I am reading Pierre Samuel's Algebraic Theory of Numbers. I get stuck at an equality within the proof of Proposition 3 of Section 2.7. The statement of the proposition is as follows:
Proposition 3. Let $\mathbb{K}$ be a field that is either finite or of characteristic zero. Let $\mathbb{L}$ be an extension of finite degree $n$ of $\mathbb{K}$, and let $\sigma_{1},\sigma_{2},\cdots,\sigma_{n}$ be the $n$ distinct $\mathbb{K}$-isomorphisms of $\mathbb{L}$ into an algebraically closed field $\mathbb{F}$ containing $\mathbb{K}$. If $(x_{1},x_{2},\cdots,x_{n})$ is a base for $\mathbb{L}$ over $\mathbb{K}$, then $$D(x_{1},x_{2},\cdots,x_{n})=\det(\sigma_{i}(x_{j}))^{2}\neq0$$ where $D(x_{1},x_{2},\cdots,x_{n})$ is the discriminant.
The proof starts with the definition of discriminant: $$D(x_{1},x_{2},\cdots,x_{n})=\det(\operatorname{Tr}_{\mathbb{L}/\mathbb{K}}(x_{i}x_{j})).$$ The following equality confuses me: $$\det(\operatorname{Tr}_{\mathbb{L}/\mathbb{K}}(x_{i}x_{j}))=\det(\sum_{k=1}^{n}\sigma_{k}(x_{i}x_{j})).$$
I want to ask the following two questions:
1) I don't really understand how we reach the equality $$\det(\operatorname{Tr}_{\mathbb{L}/\mathbb{K}}(x_{i}x_{j}))=\det(\sum_{k=1}^{n}\sigma_{k}(x_{i}x_{j})).$$
My current guess is to use the following: For each element $y\in\mathbb{L}$, let $p(x)$ be the minimal polynomial of $y$ over $\mathbb{K}$. Then $$p(x)^{\frac{n}{deg(p)}}=(x-\sigma_{1}(y))(x-\sigma_{2}(y))\cdots(x-\sigma_{n}(y))$$
Is this statment true? And why is it true or false?
If this is true, then I can get: $$\operatorname{Tr}_{\mathbb{L}/\mathbb{K}}(x_{i}x_{j})=\sum_{k=1}^{n}\sigma_{k}(x_{i}x_{j})$$ as desired.
2) In the paragraph following the proof of this proposition, the author wrote
The relation $D(x_{1},x_{2},\cdots,x_{n})\neq0$ means that the bilinear form $(x,y)\mapsto \operatorname{Tr}_{\mathbb{L}/\mathbb{K}}(xy)$ is non-degenerate.
Why is this the case?