Cartier and Weil Divisors Line Bundles and Picard Groups — Examples

The charts and all points except infinity. By [F1], ${\mathbb{P}}_k^1=U_0\cup U_{\mathrm{\infty }}$, the overlap $D(t)\subseteq U_0$ is identified with $D(u)\subseteq U_{\mathrm{\infty }}$, and $tu=1$ on it. The ideal $(u)\subseteq k[u]$ is maximal with residue field $k[u]\mathrm{/}(u)\cong k$, so every prime of $k[u]$ containing $u$ equals $(u)$; hence every point of $U_{\mathrm{\infty }}$ other than $\mathrm{\infty }=(u)$ lies in the basic open $D(u)=U_0\cap U_{\mathrm{\infty }}\subseteq U_0$. Therefore every point of ${\mathbb{P}}_k^1$ other than $\mathrm{\infty }$ lies in $U_0$, and by [F2] it corresponds to a prime $\mathfrak{p}\subseteq k[t]$ with ${\mathcal{O}}_{{\mathbb{P}}_k^1,x}=k[t{]}_{\mathfrak{p}}$.

Integrality, normality and the function field. ${\mathbb{P}}_k^1$ is an integral finite-type $k$-scheme, every local ring of it is an integrally closed domain, and its function field is $k(t)=k(u)$ with $t=u^{-1}$. $U_0$ and $U_{\mathrm{\infty }}$ are integral affine schemes because $k[t]$ and $k[u]$ are domains, and they are glued along the nonempty open subscheme $D(t)\cong D(u)$. The two irreducible open charts have nonempty overlap, which is dense in both, so their union is irreducible, and stalks of the glued scheme are stalks of one of the two charts, so reducedness is inherited; hence ${\mathbb{P}}_k^1$ is integral, and it is finite type over $k$ because its affine charts are. Every local ring of ${\mathbb{P}}_k^1$ is a localisation $k[t{]}_{\mathfrak{p}}$ or $k[u{]}_{\mathfrak{q}}$ of an integrally closed domain [1.1], and localisations of integrally closed domains are integrally closed: if $x\in \mathrm{Frac}(A)$ is integral over $S^{-1}A$ with monic equation $x^n+{\sum }_{i<n}(a_i\mathrm{/}{s}_i)x^i=0$, then with $s={\prod }_{i<n}{s}_i$ the element $sx$ satisfies the monic equation $(sx{)}^n+{\sum }_{i<n}{a}_i(s^{n-i}\mathrm{/}{s}_i)(sx{)}^i=0$ over $A$, obtained by multiplying the equation of $x$ by $s^n$, whose coefficients lie in $A$; so $sx\in A$ and $x\in S^{-1}A$. The rings $k[t]$, $k[u]$ are Noetherian [F3], so ${\mathbb{P}}_k^1$ is a normal Noetherian scheme [F4]. By [F5] the function field is $\mathrm{Frac}k[t]=k(t)$, and $k(t)=k(u)$ with $t=u^{-1}$ on the overlap; in particular $f=t^2\mathrm{/}(t-1)$ is a global meromorphic unit.

The coordinates are uniformisers. The local ring at $0=(t)$ is $k[t{]}_{(t)}$, with maximal ideal generated by $t$; similarly $t-1$ generates the maximal ideal of $k[t{]}_{(t-1)}$ at $1$, and $u$ generates the maximal ideal of $k[u{]}_{(u)}$ at $\mathrm{\infty }$. Since $\dim k[t]=\dim k[u]=1$ and the ideals $(t)$, $(t-1)$, $(u)$ are nonzero, each of these local rings has dimension one and is therefore a discrete valuation ring [F4]; in each of them the displayed generator is a uniformiser, so $v_{[0]}(t)=v_{[1]}(t-1)=v_{\mathrm{\infty }}(u)=1$ and $v_{[0]}(t-1)=v_{[1]}(t)=v_{\mathrm{\infty }}(1-u)=0$, because $t-1\mathrm{\notin }(t)$, $t\mathrm{\notin }(t-1)$ and $1-u\mathrm{\notin }(u)$ are units.

Orders at the origin and at one. In $k(t)$ one has $f=t^2\cdot (t-1{)}^{-1}$, and ${\mathrm{ord}}_{[0]}(f)=2$, ${\mathrm{ord}}_{[1]}(f)=-1$. By additivity of the order, ${\mathrm{ord}}_{[0]}(f)=2v_{[0]}(t)-v_{[0]}(t-1)=2\cdot 1-0=2$ and ${\mathrm{ord}}_{[1]}(f)=2v_{[1]}(t)-v_{[1]}(t-1)=2\cdot 0-1=-1$.

Order at infinity. On $U_{\mathrm{\infty }}$ the coordinate is $u=1\mathrm{/}t$, and $f=u^{-1}(1-u{)}^{-1}$ in $k(u)=k(t)$, so ${\mathrm{ord}}_{\mathrm{\infty }}(f)=-1$. Indeed $f=\frac{t^2}{t-1}=\frac{u^{-2}}{u^{-1}-1}=\frac{1}{u(1-u)}=u^{-1}(1-u{)}^{-1}$, and with the uniformiser $u$ and the unit $1-u$ of step 1.3, additivity gives ${\mathrm{ord}}_{\mathrm{\infty }}(f)=-v_{\mathrm{\infty }}(u)-v_{\mathrm{\infty }}(1-u)=-1-0=-1$.

Every other prime divisor has order zero. If $Z\mathrm{\ne }[0],[1],[\mathrm{\infty }]$ is a prime divisor with generic point $\xi$, then ${\mathrm{ord}}_Z(f)=0$. By 1.1 the point $\xi$ lies in $U_0$ and corresponds to a prime $\mathfrak{p}\subseteq k[t]$ with ${\mathcal{O}}_{{\mathbb{P}}_k^1,\xi }=k[t{]}_{\mathfrak{p}}$; since $Z$ is a prime divisor, $\dim k[t{]}_{\mathfrak{p}}=1=\mathrm{ht}(\mathfrak{p})$, so $\mathfrak{p}\mathrm{\ne }(0)$ [F4]. The maximal ideals $(t)$ and $(t-1)$ of $k[t]$ are the primes of the points $[0]$ and $[1]$; as they are maximal, $t\in \mathfrak{p}$ forces $\mathfrak{p}=(t)$, and $t-1\in \mathfrak{p}$ forces $\mathfrak{p}=(t-1)$. Hence $t,t-1\mathrm{\notin }\mathfrak{p}$, both elements are units of $k[t{]}_{\mathfrak{p}}$, and $f=t^2(t-1{)}^{-1}$ is a unit of ${\mathcal{O}}_{{\mathbb{P}}_k^1,\xi }$; by [F6] its valuation, and hence ${\mathrm{ord}}_Z(f)$, is $0$.

Conclusion. The principal divisor of $f$ is $\mathrm{div}(f)=2[0]-[1]-[\mathrm{\infty }]$.

Charts and points. By [F1], ${\mathbb{P}}_k^1=U_0\cup U_{\mathrm{\infty }}$, the overlap $D(t)\subseteq U_0$ is identified with $D(u)\subseteq U_{\mathrm{\infty }}$, and $tu=1$ on it; every point of ${\mathbb{P}}_k^1$ other than $\mathrm{\infty }=(u)$ lies in $U_0$ and corresponds to a prime $\mathfrak{p}\subseteq k[t]$ with ${\mathcal{O}}_{{\mathbb{P}}_k^1,x}=k[t{]}_{\mathfrak{p}}$. Since $(u)\subseteq k[u]$ is maximal with $k[u]\mathrm{/}(u)\cong k$, every prime of $k[u]$ containing $u$ equals $(u)$; hence $U_{\mathrm{\infty }}\setminus \{\mathrm{\infty }\}=D(u)=U_0\cap U_{\mathrm{\infty }}$.

Integrality, normality, function field, dimension. ${\mathbb{P}}_k^1$ is an integral finite-type $k$-scheme of chain dimension one, every local ring of it is an integrally closed domain, and its function field is $k(t)=k(u)$ with $t=u^{-1}$. $U_0$ and $U_{\mathrm{\infty }}$ are integral and glued along the nonempty open $D(t)\cong D(u)$, so ${\mathbb{P}}_k^1$ is integral, and finite type over $k$ because its affine charts are. Localisations of integrally closed domains are integrally closed: if $x\in \mathrm{Frac}(A)$ is integral over $S^{-1}A$ with monic equation $x^n+{\sum }_{i<n}(a_i\mathrm{/}{s}_i)x^i=0$, then with $s={\prod }_{i<n}{s}_i$ the element $sx$ satisfies the monic equation $(sx{)}^n+{\sum }_{i<n}{a}_i(s^{n-i}\mathrm{/}{s}_i)(sx{)}^i=0$ over $A$, so $sx\in A$ and $x\in S^{-1}A$. Hence every local ring is an integrally closed domain and ${\mathbb{P}}_k^1$ is a normal Noetherian scheme [F4]; by [F5] its function field is $k(t)=k(u)$. For the dimension, $U_0=\mathrm{Spec}k[t]$ is a Noetherian space of chain dimension $\dim k[t]=1$ whose proper closed subsets are finite unions of points; since $U_0$ is dense in ${\mathbb{P}}_k^1$, the only proper irreducible closed subsets of ${\mathbb{P}}_k^1$ are the points, so its chain dimension is one (Chain dimension and the empty-space convention).

Factorisations and the degree count. $p={\prod }_i{r}_i^{a_i}$ and $q={\prod }_j{s}_j^{b_j}$ with pairwise distinct monic irreducibles $r_i$ and $s_j$, no $r_i$ equal to any $s_j$, and $m={\sum }_i{a}_i\mathrm{deg}{r}_i$, $n={\sum }_j{b}_j\mathrm{deg}{s}_j$. The monic polynomial $p$ factors as a product of monic irreducibles: a factorisation $p=c\prod r_i^{a_i}$ has leading coefficient $c\prod (\text{leading coefficients})=c$ if each $r_i$ is monic, so $c=1$; the same holds for $q$. Coprimality of $p$ and $q$ says no monic irreducible divides both, so the two families are disjoint. Degrees add: $m=\mathrm{deg}p={\sum }_i{a}_i\mathrm{deg}{r}_i$ and $n=\mathrm{deg}q={\sum }_j{b}_j\mathrm{deg}{s}_j$.

Orders at the finite points. For every monic irreducible $r\in k[t]$, ${\mathrm{ord}}_{[r]}(f)$ equals $a_i$ if $r=r_i$, equals $-b_j$ if $r=s_j$, and is $0$ otherwise. Let $r$ be monic irreducible and $\mathfrak{p}=(r)$. The local ring ${\mathcal{O}}_{{\mathbb{P}}_k^1,[r]}=k[t{]}_{\mathfrak{p}}$ is a one-dimensional local domain [F4] and hence a discrete valuation ring whose maximal ideal is generated by the uniformiser $r$ [F7]. If $r=r_i$ then $r_i\mid p$ and $r_i\nmid q$, so $q$ is a unit of $k[t{]}_{\mathfrak{p}}$ and additivity gives ${\mathrm{ord}}_{[r]}(f)=a_i\cdot 1-0=a_i$; if $r=s_j$ then $p$ is a unit and $q=r^{b_j}\cdot (\text{unit})$, giving ${\mathrm{ord}}_{[r]}(f)=-b_j$; and if $r$ divides neither $p$ nor $q$, both are units and the order is $0$.

Order at infinity. ${\mathrm{ord}}_{\mathrm{\infty }}(p)=-m$, ${\mathrm{ord}}_{\mathrm{\infty }}(q)=-n$, and hence ${\mathrm{ord}}_{\mathrm{\infty }}(f)=n-m$. On $U_{\mathrm{\infty }}$ one has $t=u^{-1}$, so for a monic polynomial $g(t)=t^d+c_{d-1}{t}^{d-1}+\cdots +c_0$ of degree $d$, $g=u^{-d}(1+c_{d-1}u+\cdots +c_0{u}^d)$ with the second factor equal to $1$ at $u=0$, hence a unit of $k[u{]}_{(u)}=O_{{\mathbb{P}}_k^1,\mathrm{\infty }}$; with the uniformiser $u$ this gives ${\mathrm{ord}}_{\mathrm{\infty }}(g)=-d$. Applying this to $p$ and $q$ and using additivity yields ${\mathrm{ord}}_{\mathrm{\infty }}(f)=(-m)-(-n)=n-m$.

The divisor. $\mathrm{div}(f)={\sum }_i{a}_i[r_i]-{\sum }_j{b}_j[s_j]+(n-m)[\mathrm{\infty }]$, a finite Weil sum, and all its terms are closed points, so it is an element of $\mathrm{Div}({\mathbb{P}}_k^1)$. By steps 1.4 and 1.5 these are exactly the nonzero orders of $f$ at prime divisors; the remaining prime divisors have order zero. The support is finite, so the locally finite sum of [F7] is this finite sum, and since ${\mathbb{P}}_k^1$ has chain dimension one its prime divisors are closed points [F6].

Degree. ${\mathrm{deg}}_k\mathrm{div}(f)={\sum }_i{a}_i\mathrm{deg}{r}_i-{\sum }_j{b}_j\mathrm{deg}{s}_j+(n-m)=m-n+(n-m)=0$. By step 1.3 the finite terms have total degree ${\sum }_i{a}_i\mathrm{deg}{r}_i-{\sum }_j{b}_j\mathrm{deg}{s}_j=m-n$, by [F9] the residue degree of $[r]$ is $[\kappa ([r]):k]=\mathrm{deg}r$ and $[\kappa (\mathrm{\infty }):k]=1$, and by [F6] the degree is additive over the coefficients; adding the coefficient ${\mathrm{ord}}_{\mathrm{\infty }}(f)=n-m$ of step 1.5 gives $m-n+(n-m)=0$.

Conclusion. On ${\mathbb{P}}_k^1$ the principal divisor of $f=p\mathrm{/}q$ is ${\sum }_i{a}_i[r_i]-{\sum }_j{b}_j[s_j]+(n-m)[\mathrm{\infty }]$ and has degree zero; the Axiom of Choice is inherited from the two-affine projective-line construction [F1] and the properness theorem [F6]. The finite factorisation and valuation computation make no further choice.

The origin $Z=V(t)$ is a prime divisor of $Y$ with local equation $t$, and $t$ is a regular function on $Y$; its divisor is the Weil divisor $[0]$, whose local equation at the origin is the meromorphic unit $t$.

The structure-sheaf pullback sends $t$ to $0\in {\mathcal{O}}_X(X)=k$. This zero is a meromorphic function on $X$, but it is not a regular section: multiplication by $0$ on the nonzero ring $k$ is not injective. Since $t$ is a regular section on $Y$, $i^{\mathrm{\#}}$ fails the condition in [F4] for inducing a pullback map on meromorphic functions. Thus no pullback of the local equation as a meromorphic unit is defined, and no order of its image can be evaluated.

The set-theoretic inverse image is not a divisor of the source either: $i^{-1}(Z)=X$ by [F3], a closed subscheme of codimension zero rather than a formal sum of prime divisors, and $\mathrm{Div}(X)=0$ by [F2], so there is no nonzero Weil divisor of $X$ that could receive the class $[0]$.

The failure is not an artefact of the choice of equation: by [F5] the pulled-back equation $0$ cuts out no effective Cartier divisor on $X$, so the associated invertible-sheaf construction also has no input; the divisor $V(t)$ of the target simply has no pulled-back divisor along $i$ in the sense of pulling back its equation.

Therefore the proposed pullback of the Weil divisor $[0]$ along the morphism $i:\mathrm{Spec}k\to {\mathbb{A}}_k^1$ is undefined: the structure-sheaf image of its equation is $0$, which is not a meromorphic unit, and the source possesses no prime divisor at all. This refutes the general claim that arbitrary morphisms carry a pullback of Weil divisors defined by pulling back local equations.

The ring $A=k[ϵ]\mathrm{/}({ϵ}^2)$ is local with maximal ideal $(ϵ)$, and its units are exactly the elements $a+bϵ$ with $a\mathrm{\ne }0$. Indeed $A\mathrm{/}(ϵ)\cong k$ is a field, so $(ϵ)$ is maximal; if $a\mathrm{\ne }0$ then $(a+bϵ)(a^{-1}-b{a}^{-2}ϵ)=1+b{a}^{-1}ϵ-b{a}^{-1}ϵ=1$, so $a+bϵ$ is a unit; conversely an element $bϵ$ of $(ϵ)$ is not a unit because $A\mathrm{/}(ϵ)\cong k$ kills it.

The element $ϵ$ is a zero divisor of $A$: $ϵ\mathrm{\ne }0$ by hypothesis and $ϵ\cdot ϵ={ϵ}^2=0$, so multiplication by $ϵ$ on $A$ sends the nonzero element $ϵ$ to $0$ and is not injective. The same computation gives $aϵ\cdot ϵ=0$ for every $a\in k$.

The generators of the ideal $(ϵ)$ are exactly the elements $aϵ$ with $a\mathrm{\ne }0$: an element of $(ϵ)$ is $c+dϵ$ times $ϵ$, which equals $cϵ$ because ${ϵ}^2=0$; if $a\mathrm{\ne }0$ then $ϵ=a^{-1}(aϵ)$ lies in $(aϵ)$, so $(aϵ)=(ϵ)$, and conversely a generator $f=(c+dϵ)ϵ=cϵ$ of $(ϵ)$ satisfies $ϵ\in (f)=(cϵ)$ only if $c\mathrm{\ne }0$. By step 1.2 every such generator is a zero divisor.

The closed subscheme $Z=V(ϵ)$ is not an effective Cartier divisor on $X$. Suppose it were, say $Z=Z_D$ for an effective Cartier divisor $D$. By [F4] the ideal sheaf $I_D$ equals the ideal sheaf of $Z$, which on the single chart $X$ is $(ϵ)$; since $X$ has only one point, its only nonempty open is $X$, so an effective datum supplies an equation $f\in A={\mathcal{O}}_X(X)$; by [F4] we have $I_D{\mathrm{\mid }}_X=fA$, so $f$ generates $(ϵ)$. By step 2.1 the element $f$ is a zero divisor, so multiplication by $f$ on $A$ is not injective and $f$ is not a regular section; this contradicts the effectiveness requirement of [F3].

Therefore $Z=V(ϵ)$ is locally principal, being cut out on its unique affine chart by the principal ideal $(ϵ)$, yet it is not an effective Cartier divisor, because no generator of that ideal is a nonzerodivisor. This refutes the claim that local principality alone makes a closed subscheme an effective Cartier divisor.