
MHB Master
#1
November 9th, 2015,
20:30
I am reading P.M. Cohn's book: Introduction to Ring Theory (Springer Undergraduate Mathematics Series) ... ...
I am currently focused on Section 2.2: Chain Conditions ... which deals with Artinian and Noetherian rings and modules ... ...
I need help to get started on Exercise 3, Section 2.2, page 65 ...
Exercise 3 (Section 2.2, page 65) reads as follows:
Any help will be much appreciated ...
Peter

November 9th, 2015 20:30
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MHB Master
#2
November 9th, 2015,
21:35
Here's one counterexample for you to chew on:
The endomorphism $\Bbb Z \to \Bbb Z$ given by:
$k \mapsto 2k$ is injective, but is not an automorphism.
Here is another counterexample:
Let $G= \{z \in \Bbb C: \exists k \in \Bbb N\text{ with }z^{2^k} = 1\}$.
This is an abelian group, and thus a $\Bbb Z$module.
Verify that $z \mapsto z^2$ is a surjective endomorphism, but not an automorphism.
Exploring WHY these two examples fail may give you some insight as to how to prove the theorem stated.

MHB Master
#3
November 10th, 2015,
03:04
Thread Author
Originally Posted by
Deveno
Here's one counterexample for you to chew on:
The endomorphism $\Bbb Z \to \Bbb Z$ given by:
$k \mapsto 2k$ is injective, but is not an automorphism.
Here is another counterexample:
Let $G= \{z \in \Bbb C: \exists k \in \Bbb N\text{ with }z^{2^k} = 1\}$.
This is an abelian group, and thus a $\Bbb Z$module.
Verify that $z \mapsto z^2$ is a surjective endomorphism, but not an automorphism.
Exploring WHY these two examples fail may give you some insight as to how to prove the theorem stated.
Hi Deveno,
Thanks for the help ...
Will just give you some ... pretty simple and thoroughly inconclusive ... thoughts regarding your first counterexample ...
You write:
"... The endomorphism $\Bbb Z \to \Bbb Z$ given by:
$k \mapsto 2k$ is injective, but is not an automorphism. ... "
Now ... some thoughts ...
$ \displaystyle \mathbb{Z}$ can be described as a right (or left) module over itself ... or can be considered a $ \displaystyle \mathbb{Z}\text{module}$ ... both these options are essentially the same and give rise to the same module ... (is that correct? ... I suspect it is trivially correct ...)
So consider $ \displaystyle Z$ as a right $ \displaystyle \mathbb{Z}\text{module}$ ... ...
Then $ \displaystyle f \ : \ \mathbb{Z} \longrightarrow \mathbb{Z}$ where $ \displaystyle f$ is such that $ \displaystyle kf = 2k$ ... ... (Note: Cohn writes mappings on the right ... ) ... is a $ \displaystyle \mathbb{Z}\text{linear mapping}$ or $ \displaystyle \mathbb{Z}\text{module homomorphism}$ ... since:
$ \displaystyle (x + y) f = xf + yf$ for all $ \displaystyle x,y \in \mathbb{Z}$
and
$ \displaystyle (xf)r = (xr)f$ for all $ \displaystyle x \in \mathbb{Z}$ and $ \displaystyle r \in \mathbb{Z}$
We know this is the case, since
$ \displaystyle (x + y) f = 2(x + y) = 2x + 2y = xf + yf $
and
$ \displaystyle (xf)r = (2x)r = 2(xr) = (xr)f$
and we can see that f is clearly injective ... ... BUT ... it is clearly NOT surjective (e.g nothing maps onto 1) ... ...
... ... thinking ... ...
If our module, instead of being $ \displaystyle \mathbb{Z}$, had actually been the submodule generated by 2, namely
$ \displaystyle M = 2 \mathbb{Z} = \{ 2r \  \ r \in \mathbb{Z} \} = \{ ... \ ... \ 2, 0, 2, 4, 6, \ ... \ ... \ \}$
... then f as defined above would be an automorphism ... ...
Indeed so would any submodule $ \displaystyle n \mathbb{Z}$ for $ \displaystyle n \in \mathbb{Z}^+$ ...
... ... BUT ... ... how to use this analysis to approach/solve the exercise ... I think I need further help ...
Note that I side mystery has developed ... seems that from the above that $ \displaystyle \mathbb{Z}$ is not Artinian ... I have the feeling that this should to easy to show ... even formally and rigorously ... but how does this follow from the definition of an Artinian module ... or even from Cohn's basic Corollary on Artinian modules ...
In summary, could you critique my thinking so far ... ... ?
... indeed there are 3 questions (which I suspect are rather basic/trivial) that come out of the above ...
(1) Is it correct that considering $ \displaystyle \mathbb{Z}$ as a ring over itself is the same as considering $ \displaystyle \mathbb{Z}$ as a $ \displaystyle \mathbb{Z}\text{module}$ ... ...? (mind you I think that this is trivially the case ... but just being sure ... )
(2) Could you please indicate why $ \displaystyle \mathbb{Z}$ is not Artinian ... proceeding only from the definition and perhaps also from Cohn's Corollary 2.3 ... ... ?
(3) Could you please indicate what I am missing regarding how the counterexample can lead to solving the exercise ... ... ?
Peter
*** EDIT ***
The above post refers to Cohn Corollary 2.3 on the characteristics of Artinian modules ... so I am providing that Corollary, as follow:
So that MHB readers can appreciate the definitions and context of Exercise 3, including Cohn's definition of an Artinian module, I am providing the Cohn's introduction to Section 2.2: Chain Conditions ...
Last edited by Peter; November 10th, 2015 at 03:13.

MHB Master
#4
November 10th, 2015,
08:20
No, even if we began with $2\Bbb Z$ the endomorphism $kf = 2k$ would not be surjective, since, for example, $6$ would not be in the image.
The reason is indeed that $\Bbb Z$ is nonArtinian.
You see, any nontrivial $\Bbb Z$submodule of $\Bbb Z$ (which is to say, any ideal, or in this case, subgroup, since we can regard $\Bbb Z$ as either a ring (and thus a $\Bbb Z$module over itself) or an abelian group (with a natural $\Bbb Z$action)) is infinite, and so the chain of submodules:
$\Bbb Z \supsetneq 2\Bbb Z \supsetneq 4\Bbb Z \supsetneq 8\Bbb Z \supsetneq \cdots$
just keeps "going and going", it never stabilizes.
This is always a possibility with "infinite things", you can't "exhaust them" in this way. One millionth of infinity is *still* infinity.
Now, suppose $M$ is Artinian, and $f: M \to M$ is an injective endomorphism.
What can you say about the chain $M \supseteq f(M) \supseteq f^2(M) \supseteq \cdots$?
(Hint: if $f$ is NOT surjective, this chain of inclusions is *strict*, by the injectivity of $f$....)
The Noetherian case is "dual" to this, and requires a similar trick (hint: look at the kernels instead of the images).
I don't want to be more "explicit" about this, because half the fun is the joy of discovery.

MHB Master
#5
November 11th, 2015,
03:08
Thread Author
Originally Posted by
Deveno
No, even if we began with $2\Bbb Z$ the endomorphism $kf = 2k$ would not be surjective, since, for example, $6$ would not be in the image.
The reason is indeed that $\Bbb Z$ is nonArtinian.
You see, any nontrivial $\Bbb Z$submodule of $\Bbb Z$ (which is to say, any ideal, or in this case, subgroup, since we can regard $\Bbb Z$ as either a ring (and thus a $\Bbb Z$module over itself) or an abelian group (with a natural $\Bbb Z$action)) is infinite, and so the chain of submodules:
$\Bbb Z \supsetneq 2\Bbb Z \supsetneq 4\Bbb Z \supsetneq 8\Bbb Z \supsetneq \cdots$
just keeps "going and going", it never stabilizes.
This is always a possibility with "infinite things", you can't "exhaust them" in this way. One millionth of infinity is *still* infinity.
Now, suppose $M$ is Artinian, and $f: M \to M$ is an injective endomorphism.
What can you say about the chain $M \supseteq f(M) \supseteq f^2(M) \supseteq \cdots$?
(Hint: if $f$ is NOT surjective, this chain of inclusions is *strict*, by the injectivity of $f$....)
The Noetherian case is "dual" to this, and requires a similar trick (hint: look at the kernels instead of the images).
I don't want to be more "explicit" about this, because half the fun is the joy of discovery.
Thanks for the help, Deveno ...
Yes ... of course ... you are correct regarding $ \displaystyle 2$$ \displaystyle \mathbb{Z}$ ... ... hmmm ... careless of me ... ... ...
Now, I take it that you mean $ \displaystyle f \circ f$ when you write $ \displaystyle f^2$ ... ...
You write:
" ... ... Now, suppose $M$ is Artinian, and $f: M \to M$ is an injective endomorphism.
What can you say about the chain $M \supseteq f(M) \supseteq f^2(M) \supseteq \cdots$? ... ... "
... hmmm ... reflecting ...
... not completely sure ... but since the image of an Rmodule homomorphism is a submodule, we have a descending chain of submodules ... :
$M \supseteq f(M) \supseteq f^2(M) \supseteq \cdots$ ... ... (*)
where $ \displaystyle f$ is an injective endomorphism ... and we have to show that f is surjective ... that is an automorphism ...
But ... since $ \displaystyle M$ is Artinian the descending chain of submodules (*) is either finite or becomes stationary at some point ... ...
Now if $ \displaystyle f$ is NOT surjective then the chain of submodules is strictly decreasing ... BUT if it is ALWAYS strictly decreasing then it must never become stationary ... which is not possible since $ \displaystyle M$ is Artinian ... so $ \displaystyle f$ must be surjective ... but given that $ \displaystyle f$ is now a surjective and injective endomorphism ... it must be an automorphism ... QED
BUT ... what if somehow a nonsurjective endomorphism reduces the set $ \displaystyle f^n (M)$ at some point to one element, say $ \displaystyle 0$ ... then $ \displaystyle f^{n+1} (M) = f^{n+1} (M) = f^{n+1} (M) \ ... \ ... = 0$ and the chain becomes stationary ... this seems intuitively possible if $ \displaystyle M$ is finite ... but also seems possible for endomorphisms $ \displaystyle f$ that reduce the set $ \displaystyle M$ very drastically ...
Not sure how you would actually formally and rigorously prove my rather informal argument for the establishment of $ \displaystyle f$ as an automorphism ... ... can you help ...?
NOTE: I have found it difficult to try out my ideas as I do not have a good set of examples of Artinian modules ... they seem a bit hard to come by ...
Hope that my above argument/analysis makes sense ... but if you could critique what I have said ... I would be most grateful ...
Must now start thinking about the second part of the exercise ... ...
Peter
Last edited by Peter; November 11th, 2015 at 03:14.

MHB Master
#6
November 11th, 2015,
08:11
You're getting confused.
Yes, it *is* possible (even in an Artinian module) for an endomorphism to "reduce" drasticallythe point is, such endomorphisms cannot be injective.
There is nothing wrong with your "informal argument", it's one way to prove it (proof by contradiction).
As for examples of Artinian modules, here's a few:
The trivial module $\{0\}$.
Any finite abelian group ($\Bbb Z$module).
Any finitedimensional vector space ($F$module).
Any field (considered as a ring).
The ring $F[t]/\langle t^n\rangle$, for any $n \in \Bbb N$.

MHB Master
#7
November 11th, 2015,
08:21
Thread Author
Originally Posted by
Deveno
You're getting confused.
Yes, it *is* possible (even in an Artinian module) for an endomorphism to "reduce" drasticallythe point is, such endomorphisms cannot be injective.
There is nothing wrong with your "informal argument", it's one way to prove it (proof by contradiction).
As for examples of Artinian modules, here's a few:
The trivial module $\{0\}$.
Any finite abelian group ($\Bbb Z$module).
Any finitedimensional vector space ($F$module).
Any field (considered as a ring).
The ring $F[t]/\langle t^n\rangle$, for any $n \in \Bbb N$.
Hi Deveno,
Very good to know my argument is OK ...
... Yes, take your point regarding my concerns ...
Thanks for the examples ... but ... with respect to a field being an Artinian module, I am assuming the ring (field) is viewed as a module over itself ... *** EDIT *** ... so the field $ \displaystyle \mathbb{Q}$ would be regarded as a module since $ \displaystyle \mathbb{Q}$ is a ring and could be viewed as a module over itself ...
Thanks again for all your patient help ... it is much appreciated ...
Peter
Last edited by Peter; November 11th, 2015 at 08:31.

MHB Master
#8
November 11th, 2015,
22:59
Thread Author
Originally Posted by
Deveno
No, even if we began with $2\Bbb Z$ the endomorphism $kf = 2k$ would not be surjective, since, for example, $6$ would not be in the image.
The reason is indeed that $\Bbb Z$ is nonArtinian.
You see, any nontrivial $\Bbb Z$submodule of $\Bbb Z$ (which is to say, any ideal, or in this case, subgroup, since we can regard $\Bbb Z$ as either a ring (and thus a $\Bbb Z$module over itself) or an abelian group (with a natural $\Bbb Z$action)) is infinite, and so the chain of submodules:
$\Bbb Z \supsetneq 2\Bbb Z \supsetneq 4\Bbb Z \supsetneq 8\Bbb Z \supsetneq \cdots$
just keeps "going and going", it never stabilizes.
This is always a possibility with "infinite things", you can't "exhaust them" in this way. One millionth of infinity is *still* infinity.
Now, suppose $M$ is Artinian, and $f: M \to M$ is an injective endomorphism.
What can you say about the chain $M \supseteq f(M) \supseteq f^2(M) \supseteq \cdots$?
(Hint: if $f$ is NOT surjective, this chain of inclusions is *strict*, by the injectivity of $f$....)
The Noetherian case is "dual" to this, and requires a similar trick (hint: look at the kernels instead of the images).
I don't want to be more "explicit" about this, because half the fun is the joy of discovery.
Hi Deveno,
... ... just thinking about the Noetherian case ... but have a few issues/problems ... hope you can help ...
You write:
" ... ... The Noetherian case is "dual" to this, and requires a similar trick (hint: look at the kernels instead of the images). ... ... "
A "dual" argument ... I think ... wold run like this:
$ \displaystyle f$ is a surjective endomorphism
$ \displaystyle \Longrightarrow $
there exists an ascending series of submodules as follows:
$ \displaystyle \text{ ker f } \subseteq \text{ ker } f^2 \subseteq \text{ ker } f^3 \subseteq \ ... \ ... \ $ ... ... ... (1)
If $ \displaystyle f$ is NOT injective then the inclusions in (1) are strict ... ...
Thus if $ \displaystyle f$ is not injective the chain in (1) never becomes stationary ... ...
BUT ... this contradicts the assumption that M is Noetherian ...
Thus $ \displaystyle f$ must be injective ... and thus is an automorphism ...
Problems/issues
Issue (i)
A little unsure of how to demonstrate (formally and rigorously) that (1) holds ... but suspect that it is due to
(a) if $ \displaystyle f$ is injective then $ \displaystyle \text{ ker } f = \{ 0_M \}$ and so equality holds and so (1) is true ...
(b) if $ \displaystyle f$ is not injective then $ \displaystyle \text{ ker } f \neq \{ 0_M \}$ and hence (I think) strict inclusion holds and so (1) is true ...
Issue (ii)
I am unsure how to demonstrate (formally and rigorously) that the proposition:
"If $ \displaystyle f$ is NOT injective then the inclusions in (1) are strict ... ... "
is true ... seems true from really simple cases but how do you formulate a proof ...
Can you help ... ?
Peter
Last edited by Peter; November 11th, 2015 at 23:08.

MHB Master
#9
November 13th, 2015,
07:59
Here is how such an argument *starts*:
Suppose $f$ is not injective. Then there exists an $m \neq 0$, with $m \in \text{ker }f$.
Since $f$ is by supposition surjective, $m = f(m')$ for some $m'$.
Now $f^2(m') = f(f(m')) = f(m) = 0$. Thus $m ' \in \text{ker }f^2$.
But $m' \not\in \text{ker }f$, since $f(m') = m \neq 0$, so $\text{ker }f \subsetneq \text{ker }f^2$.

MHB Master
#10
November 13th, 2015,
09:04
Thread Author
Originally Posted by
Deveno
Here is how such an argument *starts*:
Suppose $f$ is not injective. Then there exists an $m \neq 0$, with $m \in \text{ker }f$.
Since $f$ is by supposition surjective, $m = f(m')$ for some $m'$.
Now $f^2(m') = f(f(m')) = f(m) = 0$. Thus $m ' \in \text{ker }f^2$.
But $m' \not\in \text{ker }f$, since $f(m') = m \neq 0$, so $\text{ker }f \subsetneq \text{ker }f^2$.
Thanks Deveno ... most helpful ...
Peter