Consider a DVR $R$ with the maximal ideal $(u)$ containing its residue field $k=R/(u)$. The exercise claims that the dimension of $\operatorname{Spec}(R[t])$ is not equal to the Krull dimension of all stalks $\mathcal{O}_p$ of the structure sheaf at closed points $p$.
As far as I understand, this is due to the fact that we have maximal ideals of different heights. In this case, the height of $(1-ut)$ is 1, while the height of $(u,t)$ is 2. But how can I show that the height of $(1-ut)$ is actually 1? And how does the requirement that $k\subset R$ enters this argument?
Best Answer
Krull's height theorem is what you're looking for: any minimal prime ideal lying over an ideal with $n$ generators is of height at most $n$. Since $(1-ut)$ is generated by one element and is prime (the quotient is $\operatorname{Frac} R$, a domain), it must be of height at most one. It's not of height zero, since that would imply it's a minimal prime of $R[t]$ - as $R[t]$ is a domain, its unique minimal prime is $(0)\neq(1-ut)$.
$k\subset R$ is not actually important to the argument here. Letting $u$ be a uniformizer for $R$, we have $R[t]/(t,u)=k$, while $R[t]/(1-ut)=\operatorname{Frac} R$, regardless of whether $k\subset R$ or not. The reason Hartshorne points this out is to emphasize that the nice conclusions about dimension really do depend on finite type over a field - just being a scheme over a field is not enough.