In the Exercise 16 of Chapter 5, the author wrote down as
$\cdots$. The generator $x_{n}$ is algebraic over $k[x_{1}, \cdots, x_{n-1}],$ hence there exists a polynomial $f\neq 0$ in $n$ variables such that $f(x_{1},\cdots, x_{n})=0$. Let $F$ be the homogeneous part of highest degree in $f$. Since $k$ is infinte, there exists $\lambda_{1},\cdots, \lambda_{n-1} \in k$ such that $F(\lambda_{1},\cdots, \lambda_{n-1},1) \neq 0$. $\cdots$
What is $F$? If we let $$f = \sum_{\alpha \in \mathbb{N}^{n}}c_{\alpha}x^{\alpha}$$ with the highest degree $q=|\alpha| = \alpha_{1}+\cdots + \alpha_{n}$,then the most plausible candidate for $F$ is $$F := \sum_{\alpha \in \mathbb{N}^{n}, |\alpha| =q}c_{\alpha}x^{\alpha}.$$
However, in this case, I don't know how to get such lambdas. Otherwise, if we regard $f$ as an element in $k[x_{1},\cdots, x_{n-1}][x_{n}],$ then $$f = \sum_{i=1}^{l}g_{i}x_{n}^{i}$$ for some $g_{i} \in k[x_{1},\cdots, x_{n-1}]$, thus we can let $$F = \sum_{\alpha \in \mathbb{N}^{d}, |\alpha| = q,\alpha_{n} = l}c_{\alpha}x^{\alpha}.$$ In this case, $F$ has such lambdas, otherwise, $F(x_{1},\cdots, x_{n}) =x_{n}^{l}F(x_{1},\cdots, x_{n-1},1) =0$ for any $x_{1},\cdots, x_{n-1}$ in field $k$,thus $f=0$, contradiction.
Which one is correct?
$$f(x_1,\ldots,x_n)=\sum_{j=0}^M f_j(x_1,\ldots,x_n)$$ where $f_j$ is homogeneous of degree $j$. So $F$ is $f_j$ for the largest $j$ such that $f_j\ne 0$.
If $F(x_1,\ldots,x_n)$ is a nonzero homogeneous polynomial, then $G(x_1,\ldots,x_{n-1})=F(x_1,\ldots,x_{n-1},1)$ is a nonzero polynomial. Over an infinite field, every nonzero polynomial has a "nonzero": there are $\lambda_j$ with $G(\lambda_1,\ldots,\lambda_{n-1})\ne0$.