What is the dimension of $R[[X]]$ where $R$ is a Noetherian ring?

Question

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 ?

Answer 1

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.

Answer 2

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.