Krull dimension of $K[[x]]$

Question

Well, I know that this is noetherian but I don't really understand how I can get a chain of prime ideals here to find the Krull dimension and assure that it is the supremum.

Answer 1

Assuming $K$ is a field, $\text{dim}\,K[[X]]=1$. More generally there is the following (Theorem 15.4 from Matsumura):

Theorem. Let $R$ be a noetherian ring. Then $$\text{dim}\,R[X_{1},\cdots,X_{n}]=\text{dim}\,R[[X_{1}\cdots,X_{n}]]=\text{dim}\,R+n$$

The proof for the power series case is as follows. By induction we can assume $n=1$. Any prime ideal of $R[[X]]$ is of the form $\mathfrak{M}=(\mathfrak{m},X)$ with $\mathfrak{m}=\mathfrak{M}\cap R$ a maximal ideal of $R$. Thus $\text{ht}\,\mathfrak{M}=\text{height}\,\mathfrak{m}+1$. Conversely if $\mathfrak{m}$ is a maximal ideal of $R$ then $\text{ht}\,(\mathfrak{m},X)=\text{ht}\,\mathfrak{m}+1$, which shows the equality.