Why is a field extension separable if and only if the discriminant of the basis of the field extension is nonzero?

Question

Let $L/K$ be a finite dimensional field extension. We define the trace function $T_{L/K}(x)$ of $L$ over $K$ as $T_{L/K}(x)=\textrm{trace}(r_x)$, where $x\in L$ and $r_x$ is matrix given by the regular representation of $L$ over $K$. Furthermore, given some basis $x_1,\ldots,x_n$ of $L$ over $K$, we define the discriminant of the basis to be $$\Delta(x_1,\ldots,x_n)=\det[T_{L/K}(x_ix_j)].$$ I am reading Janusz's Algebraic Number Fields and in a proof he claims that $L/K$ is separable if and only if $\Delta(x_1,\ldots,x_n)\neq 0$. He seems to be basing it on the fact that the bilinear form $(x,y)=T_{L/K}(xy)$ is nondegenerate if and only if $L/K$ is separable, which was previously proven. How do I prove this? Am I missing something obvious?

Answer 1

What do you mean by "was previously proven": previously proven in the book, or previously proven as assumed background knowledge?

If $x_1, \ldots, x_n$ and $y_1, \ldots, y_n$ are two $K$-bases of $L$ then $$ \Delta_{L/K}(x_1,\ldots,x_n) = D^2 \Delta_{L/K}(y_1,\ldots,y_n) $$ where $D$ is the determinant of the change-of-basis matrix expressing the $x$'s in terms of the $y$'s. Therefore one basis of $L/K$ has a nonzero discriminant if and only if every basis of $L/K$ has a nonzero discriminant.

More generally, for any bilinear form $B : V \times V \to K$ on a finite-dimensional $K$-vector space and $K$-bases $v_1, \ldots, v_n$ and $w_1, \ldots, w_n$ of $V$, $$ \det(B(v_i,v_j)) = D^2\det(B(w_i,w_j)), $$ where $D$ is the determinant of the change-of-basis matrix expressing the $v$'s in terms of the $w$'s. Since $D$ is nonzero, $\det(B(v_i,v_j))$ is nonzero for one $K$-basis $v_1, \ldots, v_n$ of $V$ if and only if it's nonzero for every $K$-basis of $V$. What you care about is a special case of this: the $K$-bilinear form $B(x,y) = {\rm Tr}_{L/K}(xy)$ on $L$ viewed as a $K$-vector space, and this $B$ is called the trace pairing on $L/K$.

For a bilinear form $B$ on a finite-dimensional $K$-vector space $V$: nondegeneracy of $B$ is equivalent to the nonvanishing of $\det(B(v_i,v_j))$ for some (equivalently, every) $K$-basis of $V$. See Theorem 3.1 and Definition 3.2 here.

Why is nondegeneracy of the trace pairing on $L/K$ equivalent to $L/K$ being separable? First assume $L/K$ is separable. Then the primitive element theorem tells us $L = K(\alpha)$ for some $\alpha$. In that case we can use as a $K$-basis of $L$ the powers $1, \alpha, \alpha^2, \ldots, \alpha^{n-1}$, where $n = [L:K]$. It can be shown that $$ \Delta_{L/K}(1,\alpha, \ldots, \alpha^{n-1}) = {\rm disc} f(T), $$ where $f(T)$ is the minimal polynomial of $\alpha$ over $K$, and ${\rm disc} f(T)$ is nonzero since $K(\alpha)/K$ is a separable extension (see Theorem 3.5 here). So some $K$-basis of $L$ has a nonzero discriminant (namely the power basis $1,\alpha,\ldots,\alpha^{n-1}$), which implies the the trace pairing on $L/K$ is nondegenerate by the above general nonsense about nondegeneracy of a bilinear form on a finite-dimensional vector space.

Conversely, suppose the the trace pairing on $L/K$ is nondegenerate. Then the trace function ${\rm Tr}_{L/K} : L \to K$ is not identically $0$: if it were identically $0$, a trace pairing matrix $({\rm Tr}_{L/K}(x_ix_j))$ would be the zero matrix and thus have determinant $0$, which is not true for the trace pairing on $L/K$ to be nondegenerate. It turns out, perhaps surprisingly, that the trace mapping $L \to K$ not being identically $0$ implies $L/K$ is separable, and in fact it's equivalent to $L/K$ being separable: see Theorem 1.1 here.