0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.