Let $A$ be a Noetherian local ring, $f:A \rightarrow A$ be a local ring morphism. Assume some power of $f$ is a flat morphism, must $f$ be flat as well?


Motivation: Kunz's theorem shows the result is true for a positive characterestic ring $A$ and its Frobenius morphism.



Iteration of a morphism and flatness
Yes. Assume $f^n$ is flat for some $n>1$. Then since $f^n$ is local, it is faithfully flat. For any $A$-module $M$, put $M\_1:=A\otimes\_{f,A} M$ and recursively $M\_i:=(M\_{i-1})\_1$. Let $u:E\to F$ be an injective $A$-module homomorphism. We need to show that $u\_1:E\_1\to F\_1$ is injective. Let $K$ be its kernel. We have a sequence $K\_{n-1}\to E\_n\to F\_n$ with zero composite, but by assumption $E\_n\to F\_n$ is injective, hence $K\_{n-1}\to E\_n$ is zero. Thus, $K\_{n}\to E\_{n+1}$ is also zero, but since $f^n$ is faithfully flat this implies that $K\to E\_{1}$ is zero. QED


