Consider any noetherian ring $A$ and the polynomial ring $A[x]$. Consider the quotient ring $A[x]/\langle x^2+1\rangle$. Is the dimension of this quotient ring equal to dimension of $A$ (i.e. dimension of $A[x] - 1$)?
If there is a chain of ideals in $A[x]$ given by $P_0[x] \subseteq P_1[x] \cdots, P_r[x] \subseteq P_r + \langle x \rangle$. This chain will still be prime in the quotient ring?
The ring extension $A\subset A[x]/(x^2+1)$ is integral ($x$ is a root of the monic polynomial $t^2+1\in A[t]$), so $\dim A=\dim A[x]/(x^2+1)$.
Remark. You can replace $x^2+1$ by any monic polynomial.