Krull dimension in finite ring extensions

Question

Let $K$ be a field and $R=K[a_1, \dots, a_n]$ a finite ring extension. Suppose that the degree of transcendence of $R$ over $K$ is $r$. Then the Krull dimension of $R$ is at most $r$. I would like to prove it is exactly $r$. If ${t_1, \dots , t_r}$ is a transcendence base of $R$ over $K$ then $S=K[t_1, \dots, t_r]$ is a polynomial ring over a field in r variables and therefore has dimension r. I cannot prove that the r prime ideals $(t_1), \dots, (t_1, \dots, t_r)$ in $S$ have proper prime extensions in $R$. I would then be done if I could prove it. It would also be nice to use a lying over theorem. R is an algebraic extension of S, but I cannot prove it is an integral extension. Once again I would be done.

Answer 1

If $A$ is a finitely generated algebra over a field $K$, then there exist $r$ elements $x_1,x_2,\cdots, x_r\in A$ which are algebraically independant over $K$ and such that $A$ is a finitely generated module over $K[x_1,x_2,\cdots, x_r]$.
The Krull dimension of $A$ is then $r$, which proves that the number $r$ is intrinsically determined by $A$ although of course the $x_i$'s may be chosen in many ways.
Moreover, if $A$ is a domain, the number $r$ is also the transcendence degree over $K$ of the fraction field of $A$ (there is no reasonable notion of transcendence degree for an arbitrary $K$-algebra $A$).

The above is Noether's normalization theorem, one of the most important results in all of commutative algebra or basic algebraic geometry.
The geometric interpretation is that if you have an $r$-dimensional affine variety $X$ over a field $K$, it can always be presented as a finite ramified covering $\pi:X\to \mathbb A^r_K$ of $r$-dimensional affine space.

Non-bibliography
Instead of giving a reference, I issue a challenge: name a commutative algebra book that does not contain Noether's normalization theorem!