Nakayama's lemma and applications

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Nakayama’s lemma and applications

Rodney Coleman, Laurent Zwald

To cite this version:

Rodney Coleman, Laurent Zwald. Nakayama’s lemma and applications. 2020. �hal-03040587�

Nakayama's lemma and applications

Rodney Coleman, Laurent Zwald December 4, 2020

Abstract

We give two versions of Nakayama's lemma in the context of commutative rings and some applications, in particular, we prove two versions of Krull's intersection theorem. To do so, we will use the Artin-Rees lemma, for which we will give a detailed proof. We will suppose that all rings are commutative with identity.

There are various closely related results all of which bear the name Nakayama's lemma. Here we will prove two such results.

Theorem 1 (Nakayama's lemma version 1) LetRbe a ring,M a nitely generated module over Nakathm1

R andI an ideal in R, with IM =M. Then there exists a∈I such that (1−a)M = 0.

proof Let x₁, . . . , x_m be generators of M over R. Since M = IM, each x_i may be written x_i =Pm

j=1z_ijx_j, withz_ij ∈I. In other words,Pm

j=1(δ_ij−z_ij)x_j= 0, whereδ_ij is the Kronecker symbol. Let d be the determinant of the m×m matrix A = (δ_ij −z_ij). We haved = 1−a, where a∈I, because all the terms of the matrix Ahave this form or are elements of I. There exists a matrixB= (bij), the complementary matrix ofA, such thatBA=AB=dIm. Writing out the terms in the productBA, we havePm

i=1bhiaij=dδhj, for allhandj. Hence

m

X

j=1 m

X

i=1

bhiaijxj=

m

X

j=1

dδhjxj =dxh,

for everyh. However, we may change the order of summation to obtain

m

X

i=1 m

X

j=1

aijxjbhi=dxh.

Given thatPm

j=1aijxj= 0, becauseaij =δij−zij, we obtain thatdxh= 0, for allh; this implies

thatdM = 0, as required. 2

In a commutative ringR the Jacobson radicalJ(R)is the intersection of all maximal ideals ofR. An element a∈Ris quasi-regular if 1−ais a unit.

Proposition 1 The Jacobson radicalJ(R)is the largest ideal inRconsisting entirely of quasi- regular elements.

proof Leta∈J(R). If1−ais not a unit, then(1−a)is a proper ideal contained in a maximal idealM. But thena and 1−a are both contained in M, which implies that1 belongs to M, which is impossible. ThusJ(R)is composed entirely of quasi-regular elements.

Now letI be an ideal composed entirely of quasi-regular elements. Ifa∈I\J(R), then for some maximal idealM, adoes not belong toM. SinceM is maximal, we haveI+M =R, so 1 =b+c, withb∈I andc∈M. Asb is quasi-regular,c= 1−b is a unit and it follows that1 belongs toM, which is impossible. HenceI⊂J(R). This concludes the proof. 2 Theorem 2 (Nakayama's lemma version 2) Let R be a ring,M a nitely generated R-module Nakathm2

andI an ideal ofR. IfI⊂J(R) andIM =M, then M ={0}.

proof From Theorem1, there is an element^Nakathm1 a∈Isuch that(1−a)M = 0. Asa∈J(R),(1−a)

is invertible, henceM = 0. 2

We give a second proof of Theorem ^Nakathm22:

proof Assume that M 6= {0} and let x1, . . . , xn be generators of M, with n minimal. Then n >0and xi6= 0, for alli. Sincexn∈M =IM, we may writexn=Pm

i=1biyi, withbi∈I and yi ∈M. Eachyi can be expressed in terms of the generatorsx1, . . . , xn: yi=Pn

j=1aijxj, with aij ∈R. Hence

xn=

m

X

i=1

bi





n

X

j=1

aijxj



=

n

X

j=1 m

X

i=1

aijbj

! xj =

n

X

j=1

cjxj,

wherecj∈I. From this we obtain

(1−cn)xn =

n−1

X

j=1

cjxj,

where 1−c_n is a unit, because I is included in the Jacobson radical. If n > 1, thenx_n is a linear combination of the other x_i, contradicting the minimality of n. Hence n= 1. But then (1−c_n)x₁= 0, which implies thatx₁ = 0, which is again a contradiction. ThereforeM ={0}. 2

Here is a rst application of Nakayama's lemma. A surjective endomorphism of a nite- dimensional vector space is always injective. We have an analogous result for a nitely generated module over a ring.

Theorem 3 Let R be a ring and M a nitely generated R-module. If f is a surjective endo- morphism ofM, thenf is injective.

proof Let R be a ring, M a nitely generated R-module and f : M −→ M a surjective endomorphism. We makeM into anR[X]-module by settingP(X)·m=P(f)(m). M is nitely generated overR, hence overR[X]: Ifm₁, . . . , m_k are generators ofM overRandm∈M, then there existr₁, . . . , r_k∈Rsuch thatm=r₁m₁+· · ·+r_km_k. IfP_i(X)is the constant polynomial of valuer_i, then

P1(X)·m1+· · ·+Pk(X)·mk =r1m1+· · ·+rkmk =m, so themi form a set of generators ofM overR[X].

LetI= (X)⊂R[X]. Then

I·M =R[X]X·M =R[f]f(M)⊂M.

However, if m ∈ M, then there exists m⁰ ∈ M such that f(m⁰) = m, because f is surjective.

Also,1∈R[X], som∈R[f]f(M)and it follows thatR[f]f(M) =M. From Theorem^Nakathm11, there exists Q(X)X ∈ I such that (1−Q(X)X)·M = 0, i.e., m = Q(f)f(m), for all m ∈ M. If f(m) = 0, then

m=Q(f)f(m) = 0,

sof is injective. 2

We now turn to Krull's intersection theorem, where once again we will use Nakayama's lemma.

We need a preliminary result.

Nakalem1 Lemma 1 (Artin-Rees) Let I be an ideal in a noetherian ring R, M a nitely generated R- module and N a submodule ofM. Then there exists a positive integer c such that, forn≥c,

IⁿM∩N =I^n−c(I^cM ∩N).

proof Consider the set

S=R⊕I⊕I²⊕ · · ·=M

n≥0

Iⁿ,

whereI⁰=R. We recall that in a direct sumS, if(a0, a1, . . .)∈S, then all but a nite number of theaiare nonzero. We dene an addition and a multiplication onSas for polynomials, using the fact that I^mIⁿ ⊂I^m+n. With these operation S is a ring. AsR is noetherian,I is nitely generated, so we may writeI = (r1, . . . , rt)⊂R. We dene a mappingφ:R[X1, . . . , Xt]−→S by

φ(X₁^e1· · ·X_t^et) =r₁^e1· · ·r^e_t^t ⊂I^e0t,

where e⁰_t = max{e1, . . . , et}. The mapping φ is clearly a ring homomorphism. However, φ is also surjective: An element a ∈ Iⁿ is a sum of products of the form a₁· · ·a_n, with a_i ∈ I. We may write a_i = Pt

j=1a_ijr_j, with a_ij ∈ R, so a₁· · ·a_n is a sum of monomials of the form br₁^ei· · ·r^e_t^t, with b ∈ R and e₁+· · ·+e_t = n. As φ(bX₁^e1· · ·X_t^et) = br₁^e1· · ·r_t^et, there exists f ∈R[X₁,· · ·, X_t] such that φ(f) =a. It now follows easily that φis surjective, so there is an isomorphism ofR[X₁, . . . , X_t]/ker(φ)ontoS. HenceS is noetherian.

We now consider the setA dened by

A=M ⊕IM⊕I²M ⊕ · · ·=M

n≥0

IⁿM.

We dene an addition on A in a natural way and a scalar multiplication by elements of S in a way analogous to that used to dene the multiplication dened on S, once again using the fact that I^mIⁿ ⊂I^m+n. With these operations A is an S-module. We claim thatA is nitely generated: Letm1, . . . , ms be a generating set of M. If u∈IⁿM, thenuis a sum of terms of the formam, wherea∈Iⁿ andm∈M.

As we saw above, the element ais a sum of monomials of the form br₁^e1· · ·r^e_t^t, with b∈ R and e1+· · ·+et = n; also, m =v1m1+· · ·+vsms, with vi ∈ R. Therefore am is a sum of elements of the form (br₁^e1· · ·r_t^et)vjmj. Collecting terms, we nd that am can be written in the formc1m1+· · ·+csms, where ci ∈Iⁿ and it follows that a termu∈IⁿM has this form.

Any element of A is a nite sum of elements of this form with dierent values ofn, so the set m₁, . . . , m_sgeneratesAoverS, thusA is nitely generated as claimed.

Our next step is to consider the subset B =L

n≥0(IⁿM ∩N)ofA. We claim thatB is an S-submodule ofA. It is sucient to consider the scalar product of a setIⁿwith a setI^mM∩N. We must show that this product is included in B. An element of Iⁿ is a sum of monomials of the form br^e₁¹· · ·r^e_t^t, withb∈R ande1+· · ·+et=n. As such monomials belong toR andN is anR-submodule, the scalar product ofbr₁^e1· · ·r_t^et with an element ofI^mM ∩N must belong toB. It follows that B is anS-submodule ofA.

Since S is a noetherian ring andA nitely generated,B must be nitely generated. Let ξj, j = 1, . . . , k, be a set of generators. By decomposing each ξj into its homogeneous parts, we may assume that eachξj belongs toI^djM∩N, for somedj. We setc= max{dj}. Suppose now thatx∈IⁿM∩N. There exist u₁, . . . , u_k∈S such thatx=u₁ξ₁+· · ·+u_kξ_k. The termu_jξ_j belongs toIⁿM ∩N =I^n−dj(I^djM ∩N), so there existsu⁰_j ∈I^n−dj andm_j ∈I^djM ∩N such thatu_jξ_j=u⁰_jm_j. Now m_j=v_j,1ξ₁+· · ·+v_j,kξ_k, with thev_j,i∈S. Sincem_j ∈I^djM ∩N, we must havemj =aξj, with a∈R, and it follows thatujξj = ˜ujξj, with u˜j ∈I^n−dj. From this we deduce thatx=Pk

j=1u˜jξj, withu˜j∈I^n−dj andξj∈I^djM∩N. However,

n−dj=n−c+c−dj =⇒I^n−dj(I^djM ∩N) =I^n−c(I^c−djI^djM∩N) =I^n−c(I^cM ∩N) and it follows thatx∈I^n−c(I^cM ∩N).

We have established that IⁿM ∩N ⊂ I^n−c(I^cM ∩N). The inclusion I^n−c(I^cM ∩N) ⊂ IⁿM∩N is trivial, therefore we have the desired equality. 2 Nakacor1 Corollary 1 Let R be a noetherian ring and I, J ideals in R. Then there exists c ∈ N^∗ such

that

Iⁿ∩J =I^n−c(I^c∩J), forn≥c.

proof It is sucient to setM =R in the Artin-Rees lemma^Nakalem11. 2 We now aim to prove two versions of Krull's intersection theorem. We will use the corollary to the Artin-Rees lemma and both versions of Nakayama's lemma.

Theorem 4 Let R be a noetherian integral domain andI a proper ideal inR. Then∩^∞_n=1Iⁿ= (0).

proof LetJ=∩^∞_n=1Iⁿ. From Corollary1 there exists^Nakacor1 c∈N^∗such that Iⁿ∩J =I^n−c(I^c∩J),

forn≥c. Settingn=c+ 1we obtain

I^c+1∩(∩^∞_n=1Iⁿ) =I(I^c∩(∩^∞_n=1Iⁿ)), i.e.,

∩^∞_n=1Iⁿ=I(∩^∞_n=1Iⁿ).

AsRis noetherian, the idealJ =∩^∞_n=1Iⁿis nitely generated. From Nakayama's lemma version 1 (Theorem ^Nakathm11), there exists a ∈ I such that (1−a)J = (0). As I is a proper ideal, we have a6= 1, so1−a6= 0. Ifx∈J, then(1−a)x= 0; asRis an integral domain, we must havex= 0.2 We now suppose that the ring R is not necessarily an integral domain but impose another condition.

Theorem 5 Let R be a noetherian ring and I an ideal included in the Jacobson radical J(R). Then∩^∞_n=1Iⁿ= (0).

proof We proceed as in the proof of the previous theorem to show that

∩^∞_n=1Iⁿ=I(∩^∞_n=1Iⁿ).

We now apply Nakayama's lemma version 2 (Theorem^Nakathm22) to obtainJ = (0). 2 Corollary 2 Let(R, M)be a noetherian local ring. Then ∩^∞_n=1Mⁿ= (0).

proof It is sucient to notice that in this case the Jacobson radical isM. 2