Finite Proper and Projective Morphisms — Examples

For $n\ge 1$, every exponent $m\ge 0$ has a unique form $m=qn+i$ with $q\ge 0$ and $0\le i<n$. Thus each polynomial in $k[t]$ has a unique expression ${\sum }_{i=0}^{n-1}{f}_i(t^n)t^i$ with $f_i\in k[s]$. Existence follows by grouping its monomials by their remainder modulo $n$; uniqueness follows because the exponents $qn+i$ are distinct for distinct pairs $(q,i)$. Taking only the $i=0$ term also shows that the ring map $k[s]\to k[t]$, $s\mapsto t^n$, is injective. Therefore $1,t,\dots ,t^{n-1}$ is a free $k[s]$-basis of rank $n$. [F1, algebra] 1.2 Suppose $\mathrm{char}(k)=p>0$ and $p\mid n$. By [F13], $p\cdot 1_k=0$, and by [F14], $p$ is prime; set $r=n\mathrm{/}p$, so $0<r<n$. In the generic extension $k(s)\subseteq k(t)$, let $\alpha =t^r$. It satisfies ${\alpha }^p=s$, hence is algebraic by [F15]. It is not in $k(s)$: if $t^r=P(t^n)\mathrm{/}Q(t^n)$ for $P,Q\in k[s]$ and $Q\mathrm{\ne }0$, then $t^{r}Q(t^n)=P(t^n)$. Every exponent on the left is congruent to $r$ modulo $n$, while every exponent on the right is divisible by $n$; since $0<r<n$, equality is impossible. Let $m_{\alpha }$ be its minimal polynomial. By [F16], $m_{\alpha }\mid X^p-s$. In an algebraic closure, $X^p-s=(X-\alpha {)}^p$ because ${\alpha }^p=s$. Since $\alpha \mathrm{\notin }k(s)$, $m_{\alpha }$ has degree at least $2$; all its roots are $\alpha$, so it has a repeated root. By [F17] and [F18], $\alpha$ is inseparable over $k(s)$, so the generic extension is not separable. [F12, F13, F14, F15, F16, F17, F18, algebra] 1.3 For $n=0$, the coordinate map is $k[s]\to k[t]$, $s\mapsto 1$. The $k[s]$-module action on $k[t]$ factors through $k[s]\mathrm{/}(s-1)\cong k$, so if it were finitely generated as a $k[s]$-module then $k[t]$ would be finite-dimensional over $k$. This is impossible because $1,t,t^2,\dots$ are linearly independent over $k$. The finite-morphism condition already fails on the whole target affine open, so the constant map is not finite. [F3, F4, algebra] 2.1 Let $U\subseteq \mathrm{Spec}k[s]$ be any affine open. Its closed complement is $V(I)$ for an ideal $I$ by [F6], and [F5] writes $I=(g)$, so $U=D(g)$ by [F7]. This includes $U=\mathrm{\varnothing }$ with $g=0$ and the whole target with $g=1$. For $g\mathrm{\ne }0$, [F2] and [F8] identify its inverse image with $D(g(t^n))=\mathrm{Spec}k[t{]}_{g(t^n)}$ and its coordinate map with $k[s{]}_g\to k[t{]}_{g(t^n)}$. The image $g(t^n)$ is nonzero by step 1.1. Every localized element is $h(t)\mathrm{/}g(t^n{)}^N$ by [F9]; writing $h$ in the basis from step 1.1 shows that the localized basis spans over $k[s{]}_g$. To prove independence, clear the coefficient denominators in a relation. By [F11], some power of $g(t^n)$ then kills the resulting numerator. The ring $k[t]$ is a domain, since leading coefficients of nonzero polynomials over $k$ multiply to a nonzero coefficient, so this power can be cancelled. The original basis independence then makes every coefficient zero. Thus the localized algebra is module-finite over $k[s{]}_g$. If $g=0$, [F10] gives the zero coordinate ring on the empty inverse image and empty target open, and the zero module is finite. By [F3] and [F4], these checks on every affine target open prove finiteness. [F2, F3, F4, F5, F6, F7, F8, F9, F10, F11, step 1.1, algebra] 3.1 At $n=1$, the basis is $\{1\}$ and the map is the identity; the generic extension is $k(t)\mathrm{/}k(t)$ and is separable. The proof uses no choice: the basis is explicit, an arbitrary affine target open is handled one at a time, and the characteristic argument uses one explicit element. The empty target open is handled in step 2.1, and there is no empty-source case or interval endpoint. The source is nonempty because $k[t]$ is a domain and $(0)$ is prime. [F12, step 1.1, step 2.1, step 1.2, step 1.3] $\mathrm{\square }$

By the converse clause of [F1] applied to the quotient map $\pi :A\to A\mathrm{/}I$, the induced morphism $i$ is a closed immersion. Now fix an affine open $U=\mathrm{Spec}{A}^{\mathrm{\prime }}\subseteq \mathrm{Spec}A$. By the first clause of [F1] applied to the closed immersion $i$ and this $U$, there is a unique ideal $I^{\mathrm{\prime }}\subseteq A^{\mathrm{\prime }}$ with $i^{-1}(U)\cong \mathrm{Spec}(A^{\mathrm{\prime }}\mathrm{/}{I}^{\mathrm{\prime }})$ over $U$; in particular the inverse image $i^{-1}(U)$ is affine.

The quotient ring $A^{\mathrm{\prime }}\mathrm{/}{I}^{\mathrm{\prime }}$ is a cyclic $A^{\mathrm{\prime }}$-module: the class $1+I^{\mathrm{\prime }}$ generates it, because every $a+I^{\mathrm{\prime }}=a\cdot (1+I^{\mathrm{\prime }})$ lies in the submodule generated by $1+I^{\mathrm{\prime }}$, and that submodule is contained in $A^{\mathrm{\prime }}\mathrm{/}{I}^{\mathrm{\prime }}$. If $I^{\mathrm{\prime }}=A^{\mathrm{\prime }}$ then $A^{\mathrm{\prime }}\mathrm{/}{I}^{\mathrm{\prime }}$ is the zero module, which is generated by the empty family. Hence $A^{\mathrm{\prime }}\mathrm{/}{I}^{\mathrm{\prime }}$ is finitely generated, equivalently module-finite, over $A^{\mathrm{\prime }}$ by [F4] and [F5].

Steps 1.1 and 1.2 verify the condition of [F3] on the affine open $U$: the inverse image $i^{-1}(U)=\mathrm{Spec}(A^{\mathrm{\prime }}\mathrm{/}{I}^{\mathrm{\prime }})$ is affine, and its coordinate algebra is module-finite over the coordinate algebra $A^{\mathrm{\prime }}$ of $U$. Since $U$ was an arbitrary affine open of $\mathrm{Spec}A$, the morphism $i$ is finite.

By step 1.1 the morphism $i$ is a closed immersion, so the AC-qualified [F2] shows that $i$ is proper. Combining with step 2.1, the morphism $\mathrm{Spec}(A\mathrm{/}I)\to \mathrm{Spec}A$ is both finite and proper.

Extreme cases. If $I=A$ then $A\mathrm{/}I=0$ and the source is $\mathrm{Spec}0=\mathrm{\varnothing }$, the empty closed immersion, to which [F1] and [F2] apply with the empty case included; if $I=0$ then $A\mathrm{/}I\cong A$ canonically and $i$ is the identity morphism, a closed immersion by [F1] and proper by [F2]; if $A=0$ then source and target are both empty and $i$ is again the identity of the empty scheme, covered by the same argument. The Axiom of Choice [F6] is assumed and is used only through the AC-qualified suppliers [F1] and [F2]. No Noetherian, field, reducedness or nonemptiness hypothesis is imposed.

Since not all $a_j$ are zero, fix an index $i$ with $a_i\mathrm{\ne }0$. By [F8], applied to the inclusion $R\hookrightarrow K$ and the elements $a_{\mathrm{\ell }}\mathrm{/}{a}_i\in K$ for $\mathrm{\ell }\mathrm{\ne }i$, there is a unique $R$-algebra map $\psi :R[x_{\mathrm{\ell }}^{(i)}:\mathrm{\ell }\mathrm{\ne }i]\to K$ with $\psi (x_{\mathrm{\ell }}^{(i)})=a_{\mathrm{\ell }}\mathrm{/}{a}_i$; by [F3] it corresponds to a morphism $\mathrm{Spec}K\to U_i^R\subseteq {\mathbb{P}}_R^n$, which is the $K$-point $[a_0:\cdots :a_n]$ of the tuple and is a morphism over $\mathrm{Spec}R$ because the composite $R\to R[x_{\mathrm{\ell }}^{(i)}]\stackrel{\psi }{\to }K$ is the structure map $R\hookrightarrow K$ of [F2]. Replacing the tuple by $(\lambda a_j)$ with $\lambda \in K^{\times }$ multiplies each ratio $a_{\mathrm{\ell }}\mathrm{/}{a}_i$ by $\lambda {\lambda }^{-1}=1$, hence gives the same $\psi$ and the same point.

There is an index $m$ with $a_m\mathrm{\ne }0$ and $a_j\mathrm{/}{a}_m\in R$ for every $j$. Start with $m:=i$, so that $a_m=a_i\mathrm{\ne }0$; process the indices $j\mathrm{\ne }i$ one at a time, maintaining the invariant that $a_m\mathrm{\ne }0$ and $a_{j^{\mathrm{\prime }}}\mathrm{/}{a}_m\in R$ for every already processed $j^{\mathrm{\prime }}$. If $a_j=0$, then $a_j\mathrm{/}{a}_m=0\in R$ and $m$ is kept. If $a_j\mathrm{\ne }0$, apply [F2] to $x=a_j\mathrm{/}{a}_m\in K^{\times }$: either $a_j\mathrm{/}{a}_m\in R$ and $m$ is kept, or $a_m\mathrm{/}{a}_j\in R$ and we replace $m$ by $j$; in the second case $a_{j^{\mathrm{\prime }}}\mathrm{/}{a}_j=(a_{j^{\mathrm{\prime }}}\mathrm{/}{a}_m)(a_m\mathrm{/}{a}_j)\in R$ for every processed $j^{\mathrm{\prime }}$, while $a_j\mathrm{/}{a}_j=1$, so the invariant is preserved. After the finitely many indices have been processed, $a_j\mathrm{/}{a}_m\in R$ for all $j$ and $a_m\mathrm{\ne }0$.

Conversely every morphism $p:\mathrm{Spec}K\to {\mathbb{P}}_R^n$ over $\mathrm{Spec}R$ arises from such a tuple: by [F1] the charts cover ${\mathbb{P}}_R^n$, so the image of the unique point of $\mathrm{Spec}K$ lies in some chart $U_i^R$, and $p$ factors through the open immersion $U_i^R\hookrightarrow {\mathbb{P}}_R^n$ by [F4]. The factorisation corresponds by [F3] to an $R$-algebra map $\psi :R[x_{\mathrm{\ell }}^{(i)}:\mathrm{\ell }\mathrm{\ne }i]\to K$, and putting $a_{\mathrm{\ell }}:=\psi (x_{\mathrm{\ell }}^{(i)})$ for $\mathrm{\ell }\mathrm{\ne }i$ and $a_i:=1$ gives a tuple with $a_i\mathrm{\ne }0$ that determines $p$ by step 1.1.

Since $\psi (x_m^{(i)})=a_m\mathrm{/}{a}_i\mathrm{\ne }0$ by step 1.2, the image point of $p$ lies in the distinguished open $D(x_m^{(i)})=U_i^R\cap U_m^R$ of [F1]; hence $p$ factors through the open subscheme $U_m^R$ by [F4]. By the transition formula of [F1] the factorisation corresponds by [F3] to the $R$-algebra map $R[x_{\mathrm{\ell }}^{(m)}:\mathrm{\ell }\mathrm{\ne }m]\to K$ sending $x_{\mathrm{\ell }}^{(m)}$ to $c_{\mathrm{\ell }}:=a_{\mathrm{\ell }}\mathrm{/}{a}_m$ for $\mathrm{\ell }\mathrm{\ne }m$, and $c_{\mathrm{\ell }}\in R$ for every $\mathrm{\ell }\mathrm{\ne }m$ by step 1.2.

By [F8], applied to the inclusion $R\hookrightarrow R$ and the elements $c_{\mathrm{\ell }}\in R$ of step 2.2, there is a unique $R$-algebra map $\phi :R[x_{\mathrm{\ell }}^{(m)}:\mathrm{\ell }\mathrm{\ne }m]\to R$ with $\phi (x_{\mathrm{\ell }}^{(m)})=c_{\mathrm{\ell }}$; by [F3] it corresponds to a morphism $q:\mathrm{Spec}R\to U_m^R\subseteq {\mathbb{P}}_R^n$.

The composite $\pi \circ q$ corresponds by [F3] to the ring map $R\to R[x_{\mathrm{\ell }}^{(m)}]\stackrel{\phi }{\to }R$, which is the identity of $R$; by the affine anti-equivalence [F3], this identity of ring maps gives $\pi \circ q={\mathrm{id}}_{\mathrm{Spec}R}$.

The composite $q\circ j$ corresponds by [F3] to the ring map $R[x_{\mathrm{\ell }}^{(m)}]\stackrel{\phi }{\to }R\hookrightarrow K$, which sends $x_{\mathrm{\ell }}^{(m)}$ to $c_{\mathrm{\ell }}$; by step 2.2 this is the same $R$-algebra map $R[x_{\mathrm{\ell }}^{(m)}]\to K$ that describes the factorisation of $p$ through $U_m^R$, so $q\circ j=p$.

Let $q^{\mathrm{\prime }}:\mathrm{Spec}R\to {\mathbb{P}}_R^n$ be any morphism over $\mathrm{Spec}R$ with $q^{\mathrm{\prime }}\circ j=p$. Then $q$ and $q^{\mathrm{\prime }}$ are both lifts of the valuative diagram of [F6] consisting of $j$, the generic map $p$ and the base morphism ${\mathrm{id}}_{\mathrm{Spec}R}$: indeed $\pi q=\mathrm{id}$ and $\pi q^{\mathrm{\prime }}=\mathrm{id}$ and $qj=p=q^{\mathrm{\prime }}j$. Since $\pi$ is separated by [F5], [F7] gives $q^{\mathrm{\prime }}=q$. Moreover the condition of being over $\mathrm{Spec}R$ is automatic for a morphism $q^{\mathrm{\prime }}$ with $q^{\mathrm{\prime }}\circ j=p$: let $u:R\to R$ be the ring map corresponding to $\pi \circ q^{\mathrm{\prime }}$ by [F3]. Since $(\pi q^{\mathrm{\prime }})j=\pi p=j$, the composite $R\stackrel{u}{\to }R\hookrightarrow K$ equals the given inclusion $R\hookrightarrow K$. The inclusion is injective, hence $u={\mathrm{id}}_R$ and $\pi q^{\mathrm{\prime }}={\mathrm{id}}_{\mathrm{Spec}R}$ by [F3]. Thus no extension of $p$ to $\mathrm{Spec}R$ other than $q$ exists, and the extension is unique.

This completes the example: the index $i$ of the statement is the index constructed in step 1.2, the ratios $a_j\mathrm{/}{a}_i$ are the elements $c_{\mathrm{\ell }}$ of step 2.2, and step 2.2, step 3.1, step 4.1, step 4.2 and step 5.1 exhibit them as the unique extension $\mathrm{Spec}R\to {\mathbb{P}}_R^n$ of the point. The construction is explicit and uses no choice principle: only finitely many coordinates are inspected and the index is updated by the dichotomy [F2], so the case of zero coordinates is included, the case $n=0$ has the single chart $U_0^R=\mathrm{Spec}R$ with no variables and the structure maps as $\psi$ and $\phi$, and the case $R=K$ is included as a valuation ring that is a field.

The multiplicative set $S$ contains no zero divisors of $k[t]$, and $k[t]$ is a domain by [F6], so [F5] shows that the localisation map $k[t]\to R$ is injective; hence $R$ is a domain and $(0)$ is a prime of $R$. Since $(t)\subseteq k[t]$ is a prime disjoint from $S$, [F4] shows that $R$ is a local ring with unique maximal ideal $\mathfrak{m}=tR$, and $t\in \mathfrak{m}$ is therefore not a unit of $R$. In particular $\mathrm{Spec}R$ contains the two points $(0)$ and $\mathfrak{m}$.

By [F9], ${\mathbf{A}}_k^1=\mathrm{Spec}k[x]$ over $S_0=\mathrm{Spec}k$, and $\mathrm{Spec}R\to S_0$ is the morphism induced by the field map $k\to R$. The canonical map $k[x]{\otimes }_{k}R\to R[x]$, $x^i\otimes r\mapsto r{x}^i$, is an isomorphism of $R$-algebras because $k[x]$ is the free $k$-module with basis $x^i$; hence [F7] identifies the base change ${\mathbf{A}}_k^1{\times }_{S_0}\mathrm{Spec}R$ with $\mathrm{Spec}R[x]$, with projection $q:\mathrm{Spec}R[x]\to \mathrm{Spec}R$ induced by $R\hookrightarrow R[x]$. By [F8] this $q$ is exactly the base change of ${\mathbf{A}}_k^1\to \mathrm{Spec}k$ along $\mathrm{Spec}R\to \mathrm{Spec}k$.

Let $Z=V(tx-1)\subseteq \mathrm{Spec}R[x]$ be the closed subset defined by the element $tx-1$. We compute its image. If $P\in Z$ is a prime containing $tx-1$ and if $t\in P\cap R$, then $tx-(tx-1)=1\in P$, a contradiction; so $t\mathrm{\notin }P\cap R$ and $P\cap R\in D(t)$. Conversely, let $\mathfrak{q}\in D(t)$ be a prime of $R$ and let $\psi :R[x]\to \mathrm{Frac}(R\mathrm{/}\mathfrak{q})$ be the $R$-algebra map sending $x$ to the inverse of the image of $t$, which is legitimate because $t\mathrm{\notin }\mathfrak{q}$. Its kernel $P$ is prime, contains $tx-1$, and satisfies $P\cap R=\mathfrak{q}$ because the composite $R\to \mathrm{Frac}(R\mathrm{/}\mathfrak{q})$ has kernel $\mathfrak{q}$. Hence the image of $Z$ under the base-changed projection $q$ of step 1.2 is exactly the basic open $D(t)\subseteq \mathrm{Spec}R$.

The subset $D(t)$ is open but not closed. It is nonempty because $(0)\in D(t)$ by step 1.1, and it is not all of $\mathrm{Spec}R$ because the maximal ideal $\mathfrak{m}=tR$ contains $t$ and so lies outside $D(t)$. If $D(t)$ were closed, then by [F3] it would equal $V(I)$ for the radical ideal $I$; from $(0)\in V(I)$ we get $I\subseteq (0)$, hence $I=0$, hence $V(I)=\mathrm{Spec}R\mathrm{\ne }D(t)$, a contradiction. Therefore $q$, whose image of the closed subset $Z$ is $D(t)$ by step 2.1, is not a closed map.

By [F2] and step 1.2, a nonclosed base-changed projection exhibits a failure of universal closedness of ${\mathbf{A}}_k^1\to \mathrm{Spec}k$, so that morphism is not universally closed and hence not proper by [F1]. The morphism is nevertheless of finite type and separated, being affine, so the failure is exactly in universal closedness. No choice principle is used: the prime witnessing nonclosedness is the maximal ideal $tR$, and the prime making $D(t)$ nonempty is $(0)$. Equivalently, the point $x=1\mathrm{/}t\in K$ has no $R$-lift, since a compatible $R$-point would give a $k$-algebra map $k[x]\to R$ with $x\mapsto 1\mathrm{/}t$ while $1\mathrm{/}t\mathrm{\notin }R$ because $t$ is not a unit of $R$ by step 1.1.

The ring $R=k[t]$ is an integral domain by [F5], so $(0)$ is a prime ideal and is a point of $X=\mathrm{Spec}R={\mathbf{A}}_k^1$. The indeterminate satisfies $t\mathrm{\ne }0$ in $k[t]$, so $t\mathrm{\notin }(0)$ and therefore $(0)\in D(t)$.

The principal ideal $(t)$ equals $\{f\in k[t]:f(0)=0\}$. For $f=tg$ one has $f(0)=0\cdot g(0)=0$, so $(t)$ is contained in that set and in particular $t\in (t)$ by [F4]; conversely, if $f(0)=0$ then the constant coefficient of $f$ vanishes, so writing $f$ as its finite coefficient sum gives $f=t\cdot ({\sum }_{i\ge 1}{a}_i{t}^{i-1})\in (t)$. Hence $(t)$ is a prime ideal: it is proper because $1\mathrm{\notin }(t)$, as $\mathrm{ev}(1)=1\mathrm{\ne }0$; and if $fg\in (t)$ then $f(0)g(0)=\mathrm{ev}(f)\mathrm{ev}(g)=\mathrm{ev}(fg)=0$, so $f(0)=0$ or $g(0)=0$ because $k$ is an integral domain, and then $f\in (t)$ or $g\in (t)$. Thus $(t)$ is a point of $X$ and $t\in (t)$, so $(t)\mathrm{\notin }D(t)$.

The set $D(t)=X\setminus V((t))$ is open in $X$ by [F2], and the inclusion $j:D(t)\to X$ of this open subscheme is an open immersion by [F7] whose image is exactly the subset $D(t)\subseteq X$. By steps 1.1 and 1.2 that image contains $(0)$ and omits $(t)$, so it is a nonempty proper subset of $X$.

The image $D(t)$ is not closed in $X$. If it were closed, then, the closed subsets of $X$ being the vanishing sets [F2], there would be an ideal $I\subseteq k[t]$ with $D(t)=V(I)$; by [F5] the ideal is principal, $I=(g)$, so $D(t)=V(g)$. Since $(0)\in D(t)$ we get $g\in (0)$ by [F3], that is $g=0$, and then $D(t)=V((0))=X$ by [F3], contradicting the properness of the image recorded in step 2.1.

It follows that $j$ is not a closed map of topological spaces: the whole source $D(t)$ is a closed subset of the source, and its image under $j$ is $D(t)$, which is not closed in $X$ by step 3.1.

Suppose $j$ were universally closed. Taking the identity morphism $X\to X$ as test morphism, the definition [F8] would make the base-changed projection $D(t){\times }_{X}X\to X$ a closed map. By [F9] the scheme $D(t){\times }_{X}X$ is projection-compatibly isomorphic to $D(t)$, and under this isomorphism the projection corresponds to $j$, so $j$ would be a closed map, contradicting step 4.1. Therefore $j$ is not universally closed.

Since properness requires universal closedness by [F8], the open immersion $j$ is not proper; directly, if $j$ were proper then [F8] would make its image closed, contradicting step 3.1. So the claim that every open immersion is proper is refuted. The argument applies to every field, including finite fields and fields of characteristic two, since the two points $(0)$ and $(t)$ of $X$ exist for every field; both are exhibited explicitly and no choice principle is used, the only inputs being the definitional closedness description of the Zariski topology, the principal-ideal property of $k[t]$, and the elementary computation $(t)=\{f:f(0)=0\}$ of step 1.2.