Intereting Posts

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Let $G=\{\sigma_1=1, \sigma_2, \dots,\sigma_n\}$ be a subgroup of automorphisms of a field $K$ and let $F$ be the fixed field. Then $[K:F]=|G|$

Why is this always true? I thought this would be true only if $K$ was the Galois extension of $F$ and $G$ was $\text{Aut}(K/F)$

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- showing that $n$th cyclotomic polynomial $\Phi_n(x)$ is irreducible over $\mathbb{Q}$
- Galois group of the product of two fields being the product of Galois groups?
- How to determine the density of the set of completely splitting primes for a finite extension?

- Where can I find the paper by Shafarevich on the result of the realization of solvable groups as Galois groups over $\mathbb{Q}$?
- Finding a unique subfield of $\mathbb{Q}(\zeta)$ of degree $2$?
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- Irreducible solvable equation of prime degree
- Why does the elliptic curve for $a+b+c = abc = 6$ involve a solvable nonic?
- Finding Galois group of $x^6 - 3x^3 + 2$
- Why does an irreducible polynomial split into irreducible factors of equal degree over a Galois extension?
- Irreducible factors of a polynomial in a Galois extension
- Resolvent of the Quintic…Functions of the roots

Let $E$ be a field, $G$ be a finite subgroup of $\textrm{Aut}(E)$ and $F:=\textrm{Fix}(G)$. The result you are trying to prove is usually called fixed subfield formula and relies on the Artin’s lemma, that is:

Lemma 1.$[E:F]\leqslant |G|$.

**Proof.** A proof can be found in the Jacobson’s *Basic Algebra I*:

$\Box$

To establish the fixed subfield formula, one has now to show that:

Lemma 2.$[E:F]\geqslant|G|$.

**Proof.** Let $m:=\left[E:F\right]$, $n:=|G|$ and assume by contradiction that $m<n$. Let $(x_i)_{i\in\{1,\ldots,m\}}\in E^m$ be a $F$-basis of $E$ and $\sigma_1,\ldots,\sigma_n$ be the distinct elements of $G$. Since $m<n$ there exists $(y_i)_{i\in\{1,\ldots,n\}}\in E^n$ a nonzero solution to the following homogeneous system: $$\forall i\in\{1,\ldots,m\},\sum_{i=1}^n\sigma_j(x_i)Y_i=0_E.$$

Let $x\in E$, there exists $(\lambda_i)_{i\in\{1,\ldots,m\}}\in(E^G)^m$ 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: $$\sum_{i=1}^n\sigma y_i=0_{\textrm{Map}(G,E)}.$$

Since $(y_i)_{i\in\{1,\ldots,n\}}$ is nonzero and $\sigma_1,\ldots,\sigma_n$ are pairwise distinct, here is a contradiction to Dedekind’s lemma (see Lemma $1.1.$ at p$.1$). $\Box$

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