Theorem 13.1 in Chapter 6 of Lang's Algebra (Revised Third Edition) states the following:
Theorem. Let $K:k$ be a finite Galois extension of degree $n$. Let $\sigma_1, \ldots, \sigma_n$ be all the elements of the Galois group of $G$ of $K:k$. Then there is an element $w\in K$ such that $\sigma_1(w), \ldots, \sigma_n(w)$ forms a $k$-basis for $K$.
The proof given is as follows:
Assume that $k$ is infinite. For each $\sigma\in G$, let $X_\sigma$ be a variable, and let $t_{\sigma, \tau}=X_{\sigma^{-1}\tau}$. Let $X_i=X_{\sigma_i}$. Define a polynomial $f(X_1, \ldots, X_n)\in K[X_1, \ldots, X_n]$ as $$f(X_1, \ldots, X_n) = \det(t_{\sigma_i, \sigma_j})$$ Then $f$ is not identically zero, since substituting $1$ in place if $X_{\text{id}}$ and $0$ in place of all other variables yields a non-zero value.
Now here is what I am unable to understand:
Hence $f(\sigma_1(x), \ldots, \sigma_n(x))$ will not vanish for all $x\in K$. (That is, there exists $x\in K$ for which this does not vanish).
Can somebody please explain how the last line is inferred? Thanks.
If the 3rd edition is like the 1st, then you will see that the preceding theorem establishes that, over an infinite field $K$, any finite group $\{\sigma_1(x),\ldots,\sigma_n(x)\}$ of distinct automorphisms of $K$ consists of functions that are algebraically independent over $K$. Thus if $f(x_1,\ldots,x_n)\in K[x_1,\ldots,x_n]$ is not the zero polynomial, then $f(\sigma_1(x),\ldots,\sigma_n(x))$ will not be the zero function.