Exercise 14, § 1

Let $V$ be a discrete valuation ring, and let $\pi$ be a uniformizer of $V$. In the ring $V[T]$, the ideal $m_1 = (\pi T - 1)$ is maximal and has height 1, the ideal $m_2 = (\pi) + (T)$ is maximal and has height 2. The fields $V[T]/m_1$ and $V[T]/m_2$ are isomorphic to the field of fractions of V and the residue field of V respectively. We have $\dim(V[T]) = 2$, but $\dim(V[T]/m_1) + \dim(V[T]_{m_1}) = 1$ and $\dim(\mathrm{gr}_{m_1}(V[T])) = 1$.