The Stacks project

[Example 3.1.9, SV]

Example 62.7.2. There exist relative $r$-cycles which are not equidimensional. Namely, let $k$ be a field and let $X = \mathop{\mathrm{Spec}}(k[x, y, t])$ over $S = \mathop{\mathrm{Spec}}(k[x, y])$. Let $s$ be a point of $S$ and denote $a, b \in \kappa (s)$ the images of $x$ and $y$. Consider the family $\alpha $ of $0$-cycles on $X/S$ defined by

  1. $\alpha _ s = 0$ if $b = 0$ and otherwise

  2. $\alpha _ s = [p] - [q]$ where $p$, resp. $q$ is the $\kappa (s)$-rational point of $\mathop{\mathrm{Spec}}(\kappa (s)[t])$ with $t = a/b$, resp. $t = (a + b^2)/b$.

We leave it to the reader to show that this is compatible with specializations; the idea is that $a/b$ and $(a + b^2)/b = a/b + b$ limit to the same point in $\mathbf{P}^1$ over the residue field of any valuation $v$ on $\kappa (s)$ with $v(b) > 0$. On the other hand, the closure of the support of $\alpha $ containes the whole fibre over $(0, 0)$.

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