# A decreasing sequence of images of an endomorphisme

• A
Let ##M## be a left R-module and ##f:M \to M## an R-endomorphism.

Consider this infinite descending sequence of submodules of ##M##

##M \supseteq f(M) \supseteq f^2(M) \supseteq f^3(M) \supseteq \cdots (1)##

Can anybody show that the sequence (1) is strictly descending if ##f## is injective and ##f## is NOT surjective ?

So you have to prove that ## f^n(M) \neq f^{n+1}(M)## for all ##n > 0##, given that ##f## is injective and ##f## is not surjective.
Of course ##M \neq f(M)##.

Can someone help me with this?

fresh_42
Mentor
2021 Award
Let ##M## be a left R-module and ##f:M \to M## an R-endomorphism.

Consider this infinite descending sequence of submodules of ##M##

##M \supseteq f(M) \supseteq f^2(M) \supseteq f^3(M) \supseteq \cdots (1)##

Can anybody show that the sequence (1) is strictly descending if ##f## is injective and ##f## is NOT surjective ?

So you have to prove that ## f^n(M) \neq f^{n+1}(M)## for all ##n > 0##, given that ##f## is injective and ##f## is not surjective.
Of course ##M \neq f(M)##.

Can someone help me with this?
You could prove the alternative definition of injectivity:
##f\, : \,M \longrightarrow N## is injective if and only if for all sets ##M'## and all ##g,h\, : \, M' \longrightarrow M## holds ##f\circ g = f \circ h## implies ##g=h##.
and apply it to the given situation.

Thank you for your suggestion, but I do not see how to apply this. Can you give me one more hint, please?

fresh_42
Mentor
2021 Award
Usually we have ##f(x)=f(y) \Longrightarrow x=y##, i.e. a left cancellation of ##f##. The definition I gave says, this is also true for mappings. Given any functions of sets, then ##f\,g=f\,h \Longrightarrow g=h## is the same as injectivity.

Now assume ##f^n(M)=f^{n+1}(M)##. This means ##f\circ f^{n-1} = f \circ f^n ## and by the alternative definition ##f^{n-1} = f^n.## You can either assume ##n## to be minimal, proceed by induction or simply say etc. to achieve a contradiction, because you end up with ##\operatorname{id}_M=f## which is surjective.

So all you have to show is, that the two definitions are equivalent.

Sorry, see the next post.

Last edited:
I am sorry. fresh_42, but i found a glitch. I think your answer is not correct. ##f^n(M)=f^{n+1}(M)## does not mean that ##f^n(x)=f^{n+1}(x)## for all ##x \in M##. It means that the two sets ##f^n(M)=\{f^n(x) | x \in M \}## and ##f^{n+1}(M)=\{f^{n+1}(x) | x \in M \}## are equal. So, one cannot conclude that ##f^n=f^{n+1}##.
Sorry I was too quickly.

fresh_42
Mentor
2021 Award
I am sorry. fresh_42, but i found a glitch. I think your answer is not correct. ##f^n(M)=f^{n+1}(M)## does not mean that ##f^n(x)=f^{n+1}(x)## for all ##x \in M##. It means that the two sets ##f^n(M)=\{f^n(x) | x \in M \}## and ##f^{n+1}(M)=\{f^{n+1}(x) | x \in M \}## are equal. So, one cannot conclude that ##f^n=f^{n+1}##.
Sorry I was too quickly.
But isn't the idea the same? Assume ##f^n(M)=f^{n+1}(M)## as sets. Then for all ##x \in M## there is an ##y\in M## such that ##f^n(x)=f^{n+1}(y)##. This leads to ##x=f(y)## and again surjectivity.

I do not see it. ##f^n(x)=f^{n+1}(y)## does not lead to ##f\circ f^{n-1} = f \circ f^n## and so on.

You mean, because ##f^n## is injective ? Then ##f^n(x)=f^nf(y)## and ##x=f(y)##.

fresh_42
Mentor
2021 Award
I do not see it. ##f^n(x)=f^{n+1}(y)## does not lead to ##f\circ f^{n-1} = f \circ f^n## and so on. Can you be more specific, please?
Answer the following questions, one by one.

What does ##f^n(M)=f^{n+1}(M)## mean in terms of elements?
If we consider this elementwise equation, can we cancel an ##f\,##?
Is this cancellation repeatable?
With which equation will we end up?
What does this last equation express?

Our posts crossed, I edited my former post.

fresh_42
Mentor
2021 Award
Our posts crossed, I edited my former post.
Yes, but I do not mean that ##f^n## is injective. This would have to be proven first. But we know that ##f## is injective, so we can write it the other way around: ##f^n(x)=f(f^{n-1}(x))## and use injectivity of ##f^1##.

I am interested in what you meant.

But ##f^n## is injective is easy to prove, if ##h:A \to B## is injective and ##g:B \to C## in injective, then ##g \circ h:A \to C## is injective.
If it is this easy, why didn't I see it ?

fresh_42
Mentor
2021 Award
If it is this easy, why didn't I see it ?
I assume because you were caught in the ring theoretic view of things and the other definition forced you to look more generally on it. Btw., do you know an example of such a chain?

##\mathbb{Z}## is not artinian.

Define ##f:\mathbb{Z} \to \mathbb{Z}:n \mapsto 2n##.

Then the sequence becomes
##(2) \supset (4) \supset (8) \supset \cdots \supset (2^n) \supset \cdots##

decreasing and not terminating

Thanks for showing me the way.

• fresh_42
mathwonk
Homework Helper
If f:S-->T is injective, and U is a proper subset of S, then f(U) is strictly smaller than f(S). QED.

Proof ?

mathwonk
Homework Helper
Lemma: If f:S-->T is injective, and U is a proper subset of S, then f(U) is a proper subset of f(S).

proof: by definition of injectivity, if U is a subset of S, then f(S) is the disjoint union of f(U) and f(S-U), and since U is a proper subset of S, both S-U and f(S-U) are non empty, hence f(U) is a proper subset of f(S), since it is missing the points of f(S-U).

More simply perhaps, if x is a point of S which is not in U, then f(x) is a point of f(S) which is not in f(U). I.e. since f is injective, and since all points of U are different from x, no point of U can map to the point f(x). I hope it is clear this follows immediately from the meaning of injectivity.

Corollary: Since by hypothesis in your problem we have f not surjective, hence f(M) is a proper subset of M, it follows that f(f(M)) is a proper subset of f(M). Continuing, we have the result you want, i.e. then also f(f(f(M))) is a proper subset of f(f(M)), .......etc.

Last edited:
"by definition of injectivity, if U is a subset of S, then f(S) is the disjoint union of f(U) and f(S-U), and since U is a proper subset of S, both S-U and f(S-U) are non empty, hence f(U) is a proper subset of f(S), since it is missing the points of f(S-U)"
I really do not understand what you are saying here.Your second proof is easier, but why not write it down like this:

##U \subset S## proper, so there is an ##x \in S## such that ##x \notin U##. Suppose ##fx \in fU##, then there is a ##y \in U## such that ##fx = fy##, by the injectivity of ##f## we have: ##x = y \in U##, contradiction, so ##fU \subset fS## proper

It is the same but much easier to read.

mathwonk