Noether normalization lemma proof I would like to prove the following statement without using Noether normalization lemma (cause it is actually the base case in the induction process of the proof of this lemma).

Let $k$ a field with an infinity of elements, and $A=k[a_1]$ a finitely generated $k$-algebra. Then there exist $b_1\in A$ such that

*

*$\lbrace b_1 \rbrace$ is algebraically independent over $k$; and

*$A$ is a finite $k[b_1]$-module.


Let $\varphi : k[X_1]\to k[a_1]$ defined by $\varphi (X_1)=a_1$. $\varphi$ is surjective so using the isomorphism theorem we have
$$k[X_1]/\ker \varphi \cong k[a_1].$$
$\bullet$ If $\ker\varphi =\lbrace 0\rbrace$ then $b_1=a_1$ suit.
$\bullet$ If not there exist $P\in\ker\varphi$, since $k[X_1]$ is principal and $k$ is a field we can assume that $P$ is a monic polynomial with $\deg P \geq 1$ and $\ker\varphi = <P>$.
To prove $2.$ I note that $k\subset k[a_1]$ and by defition of $\ker\varphi$, there exist a monic polynomial $P$ such $P(a_1)=0$, so $a_1$ is algebraic integer over $k$. And that implies that A si a $k$-module. So here, I would choose $b_1\in k^*$ to have $k[b_1]=k$ and $2.$. But if I do so $1.$ is not true.
To prove $1.$, $\lbrace b_1 \rbrace$ is algebraically independent over $k$ means that $\varphi : k[X_1]\to k[a_1]$ defined by $\varphi (X_1)=b_1$ is injective, e.g. if $P\in k[X_1]$, $\varphi (P)=0\implies P=0$. Here I d'on't know how to choose $b_1$.
Any help will be greatly appreciate.
 A: Noether normalization says:

Let $k$ be an infinite field, $A = k[a_1, ... , a_n]$ a finitely generated $k$-algebra.  Then for some $0 \leq r \leq n$, there exist $r$ elements $b_1, ... , b_r \in A$, algebraically independent over $k$, such that $A$ is finitely generated as a module over $k[b_1, ... , b_r]$.

If $S \subseteq A$, the ring $k[S]$ is by definition the intersection of all subrings of $A$ containing $k$ and $S$.  If it happens that $r = 0$, then $k[b_1, ... , b_r]$ just means $k[\emptyset] = k$, so Noether normalization just says that $A$ is already finitely generated as a module over $k$.  You need to consider the possibility that $r = 0$ when you formulate what Noether normalization is saying in the case $n = 1$.
A: Your condition 1 is either vacuous or false depending on what it's supposed to mean. You should just remove it altogether to get a true statement.
Indeed, $A$ can be finite-dimensional algebra over $k$, and in this case you can't take $b_1$ transcendent.
The mistake is that in Noether's lemma, if $A = k[a_1,\dots,a_n]$ then $A$ is a finite extension of some $B = k[b_1\dots,b_r]$ with the $b_i$ algebraically independent, but $r$ is not always equal to $n$. So here you have $n=1$ but you may have $r=0$.
