Editing Normal basis (section)

== Proof for the case of infinite fields ==
Suppose <math>K/F</math> is a finite Galois extension of the infinite field ''F''. Let {{nowrap|1=[''K'' : ''F''] = ''n''}}, <math>\text{Gal}(K/F) = G =\{\sigma_1...\sigma_n\}</math>, where <math>\sigma_1 = \text{Id}</math>. By the [[primitive element theorem]] there exists <math>\alpha \in K</math> such <math>i\ne j\implies \sigma_i(\alpha)\ne\sigma_j(\alpha)</math> and <math>K=F[\alpha]</math>. Let us write <math>\alpha_i = \sigma_i(\alpha)</math>. <math>\alpha</math>'s (monic) minimal polynomial ''f'' over ''K'' is the irreducible degree ''n'' polynomial given by the formula
<math display="block">\begin {align}
f(X) &= \prod_{i=1}^n(X - \alpha_i)
\end {align}</math>
Since ''f'' is separable (it has simple roots) we may define 
<math display="block">
\begin {align}
g(X) &= \ \frac{f(X)}{(X-\alpha)f'(\alpha)}\\
g_i(X) &= \ \frac{f(X)}{(X-\alpha_i) f'(\alpha_i)} =\ \sigma_i(g(X)).
\end {align}
</math>
In other words,
<math display="block">\begin {align}
g_i(X)&= \prod_{\begin {array}{c}1 \le j \le n \\ j\ne i\end {array}}\frac{X-\alpha_j}{\alpha_i - \alpha_j}\\
g(X)&= g_1(X).
\end {align}</math>
Note that <math>g(\alpha)=1</math> and <math>g_i(\alpha)=0</math> for <math>i \ne 1</math>. Next, define an <math>n \times n</math> matrix ''A'' of polynomials over ''K'' and a polynomial ''D'' by
<math display="block">
\begin {align}
A_{ij}(X) &= \sigma_i(\sigma_j(g(X)) = \sigma_i(g_j(X))\\
D(X) &= \det A(X).
\end {align}</math>
Observe that <math>A_{ij}(X) = g_k(X)</math>, where ''k'' is determined by <math>\sigma_k = \sigma_i \cdot \sigma_j</math>; in particular <math>k=1</math> iff <math>\sigma_i = \sigma_j^{-1}</math>. It follows that <math>A(\alpha)</math> is the permutation matrix corresponding to the permutation of ''G'' which sends each <math>\sigma_i</math> to <math>\sigma_i^{-1}</math>. (We denote by <math>A(\alpha)</math> the matrix obtained by evaluating <math>A(X)</math> at <math>x=\alpha</math>.) Therefore, <math>D(\alpha) = \det A(\alpha) = \pm 1</math>. We see that ''D'' is a non-zero polynomial, and therefore it has only a finite number of roots. Since we assumed ''F'' is infinite, we can find <math>a\in F</math> such that <math>D(a)\ne 0</math>. Define
<math display="block">
\begin {align}
\beta &=  g(a) \\ 
\beta_i &= g_i(a) = \sigma_i(\beta).
\end {align}
</math>
We claim that <math>\{\beta_1, \ldots, \beta_n\}</math> is a normal basis. We only have to show that <math>\beta_1, \ldots,\beta_n</math> are linearly independent over ''F'', so suppose <math display="inline">\sum_{j=1}^n x_j \beta_j = 0</math> for some <math>x_1...x_n\in F</math>. Applying the automorphism <math>\sigma_i</math> yields <math display="inline">\sum_{j=1}^n x_j \sigma_i(g_j(a)) = 0</math> for all ''i''. In other words, <math>A(a) \cdot \overline {x} = \overline {0}</math>. Since <math>\det A(a) = D(a) \ne 0</math>, we conclude that <math>\overline x = \overline 0</math>, which completes the proof.

It is tempting to take <math>a=\alpha</math> because <math>D(\alpha)\neq0</math>. But this is impermissible because we used the fact that <math>a \in F</math> to conclude that for any ''F''-automorphism <math>\sigma</math> and polynomial <math>h(X)</math> over <math>K</math> the value of the polynomial <math>\sigma(h(X))</math> at ''a'' equals <math>\sigma(h(a))</math>.