A decreasing sequence of images of an endomorphisme

[steenis]

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]

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.

 

[steenis]

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

 

[fresh_42]

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.

 

[steenis]

Sorry, see the next post.

 

[steenis]

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]

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.

 

[steenis]

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]

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?

 

[steenis]

Our posts crossed, I edited my former post.

 

[fresh_42]

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 $f^1$.

 

[steenis]

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]

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?

 

[steenis]

$\mathbb{Z}$ 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.

 

[mathwonk]

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

 

[steenis]

Proof ?

 

[mathwonk]

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.

 

[steenis]

"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]

well done. but i am puzzled why you had to ask how to prove this, since you did it so nicely. did you try it before asking?

edit: I have had an idea as to why this proof was not obvious and I suggest the following:

lemma: a function f:S-->T is injective if and only if whenever U,V are disjoint subsets of S, then f(U) and f(V) are disjoint subsets of T.

Try proving this. I am sure you can do it easily. This statement seemed obvious to me, and I more or less took it for granted in giving a brief solution to this problem, but for someone just meeting the notion of injectivity, it might not be. It may help to draw a picture of an injective function, which just faithfully embeds the domain into the range. I think this more global view of injectivity is more useful then the bare definition involving only two (arbitrary) domain points.