# trouble understanding lemma for normal basis theorem

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?

• Can we not use $a$ and $\alpha$ together... – Kenny Lau Jul 1 '18 at 12:32
• 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 $x_i$. – zinR Jul 1 '18 at 12:37
• 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). – user1551 Jul 1 '18 at 12:55

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.
• $(\sigma_i(x_1))_{1\leq i \leq n} , ..., (\sigma_i(x_n))_{1\leq i \leq n}$ 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? – zinR Jul 1 '18 at 13:31