Let $R$ be a Noetherian ring, if $I$ is an ideal of ring $R$, then the blowup algebra of $R$ is the subalgebra of $R[x]$ given by $B_I(R):=R\bigoplus Ix\bigoplus I^2x^2\bigoplus \dots \cong R[Ix]$. Show that the dimension of $B_I(R)$ is the maximum of the numbers $\dim R/P$, where $P$ ranges over minimal primes containing $I$, and $1+\dim R/Q$ where $Q$ ranges over minimal primes not containing $I$. (Eisenbud, Exercise 13.8 (2).)
I am able to show the minimal primes of $B_I(R)$ are the ideals of the form $PR[x]\bigcap B_I(R)$, where $P$ is a minimal ideal of $R$. Thus we can reduce to the case where $R$ is an integral domain by factoring out some minimal prime. However, then I am not sure how to proceed.
I think I somehow found a way like following. First, show that if $P$ is a prime of $R$, then $PR[x]\bigcap B_I(R)$ is a prime of blowup algebra. Then, we can prove $B_I(R)/PR[x]\bigcap B_I(R)\cong B_{(I+P)/P}(R/P)$. If $P$ contains $I$, the dimension is obviously maximal among $\dim R/P$; otherwise, we can prove if $P_1\supset P_2\supset\dots$ is a chain of distinct primes in $R$, then $P_1R[x]\bigcap B_I(R)\supset P_2R[x]\bigcap B_I(R)\supset \dots$ is a chain of distinct prime of $B_I(R)$ and we know the correspondence between their minimal ideals, thus we can create a chain of primes in $B_{(I+Q)/Q}(R/Q)$ which has length $\dim R/Q+1$(since $(I+Q)/Q$ is nonzero), here $Q$ is a minimal prime not containing $I$. And the other side of dimension equality is obvious (i.e. by Nagata's Altitude Formula).