# 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

1. $$\lbrace b_1 \rbrace$$ is algebraically independent over $$k$$; and
2. $$A$$ 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 =

$$

.

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.

• I've just edited the proof from the Wikipedia's article adding this case. Commented Apr 12 at 14:29

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 $n = 1$.

• Also, the empty set is a subset of $A$ which is algebraically independent over $k$.
– D_S
Commented Apr 13, 2016 at 14:04

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 $r=0$.

• Is that means that I have to replace my stratement by or $r=1$ in this case 1. and 2. are correct or $r=0$ and I replace 1. by for all $b_1\in A$, $b_1$ is transcendent over $k$ and I delete 2. ? Or should I replace 2. by $A$ is a finite $k$-module ? Commented Apr 12, 2016 at 20:03
• But I still have to prove that if $r=0$ all $b_1\in A$ are transcendent, otherwise the theorem is vacuous (since we accept $r=0$), right ? Commented Apr 12, 2016 at 20:04
• In any case, $A$ is finite over $k[b_1]$ for some $b_1$. If $b_1$ is algebraic over $k$ then this corresponds to $r=0$ and actually $A$ is finite over $k$ (and you can take $b_1=1$). If $b_1$ is transcendent, this is the case $r=1$. But in all cases, the statement 1 does not matter. Commented Apr 12, 2016 at 20:08
• But I don't understand what means this theorem then. In it's general form the Noether normalisation lemma : Let $k$ be an infinite field, and $A = k[a_1,\cdots ,a_n]$ a finitely generated $k$-algebra. Then there exist $r\leq n$ and $b_1,\cdots , b_r$ in $A$ such that : 1 .$b_1,\cdots , b_r$ are algebraically independent over $k$; and 2. $A$ is a finite $k[b_1,\cdots ,b_r]$-module. If I Always condider $r=0$ then $1.$ does not make any sense and $2.$ is always true. Commented Apr 12, 2016 at 21:10
• On my last comment I think I made a mistake. With $r=0$, 2. is true if and only if $a_1,\cdots ,a_n$ are algebraic elements over $k$. Commented Apr 12, 2016 at 21:22