Let $L$ be perfectoid field of characteristic $p$ and $L'$ be a finite extension of $L$. Then I want to prove the trace map $\text{Tr}\_{L'/L}: m\_{L'}\rightarrow m\_L$ is surjective. I find a proof in Kedlaya' paper "On categories of $(\phi,\Gamma)$-modules", but I can't understand it. The proof is as follows:


**The surjectivity is equivalent to the fact that the cokernel is annihilated by all of $m\_L$.(Why?)**


**Since $L'/L$ is finite separable extension, then the annihilator of the cokernel is nonzero.(Why?)**


**Since the annihilator is nonzero and closed under taking $p$-th roots, so the cokernel is annihilated by all of $m\_L$, so this map is surjective.(This is easy to see).**


Could someone help me figure out these two questions in this proof?


Thanks!



perfectoid field of characteristic $p$
Write $I$ for the image of $\operatorname{Tr}\_{L'/L} \colon \mathfrak m' \to \mathfrak m$, which is an ideal because $\operatorname{Tr}\_{L'/L}$ is $\mathcal O\_L$-linear.


(1) The first sentence is just the non-discreteness of the valuation. This forces $\mathfrak m^2 = \mathfrak m$, so if $I \subseteq \mathfrak m$ with $\mathfrak m^2 \subseteq I$, we must have $I = \mathfrak m$.


(2) The second sentence follows since $\operatorname{Tr}\_{L'/L} \colon L' \to L$ is nonzero if (and only if) $L'/L$ is separable (see e.g. Tag [0BIL](https://stacks.math.columbia.edu/tag/0BIL)). Then the same goes for $\operatorname{Tr}\_{L'/L} \colon \mathfrak m' \to \mathfrak m$, so there exists a nonzero $x \in I$. Then $x\mathfrak m \subseteq (x) \subseteq I$, i.e. $x \in \operatorname{Ann}(\mathfrak m/I)$.


