Noether normalization theorem I'm reading: Hulek, "Elementary Algebraic Geometry", i can't understand a comment he does about Noether normalization theorem, which tells:
Le $k$ be a field with infinitely many elements, let $A=k[a_1,\ldots,a_n]$ be a finitely generated $k$-algebra, then we can find $y_1,\ldots,y_m$ elements of $A$, with $m\leq n$ such that $y_1,\ldots,y_m$ are algebraically indipendent over $k$ and $A$ is a finitely generated $k[y_1,\ldots,y_m]$-module
In the comments after the proof, Hulek says:"....we see that $y_1,\ldots,y_m$ can be taken to be any "general" choice of linear forms in $a_1,\ldots,a_n$....."
Why linear forms? $y_1,\ldots,y_m$ are elements of $k[a_1,\ldots,a_n]$ so $y_i's$ should be polynomials of arbitrary degree in $a_1,\ldots,a_m$....what does it means linear form in this contest?
 A: I was confused a bit as well when I saw this statement for the first time.
The induction, in some sense, "hides" this result. 
1.
If $a_1, \dots, a_n$ are algebraically independent, then one can take $y_1 = a_1, \dots , y_n = a_n$ or any other algebraically independent linear forms in $a_1, \dots, a_n$.
2.
Suppose $a_1, \dots, a_n$ are algebraically dependent over $k$. Then there is a nonzero polynomial $f \in k[x_1, \dots, x_n]$ such that $f(a_1, \dots, a_n) = 0$.
For $i = 1, \dots, n-1$ set $a_i' = a_i - \alpha_i a_n$, where $\alpha_i \in k$. When the field $k$ is infinite, we can choose $\alpha_i$'s and a constant $c \in k$ such that 
$$
  \frac{1}{c} f(a'_1 + \alpha_1 a_n,  \dots, a'_{n-1} + \alpha_{n-1} a_n, a_n) = 0
$$
is a monic relation for $a_n$ over $A' = k[a_1', \dots, a_{n-1}']$. Therefore $A = A'[a_n]$ is finite over $A'$.
Now we repeat the same argument with $A = A'$. If $a_1' , \dots, a_{n-1}'$  are algebraically independent, we can take $y_1 = a_1', \dots, y_{n-1} = a_{n-1}'$, i.e. $y_1 = a_1 + \alpha_1 a_n, \dots, y_{n-1} = a_{n-1} + \alpha_{n-1} a_n$ and so on.
This shows that $y_i$'s can be chosen to be linear forms in $a_1, \dots, a_n$. 
Note that to choose $\alpha_i$'s we do the following: We consider the expression
$$f(a'_1 + \alpha_1 x_n,  \dots, a'_{n-1} + \alpha_{n-1} x_n, x_n) = g(\alpha_1, \dots, \alpha_{n-1})x_n^{\deg(f)} + \text{polynomial in } x_n \text{ of a lower degree}$$
where $g \neq 0$ and take any $\alpha_i$'s such that $g(\alpha_1, \dots, \alpha_{n-1}) \neq 0$. The set $\{P \in \mathbb{A}^{n-1}_k: g(P) \neq 0 \}$ is nonempty (because $k$ is infinite) Zariski open and hence dense. In other words almost any choice will work fine. I assume it is what the author means with "general choice".
