Maximal Ideals of Different Heights in a domain In continue of this question: Maximal Ideals of Different Heights ; and regarding  @rschwieb 's comment,
Now I want to construct other cases with the following property:
1- The ring $R$ is a domain and has some maximal ideals, one with height $3$ and one with height $2$.
2- The ring $R$ is a domain and has infinitely many maximal ideals, one with height $3$ and one with height $2$. (I dont want use localizing in 2 primes of different heights).

Any other case of rings with  maximal ideals of different heights are welcomed.
 A: Partial answer:
The paper Hilbert integral domains with maximal ideals of preassigned height by William Heinzer has an example, but it is not simple.
It is available at "Journal of Algebra, Volume 29, Issue 2, May 1974, Pages 229-231".
Also, you can find another example as reference 5 of the above paper.
A: Let $R$ be a valuation domain of dimension $2$ and assume $m=tR$ for its maximal ideal and $p R_p=s R_p$ for the other non-zero prime ideal. Consider the polynomial ring $R[X]$. Its Krull dimension is $3$: according to Theorem 2 in
M. Nagata, Finitely generated rings over a valuation ring, J. Math. Kyoto Univ. 5-2 (1966), 163-169
every maximal saturated chain
$0\subset p_1\subset\ldots\subset p_\ell$
of prime ideals in $R[X]$ has a length equal to
$\ell=\mathrm{height}(p_\ell\cap R)+1-\mathrm{trdeg}(R[X]/p_\ell|R/(p_\ell\cap R))$.
The right-hand side becomes maximal, if one chooses $p_\ell$ such that for the transcendence degree $\mathrm{trdeg}(R[X]/p_\ell|R/(p_\ell\cap R))=0$ holds and $p_\ell\cap R$ is maximal. An example of a maximal saturated chain of prime ideals is $0\subset p[X]\subset m[X]\subset m[X]+XR[X]$.
The ideal $(tX-1)R[X]$ is a prime ideal of height $1$, since $(tX-1)K[X]$ is a prime ideal, where $K$ is the field of fractions of $R$.
We have $R[X]/(tX-1)R[X]=R[\frac{1}{t}]=R_p$: the kernel of the evaluation homomorphisms $R[X]\rightarrow R[\frac{1}{t}]$ contains $(tX-1)R[X]$. Now let $f\in R[X]$ be a polynomial with $f(\frac{1}{t})=0$. We may assume that one of its coefficients is a unit of $R$. In $K[X]$ we have the factorization $f=q(tX-1)$. Gauss's lemma then yields $q\in R[X]$.
$R[\frac{1}{t}]$ is a proper overring of $R$, hence it is either equal to $R_p$ or to $K$. However $\frac{1}{s}\not\in R[\frac{1}{t}]$: if it were, we had a relation
$
\frac{1}{s}=\sum\limits_{k=0}^nr_k\frac{1}{t^k}, r_k\in R.
$
Multiplikation by $s$ yields
$
1=\sum\limits_{k=0}^nr_k\frac{s}{t^k},
$
which is a contradiction, because $\frac{s}{t^k}\in m$ for all $k$.
The valuation domain $R_p$ has dimension $1$, so that we obtain a maximal chain of primes $0\subset (tX-1)\subset q$, where $q/(tX-1)R[X]=pR_p$.
According to Corollary 3.4 of
Bouvier, Dobbs, Fontana, Universally catenarian integral domains, Advances in Mathematics 72,
$R[X]$ is catenarian, hence $q$ has height $2$. Actually Theorem 1 of
Bouvier, Dobbs, Fontana, TWO SUFFICIENT CONDITIONS FOR UNIVERSAL CATENARITY, COMMUNICATIONS IN ALGEBRA, 15(4), 861-872  (1987)
is the more adequate reference. The first one however is available online.
