Importance of Noether normalisation lemma

The Noether normalization lemma states that if $k$ is a field, and $A$ a finitely generated $k$-algebra, then there exist elements $z_1,...,z_m \in A$ such that

(i) $z_1,...,z_m$ are algebraically independent over $k$

(ii) $A$ is finite over $B=k[z_1,...,z_m]$.

I understand the theorem and its proof, but I am having troubles seeing its importance. What does this actually tell us? Most importantly, are there any strong statements that suddenly becomes trivial consequences of the theorem? At the moment it's just a statement to me; I don't see it in a context. Any help?

If you have found the Noether Normalization Lemma in a commutative algebra book, just read on. You will see many applications, for example in dimension theory. The Lemma implies for example the fundamental formula $\dim(X)=\mathrm{trdeg}(K(X)/k)$ for affine varieties $X$ over a field $k$, and that $\dim(X \times_k Y) = \dim(X) + \dim(Y)$ if $X,Y$ are affine varieties over $k$. The Lemma is the main ingredient in the proof of Zariski's Lemma, which in turn implies Hilbert's Nullstellensatz.

By the way, the Lemma has a nice geometric interpretation: Every affine variety over a field has a finite map to some affine space $\mathbb{A}^n$ (and this $n$ is the dimension of the variety). See SE/986279 for a specific example. In Eisenbud's book on commutative algebra you will also find a finer version which starts with a sequence of subvarieties which then corresponds to the sequence $\mathbb{A}^0 \subseteq \mathbb{A}^1 \subseteq \dotsc \subseteq \mathbb{A}^n$.

Another nice application is the generic smoothness of geometrically reduced finite type $k$-schemes. The intuition is very simple: As Martin explained, Noether normalization says that any affine finite type $k$-scheme $X$ admits a finite surjective map $\pi: X\to {\mathbb A}_k^n$ over some affine space ${\mathbb A}_k^n$. You can imagine such $\pi$ as some kind of branched covering of ${\mathbb A}_k^n$; in particular, intuitively one should get an honest covering by removing from ${\mathbb A}_k^n$ the closed set $Z$ of ramification points, and then $X\setminus\pi^{-1}(Z)$ should inherit the smoothness of ${\mathbb A}_k^n\setminus Z$.

This was very vague and informal, but can actually be made precise: First, if $X=\text{Spec}(A)$ is geometrically reduced and integral (we may assume this without loss of generality) and finite surjective over ${\mathbb A}_k^n=\text{Spec}(k[X_1,...,X_n])$ via some $\pi: X\to {\mathbb A}_k^n$, generic flatness tells us that there is some non-vanishing $f\in k[X_1,...,X_n]$ such that $A_f$ is free/flat over $k[X_1,...,X_n]_f$, so we may assume right from the beginning that $X$ admits a finite and flat morphism $\pi: X\to O$ to an open subscheme of ${\mathbb A}_k^n$. In this situation, the discriminant $\text{disc}(\pi)\in{\mathscr O}(O)$ of $\pi$ describes its ramification points. At the generic point, $\text{disc}(\pi)\in k(O)$ measures the separability of the finite extension $k(X)/k(O)$, which is assured by our assumption of geometric reducedness of $X$. Hence, $\text{disc}(\pi)\neq 0$ and the ramificiation points form a proper closed subscheme $Z$ of $O$. Removing this subscheme, we obtain a finite, flat and unramified morphism, i.e. a finite étale cover, $\pi: X\setminus\pi^{-1}(Z)\to O\setminus Z$, and the smoothness of $O\setminus Z$ is inherited to $X\setminus\pi^{-1}(Z)$.