Transcendence degree and Krull dimension for homomorphic images of power series rings

Question

Let $k$ be a field of characteristic zero and $I$ be a radical ideal of $k[[x_1,\ldots, x_n]]$. Let $P$ be a minimal prime ideal of the reduced ring $R:=k[[x_1,\ldots, x_n]]/I$. Then, $R_P$ is a field extension of $k$. My question is: Is it true that $\dim(R/P)$ equals the transcendence degree of $R_P$ over $k$ ?

Answer 1

Let $k$ be countable and let $I=(0)$. Then $R_P$ is the field $k((x_1,\dots,x_n))$, which is uncountable provided $n>0$, and therefore of uncountable transcendence degree over $k$, since an algebraic extension of a countable field is countable. The Krull dimension, on the other hand, is $n$.