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I'm having trouble understanding the following proof:

Lemma: Let $K \subseteq L$ be a Galois extension and the Galois group $G(L/K) = \{ \sigma_1, ..., \sigma _n \}$. Elements $x_1, ..., x_n \in L$ form a $K$-basis of $L$ if and only if $\det [\sigma_i(x_j)] \neq 0.$

Proof: The given elements are linearly dependent if and only if there are elements $a_1,...,a_n\in K$ not all equal to $0$ such that $a_1 x_1 + \dots + a_n x_n =0$. Letting all $\sigma _i$ act on this equality, we get \begin{align*} a_1 \sigma_1 (x_1) + \dots + a_n \sigma_1(x_n) &= 0 \\ a_1 \sigma_2 (x_1) + \dots + a_n \sigma_2(x_n) &= 0 \\ \dots \\ a_1 \sigma_n (x_1) + \dots + a_n \sigma_n(x_n) &= 0. \end{align*} The above system of linear equations has a nonzero solution $a_1,...,a_n$ if and only if the determinant of the coefficient matrix (consisting of $\sigma_i(x_j) $) equals $0$.

I understand that if there is a nonzero solution $a_1,...,a_n \in K$, then this determinant must be zero.

Conversely, I do not understand why the determinant being zero implies there is a nonzero solution $a_1,...,a_n \in K$. Since the coefficients are in $L$, the only thing I should be able to deduce is that there is a nonzero solution $a_1,...,a_n \in L$, while in general $K \neq L$.

What am I missing?

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    $\begingroup$ Can we not use $a$ and $\alpha$ together... $\endgroup$ – Kenny Lau Jul 1 '18 at 12:32
  • $\begingroup$ I am sorry. I copied that from the book I found the proof in. (Which by the way is Brzezinski: Galois Theory Through Exercises.) If you wish I'll replace $\alpha_i$ by $x_i$. $\endgroup$ – zinR Jul 1 '18 at 12:37
  • $\begingroup$ This is off-topic, but this reminds me of Bak and Newman's Complex Analysis, which also uses $a$ and $\alpha$ together in proofs. This is just horrible (although the book itself is good). $\endgroup$ – user1551 Jul 1 '18 at 12:55
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My justification might be circular in your context since you are developing Galois theory, but if we forget this context and just view it from the outside (Galois theory can definitely be proved without your theorem), then I can justify it using Galois descent.

If the system of linear equations has a non-trivial $L$-solution, then the corresponding linear transformation has non-trivial $L$-kernel, which forms a vector space $V$. By Galois descent, $V^\Gamma = V \cap K^n$ is non-trivial, so the system of equations has a non-trivial $K$-solution.

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    $\begingroup$ This is not easy to follow, since I am not at all familiar with Galois descent. (Your source seems like a lot to digest.) Is it possible to translate this into a more elementary proof? If not, how come that the author of the book I found that proof in did not notice this big of a gap? $\endgroup$ – zinR Jul 1 '18 at 13:12
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    $\begingroup$ $(\sigma_i(x_1))_{1\leq i \leq n} , ..., (\sigma_i(x_n))_{1\leq i \leq n}$ is a $L$-Basis of $L^n$ $\Leftrightarrow$ the determinant of the above matrix is not 0 $\Leftrightarrow$ there is no non-trivial $L$-solution. Maybe we can use Dedekind's lemma somehow? $\endgroup$ – zinR Jul 1 '18 at 13:31
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    $\begingroup$ Yes, it seems to me this follows immediately from Dedekind. @zinR $\endgroup$ – Pierre-Yves Gaillard Jul 1 '18 at 15:38
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    $\begingroup$ Here (Lemma 3.4) it is proven with Dedekind. Is this what you had in mind? Actually quite a simple proof, and appearently from the same author whose text on Galois descent was linked above. $\endgroup$ – zinR Jul 1 '18 at 15:41
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    $\begingroup$ @Pierre-YvesGaillard I see, thank you (also for explaining to me how to use the @ symbol). $\endgroup$ – zinR Jul 7 '18 at 16:54

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