Krull dimension of a local ring and completion Let $A$ be a local ring (not noetherian) of finite Krull dimension such that its maximal ideal $\mathfrak{m}$ is of finite type.
Let $\hat{A}$ be its $\mathfrak{m}$-adic completion.
Do we have that $\dim(A)=\dim(\hat{A})$?
 A: There exist examples of commutative local rings of finite dimension whose completions have infinite dimension (all dimensions are Krull, of course). Take $A$ to be a complete valuation ring, whose valuation of non-discrete but of rank 1 (i.e., the value group can be embedded in $\mathbb{R}$, but is not cyclic), and consider the ring $A[X]$. Since $A$ has dimension 1, the dimension of $A[X]$ is at most 3 (since $\dim A[X]$ is bounded by $2\dim A+1$ in general). On the other hand, the completion of $A[X]$ in any ideal containing X is the ring of formal power series $A[\![X]\!]$ (since A itself is already complete). But in this case the dimension of $A[\![X]\!]$ is infinite --- see Krull dimension in power series rings'' by Arnold, Trans of the AMS, 1973. If byideal of finite type'' you mean an ideal generating the same topology as a finitely generated ideal (since here the maximal ideals themselves are not finitely generated), then this answers your question fully, and also shows that the inequality from the remark to the answer does not always hold.
A: Let $\Gamma$ be the group $\mathbb{Z}\times\mathbb{Z}$ with the lexicographic ordering. Then any valuation ring $A$ with value group $\Gamma$  is a counterexample.
Indeed, the maximal ideal of $A$ is principal, the Krull dimension of $A$ is $2$ (the rank of $\Gamma$), and the completion of $A$ is a discrete valuation ring, so its Krull dimension is $1$. (Of course, $A\to \hat{A}$ is not injective in this case.)
