0−−−→M −→ 0 (3.9)
splits, so Ker(α0) = Im(α1) is projective. Hence pd(M) = 0. SupposeM is not projective, thenthe short exact sequence (3.9) does not split. Therefore Im(α1) = Ker(α0) is not projective.For l = 2, the exact sequence (3.6) does not split, therefore Im(α2) = Ker(α1) is notprojective. Continuing this process Im(α1), Im(α2), . . . , Im(αl−1) are not projective and Im(αl)is projective. In this way after performing a finite number of steps, we obtain the minimumlength of free resolution of M which is the projective dimension of M such that pd(M) = lsee Algorithm 1.
Remark 3.2. Nowwe give an algorithm to calculate the projective dimension of a module overthe Laurent polynomial ring. We follow the technique discussed in [16] over polynomial ring.
Example 3.3. Let
A = R[x, x−1, y, y−1
] ∼= R[x, y, p, q
]
(xp − 1, yq − 1
) (3.10)
ringR = 0, (x, y, p, q), dp;
ideal I = xp − 1, yq − 1;
I = std(I);
I;
I[1] = yq − 1;
6 ISRN Algebra
Let
A = R[x±11 , x
±12 , . . . , x
±1n ] ∼= R[x1, y1, x2, y2, . . . , xn, yn]
(x1y1 − 1, x2y2 − 1, . . . , xnyn − 1).
Objective: Computation of the projective dimension of a module.Input: A finitely generated A-moduleM =< f1, f2, . . . , fm >⊂ At anda positive integer l.Output: A projective dimension ofM and a list of matrices α1, α2, . . . , αlwith αi ∈ Mat(li−1 × li, A), where i = 1, 2, . . . , l, such that
℘ : 0Plαl−−→ Pl−1
αl−1−−−→ P1α1−−→ P0
α0−−→ P0/M −→ 0is a free resolution of P0/M. If pd(M) = 0, thenM is projective. Thealgorithm returnsM ⊕Ar ∼= Al.START:initialize i = 1;if (α1 does not split)α1 = matrix(f1, f2, . . . , fm) ∈ Mat(l ×m,A) andpd(M) = 1;
elseLet β1 be the split of α1. Then pd(M) = 0 andP1 ⊕M ∼= P0 ∼= ker(β1) ⊕ P1;
end ifif (i = l)
return pd(M);elsewhile (i < l) doi + +;αi = syz(αi−1);
end loopend ifif (αl does not split)
pd(M) = l;else
Let βl be the split of αl. Then pd(M) = l − 1;end if
return pd(M);STOP.
Algorithm 1
I[2] = xp − 1;
qringA = I;
A;∗∗quotient ring from ideal∗∗
[1] = yq − 1;
[2] = xp − 1;
Poly f1 = x3p + (x3p)−1 + y4q;
Poly f2 = xp + (yq)−1 + yq;
IdealM = (f1, f2);
Resolution L =mres(M, 0);
ISRN Algebra 7
We find a resolution of P0/M over Laurent polynomial ring:
0 → P0α0−−→ P0/M → 0.
Hence, projective dimension ofM is 0.
Example 3.4. Let
A = R[x, x−1, y, y−1, z, z−1, s, s−1
]
∼= R[x, y, z, p, q, u, s, t
]
(xp − 1, yq − 1, zu − 1, st − 1
)
(3.11)
ringR = 0, (x, y, z, p, q, u, s, t), dp;
ideal I = xp − 1, yq − 1, zu − 1, st − 1;
I = std(I);
I;
I[1] = st − 1
I[2] = zu − 1
I[3] = yq − 1
I[4] = xp − 1
qringA = I;
A;∗∗quotient ring from ideal∗∗
[1] = st − 1
[2] = zu − 1
[3] = yq − 1
[4] = xp − 1
poly f = x3p + (x3p)−1 + y4q + zu2;
poly g = x6p2 + (yq)−1 + yq + s3t + yq;
idealM = (f, g);
resolution L = mres(M, 0).
We find a resolution of P0/M over Laurent polynomial ring:
0 → P 332 ⊕ P 6
33α11−−→ P30 ⊕ P 5
31α10−−→ P 2
28 ⊕ P 329
α9−−→ P 226 ⊕ P 2
27α8−−→ P−1
20 ⊕ P21 ⊕ P 322 ⊕ P23 ⊕ P24 ⊕ P25
α7−−→P 215⊕P 2
16⊕P 217⊕P18⊕P19
α6−−→ P 311⊕P 2
12⊕P 213⊕P14
α5−−→ P 48 ⊕P 2
9 ⊕P 210
α4−−→ P 35 ⊕P6⊕P7
α3−−→ P−12 ⊕P3⊕P4 α2−−→
P1α1−−→ P 0
0α0−−→ P0/M → 0.
Hence, projective dimension ofM is 11.
Remark 3.5. Examples 3.3 and 3.4 are verified using the singular software [17].
8 ISRN Algebra
4. A Cancellation Theorem for Projective Modules over LaurentPolynomial Ring
Proposition 4.1. Let P and P ′ be projective R[X,X−1]-modules and let ϕ : P ′ → P , ψ : P → P ′
be injective homomorphisms. If R[X,X−1], P/ϕψP , and P ′/ψϕP ′ are projective over R, then P/ϕP ′
and P ′/ψP are also projective over R.
Proof. Since P and P ′ are R[X,X−1]-projective and R[X,X−1] is R-projective, then we have
P ⊕ P1 ∼= R[X,X−1
]n
P ′ ⊕ P ′1∼= R
[X,X−1
]m
R[X,X−1
]⊕Q ∼=
∑R
...
R[X,X−1
]n ⊕Qn ∼=∑
R
P ⊕ P1 ⊕Qn ∼=∑
R
P ⊕Qn1∼=∑
R,
(4.1)
where P1 ⊕Qn = Qn1 . Therefore, P is a R-projective. Similarly P ′ is also a R-projective. Since
0 −→ Pβ−−→ P ′ −→ P ′/ψP −→ 0, (4.2)
pd(P ′/ψP) ≤ 1 and similarly pd(P/ϕP ′) ≤ 1. Now isomorphism between P and ψP inducesP/ϕP ′ ∼= ψP/ψϕP ′. We have the exact sequences
0 P P P /ψP 0
0 ψP/ψϕP P /ψϕP P /ψP 0
(4.3)
By Schanuel’s lemma, P ′/ψϕP ′ ⊕ P ∼= ψP/ψϕP ′ ⊕ P ′ so that P ′/ψϕP ′ ⊕ P ∼= P/ϕP ′ ⊕ P ′. SinceP ′/ψϕP ′, P are R-projective and direct sum of projective module is also projective module,then P ′/ψϕP ′ ⊕ P is R-projective. By the isomorphism, P/ϕP ′ ⊕ P ′ is an R-projective. P/ϕP ′ ⊕P ′ ⊕P0 ∼= Rn follows from the definition of R-projective module, where P0 is an R-module. LetP = P ′ ⊕ P0 be an R-module. Then P/ϕP ′ ⊕ P ∼= Rn. Hence P/ϕP ′ is an R-projective. SimilarlyP ′/ψP is also an R-projective.
Corollary 4.2. Let P and P ′ be projective R[X,X−1]-modules with P ⊃ P ′ ⊃ fP , where f is adoubly monic polynomial of Laurent polynomial ring. If R[X,X−1] and R[X,X−1]/fR[X,X−1] areR-projective, then P/P ′ is also R-projective.
ISRN Algebra 9
Proof. Take ϕ the inclusion map P ′ ↪→ P and ψ the multiplication by f from P → P ′. ThenP/ϕψP = P/fP and P ′/ψϕP ′ = P ′/fP ′. Since P , P ′ are R[X,X−1]-projective and R[X,X−1],R[X,X−1]/fR[X,X−1] are R-projective. Then P/fP and P ′/fP ′ are R[X,X−1]/fR[X,X−1]-projective. Therefore P/fP and P ′/fP ′ are R-projective. Hence P/ϕψP and P ′/ψϕP ′ are alsoR-projective. By Proposition 4.1, module P/P ′(= P/ϕP ′) is R-projective.
Proposition 4.3. LetM be an R-module and f ∈ EndR(M). The characteristic sequence of f
0 −→M[X,X−1
] X.1M[X,X−1]−f[X,X−1]−−−−−−−−−−−−−−−−−−→M[X,X−1
] ϕf−−−→ fM −→ 0 (4.4)
is an exact sequence in R[X,X−1]-module.
Proof. Clearly ϕf is surjective, and we have
ϕf(X.1M[X,X−1] − f
[X,X−1
])(∑mnX
n)= ϕf
(∑(mnX
n+1 − f(mn)Xn))
=∑(
fn+1(mn) − fn(f(mn)
))
=∑(
fn+1(mn) − fn+1(mn))
= 0.
(4.5)
Im(X.1M[X,X−1]−f[X,X−1]) ⊆ Ker(ϕf). Since,X.1M[X,X−1]−f[X,X−1] raises degree by one, andpreserves leading coefficients, it is monomorphism. Finally, suppose
z =∑
mnXn ∈ Ker
(ϕf
)=⇒ ϕf
(∑mnX
n)= 0, that is,
∑fn(mn) = 0. (4.6)
Then
z = z −∑
fn(mn)
=∑(
mnXn − fn(mn)
)
=∑(
Xn.1M[X,X−1] − fn)(mn)
=(X.1M[X,X−1] − f
[X,X−1
])[
· · · − 1X − f
Xn − fnXnfn
− 1X − f
Xn−1 − fn−1Xn−1fn−1
· · ·
− 1X − f
X − fXf
+ 0 +X − fX − f +
X2 − f2
X − f + · · ·
+Xn − fnX − f + · · ·
]
(mn)
10 ISRN Algebra
=(X.1M[X,X−1] − f
[X,X−1
])∑hn(mn)
=⇒ z ∈ Im(X.1M[X,X−1] − f
[X,X−1
]).
(4.7)
Therefore Ker(ϕf) ⊆ Im(X.1M[X,X−1] − f[X,X−1]).Hence Im(X.1M[X,X−1] − f[X,X−1]) = Ker(ϕf).
Theorem 4.4. Let P and P ′ be finitely generated projective modules over R[X,X−1]. Suppose thatP ⊃ P ′ ⊃ fP for some doubly monic polynomial f ∈ R[X,X−1]. Then P and P ′ are stably isomorphic.In particular, if Pf ∼= P ′
f, then P and P ′ are stably isomorphic.
Proof. Take M = P/P ′. Since f is doubly monic polynomial, R[X,X−1]/(f) is free as an R-module. Therefore P/fP is R-projective, whenceM is R-projective by Corollary 4.2. We haveexact sequences of R[X,X−1]-modules
0 M 0
0 M [X,X−1] M [X, X−1] M 0
P P
(4.8)
where M[X,X−1] = M⊗RR[X,X−1]. Since P ′ ⊂ P and M = P/P ′, the inclusion map i fromP ′ to P and the surjective map π from P to M makes the first sequence an exact sequence.By Proposition 4.3, second sequence is also an exact sequence. Since M is an R-projective,M[X,X−1] is an R[X,X−1]-projective. Therefore by the Schanuel’s lemma, P ⊕M[X,X−1] ∼=P ′ ⊕M[X,X−1]. Hence P , P ′ are stably isomorphic.
Theorem 4.5. Let F be a free R[X,X−1]-module of rank 2, P a projective R[X,X−1]-module andF ⊃ P ⊃ XF. Then P ∼= F.
Proof. Let {f1, f2} be a free R[X,X−1]-basis for F, and let F1 = {a1f1 + a2f2 | a1, a2 ∈ R}. ThenF1 is free over R of rank 2.
Claim. We show that each element of F1 ⊕XF ⊕ (1/X)F can be written as elements of F.
F1 ⊕XF ⊕ 1XF ⇐⇒ a1f1 + a2f2 +X
(g1f1 + g2f2
)+
1X
(h1f1 + h2f2
)
⇐⇒(
a1 +Xg1 +1Xh1
)
f1 +(
a2 +Xg2 +1Xh2
)
f2
⇐⇒ Uf1 + Vf2
⇐⇒ F(direct sum as R-module),
(4.9)
where g1, g2, h1, h2, U, and V are the elements of R[X,X−1]. Thus with P1 = P ∩ F1, P =P1⊕XF⊕(1/X)F (direct sum asR-module). By Corollary 4.2, F1/P1 = F/P is a projective over
ISRN Algebra 11
R, so F1 = P1⊕P ′1 for some P ′
1. Now F = P1[X,X−1]⊕P ′1[X,X
−1] and P = P1⊕XF⊕(1/X)F = P1⊕X(P1[X,X−1]⊕P ′
1[X,X−1])⊕(1/X)(P1[X,X−1]⊕P ′
1[X,X−1]) = P1[X,X−1]⊕P ′
1[X,X−1] ∼= F.
5. A Qualitative Version of Horrocks’ Theorem
Theorem 5.1. Let R be a local ring and P be a projective R[X,X−1]-module. Then, for any doublymonic polynomial f in R[X,X−1], the R[X,X−1]-module P and the R[X,X−1]f -module Pf have thesame minimal number of generators, that is, μ(P) = μ(Pf).
Proof. Let f be a doubly monic polynomial in R[X,X−1]. Since there is a natural surjectivemap φ : P → Pf , we have μ(Pf) ≤ μ(P). To complete the proof it is enough to show μ(Pf) ≥μ(P). We have the following two cases.
Case 1. Pf is a free module over R[X,X−1]f . By Theorem 2.2, P is free and the result follows.
Case 2. Suppose Pf is not free over R[X,X−1]f . Then Rank(Pf) < μ(Pf). Let μ(Pf) = n. We canwrite Pf as a direct summand of (R[X,X−1]f)
n ∼= (R[X,X−1]n)f . Since Rank(P) = Rank(Pf) <Rank(R[X,X−1]n). By Theorem 2.3, there exists an R[X,X−1]-module P ′ such that P ⊕ P ′ ∼=R[X,X−1]n. Therefore μ(P) ≤ n = μ(Pf).
References
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12 ISRN Algebra
[16] G.-M. Greuel and G. Pfister,A Singular Introduction to Commutative Algebra, Springer, Berlin, Germany,2008.
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