Again I am stuck trying to solve an exercise in Bosch's Algebraic Geometry. I apologize for this rather lengthy post.
For a discrete valuation ring $R$, consider the scheme $S=\rm{Spec}(R)$ . Determine the dimension $\rm{dim}(S)$ as well as the local dimension $\rm{dim}_s(S)$ for all points $s\in S$. Do the same for the affine $n$ - space $\mathbb{A}^n_S$ and the projective space $\mathbb{P}^n_S$ over $S$. Hint: View $\mathbb{A}^n_S$ and $\mathbb{P}^n_S$ as relative schemes over $S$ and look at their fibers.
Here is what I have: Let $\mathcal{m}$ be the maximal ideal in $R$. $\rm{dim}(S)=1$ since $(0)\subset \mathcal{m}$ is the longest chain of primes in $R$. For $s_0 = (0)$ the generic point we have $\rm{dim}_{s_0}(S) = \rm{dim}(R_{(0)}) = \rm{dim}(K)=0$, where $K$ is the field of fractions of $R$. And for the closed point $s_1 = \mathcal{m}$, we have $\rm{dim}_{s_1}(S) = \rm{dim}(R_{\mathcal{m}}) = \rm{dim}(R) =1$.
Now cosider $\mathbb{A}_S^n = \rm{Spec}(A)$, with $A:= R[t_1,...,t_n]$ and the natural morphism $\phi:\mathbb{A}_S^n\rightarrow S$. Using the hint I can compute the fibers $$ \phi^{-1}((0)) = \rm{Spec}(K[t_1,...,t_n]);\quad \phi^{-1}(\mathcal{m}) = \rm{Spec}(R/\mathcal{m}[t_1,...,t_n]). $$
Since $R$ is Noetherian, we have $\rm{dim}(\mathbb{A}^n_S)=n+\rm{dim}(R)=n+1$. The fibers are $n$ - dimensional. But how do I compute the local dimensions of all points in the fibers?