Let $R$ be a commutative Noetherian ring, and $\phi: R \to R$ be a ring homomorphism. For an $R$-module $M$, let $^{\phi}M$ be the $R$-module defined via restriction of scalars via $\phi$, i.e., as abelian groups, $M$ and $^{\phi}M$ are the same, but the $R$-action on $^{\phi}M$ is given by $r\*x:=\phi(r)x, \forall r \in R, x \in M$. For any $R$-modules $M,N$, it is clear that $\text{Hom}\_R(M,N)\subseteq \text{Hom}\_R(^{\phi}M, ^{\phi}N)$. In particular, any exact sequence of $R$-modules $M\xrightarrow{f} N \xrightarrow{g} L$ gives rise to an exact sequence of $R$-modules $^{\phi}M\xrightarrow{f} \space^{\phi}N \xrightarrow{g} \space ^{\phi}L$.


Now I want to apply the above scenario to the case where $(R,\mathfrak m,k)$ is a prime characteristic, $F$-finite, local Cohen--Macaulay ring of dimension $1$, $\phi$ is some $e$-th iteration of the Frobenius map, $M=R$, and $f$ is the multiplication by some non-zero-divisor $x\in \mathfrak m$. If I am not mistaken, then $^{\phi}R$ is exactly what is usually written as $F^e\_\*R$. The exact sequence $0\to R \xrightarrow{x} R\to R/xR\to 0$ then gives rise to exact sequence $0\to F^e\_\* R \xrightarrow{x} F^e\_\* R \to F^e\_\*(R/xR)\to 0$ implying $F^e\_\*R/xF^e\_\*R \cong F^e\_\*(R/xR)$. Now, $R/xR$ is an Artinian local ring, so $\mathfrak m^{p^e}(R/xR)=0$ for all large enough $e$, hence $\mathfrak m F^e\_\*(R/xR)=0$ for all large enough $e$, hence $F^e\_\* (R/xR)$ is a finite dimensional $k$-vector space. So, $F^e\_\*R/xF^e\_\*R \cong F^e\_\*(R/xR)$ is a finite dimensional $k$-vector space for large enough $e$. Now, take $R$ to be $F$-split, for instance take $R=\mathbb F\_p[[u,v]]/(uv)$, and $x:=u+v$ a non-zero-divisor on $R$. Then, $R$ is a direct summand of $F^e\_\*R$ for all $e>0$, hence $R/xR$ is a direct summand of $F^e\_\*R/xF^e\_\*R \cong F^e\_\*(R/xR)$, but this is a finite dimensional $k$-vector space for large enough $e$, hence $R/xR$ is annihilated by $\mathfrak m$, i.e., $\mathfrak m \subseteq xR$, hence $\mathfrak m=xR$, so $R$ must be regular! But clearly, $R$ is not regular!!


My question is: Where am I going wrong in the above argument?



Finding the mistake in an argument concerning $F$-finite $F$-split local Cohen--Macaulay ring of dimension $1$
@uno, you are correct in the comments above, $F^e\_\*(-)$ is not $R$-linear. In particular, the mistake is the line right here:



> 
> The exact sequence $0\to R \xrightarrow{x} R\to R/xR\to 0$ then gives rise to exact sequence $0\to F^e\_\* R \xrightarrow{x} F^e\_\* R \to F^e\_\*(R/xR)\to 0$ implying $F^e\_\*R/xF^e\_\*R \cong F^e\_\*(R/xR)$.
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> 
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You have to apply the functor $F^e\_\*$ to the map $\cdot x$ as well. In particular, what you get is an exact sequence:


$$0\to F^e\_\* R \xrightarrow{F^e\_\* (\cdot x)} F^e\_\* R \to F^e\_\*(R/xR)\to 0.$$


The $F^e\_\* (\cdot x)$-map is not the same as multiplication by $x$. Multiplication by $x$ is the same as the $F^e\_\* (\cdot x^{p^e})$-map though.


