What is the dimension of $R[[X]]$ where $R$ is a Noetherian ring? If $R$ is a Noetherian ring then we know that the Krull dimension of the polynomial ring $R[X]$ is $\rm dim(R)+1.$ Is there any formula for the Krull dimension of the power series ring $R[[X]] $ When $R$ is a Noetherian ring ?
 A: When $R$ is Noetherian, then the same formula holds for polynomials and for power series:
$$
 \dim R[X] = \dim R[[X]] = \dim R + 1.
$$
A reference was given by user26857 in the comments; another one would be Matsumura's Commutative ring theory, Theorem 15.4.
It may be interesting to point out that this is not true anymore if $R$ is not required to be Noetherian.  In his 1973 paper Krull Dimension in Power Series Rings, J.T.Arnold gives examples of such rings where $\dim R$ is finite, but $\dim R[[X]]$ is infinite.
A: Note that the Chevalley dimension $s(R)$ of a local ring $(R, m)$ is the smallest
integer $r$ for which there exist $a_1, a_2,..., a_r \in m$ with
$lenght(R/(a_1, ...., a_r)M) <\infty$.
Fact1: Let $(R, m)$ be a Noetherian local ring. Then clearly
$R[[x]]$ is is local ring with maximal ideal $m'=<m,x>$.
Fact2: For a Noetherian local ring $(R,m)$ $dim (R)= s(R)$.
Fact3: Clearly $s(R)+1=s(R[[x]])$.
Fact4: For an arbitrary (non-local) commutative ring $R$,
$dim(R)=sup\{dim(R_m)| m$ is a maximal ideal of $R\}$.
Fact5: For a local ring $(R,m)$, we have $R[[x]]_{<m,x>}\cong
R_m[[x]]$.
By combining above facts we have:
$dim(R[[x]])=sup\{R[[x]]_{<m,x>}| m$ is a maximal ideal of
$R\}=\\
sup\{R_m[[x]]| m$ is a maximal ideal of $R\}=\\
sup\{R_m+1| m$ is a maximal ideal of $R\}=\\
sup\{R_m| m$ is a maximal ideal of $R}+1=dim(R)+1.
For more details see commutative algebra, by Gopalakrishnan,chapter 8.
