In a homework dealing with Galois' theory, I am asked to prove the following standard statement, known as the fixed subfield formula:

Theorem. Let $L$ be a field and $G$ be a finite subgroup of $\textrm{Aut}(L)$, then: $$\left[L:L^G\right]=|G|\textrm{ and }G=\textrm{Aut}_{L^G}(L).$$

However, I am bound to use very specific tools that I shall introduce to you. First, let me give you a:

Definition 1. Let $L$ be a field, $G$ be a finite subgroup of $\textrm{Aut}(L)$ and $V$ be a $L$-vector space. One says that a map $\cdot:G\times V\mapsto V$ is a $L^G$-structure of $V$ if and only if:

(i) $\cdot:G\times V\rightarrow V$ is a group action of $G$ on $V$.

(ii) For all $\sigma\in G$,$V\ni v\mapsto\sigma\cdot v\in V$ is a $L^G$-linear map.

(iii) For all $(\lambda,v)\in L\times V$, $\sigma\cdot\lambda v=\sigma(\lambda)(\sigma\cdot v)$.

From there, I have shown the:

Proposition 1. Let $L$ be a field, $G$ be a finite subgroup of $\textrm{Aut}(L)$ and $V$ be a $L$-vector space.

(i) $V^G$ is a $L^G$-subvector space of $V$.

(ii) Let $(v_i)_{i\in\{1,\ldots,n\}}\in(V^G)^n$ be $L^G$-linearly independent in $V^G$, then $(v_i)_{i\in\{1,\ldots,n\}}$ is $L$-linearly independent in $V$.

Besides, I have also derived the two following constructions:

Proposition 2. Let $L$ be a field and $G$ be a finite subgroup of $\textrm{Aut}(L)$.

(i) Let $V$ be a finite-dimensional vector space over $L$ and $(v_i)_{i\in I}$ a $L$-basis of $V$, then $V$ is equipped with a $L^G$-structure given by: $$\left\{\begin{array}{ccc}G\times V & \rightarrow & V\\\displaystyle\left(\sigma,\sum_{i\in I}\lambda_iv_i\right) & \mapsto & \displaystyle\sum_{i\in I}\sigma(\lambda_i)v_i\end{array}\right..$$

(ii) The $L$-vector space $\textrm{Map}(G,L)$ of the maps from $G$ to $L$ is equipped with a $L^G$-structure given by: $$\left\{\begin{array}{ccc}G\times\textrm{Map}(G,L) & \rightarrow & \textrm{Map}(G,L)\\(\sigma,\Phi) & \mapsto & G\ni\tau\mapsto\sigma(\Phi(\sigma^{-1}\circ\tau))\end{array}\right..$$

Thus far, here my:

Theorem's proof: Let $m:=\left[L:L^G\right]$, $n:=|G|$ and assume by contradiction that $m<n$. Let $(x_i)_{i\in\{1,\ldots,m\}}\in L^m$ be a $L^G$-basis of $L$ and $\sigma_1,\ldots,\sigma_n$ be the distinct elements of $G$. Since $m<n$ there exists $(y_i)_{i\in\{1,\ldots,n\}}\in L^n$ a nonzero solution to the following homogeneous system: $$\forall i\in\{1,\ldots,m\},\sum_{i=1}^n\sigma_j(x_i)Y_i=0_L.$$ Let $x\in L$, there exists $(\lambda_i)_{i\in\{1,\ldots,m\}}\in(L^G)^m$ such that: $$x=\sum_{i=1}^m\lambda_ix_i.$$ One derives from the construction of $(y_i)_{i\in\{1,\ldots,n\}}$: $$\sum_{i=1}^n\sigma(x)y_i=\sum_{i=1}^m\lambda_i\sum_{j=1}^n\sigma_i(x_j)y_i=0_L.$$ Therefore, one has: $$\sum_{i=1}^n\sigma y_i=0_{\textrm{Map}(G,L)}.$$ Since $(y_i)_{i\in\{1,\ldots,n\}}$ is nonzero and $\sigma_1,\ldots,\sigma_n$ are pairwise distinct, here a contradiction to Dedekind's lemma. Therefore, one has $m\geqslant n$. $\Box$

I struggle to prove that $m\leqslant n$, I tried to use the $L^G$-structure on $\textrm{Map}(L,G)$ from proposition $2.$ (ii) along with the proposition $1.$ (ii) but I didn't manage to succeed. Any hints will be appreciated.

  • $\begingroup$ Please correct the Theorem. Should the "and" be an "iff", or the $G$ not a general subgroup? $\endgroup$ – darij grinberg Nov 28 '15 at 1:27
  • $\begingroup$ Something is wrong. I can take $G$ to be any finite subgroup of $\mathrm{Aut}_K(L)$ including the trivial group. $\endgroup$ – Batominovski Nov 28 '15 at 1:34
  • 1
    $\begingroup$ It should be $[L : Fix(G)] = |G|$ I think $\endgroup$ – Adam Jaffe Nov 28 '15 at 1:44
  • $\begingroup$ My bad, $K$ isn't any field, I'm going to fix that. Thank you! $\endgroup$ – C. Falcon Nov 28 '15 at 1:44

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