Some hours ago, a question was posted, asking (citation by heart, not literally)

Let $R$ be a commutative ring (with unit). What can be said about the Krull dimension of an algebraic, non-integral extension $S = R[x]$ of $R$ ?

I don't know why the question disappeared in the meanwhile. Nevertheless I find it interesting and repost it therefore.

Let $R$ be noetherian (with finite dimension). Since $S = R[x]$ is a quotient of a polynomial ring $R[T]$, there are epimorphisms $R[T] \to S \to R$ showing $\dim R + 1 = \dim R[T] \ge \dim S \ge \dim R$. Thus $\dim S = \dim R$ or $\dim S = \dim R +1$.

Both cases can occur:

1) Let $R = \mathbb{Z}$ and $S=R[T]/(pT) = R[x]$ with $x = T + (pT)$. Since $x$ is annulated by the polynomial $pX$, the extension is (non-integral) algebraic and since $pT \in R[T]$ is homogenous and regular, $\dim S = \dim R[T] -1 = \dim R = 1$.

2) Let $R =\mathbb{Z}/p^2$ and $S = R[T]/(pT)$. As above the extension is (non-integral) algebraic and since $pT$ is nilpotent $\dim S = \dim R[T] = \dim R + 1 \;(= 1)$.