 Solution

Since $A$ is a regular local ring, $\dim A=d$ is finite. Choose a chain of prime ideals of length $d$,Such a chain exists because the supremum defining this finite integer Krull dimension is attained; a maximal-length chain in a local ring ends at its maximal ideal.

For each $i$, the ideal ${\mathfrak{p}}_i$ is finitely generated because $A$ is Noetherian. This ensures that its extension ${\mathfrak{p}}_{i}B$ consists precisely of series all of whose coefficients lie in ${\mathfrak{p}}_i$: if ${\mathfrak{p}}_i=(b_1,\dots ,b_r)$, write each coefficient as ${\sum }_j{b}_j{c}_{jn}$ and gather them into the finitely many series ${\sum }_n{c}_{jn}{x}^n$. Thus coefficient reduction givesThe coefficient ring $A/{\mathfrak{p}}_i$ is an integral domain, and so is its formal power series ring: the first nonzero coefficient of a product is the product of the first nonzero coefficients of its factors. Hence each ${\mathfrak{p}}_{i}B$ is prime. Also ${\mathfrak{p}}_{i}B\cap A={\mathfrak{p}}_i$, so the strict inclusions persist upon extension. Finally, $x\notin \mathfrak{m}B$, since its image in $B/\mathfrak{m}B\cong k[[x]]$ is nonzero. We obtain the prime chainof length $d+1$. Therefore $\dim B\ge d+1$. Together with part (b),Its embedding dimension is also $d+1$, by part (b), so $B$ is itself a regular local ring. This proves that formal power series preserve regular local rings.