In my commutative algebra class we proved the following theorem:
Contractions of maximal ideals: Let $A\xrightarrow{\ \ f\ \ }B$ be a homomorphism of $K$-algebras. Suppose that $B$ is of finite type. Then for every maximal ideal $\mathfrak{m}\trianglelefteq B$ the contraction $f^{-1}\mathfrak{m}\trianglelefteq A$ is also maximal.
We did this, when we had not yet developed a lot of machinary and the proof was rather messy and technical. We used this theorem to prove Hilbert's Nullstellensatz. Later we gave a simpler proof of Hilbert's Satz using Noether normalization. I was wondering:
Is it possible to give a simple proof for the above theorem using Noether normalization?
We had two (very closely related) theorems, called Noether normalization:
Noether 1: Let $K$ be a field, $0\not=f\in K[x_1,\dots,x_n]$. Then there exists an isomorphism $$K[x_1,\dots,x_n]\xrightarrow{\ \ \varphi\ \ }K[x_1,\dots,x_{n-1}][x_n]$$ of $K$-algebras, such that for $\displaystyle \varphi(f)=\sum_{i=0}^ma_ix_n^i$ with $a_i\in K[x_1,\dots, x_{n-1}]$ we have $a_m=1$.
Noether 2: Let $B$ be a finite type $K$-algebra. Then $K\hookrightarrow B$ factors as $K\hookrightarrow K[x_1,\dots,x_n]\hookrightarrow B$, where the second arrow is integral.
I think, I found a way to give a more conceptual proof. Consider the map $K\hookrightarrow B\twoheadrightarrow B/\mathfrak{m}$. By Noether normalization, this factors as $K\hookrightarrow K[x_1,\dots,x_n]\hookrightarrow B/\mathfrak{m}$, where the second arrow is an integral extension. Using a little bit of dimension theory, $$n=\dim K[x_1,\dots,x_n]=\dim B/\mathfrak m=0\,,$$ so $B/\mathfrak{m}$ is integral over $K$. Now $\mathfrak{n}$ is the kernel of $A\xrightarrow{\ \ f\ \ }B\twoheadrightarrow B/\mathfrak{m}$. The homomorphism theorem yields an embedding $A/\mathfrak{n}\hookrightarrow B/\mathfrak{m}$, so each element of $A/\mathfrak{n}$ is integral over $K$. But for two rings, one integral over the other, one of them is a field if and only if the other one is. Thus, $A/\mathfrak{n}$ is a field and the result follows.