Let $k$ be an infinite field and $f(x_1,x_2,\cdots,x_n)$ be a non-constant polynomial over $k$. Then it is known that -
There exists $c_1,c_2,\cdots,c_n\in k$ such that $f(c_1,c_2,\cdots,c_n)\neq 0$.
Q. Let $K$ be a field extension of $k$ of finite degree. Suppose $f(x_1,x_2,\cdots,x_n)$ is a non-constant polynomial over bigger field $K$. Can we say that there exists $c_1,\cdots,c_n$ in smaller field $k$ such that $f(c_1,c_2,\cdots,c_n)\neq 0$?
The fact stated before question I saw in the book of P. M. Cohn on Basic Algebra (p. 228). But in the primitive element theorem, he considers a non-constant polynomial $f(x_1,x_2,\cdots,x_n)$ over bigger field $K$ and asserts that there exists $c_1,\cdots,c_n$ from smaller field $k$ such that $f(c_1,\cdots,c_n)\neq 0$. I didn't get this jumping.
Let $k$ be an infinite field, $K/k$ any field extension, and $f\in K[X_1,\ldots, X_n]$ a non-constant polynomial. Let $S$ be the set of all $m\in\{0,1,\ldots,n\}$ such that there exist $c_1,\ldots, c_n\in K$ such that $f(c_1,\ldots, c_n)\ne 0$ and $c_i\in k$ for $i\le m$. We already know that $0\in S$. Let $s=\max S$. So there exist $c_1,\ldots, c_n\in K$ such that $f(c_1,\ldots, c_n)\ne 0$ and $c_i\in k$ for $i\le s$.
Assume $s<n$ and let $g(X)=f(c_1,\ldots, c_s,X,c_{s+2},\ldots, c_n)\in K[X]$. By maximality of $s$, $g(x)=0$ for all $x\in k$. But then $g$ has infinitely many roots and must be the zero polynomial. On the other hand, $g(c_{s+1})\ne0$, contradiction. We conclude that $s=n$, i.e., there exist $c_1,\ldots, c_n\in k$ such that $f(c_1,\ldots, c_n)\ne 0$.