# Surjective ring morphism $f:R\to R$ satisfies Ker$(f^{n+1})\subset$ Ker$(f^n)$ then $f$ is injective.

As in the title, the set-up of the problem is as follows: $$f: R\to R$$ is a surjective ring homomorphism and $$R$$ is a commutative ring. Suppose that for some $$m\in \mathbb{N}$$, Ker$$(f^{m+1})\subset$$ Ker$$(f^m)$$. Prove that $$f$$ is injective.

Here are my thoughts so far: We know by the first isomorphism theorem (since $$f$$ is surjective) that there is an isomorphism $$\phi_n: R \to R/\text{Ker}(f^n)$$ for any $$n\in \mathbb{N}$$. Now consider the map $$R \longrightarrow^{f^{m+1}} R \longrightarrow^{\pi_m}R/\text{Ker}(f^n)\longrightarrow^{\phi_m^{-1}} R.$$ Now the above composition is an isomorphism by our hypotheses and I would like to conclude that the map agrees with $$f$$ but I don't see why this should be true (in fact I know that it shouldn't be in general but I feel that I am on the right path).

I would appreciate a hint or some guidance on how to make my solution more complete. This is not a HW question (it is a problem on this practice qualifying exam).

I think it might be more straightforward to work from first principles. While I think your solution could work, I find it easier just to use the definition of the kernel. The key is that your assumption gives you an inclusion $$\text{ker}(f^{m+1})\subset\text{ker}(f^m)$$, but there is always the reverse inclusion $$\text{ker}(f^m)\subset\text{ker}(f^{m+1})$$ since $$f$$ is a ring homomorphism and must map 0 to 0. When you have these two inclusions together, you get that $$f^m(R)\xrightarrow{f}f^{m+1}(R)$$ is not just surjective but also injective. Once you convince yourself of this, try to conclude that if this restriction of $$f$$ is injective then so is $$f$$ itself.
By assumption, $$\text{ker}(f^{m+1}) = \text{ker}(f^m)$$, so the restriction $$f|_{f^m(R)}$$ is injective since $$f\circ f^m(x) = 0$$ implies that $$f^m(x) = 0$$ for every $$x\in R$$. However, $$f$$ is surjective, so all of its iterates must be surjective as well. Then the restriction $$f|_{f^m(R)}$$ is actually just $$f$$ itself. Therefore, $$f$$ is injective.