# Clarifications on Noether Normalization

I finished reading Noether Normalization but given that I have almost no prior algebra training I am concerned that my understanding is wrong. (Starting masters in Mathematics but previously was an engineer) The reference I took is as follows:

Theorem (Noether normalisation Lemma). Let $k$ be a field and $A$ a finitely generated $k$-algebra; then there are elements $z_1, \dots , z_m \in A$ such that $B = k[z_1,\dots, z_m]$ is isomorphic to a polynomial ring in $m$-variables and $A$ is integral over $B$. In fact $A$ is a finitely generated $B$-module.

Instead of questions, I would like to make statements based on my understanding:

S1) Polynomial ring in $n$-variables: $R=k[x_1,x_2,\dots,x_n]$ where $x_i$ are indeterminates.
$\forall x\in R, x=\sum k_i{x_1}^{\alpha_1}\dots{x_n}^{\alpha_n}$
Finitely generated $k$-algebra is usually given as $A=k[a_1,a_2,\dots,a_n]$.
I saw a definition of finitely generated $k$-algebra from here , which clarifies that
(a) Finite $k$-algebra: $\forall x\in A, x=k_1a_1+k_2a_2+\dots+k_na_n$ (i.e. as a $k$-module)
(b) Finitely generated $k$-algebra: $\forall x\in A, x=\sum k_i{a_1}^{\alpha_1}\dots{a_n}^{\alpha_n}$ (i.e. like a $n$-variable polynomial ring)

S2) The elements $z_i$ are algebraically independent and hence transcendental.
i.e. $\forall f\in k[x]$, if $f[z_i]=0$, then $f=0$
Similarly, $\forall f\in k[x_1,\dots,x_m]$, if $f[z_1,\dots,z_m]=0$, then $f=0$
For any $x\in A, x\not\in\lbrace z_1,\dots,z_m\rbrace$: there is some polynomial $0\neq f\in k[x_1,\dots,x_{m+1}]$ such that $f(z_1,\dots,z_m,x)=0$
Otherwise we need to include $x$ into the algebraic independent set.
Hence this shows that $x$ can be formed by $\lbrace z_1,\dots,z_m\rbrace$ using $f(z_1,\dots,z_m,x)=0$

S3) $A$ is a finitely generated $B$-module since: $\forall x\in A: x=\sum k_i{z_1}^{\alpha_1}\dots{z_m}^{\alpha_m}\in B$
This is similar to a restatement of S2.

S4) Let $P=k(z_1,\dots,z_m)[x]$.
$P$ is the ring of polynomials in $x$ but with coefficients in $k(z_1,\dots,z_m)$, which are rational of the form $\frac{f}{g}|f,g\in k[z_1,\dots,z_m], g(z_1,\dots,z_m)\neq 0$
We would call $k(z_1,\dots,z_m)$ the field of fractions $=Quot(B)$.
Then $P$ is the ring of polynomials over $Quot(B)$.
We can "evaluate" a polynomial using $x=b\in B$

Are the statements correct?

• In part 2, you want $f(x_1,\ldots,x_{m+1})$ to be monic in $x_{m+1}.$ – Andrew Sep 15 '12 at 16:16
• Do you mean that the statement is false if $0\neq f\in k[x_1,\dots,x_{m+1}]$ but $f(x_1,\dots,x_{m+1})$ is not monic in $x_{m+1}$? – Yong Hao Ng Sep 16 '12 at 1:18
• I mean that, by definition, an element $a\in A$ is integral over $B$ if there is a (nonzero) monic polynomial $f\in B[x]$ such that $f(a)=0,$ so your statement 2 is true, but something slightly stronger holds. – Andrew Sep 16 '12 at 5:27
• Ah! I got it, thanks =D – Yong Hao Ng Sep 16 '12 at 9:05