The doubled-origin diagonal is not closed

By [F3] the product $D{\times }_{k}D$ is covered by the four open subschemes $U{\times }_{k}U$, $U{\times }_{k}V$, $V{\times }_{k}U$, $V{\times }_{k}V$, of which the cross term is $U{\times }_{k}V=\mathrm{Spec}(k[x]{\otimes }_{k}k[y])=\mathrm{Spec}k[x,y]$, with ${\mathrm{pr}}_1$ the first projection and ${\mathrm{pr}}_2$ the second.

By [F2] the inverse image of $U{\times }_{k}V$ under $\mathrm{\Delta }$ is $U\cap V$, and the restriction of $\mathrm{\Delta }$ to it is the morphism $U\cap V\to U{\times }_{k}V$ whose composites with ${\mathrm{pr}}_1$ and ${\mathrm{pr}}_2$ are the two inclusions; under the identification of step 1.1 it is the morphism of affine schemes corresponding to the ring map $k[x,y]\to \mathrm{\Gamma }(U\cap V,{\mathcal{O}}_D)$ with $x\mapsto t$, $y\mapsto t$, where $U\cap V$ is the glued ${\mathbb{G}}_m=\mathrm{Spec}k[t,t^{-1}]$.

The point $(0,0)\in U{\times }_{k}V$ is not in $\mathrm{\Delta }(D)$: its first projection is the closed point $0_1$ of $U$ and its second projection is the closed point $0_2$ of $V$, and these are distinct points of $D$ by [F1]. Were $(0,0)=\mathrm{\Delta }(d)$ for some $d\in D$, then ${\mathrm{pr}}_1\mathrm{\Delta }(d)=0_1$ and ${\mathrm{pr}}_2\mathrm{\Delta }(d)=0_2$ would force $0_1=d=0_2$ by [F2], a contradiction.

The image of the morphism of step 2.1 is the set $V(x-y)\setminus \{(0,0)\}$: a point of $U\cap V$ has coordinate $t\mathrm{\ne }0$, so its image satisfies $x=y\mathrm{\ne }0$, and conversely a point of $\mathrm{Spec}k[x,y]$ with $x=y\mathrm{\ne }0$ lies in $D(xy)$ and is the image of the corresponding nonzero value of $t$.

The set $V(x-y)\setminus \{(0,0)\}$ is dense in $V(x-y)$: the line $V(x-y)\cong \mathrm{Spec}k[u]$ is irreducible, so removing the single closed point given by $x=y=0$ leaves a nonempty open subset, which is dense. Hence $(0,0)$, the maximal ideal $(x,y)$ of $k[x,y]$, lies in the closure of the image of $\mathrm{\Delta }$ inside the chart $U{\times }_{k}V$.

By steps 4.1 and 2.2 the diagonal image accumulates at a point of the chart $U{\times }_{k}V\subseteq D{\times }_{k}D$ that does not belong to it, so ${\mathrm{\Delta }}_{D\mathrm{/}k}(D)$ is not closed in $D{\times }_{k}D$. This is the concrete form of the failure of separatedness recorded in [F1]: for the doubled-origin line the diagonal is a locally closed subscheme whose closure is strictly larger than its image.