Every scheme diagonal is an immersion

Choose an open cover $X={\bigcup }_i{U}_i$ with $U_i=\mathrm{Spec}{B}_i$ affine and $f(U_i)\subseteq V_i=\mathrm{Spec}{A}_i$ affine, and put $Q_i={\mathrm{pr}}_1^{-1}(U_i)\cap {\mathrm{pr}}_2^{-1}(U_i)$. By [F6] the open subscheme $Q_i$ represents $U_i{\times }_{V_i}{U}_i$, so [F4] identifies $Q_i$ with $\mathrm{Spec}(B_i{\otimes }_{A_i}{B}_i)$, an open subscheme of $X{\times }_{S}X$.

For each $x\in X$, choose $i$ with $x\in U_i$. Since both projections of ${\mathrm{\Delta }}_{X\mathrm{/}S}(x)$ are $x$ by [F1], this point lies in $Q_i$. Thus ${\mathrm{\Delta }}_{X\mathrm{/}S}(X)\subseteq Q:={\bigcup }_i{Q}_i$.

If $z={\mathrm{\Delta }}_{X\mathrm{/}S}(x)$, then ${\mathrm{pr}}_1(z)={\mathrm{pr}}_2(z)=x$ by [F1], and the residue-field map induced by either projection is the inverse of the isomorphism induced by ${\mathrm{\Delta }}_{X\mathrm{/}S}$, because the composites ${\mathrm{pr}}_j{\mathrm{\Delta }}_{X\mathrm{/}S}={\mathrm{id}}_X$ induce identity maps on residue fields; in particular the two projections induce the same map $\kappa (x)\to \kappa (z)$.

Each $Q_i$ is open by step 1.1, so $Q$ is an open subscheme of $X{\times }_{S}X$ containing the image of ${\mathrm{\Delta }}_{X\mathrm{/}S}$ by step 1.2.

The inverse image ${\mathrm{\Delta }}_{X\mathrm{/}S}^{-1}(Q_i)$ equals $U_i$, and the restricted morphism $U_i\to Q_i$ is, under the identifications of step 1.1, the morphism $\mathrm{Spec}{B}_i\to \mathrm{Spec}(B_i{\otimes }_{A_i}{B}_i)$ corresponding to the multiplication map $m_i:B_i{\otimes }_{A_i}{B}_i\to B_i$, $b\otimes b^{\mathrm{\prime }}\mapsto b{b}^{\mathrm{\prime }}$. This $m_i$ is surjective, since $b=m_i(b\otimes 1)$, and then [F5] exhibits the restricted morphism as a closed immersion; the cases $B_i=0$ and $A_i=0$ of the zero ring are included, with $m_i$ again surjective. Consequently ${\mathrm{\Delta }}_{X\mathrm{/}S}^{-1}(Q)={\bigcup }_i{\mathrm{\Delta }}_{X\mathrm{/}S}^{-1}(Q_i)=X$.

Conversely let $z$ satisfy ${\mathrm{pr}}_1(z)={\mathrm{pr}}_2(z)=x$ with a common induced residue map $\psi :\kappa (x)\to \kappa (z)$. By [F7] the point $z$ corresponds to a quadruple $(x,x,s,\mathfrak{r})$ with $\kappa (z)=\kappa (\mathfrak{r})$, and the ring map $\phi :\kappa (x){\otimes }_{\kappa (s)}\kappa (x)\to \kappa (z)$ with kernel $\mathfrak{r}$ restricts to $\psi$ on each tensor factor. Since these restrictions agree, the universal property of the tensor product factors $\phi$ through the multiplication $\kappa (x){\otimes }_{\kappa (s)}\kappa (x)\to \kappa (x)$, whose kernel ${\mathfrak{m}}_0$ is therefore contained in $\mathfrak{r}$. That multiplication maps onto the field $\kappa (x)$, so ${\mathfrak{m}}_0$ is maximal and proper, and the prime $\mathfrak{r}$ containing it is equal to it. By step 1.3 the point ${\mathrm{\Delta }}_{X\mathrm{/}S}(x)$ is the point of the product whose quadruple is $(x,x,s,{\mathfrak{m}}_0)$, so injectivity of the bijection of [F7] gives $z={\mathrm{\Delta }}_{X\mathrm{/}S}(x)$.

Steps 1.2 and 2.1 show that the open subscheme $Q$ of $X{\times }_{S}X$ contains ${\mathrm{\Delta }}_{X\mathrm{/}S}(X)$, and step 2.2 gives ${\mathrm{\Delta }}_{X\mathrm{/}S}^{-1}(Q)=X$ and shows that the restriction of ${\mathrm{\Delta }}_{X\mathrm{/}S}$ to each member of the open cover $\{Q_i\}$ of $Q$ is a closed immersion. By [F3] the morphism ${\mathrm{\Delta }}_{X\mathrm{/}S}:X\to Q$ is a closed immersion.

Composing the closed immersion ${\mathrm{\Delta }}_{X\mathrm{/}S}:X\to Q$ of step 3.1 with the open immersion $Q\hookrightarrow X{\times }_{S}X$ exhibits ${\mathrm{\Delta }}_{X\mathrm{/}S}$ as an immersion by [F2]; explicitly, $X$ is identified with the closed subscheme ${\mathrm{\Delta }}_{X\mathrm{/}S}(X)$ of $Q$, cut out on the chart $Q_i$ by the kernel of $m_i$ of step 2.2.

Steps 4.1 and 2.3 prove that ${\mathrm{\Delta }}_{X\mathrm{/}S}$ is an immersion and that a point of $X{\times }_{S}X$ lies in its image exactly when the two projections carry it to one point $x$ of $X$ and induce one and the same map $\kappa (x)\to \kappa (z)$.