Dimension of $R\left/aR\right.$. In my algebra course I was asked to solve the following problem:
Let $R$ a finite type $K$-algebra and suppose $R$ is an integral domain. If $0\neq a\in R$ is not invertible show that $\dim (R\left/aR\right.)=\dim R-1$.
($\dim$ means Krull dimension) I was able to solve it by elementary methods, however the teacher gaves us a ''hint'' to solve this problem:
If $B$ is a finite type $K$-algebra and $Q$ is a maximal ideal then:
$$\dim(B)=\limsup_{n\to\infty}\frac{\log_2\dim_K(B/Q^n)}{log_2(n)}$$
So I would like to know how to solve this problem with this hint. Somehow this was suposed to help, but it confused me more than helped, since my solution has nothing to do with this weird formula.
Since $(M\left/ Ra\right.)^n=(M^n+Ra)\left/ Ra\right.$ holds in general, we can take $Q=M/aR$ a maximal ideal of $R\left/aR\right.$. Then:
$$(R/aR)\left/ Q^n\right.=(R/aR)\left/ ((M^n+Ra)\left/ Ra\right.)\right.=R/(M^n+Ra)$$
But I dont know how to continue after this, since I dont know how to relate $\dim(R/(M^n+Ra))$ with $\dim(R/M^n)$.
 A: Consider the map $f: R \to R/M^n$ given by $f(r) = ra+M^n$, that is, multiplication by $a$. Its cokernel is $(R/M^n)/(Ra/M^n) = R/(Ra+M^n)$. In other words, we have the exact sequence of $K$-algebras
$$0 \to R/\ker(f) \xrightarrow{\cdot a} R/M^n \to R/(Ra+M^n) \to 0.$$
This is in particular an exact secuence of $K$-vector spaces, hence $\dim_K R/M^n = \dim_K R/(Ra+M^n) + \dim_K R/\ker(f)$. This already relates the two dimensions in a way that is useful, since one can observe that $M^{n-1} \subseteq \ker f$, for $aM^{n-1} \subseteq M^n$ (we need to use a maximal ideal $M$ that contains $a$, which exists since $a$ is not invertible). Hence $\dim_K R/\ker(f) \le \dim_K R/M^{n-1}$, so we reach
$$\dim_K R/(Ra+M^n) \ge \dim_K R/M^n - \dim_K R/M^{n-1}.$$
We know that $p(n) := \dim_K R/(Ra+M^n)$ and $l(n) := \dim_K R/M^n$ are polynomials for large enough $n$ (by Hilbert-Samuel theory), and then the relationship $p(n) \ge l(n) - l(n-1)$ that we just proved shows that $\deg p(n) \ge \deg l(n) - 1$. Using the theorem you stated that allows to compare this degree with the dimension, the result follows (since the other inequality is immediate using chains of primes).
