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 $\dim(R/M^n)$.