Let $K\subset L$ be a finite cyclic Galois extension with Galois group $G$. suppose $G$ is generated by $\sigma$ and the order of $G$ is $m$. Denote $\operatorname{tr}(x)=x+\sigma x+\cdots+\sigma^{m-1}x$ where $x\in L$. Then there is an exact sequence $L\xrightarrow {\sigma-1}L\xrightarrow {\operatorname{tr}}K \rightarrow 0$

There is an isomorphism $L\cong \mathbb{Z}G\otimes _{\mathbb{Z}}K$ as $G$ module. for free resolution $\mathbb{Z}G\xrightarrow {\sigma -1}\mathbb{Z}G\rightarrow \mathbb{Z}\rightarrow 0$ by tensor L and use the isomorphism,we have exact sequence $L\xrightarrow {\sigma-1}L\xrightarrow f K\rightarrow 0$.

I can't prove $f$ can be replaced by $\operatorname{tr}$. What I know is $\ker(\operatorname{tr})=\operatorname{Im}(\sigma -1)$ by Hilbert's Theorem 90. So how to show $\operatorname{tr}$ is epic?

Thank you in advance.


The trace map is non-zero (and so onto) due to Dedekind's lemma on linear independence of automorphisms. The automorphisms $\sigma^j$ for $0\le j\le m-1$ are linearly independent over $L$ and in particular $x\mapsto\sum_{j=0}^{m-1}\sigma^j(x)$ cannot be identically zero.


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