Why is : $ [ K \ : \ K^G ] = | G | $?

Question

Let $ G $ be a finite group acting as automorphisms of a field $ K $.

Why is : $$ [ K \ : \ K^G ] = | G | $$ ?

Thanks in advance for the help.

Answer 1

You can get this by proving inequalities in both directions. Let $ K^G = F $ throughout the post.

Lemma 1. $ [K : F] \leq |G| $.

Proof. Let $ |G| = n $. Pick any $ n+1 $ elements of $ K $ as $ \alpha_k $, and consider the $ n \times (n+1) $ matrix $ A $ given by $ A_{ij} = \sigma_i(\alpha_j) $, where the $ \sigma_i $ are all automorphisms in $ G $. This matrix has more columns than rows, therefore its columns are linearly dependent over $ K $. We will show that we can refine this linear dependence relation to be over $ F $ as follows. Among the nonzero vectors in the kernel of $ A $, pick one with minimal number of nonzero entries, say $ v $. We may assume that the leading entry of $ v $ is $ 1 $ by permuting and multiplying by an element of $ K $ if necessary. Since $ G $ acts on $ A $ by permuting its rows, it is easy to see that $ \sigma(v) $ is also in the nullspace for any $ \sigma \in G $, thus so is $ \sigma(v) - v $. But this last vector has less nonzero entries than $ v $, thus this vector is actually zero and $ \sigma(v) = v $. Since this holds for every $ \sigma \in G $, the entries of $ v $ are in $ K^G = F $, concluding the proof.

Lemma 2. $ [K : F] \geq |G| $.

Proof. Let $ \beta_i $ be a basis of $ K/F $, we know that a finite basis exists thanks to Lemma 1. Consider the matrix $ A_{ij} = \sigma_i(\beta_j) $, where the $ \sigma_i $ are all automorphisms in $ G $. By linear independence of characters, the rows of this matrix are linearly independent, therefore this matrix must have more columns than rows. The result follows.