Let R be a commutative ring and let R[X] be the polynomial ring over R. For an R-module M, let M[X] be the set of polynomials in an indeterminate X with coefficients in M. Then M[X] is an R[X]-module under the usual addition and the scalar multiplication as follows: For f=∑i=0maiXi,g=∑i=0nbiXi∈M[X] with m≥n and h=∑i=0ℓriXi∈R[X],

and

where ci=∑k=0irkai-k for all i=0,…,ℓ+m (cf. [2, Chapter 2, Exercise 6]). We call M[X] the polynomial R[X]-module.

Let R be a commutative ring and let M be an R-module. Recall that M is a Noetherian module if it satisfies the ascending chain condition on R-submodules of M (or equivalently, every R-submodule of M is finitely generated); and R is a Noetherian ring if R is a Noetherian R-module. It is well known as Hilbert basis theorem for Noetherian modules that if R is a commutative ring with identity and M is a Noetherian unitary R-module, then M[X] is a Noetherian R[X]-module [2, Chapter 7, Exercise 10]. When M=R, it recovers the well-known Hilbert basis theorem which states that if R is a Noetherian ring, then R[X] is also a Noetherian ring [2, Chapter 7, Theorem 7.5] (or [4, Theorem 69]).

In this note, we study Hilbert basis theorem for Noetherian modules. We first give a brief proof of Hilbert basis theorem for Noetherian modules. We next consider the converse of Hilbert basis theorem for Noetherian modules. More precisely, we show that if M[X] is a Noetherian R[X]-module, then M is a Noetherian R-module and there exists an element e∈R such that em=m for all m∈M. We also prove that if M[X] is a Noetherian R[X]-module and annR(M)=(0), then R contains the identity and M is a unitary R-module.

We start this section with Hilbert basis theorem for Noetherian modules. While the result appears in [2, Chapter 7, Exercise 10], we insert a brief proof for the sake of reader's easy understanding.

Theorem 1. Let R be a commutative ring with identity and let M be a unitary R-module. If M is a Noetherian R-module, then M[X] is a Noetherian R[X]-module.

Proof. Suppose to the contrary that M[X] is not a Noetherian R[X]-module. Then there exists an R[X]-submodule N of M[X] which is not finitely generated. Let f1 be a nonzero element of least degree in N. Then 〈f1〉⊊N. Let f2 be an element of least degree in N∖〈f1〉. Then 〈f1,f2〉⊊N. By repeating this process, for all integers k≥1, there exists an element fk+1 of least degree in N∖〈f1,…,fk〉. For each integer i≥1, let ai be the leading coefficient of fi and let di be the degree of fi. Then di≤di+1 for all integers i≥1. Since M is a Noetherian R-module, the ascending chain 〈a1〉⊆〈a1,a2〉⊆⋯ of R-submodules of M is stationary; so there exists an integer ℓ≥1 such that 〈a1,…,aℓ〉=〈a1,…,aℓ+1〉. Therefore aℓ+1∈〈a1,…,aℓ〉. Write aℓ+1=r1a1+⋯+rℓaℓ for some r1,…,rℓ∈R. Let g=fℓ+1-(r1Xdℓ+1-d1)f1+⋯+(rℓXdℓ+1-dℓ)fℓ. Then g∈N∖〈f1,…,fℓ〉 and deg(g)<deg(fℓ+1). This is a contradiction to the choice of fℓ+1. Thus M[X] is a Noetherian R[X]-module.

We next consider the converse of Hilbert basis theorem for Noetherian modules. In order to study the converse of Hilbert basis theorem for Noetherian modules, we give a result which is required to prove the main theorem.

Lemma 2. Let R be a commutative ring and let M be an R-module. If M[X] is a Noetherian R[X]-module, then the following assertions hold.

• (1)

  M is a Noetherian R-module.

• (2)

  For each m∈M, there exists an element r∈R

Proof. (1) Let N be an R-submodule of M. Then N[X] is an R[X]-submodule of M[X]. Since M[X] is a Noetherian R[X]-module, there exist g1,…,gk∈N[X] such that N[X]=〈g1,…,gk〉. Let n∈N. Then we obtain

for some f1,…,fk∈R[X] and ℓ1,…,ℓk∈Z. For each i=1,…,k, write fi=∑j=0αirijXj and gi=∑j=0βinijXj. Then we have

so N=〈n10,…,nk0〉. Hence N is a finitely generated R-submodule of M. Thus M is a Noetherian R-module.

(2) Let m∈M. Then 〈m〉⊆〈m,mX〉⊆〈m,mX,mX2〉⊆⋯ is an ascending chain of R[X]-submodules of M[X]. Since M[X] is a Noetherian R[X]-module, there exists a nonnegative integer q such that 〈m,…,mXq〉=〈m,…,mXq+1〉; so mXq+1∈〈m,…,mXq〉. Therefore we have

for some f0,…,fq∈R[X] and ℓ0,…,ℓq∈Z. For each i=0,…,q, write fi=∑j=0αirijXj. By comparing the coefficients of Xq+1 in both sides, we obtain

Note that r0q+1+⋯+rq1∈R. Thus the proof is done.

Let R be a commutative ring and let M be an R-module. Then annR(M):={r∈R|rM={0}} is an ideal of R and is called the annihilator of M in R. We are now ready to give the main result in this note.

Theorem 3. Let R be a commutative ring and let M be an R-module. If M[X] is a Noetherian R[X]-module, then the following conditions hold.

• (1)

  There exists an element e∈R such that

• (2)

  If annR(M)=(0),

Proof. (1) Suppose that M[X] is a Noetherian R[X]-module. Then by Lemma 2.1(1), M is a Noetherian R-module; so M=〈m1,…,mn〉 for some m1,…,mn∈M. By Lemma 2.1(2), for each i=1,…,n, there exists an element ri∈R such that rimi=mi. Let e1=r1+r2-r1r2 and for each i∈{2,…,n-1}, let ei=ei-1+ri+1-ei-1ri+1. Then e1mi=mi for all i=1,2. Suppose that there exists an index k∈{1,…,n-2} such that ekmi=mi for all i∈{1,…,k+1}. Then we have

for all i=1,…,k+1. Also, we obtain

Therefore ek+1mi=mi for all i=1,…,k+2. Hence by the induction, en-1mi=mi for all i=1,…,n. Let m∈M. Then we have

for some s1,…,sn∈R and ℓ1,…,ℓn∈Z. Thus we obtain

(2) Let a be any element of R. By (1), there exists an element e∈R such that em=m for all m∈M; so aem=am for all m∈M, which indicates that (ae-a)m=0 for all m∈M. Therefore (ae-a)M={0}. Since annR(M)=(0), ae-a=0. Hence ae=a. Thus e is the identity of R, which also shows that M is a unitary R-module.

As an immediate consequence of Theorems 1 and 2(2) and Lemma 2.1(1), we have

Corollary 4. Let R be a commutative ring and let M be an R-module with annR(M)=(0). Then the following conditions are equivalent.

• [(1)] R contains the identity and M is a Noetherian unitary R-module.

• [(2)] M[X] is a Noetherian R[X]-module.

By applying M=R to Theorem 2(1), we regain

Corollary 5. ([3, Theorem]) Let R be a commutative ring. If R[X] is a Noetherian ring, then R contains the identity.

We next give two examples which show that the converse of Theorem 2(2) is not true in general and the annihilator condition in Theorem 2(2) is essential.

Example 6. Let M={0,2,4} be a Z6-submodule of Z6. Then Z6 contains the identity and M is a unitary Z6-module. Note that Z6[X] is a Noetherian Z6[X]-module by Theorem 1 (or [2, Chapter 7, Exercise 10]) and M[X] is a Z6[X]-submodule of Z6[X]. Hence M[X] is a Noetherian Z6[X]-module. However, annZ6(M)={0,3}.

Example 7. Let R={0,2,4,6,8,10} be a subring of Z12 and let M={0,4,8} be an R-submodule of Z12. Then R is a commutative ring without identity and M is a nonunitary R-module with annR(M)={0,6}. Also, 4m=m for all m∈M. Finally, M[X] is a Noetherian R[X]-module. (To see this, suppose to the contrary that M[X] is not a Noetherian R[X]-module. Then there exists an R[X]-submodule N of M[X] which is not finitely generated. Let f1 be a nonzero element of least degree in N. Then 〈f1〉⊊N; so there exists an element f2 of least degree in N∖〈f1〉. For all i=1,2, let ai be the leading coefficient of fi and let di be the degree of fi. Then d1≤d2. If a1=a2=4 or a1=a2=8, then we let g=f2-(4Xd2-d1)f1. Then g∈N∖〈f1〉 and deg(g)<deg(f2). This is a contradiction to the choice of f2. If a1=4 and a2=8, then we let g=f2-(2Xd2-d1)f1. Then g∈N∖〈f1〉 and deg(g)<deg(f2). This is impossible because of the choice of f2. Similarly, if a1=8 and a2=4, then we let g=f2-(2Xd2-d1)f1. Then g∈N∖〈f1〉 and deg(g)<deg(f2). This is also absurd because of the minimality of the degree of f2 in N∖〈f1〉. Hence M[X] is a Noetherian R[X]-module.)

Let R be a commutative ring and let M be an R-module. Then the Nagata's idealization of M in R (or the trivial extension of R by M) is a commutative ring

with usual addition and multiplication defined by (r1,m1)(r2,m2)=(r1r2,r1m2+r2m1) for all (r1,m1),(r2,m2)∈R(+)M. It is routine to check that (R(+)M)[X] is isomorphic to R[X](+)M[X] (cf. [1, Corollary 4.6(1)]). It was shown that if R contains the identity and M is a unitary R-module, then R(+)M is a Noetherian ring if and only if R is a Noetherian ring and M is finitely generated [1, Theorem 4.8] (or [5, Corollary 3.9]).

Corollary 8. Let R be a commutative ring and let M be an R-module. If R[X](+)M[X] is a Noetherian ring, then R contains the identity and M is a unitary R-module.

Proof. Suppose that R[X](+)M[X] is a Noetherian ring. Since R[X](+)M[X] is isomorphic to (R(+)M)[X], (R(+)M)[X] is a Noetherian ring. Hence by Corollary 5, R(+)M contains the identity.

Let (a,n) be the identity of R(+)M. For each r∈R, (a,n)(r,0)=(r,0); so ar=r. Thus a is the identity of R. Let m∈M. Then (a,n)(0,m)=(0,m); so am=m. Thus M is a unitary R-module.

The author sincerely thanks the referee for several valuable comments. This research was supported by Kyungpook National University Research Fund, 2023.