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?



Example of a ring $R$ such that $\dim(R[[X]])<\dim(R[X])$
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. 


