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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?

Federica Maggioni
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    Linear form does mean degree 1 here. In the case where k is infinite, the theorem can be strengthened to require that the $y_i$ are linear (and in fact almost every choice of linear forms is indeed a basis) – Nehsb Mar 03 '13 at 17:50
  • @Nehsb do u mean a basis for $k[a_1,\ldots,a_n]$ as a $k$-vector space? – Federica Maggioni Mar 03 '13 at 18:26
  • Sorry, by basis I mean such that $A$ is a finitely generated $k[y_1,\cdots,y_m]$ module. – Nehsb Mar 03 '13 at 20:39
  • @Nehsb can u suggest me how to prove that or where i can find a proof – Federica Maggioni Mar 03 '13 at 20:45
  • The comment in the book seems to imply that the given proof also shows this stronger version.

    Here's a source for a proof though: http://math.stanford.edu/~vakil/216blog/ See 11.2.3

     – Nehsb Mar 03 '13 at 21:55

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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".