I am reading "A Course in Ring Theory" by Passman.
I was reading Theorem $7.8$ which state
"Let $R$ be a Dedekind domain with field of fractions $F$ and let $K$ be a finite degree separable field extension of $F$. If $S$ is the set of elements of $K$ that are integral over $R$, then $S$ is a Dedekind domain."
The proof then starts of with:
"To start we know that $S$ is a subring of $K$, having $K$ as its field of fractions. Indeed, if $\alpha \in K$, then there exists $0 \neq r \in R$ with $r\alpha \in S$. Furthermore, $S \cap F = R$, since $R$ is integrally closed. It remains to verify conditions $(1), (2),$ and $(3)$ of the definition of a Dedekind domain.
$(1)$ In view of the preceding remarks, we can let $\{\alpha_1, \alpha_2, \cdots, \alpha_n \}$ be an $F-$basis for $K$ with each $\alpha_i \in S$. Notice that multiplication by any $\beta \in K$ determines an $F$-linear transformation of the $F-$vector space $K$. We can therefore let tr$\beta$ denote the trace of this linear transformation. In other words, if $\beta \alpha_i = \sum_{j = 1}^n f_{i,j}\alpha_j$ with $f_{i,j} \in F$, then tr$\beta = \sum_{i = 1}^n f_{i,j}$. Clearly, tr$: K \rightarrow F$ is an $F$-linear functional and it is not hard to see that if $\beta \in S$, then tr$\beta \in S \cap F = R$. Furthermore, since $F/K$ is separable, we know that the $n \times n$ matrix (tr$\alpha_i \alpha_j$) is non singular. "
*** My problems: I don't really follow, why if $\beta \in S$ then tr$\beta \in S$, or why the $n \times n$ matrix (tr$\alpha_i \alpha_j$) is non singular.
For if $\beta \in S$ then tr$\beta \in S$, I know since $S$ is a ring and $\alpha_i \in S$ we have $\beta\alpha_i \in S$. Also since $S$ is a ring, if each $f_{i,i} \in S$ then tr$\beta \in S$, so I have been trying to prove each $f_{i,i} \in S$, but have not had any luck.
Sadly I haven't made any progress at all on why he $n \times n$ matrix (tr$\alpha_i \alpha_j$) is non singular.
Non-degeneracy of a bilinear map $\beta:V\times W\to k$ for $k$ vectorspaces $V,W$ means that, for $0\not=v\in V$, there is $w\in W$ such that $\beta(v,w)\not=0$, and symmetrically.
The fact you are wanting to prove is apparently the (well-known) non-degeneracy of the trace pairing $\beta(x,y)=\hbox{tr}^K_k(xy)$ from a finite, separable field extension $K$ of a field $k$.
To write things in terms of matrices and so on is not helpful, I think.
Rather, existence of a primitive element, a zero of a polynomial with no repeated roots (so the derivative of the polynomial does not vanish at that element) is what you need.
This is quite standard, and eminently google-able. It is certainly in my own algebra notes, among nearly-infinitely-many other on-line sources.
I understand that "non-degeneracy of a pairing" might be unfamiliar, but it is the correct notion here. For example, "positive-definiteness", a cousin of this notion, is not quite applicable.