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Let $L|K$ be a finite and separable field extension of degree $n$, and $\sigma_i:L\to\bar{K},\;i=1,\ldots,n$ the (distinct) $n$ $K$-embeddings of $L$ into some algebraic closure $\bar{K}$ of $K$. By definition, if $\alpha\in K$, then $\sigma_i(\alpha)=\alpha,\;\forall\; i$. Now, suppose that $\sigma_i(\lambda)=\lambda,\;\forall\; i$ for some $\lambda\in L$. Then, is it true that $\lambda\in K$?

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The subextension of the algebraic closure generated by $\lambda$ and its images under all embeddings is a finite, normal and separable extension of K, so this follows from the usual theory of Galois extensions.

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  • $\begingroup$ Let me see if I understood. Denoting by $\lambda_i$ the image of $\lambda$ under $\sigma_i$, we have $E=K(\lambda_1,\ldots,\lambda_n)\subset\bar{K}$ and $E|K$ is a Galois extension. On the other hand $\lambda_i=\lambda,\;\forall\; i$, so $E=K(\lambda)\subset L$ and $E^G=K$, where $G=\text{Aut}(E|K)\subset\{\sigma_1,\ldots,\sigma_n\}$. And since $\lambda\in E^G$, we have $\lambda\in K$ and therefore $E=K$. Is everything right? $\endgroup$ – Larara Feb 11 '15 at 5:14
  • $\begingroup$ Yes. You do have to check all the claims you made, of course :-) $\endgroup$ – Mariano Suárez-Álvarez Feb 11 '15 at 5:32
  • $\begingroup$ By the way, nothing, nothing, nothing is gained from writing $\forall$ instead of simply «for all» and it really comes out much more pro to spell it out (except in very specific circumstances, none of which apply to the usage of that symbol in this page) :-) $\endgroup$ – Mariano Suárez-Álvarez Feb 11 '15 at 5:36
  • $\begingroup$ Good to know, and thank you! $\endgroup$ – Larara Feb 11 '15 at 5:45

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