In this paper we study the transfer of the property of Hopfian modules between the right module and some of its extension classes. Namely, under certain conditions, we show that: is a Hopfian right module if and only if the skew generalized power series module is a Hopfian right module.
Hopfian module; Hopfian ring; Skew generalized power series module
primary, 06F05, 16W60; secondary, 13E10
Throughout this paper denotes an associative ring not necessarily commutative with the identity and a unitary right module. As it has been noted by Hiremath [3], the concept of Hopfian groups was introduced by Baumslag [1]. In fact, the study of endomorphism rings of various rings and modules has been a topic of keen interest since the end of the nineteen sixties when injectivity and its variants began to flourish. In 1986, Hiremath introduced the concept of the Hopfian module as follows: A right module is called Hopfian if any surjective endomorphism of is an isomorphism. The term “Hopfian” is said to be in honor of Heinz Hopf and his use of the concept of the Hopfian group in his work on fundamental groups of surfaces. Any noetherian module is Hopfian and if is a right noetherian ring, then every finitely generated module is a Hopfian module. Also, a simple ring is Hopfian, since the kernel of any endomorphism is an ideal, which is necessarily zero in a simple ring.
The module is Hopfian if and only if is a directly finite ring. Symmetrically, these two are also equivalent to the left module being Hopfian. The full linear ring of a countable dimensional vector space is a Hopfian ring which is not Hopfian as a module, since it only has three ideals, but it is not directly finite.
Varadarajan [5] and [6] showed that the right module is Hopfian if and only if the right module is Hopfian.
The motivation of this paper is to investigate how the property of Hopfian modules behaves under passage to the skew generalized power series modules.
In this section we extend the results of [9] to the skew generalized power series modules.
Let be an ordered commutative monoid. Unless stated otherwise, the operation of will be denoted additively, and the identity element by 0. Recall that is artinian if every strictly decreasing sequence of elements of is finite and that is narrow if every subset of pairwise orderincomparable elements of is finite. The following construction is due to Zhongkui [10]:
Let be a strictly ordered monoid (that is, if and , then ), a ring and a monoid homomorphism. Consider the set of all maps whose support () is artinian and narrow.
For every and , let

It follows from ([4], 4.1) that is a finite set.
This fact allows us to define the operation of multiplication (convolution) as follows:

and if . With this operation and pointwise addition becomes a ring, which is called the ring of skew generalized power series with coefficients in and exponents in .
In [8], Zhao and Jiao generalized this construction to obtain the skew generalized power series modules over skew generalized power series rings, as follows:
Let be a right module, let be the set of all maps such that supp is artinian and narrow. With pointwise addition, is an abelian additive group. For each and , the set

is finite (see [9], Lemma 1). This allows us to define the scalar multiplication of the elements of by scalars from as follows:

and if . With this operation and pointwise addition, one can easily show that is a right module, which is called the module of skew generalized power series with coefficients in and exponents in .
For every if we set , the identity map of , then is the ring of generalized power series in the sense of Ribenboim [4] and is the untwisted module of generalized power series in the sense of [7].
For any we associated the map defined by:

For any and , we define a map by:

If is a strictly totally ordered monoid, then is a nonempty wellordered subset of , for every , and we denote by the smallest element of . Also, is a nonempty wellordered subset of , for every , and we denote by the smallest element of .
The following are required in the sequel.
A monoid is called finitely generated if there exists a finite subset of such that .
If is a finitely generated monoid, then every ideal of is finitely generated, so every strictly increasing sequence of ideals is finite.
The following lemma is crucial in developing the proof of the main result.
Let be a strictly totally ordered monoid which is finitely generated and satisfies the condition that for every . Assume that for each . Then where and .
For each element , we denote by the subset . Set and . Suppose that . Define via

Set . If , then . Thus

For any with , we have

For each pair , we have . Then and . Thus and . Hence . It follows that , a contradiction. So . Since , we have

It follows that

If , then , a contradiction with the assumption that . Thus .
Denote and . Then . Since

we have

It is clear that . If , then there exists such that . Thus

a contradiction. Hence and . Then . Suppose that for a positive integer , we have found such that

where

We set and . Define via

Let . Hence

If , then

Thus

which implies that there exists such that . Thus , a contradiction. Now, suppose that . Denote . Since , we have

For any with and for every , we have . Then and , and hence . It follows that

So . Thus

Suppose that . Then there exists such that

Let for some . Then

Note that, since and , we see that

and

Thus

which implies that , a contradiction. Hence . Now, we have the infinite strictly increasing sequence of ideals of

a contradiction with Lemma 1. Therefore . ■
Now, we are able to prove the main result of this paper.
Suppose that is a strictly totally ordered monoid which is finitely generated and satisfies the condition that for every . Assume that for each . Then is a Hopfian right module if and only if is a Hopfian right module.
Suppose that is a Hopfian right module. Let be any surjective homomorphism. We want to prove that is injective to be an isomorphism. Define via . Now, define via .
(1) is an homomorphism: For any and , we have

Since for every , we get and so

(2) is a surjective map: For any , there exists such that since is surjective. Let . Then

So and using Lemma 2, . Hence

Hence is a surjective homomorphism which must be an isomorphism, since is a Hopfian right module. To prove that is injective, let be such that . It follows that . Then , since is an isomorphism. Suppose that and for any with . We show that . Define via

Thus , for any , and it follows that . Using Lemma 2, we have . Hence

Consider the following computation, for any

It follows that . Thus

which implies that , since is an isomorphism. Hence for any , and so . Therefore is an isomorphism and is a Hopfian right module.
Conversely, suppose that is a Hopfian right module. Let be any surjective homomorphism. We want to prove that is injective.
Define via for any and . We show that is an isomorphism.
(1) is an homomorphism: For any and , we set

Then clearly . Consider the following computation, for any

Thus .
(2) is a surjective map: For any and any , there exists an element such that , since is a surjective map.
Define via

Clearly, , which implies that is an artinian and narrow subset of and thus .
If , then

If , then

Thus . Hence is a surjective homomorphism which must be an isomorphism, since is a Hopfian right module. To prove that is injective, let be such that . Then for any ,

Thus and so , since is an isomorphism. Therefore is an isomorphism and is a Hopfian right module. ■
If we set , for every , we get the following result as a corollary.
Suppose that is a strictly totally ordered monoid which is finitely generated and satisfies the condition that for every . Then is a Hopfian right module if and only if is a Hopfian right module.
Published on 07/10/16
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