Example of a ring $R$ such that $\dim(R[[X]])<\dim(R[X])$ Dimension refers to the Krull dimension of a commutative ring. 
In the paper "Prime ideals in power series rings" J. Arnold gives an example of such a ring:  
Let $k$ be a field and $K=k(t)$ a simple transcendental extension of $k$. Suppose that $V=K+M$ is a discrete valuation ring with maximal ideal $M$. Let $D=k+M$. Then $\dim(D)=1$ and $\dim(D[X])=3$. Here is a proof:
$M$ is the the only nonzero prime ideal of $D$. Thus $dim(D)=dim(D_M)=1$. For a $1$-dimensional integral domain $R$, $R[X]$ is $2$-dimensional iff every localization of the integral closure of $R$ is a valuation ring. Since $D$ is integrally closed and $D_M$ is not a valuation ring, $dim(D[X])=3$.
Arnold claims that $\dim(D[[X]])=2$. If fail to see why this is true.
Are there any other examples of such rings?
 A: The primes of $D$ are $(0), P=(M*f(t):f(t)\in k(t))$. ($P$ is not finitely generated.)  
The primes of $D[X]$ are $(0), (M*f(t)*F(X,t):f(t)\in k(t))$ (a prime for each irreducible $F(X,t)$ mod $k(t)^*$ where $F(X,t) \in k(t)[X]$) (note $P*D[X]$ corresponds to $F(X,t)=1$), and $P*D[X] + (h(X))$ (a prime for each irreducible $h(X)$ mod $k^*$ where $h(X) \in k[X]$).  For example, $Q:=(M*f(t)*(1+t^2-X^3):f(t)\in k(t))$ is a prime of $D[X]$, and this prime is in the maximal chain $(0) \subset Q \subset P*D[X] \subset P*D[X]+(X)$.
In $D[[X]]$, $P*D[[X]]$ is the only prime over all primes of $D[X]$ of the form $(M*f(t)*F(X,t):f(t)\in k(t))$.  For example, a prime of $D[[X]]$ that contains $M*(1+t^2-X^3)$ also contains $M*(1+t^2)*M*(1-X^3/(1+t^2))$ and therefore also contains $P$ since $(1-X^3/(1+t^2))$ is invertible in $D[[X]]$.  This implies that $D[[X]]$ has Krull dimension=2.
A: The ring $D$ given in the question indeed has $\dim(D[[X]])$=2. This is proved in detail in the paper "Power series rings over Pruefer domains" by J. Arnold. 
