Diagonals Separated Morphisms and Valuative Uniqueness — Examples

By [F2] applied to $A\to A[x]$ twice, $\mathrm{Spec}A[x]{\times }_S\mathrm{Spec}A[x]\cong \mathrm{Spec}(A[x]{\otimes }_{A}A[x])$, and the latter is $\mathrm{Spec}A[x,y]$ with ${\mathrm{pr}}_1$ corresponding to $x\mapsto x\otimes 1=x$ and ${\mathrm{pr}}_2$ to $x\mapsto 1\otimes x=y$.

By [F1] the diagonal corresponds, under the identification of step 1.1, to a ring map $\mu :A[x,y]\to A[x]$ with $\mu (x)=x$ and $\mu (y)=x$, namely the multiplication $x\otimes 1\mapsto x$, $1\otimes x\mapsto x$.

The map $\mu$ is surjective, and its kernel is $(x-y)$: writing $A[x,y]=A[x][u]$ with $u=y-x$, the map $\mu$ is the $A[x]$-algebra map sending $u$ to $0$, whose kernel is the principal ideal $(u)=(x-y)$.

By [F3] the closed subscheme of $\mathrm{Spec}A[x,y]$ with ideal $(x-y)$ is presentable as $\mathrm{Spec}(A[x,y]\mathrm{/}(x-y))$, and $A[x,y]\mathrm{/}(x-y)\to A[x]$, $x\mapsto x$, $y\mapsto x$, is an isomorphism of $A$-algebras; hence the diagonal is exactly the closed subscheme $V(x-y)$ and in particular a closed immersion.

Now let $S$ be arbitrary and ${\mathbb{A}}_S^1=\mathrm{Spec}\mathbb{Z}[x]{\times }_{\mathrm{Spec}\mathbb{Z}}S$ as in [F6], and let $\mathrm{Spec}A\subseteq S$ be an affine open. Base changing along $\mathrm{Spec}A\hookrightarrow S$ produces the affine line $\mathrm{Spec}A[x]$ over $\mathrm{Spec}A$, whose diagonal is cut out by $x-y$ as computed in step 4.1, and by [F5] these local diagonal conditions are the restrictions to the open subscheme $\mathrm{Spec}A{\times }_S\mathrm{Spec}A$ of the product. Since these products over an affine open cover of $S$ cover ${\mathbb{A}}_S^1{\times }_S{\mathbb{A}}_S^1$, [F4] shows that the diagonal of ${\mathbb{A}}_S^1\to S$ is a closed immersion cut out by $x-y$ on each such piece, and by [F1] and [F3] its ideal in the chart ring $A[x,y]$ is $(x-y)$.

Over an affine base $S=\mathrm{Spec}A$ the four products of charts are affine by [F3]: $U_0{\times }_S{U}_0=\mathrm{Spec}A[x,y]$, $U_0{\times }_S{U}_1=\mathrm{Spec}A[x,z]$, $U_1{\times }_S{U}_0=\mathrm{Spec}A[v,y]$, $U_1{\times }_S{U}_1=\mathrm{Spec}A[v,z]$.

The equation $x_0{y}_1-x_1{y}_0=0$ is bihomogeneous of bidegree $(1,1)$, so its restriction to each product of charts is obtained by dividing by the two chosen coordinates; this gives the four displayed equations in the order $(U_0,V_0),(U_0,V_1),(U_1,V_0),(U_1,V_1)$.

On $U_0{\times }_S{U}_0=\mathrm{Spec}A[x,y]$ the equation becomes $y-x=0$, which is the difference of the two copies of the coordinate $x_1\mathrm{/}{x}_0$; by [F2] this is the chart form of the diagonal, and $A[x,y]\to A[x]$, $y\mapsto x$, exhibits it as a closed immersion with image the diagonal copy of $U_0$.

On $U_0{\times }_S{U}_1=\mathrm{Spec}A[x,z]$ the equation becomes $1-xz=0$, exactly [F2]'s mixed-chart generator $x_1^{(0)}{y}_0^{(1)}-1$ up to sign, with $z=y_0\mathrm{/}{y}_1$ and $x=x_1\mathrm{/}{x}_0$; the quotient is $A[x,x^{-1}]$ via $z\mapsto x^{-1}$, so the diagonal over this chart product is the closed subscheme isomorphic to the overlap $U_0\cap U_1$.

The remaining two products are obtained from steps 2.1 and 2.2 by swapping the two factors: on $U_1{\times }_S{U}_0=\mathrm{Spec}A[v,y]$ the equation becomes $vy=1$, and on $U_1{\times }_S{U}_1=\mathrm{Spec}A[v,z]$ it becomes $z-v=0$.

Both sides are compatible with base change along any $S^{\mathrm{\prime }}\to S$ by [F1], so the four chart computations glue: on every standard chart product the diagonal is the closed subscheme cut out by $x_0{y}_1-x_1{y}_0=0$, which is the assertion.

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.

Define $v(f\mathrm{/}g):={\mathrm{ord}}_0(f)-{\mathrm{ord}}_0(g)$ for nonzero $f,g\in k[t]$ and $v(0):=\mathrm{\infty }$, where ${\mathrm{ord}}_0$ is the order of vanishing at $0$. By [F4] this is a valuation: multiplicativity is clear from additivity of ${\mathrm{ord}}_0$ and the ultrametric inequality follows from the Taylor expansion of $f$ and $g$ at $0$; it is surjective onto $\mathbb{Z}$ since $v(t)=1$, so by [F3] it is a discrete valuation and $R=\{h\in K:v(h)\ge 0\}=k[t{]}_{(t)}$ is a discrete valuation ring with fraction field $K$.

Let $\mathrm{Spec}K\to D$ be the morphism with image the generic point of the shared ${\mathbb{G}}_m$, obtained by composing $k[x]\to K$, $x\mapsto t$, with the chart inclusion $U\hookrightarrow D$, and let $\mathrm{Spec}R\to \mathrm{Spec}k$ be the structure morphism. The square commutes, so this is a valuative diagram for $D\to \mathrm{Spec}k$ in the sense of [F2].

The generic point of $\mathrm{Spec}R$ lies in the shared overlap, so the composite $\mathrm{Spec}K\to U\hookrightarrow D$ coincides with $\mathrm{Spec}K\to V\hookrightarrow D$: both are given by the inclusion $k[t]\to k(t)$ read in the two charts, and $t$ is a unit in the glued ${\mathbb{G}}_m$.

The ring maps $k[x]\to R$, $x\mapsto t$, and $k[y]\to R$, $y\mapsto t$, define morphisms $u_R:\mathrm{Spec}R\to U\hookrightarrow D$ and $w_R:\mathrm{Spec}R\to V\hookrightarrow D$. Their composites to $\mathrm{Spec}k$ are the structure morphism. After restricting to $\mathrm{Spec}K$, they agree with the generic map of step 2.1 because the chart identifications on $D(x)\cong D(y)$ identify $x$ and $y$. Hence $u_R$ and $w_R$ are two lifts of the valuative diagram.

The lifts $u_R\mathrm{\ne }{w}_R$ are distinct: the closed point of $\mathrm{Spec}R$, corresponding to the maximal ideal $(t)$, is sent by $u_R$ to the origin $0_1$ of the chart $U$ and by $w_R$ to the origin $0_2$ of the chart $V$, and $0_1\mathrm{\ne }{0}_2$ by [F1].

Hence the displayed valuative diagram has two distinct lifts, refuting the claimed uniqueness for $D$. In this example the single discrete valuation ring $k[t{]}_{(t)}$ already detects the failure of uniqueness; this witness does not establish that DVR tests are insufficient for other morphisms.

The structure morphism ${\mathbb{A}}_k^n\to \mathrm{Spec}k$ is affine, hence separated by [F3], so [F2] applies to the $\mathrm{Spec}k$-morphism $g$ and the graph ${\mathrm{\Gamma }}_g$ is a closed immersion. By [F4] the product is $\mathrm{Spec}(k[x]{\otimes }_{k}k[y])=\mathrm{Spec}k[x_1,\dots ,x_m,y_1,\dots ,y_n]$, with ${\mathrm{pr}}_1$ acting as $x_i\mapsto x_i$ and ${\mathrm{pr}}_2$ as $y_j\mapsto y_j$.

Under the identification of step 1.1 the morphism ${\mathrm{\Gamma }}_g=(\mathrm{id},g)$ corresponds to the $k$-algebra map $\phi :k[x,y]\to k[x]$ with $\phi (x_i)=x_i$ and $\phi (y_j)=g_j(x)$.

The map $\phi$ is surjective and its kernel is the ideal $I=(y_1-g_1(x),\dots ,y_n-g_n(x))$: clearly $I\subseteq \ker \phi$, and conversely if $f\in \ker \phi$ then writing $f$ as a polynomial in the variables $u_j=y_j-g_j(x)$ with coefficients in $k[x]$ gives $f\equiv f(x,g(x))=0$ modulo $I$, so $f\in I$.

By [F5] the closed subscheme with ideal $I$ is, up to unique isomorphism over the product, the image of $\phi$, so the graph is the closed subscheme $V(I)$ of $\mathrm{Spec}k[x,y]$ with the displayed ideal.

By [F1] the composite ${\mathrm{pr}}_1\circ {\mathrm{\Gamma }}_g$ is the identity of ${\mathbb{A}}_k^m$, so the first projection restricts to a morphism ${\mathrm{\Gamma }}_g\to {\mathbb{A}}_k^m$ with inverse ${\mathrm{\Gamma }}_g$; hence ${\mathrm{pr}}_1{\mathrm{\mid }}_{{\mathrm{\Gamma }}_g}$ is an isomorphism. In coordinate rings this is the isomorphism $k[x,y]\mathrm{/}I\to k[x]$ inverse to $\phi$.

The degenerate cases are included: for $n=0$ the list of equations is empty, $I=0$, and the graph is the identity of ${\mathbb{A}}_k^m$; for $m=0$ the source is a single $k$-rational point and the graph is the closed point $(g_1,\dots ,g_n)\in {\mathbb{A}}_k^n$ cut out by $y_j-g_j$.

Steps 1.1, 4.1 and 4.2 show that ${\mathrm{\Gamma }}_g$ is a closed subscheme of ${\mathbb{A}}_k^m{\times }_k{\mathbb{A}}_k^n$ with ideal $(y_1-g_1(x),\dots ,y_n-g_n(x))$ whose first projection is an isomorphism onto ${\mathbb{A}}_k^m$, which is the assertion.

The affine line ${\mathbb{A}}_k^1=\mathrm{Spec}k[t]$ is affine over $\mathrm{Spec}k$, so by [F1] the structure morphism is separated. The points $(0)$ and $(t)$ of $\mathrm{Spec}k[t]$ are distinct: $(0)$ is the generic point, and the maximal ideal $(t)$ is a proper nonzero prime.

Let $U\subseteq \mathrm{Spec}k[t]$ be a nonempty open subset. Its complement is a proper closed subset, hence of the form $V(f)$ for some nonzero $f\in k[t]$; since $k[t]$ is a domain, a nonzero polynomial is not in the prime ideal $(0)$, so $(0)\mathrm{\notin }V(f)$ and therefore $(0)\in U$. Thus $(0)$ belongs to every nonempty open subset.

Let $U_1$ and $U_2$ be open neighbourhoods of the distinct points $(0)$ and $(t)$. Since $U_2$ is nonempty, step 1.2 gives $(0)\in U_2$, and $(0)\in U_1$ by definition; hence $U_1\cap U_2\ni (0)$ is nonempty.

Step 2.1 shows that no two distinct points of $\mathrm{Spec}k[t]$ have disjoint open neighbourhoods, so $\mathrm{\mid }{\mathbb{A}}_k^1\mathrm{\mid }$ is not Hausdorff, while step 1.1 shows that ${\mathbb{A}}_k^1$ is separated over $k$; the implication asserted in the statement is therefore false.

The morphism $j$ is an open immersion, so by [F1] it is separated; hence [F2] gives at most one lift for every valuative diagram for $j$.

By [F4] the ring $R=k[t{]}_{(t)}$ is a discrete valuation ring with fraction field $K=k(t)$, so $R\subseteq K$ is a valuation ring for the purposes of [F3].

Let $\mathrm{Spec}K\to D(t)$ correspond to the ring map $k[t,t^{-1}]\to K$ with $t\mapsto t$; its image is the generic point, which lies in $D(t)$. Let $\mathrm{Spec}R\to \mathrm{Spec}k[t]$ correspond to the localization $k[t]\to R$. The two composites $\mathrm{Spec}K\to \mathrm{Spec}k[t]$ agree, so this is a valuative diagram for $j$ in the sense of [F3].

Suppose there were a lift $u:\mathrm{Spec}R\to D(t)$. Then $u$ corresponds to a ring homomorphism $\psi :k[t,t^{-1}]\to R$ with $\psi (t)=t$, since composing $u$ with $j$ must give the base map, whose corresponding ring map is the localization $k[t]\to R$.

Here $t$ is a unit of $k[t,t^{-1}]$, so $\psi (t)$ must be a unit of $R$, every ring homomorphism sending units to units. But the image of $t$ under the localization $k[t]\to R$ is the element $t\in R$, which is not a unit: the maximal ideal of $R$ is $(t)$, so $t$ lies in the maximal ideal of the local ring $R$ and cannot be invertible there.

Steps 3.1 and 4.1 contradict each other, so no lift exists; combined with step 1.1, the displayed valuative diagram has exactly zero lifts although every valuative diagram for $j$ has at most one. This shows that the uniqueness part of the valuative criterion carries no existence assertion.

For each $n\ge 1$, write $u=t^{1\mathrm{/}n}$ and let ${\mathrm{ord}}_{u=0}$ be the usual order-of-vanishing discrete valuation on $k(u)$. Scale it by $1\mathrm{/}n$. These valuations are compatible under the transition maps indexed by divisibility, since $t^{1\mathrm{/}n}=(t^{1\mathrm{/}m}{)}^{m\mathrm{/}n}$ when $n\mid m$; multiplicativity and the ultrametric inequality for ${\mathrm{ord}}_{u=0}$ are preserved by this positive rescaling, so they define a valuation $v:K\to \mathbb{Q}\cup \{\mathrm{\infty }\}$ with $v(0)=\mathrm{\infty }$ [F2]. Its value group is $\mathbb{Q}$, since $v(t^{1\mathrm{/}n})=1\mathrm{/}n$. At each stage, the elements of nonnegative value are exactly $k[t^{1\mathrm{/}n}{]}_{(t^{1\mathrm{/}n})}$, so $V=\{z\in K:v(z)\ge 0\}$ is a valuation ring by [F1], with maximal ideal ${\mathfrak{m}}_V=\{z:v(z)>0\}\cup \{0\}$ and residue field $k$ (taking residues at any stage gives the same element of $k$). It is not a discrete valuation ring by [F2]. It is not Noetherian: if ${\mathfrak{m}}_V$ were generated by finitely many nonzero elements, the least positive value $q$ of those generators would bound below the value of every element they generate, whereas $t^{1\mathrm{/}n}\in {\mathfrak{m}}_V$ has value $1\mathrm{/}n<q$ for all sufficiently large $n$.

We show that every valuative diagram for $f$ over a discrete valuation ring $A$ has at most one lift. Let $A$ be a discrete valuation ring by [F2], with fraction field $L$, closed point $s$ and generic point ${\eta }_A$, and let $u,v:\mathrm{Spec}A\to X$ be two lifts of one diagram; then $u$ and $v$ agree on $\mathrm{Spec}L$, so $u({\eta }_A)=v({\eta }_A)=x$.

Every nonzero prime ideal $\mathfrak{p}$ of $V$ is ${\mathfrak{m}}_V$. Indeed, choose $0\mathrm{\ne }g\in \mathfrak{p}$; since $\mathfrak{p}$ is proper, $v(g)>0$. For any nonzero $h\in {\mathfrak{m}}_V$, choose $N$ with $Nv(h)\ge v(g)$. Then $h^N\mathrm{/}g\in V$, so $h^N\in gV\subseteq \mathfrak{p}$, and primality gives $h\in \mathfrak{p}$. Thus ${\mathfrak{m}}_V\subseteq \mathfrak{p}$, so equality holds. The only primes of $V$ are therefore $(0)$ and ${\mathfrak{m}}_V$. Fix $0\mathrm{\ne }g\in {\mathfrak{m}}_V$. For any $z\in K$, if $v(z)\ge 0$ then $z\in V$; if $v(z)<0$, choose $N$ with $v(z)+Nv(g)\ge 0$, so $z{g}^N\in V$ and $z=(z{g}^N)\mathrm{/}{g}^N\in V_g$. Hence $V_g=K$, and [F7] identifies $\mathrm{Spec}K$ with $D(g)\subseteq \mathrm{Spec}V$. Now [F3] glues $U$ and $W$ along this open to a scheme $X$ with $U\cap W=\mathrm{Spec}K$ and $U\cup W=X$, and the identity $V\to V$ on each chart induces $f:X\to S=\mathrm{Spec}V$.