---- Assume $(X, \mathcal{O}_{X})$ is a [[locally Noetherian scheme|Noetherian]], [[integral scheme|integral]] [[scheme]] [[scheme over a field|over]] a [[field]] $k$ which is [[regular scheme|regular]] in [[codimension of a closed subspace|codimension]] $1$. (This is the default for this chapter of the course.) > [!definition] Definition. ([[prime divisor in a scheme]]) > A **prime divisor** on $X$ is an [[integral scheme|integral]] (equivalently [[irreducible scheme|irreducible]] + [[reduced scheme|reduced]]) [[subscheme|closed subscheme]] of $X$ having [[codimension of a closed subspace|codimension]] $1$. > > We denote by $\text{Div}(X)$ the [[free abelian group]] generated by prime divisors. The elements of $\text{Div}(X)$ are called **(Weil) divisors**. > > For $Y \subset X$ a prime divisor, let $\xi$ denote its [[generic point of an integral scheme|generic point]].[^3] Then $\text{dim }{ \mathcal{O}_{X, \xi} }=1$, because $\text{codim}(Y, X)=1$.[^5] [[every DVR is a Noetherian local domain of dimension 1|This means]] $\mathcal{O}_{X, \xi}$ is [[regular local ring|regular]], hence a [[DVR]].[^2] So we have a [[discrete valuation|valuation]][^4] $\nu_{Y}:K(X)^{*} \to \mathbb{Z}.$ > [[discrete valuation|In particular]]: $\mathcal{O}_{X, \xi}=\{ f \in K(X)^{*}: \nu (f) \geq 0 \} \cup \{ 0 \}$ > So each prime divisor $Y$ corresponds to a [[discrete valuation|valuation]] $\nu_{Y}$ on the [[generic point of an integral scheme|function field]] $K(X)$ of $X$. We think of $\nu_{Y}(f)$ as specifying the order of zero of the [[rational function]] $f$ on $X$ along $Y$. (Or something... there is some intuition from complex analysis) > [!basicexample] > Let $X=\mathbb{A}^{1}_{k}=\text{Spec }k[x]$ be the [[affine scheme|scheme-theoretic affine line]]. Let $\mathfrak{p}=\langle x-a \rangle \subset k[x]$. Then $\mathcal{O}_{X, \mathfrak{p}}=k[x]_{\langle x-a \rangle}$ [[rational function|and]] $K(X)=k(x)$. Given $\frac{f}{g}\in K(X)$ nonzero, we may factor out $( x-a )^{m}$, $m \in \mathbb{Z}$, to get $\frac{f}{g}=\left( \frac{p}{q} \right)(x-a)^{m}$ [[greatest common divisor|with]] $\text{gcd}(p, x-a)=1=\text{gcd}(q, x-a)$. Then with $Y=V(\mathfrak{p})$ ([[correspondence between height-1 primes and prime divisors in affine schemes|recall]]), have (using that $\nu_{Y}$ is a [[group homomorphism]]) $\nu_{Y}\left( \frac{f}{g} \right)= \nu_{Y}\left( \frac{p}{q} \right)+ \sum \underbrace{ \nu_{Y}\big( x-a \big) }_{ =1, x-a \text{ uniformizer} }=m.$ > Thus, $\nu_{Y}$ counts the order of vanishing (if $m>0$) or the order of pole (if $m<0$) of the rational function $\frac{f}{g}$ at the point $x=a$. By definition of [[DVR]], $\mathcal{O}_{X, \mathfrak{p}}=\left\{ \frac{f}{g} \in K(X)^{*} : \nu_{Y}\left( \frac{f}{g} \right) \geq 0 \right\} \cup \{ 0 \}$ > collects the rational functions without poles at $x=a$. We can say that such rational functions are 'regular along the divisor $Y. > [!basicexample] > If $X$ is a 1-dimensional [[variety]] over $\text{Spec }k$, then for its irreducible closed subsets codimension is [[codimension is well-behaved for irreducible closed subspaces of varieties over a field|well-behaved]]. So the prime divisors of $X$ are its closed points. > Well, to be pedantic: *a priori*, saying 'its closed points' is underspecified: what is the scheme structure? In general, there can be lots of [[sheaf|structure sheaf]] choices making a given closed subset into a scheme... But it turns out here that there is a unique integral subscheme structure on $\{ x \} \subset X$. Namely, $x$ lives in some open affine $\text{Spec }A \subset X$; it's closed, so $\{ x \}=V( x )$, meaning $x=\mathfrak{m}$ is a [[maximal ideal]] of $A$. Then $\text{Spec }A/\mathfrak{m}=(0)$ as $A / \mathfrak{m}$ is a [[field]], and we identify $x$ with $(\text{Spec }A / \mathfrak{m}, \mathcal{O}_{\text{Spec } A/\mathfrak{m}})$. > [!basicexample] > The prime divisors in $\mathbb{P}^{n}_{k}=\text{Proj }k[X_{0},\dots,X_{n}]$ are precisely the sets $V(\langle f \rangle)$ for $f \in k[X_{0},\dots,X_{n}]$ [[homogeneous polynomial|homogeneous]] and [[irreducible element of an integral domain|irreducible]]. The prime divisors in $\text{Spec }k[X_{0},\dots,X_{n}]=\mathbb{A}^{n}_{k}$ are precisely the sets $V(\langle f \rangle)$ for $f \in k[X_{0},\dots,X_{n}]$ [[irreducible element of an integral domain|irreducible]]. > > Indeed, the closed irreducible subsets of $\mathbb{P}^{n}_{k}$ are exactly those of the form $V(\mathfrak{p})$ [[irreducible closed subspaces of Spec are precisely the vanishing of primes|(by a reformulation of this argument)]] for $\mathfrak{p}$ prime. A prime divisor is then a set $V(\mathfrak{p})$ of codimension 1; equivalently $\text{ht }\mathfrak{p}=1$. > > > > [!proof] > > **$Y$ prime divisor implies $Y=V(\langle f \rangle)$, $f$ irreducible.** Suppose $Y$ is a prime divisor. Then since $Y$ is closed + irreducible, it has the form $V(\mathfrak{p})$ for some prime $\mathfrak{p}$. $\mathfrak{p}$ contains an nonzero irreducible element $f$,[^1] giving inclusions $0 \subset \langle f \rangle \subset \mathfrak{p}$ but since $\mathfrak{p}$ is height-1, the inclusion must not be proper: $\langle f \rangle= \mathfrak{p}$. > > > > **$V\langle f \rangle$, $f$ irreducible, is a prime divisor.** Conversely, let $f$ be irreducible. Then the [[ideal]] $\langle f \rangle$ is prime ([[characterization of UFDs|since irreducible implies prime in UFDs]], so $V(\langle f \rangle)=V(\mathfrak{p})$ is closed + irreducible. It has codimension equal to the height of $\langle f \rangle$; by [[dimension theorem for Noetherian local rings|Krull's Height Theorem]] $\text{ht } \langle f \rangle \leq 1$ and we know $\text{ht }\langle f \rangle \neq 0$ so $\text{ht }\langle f \rangle=1$. > (Same deal for spec, maybe there are some scheme-y things to say) > > [^1]: Just because since $\mathfrak{p}$ is prime, if we factor $f$ into irreducible elements then at least one of the irreducible elements must live in $\mathfrak{p}$. [^2]: [[characterization of DVRs|Recall]]: [[regular local ring|regular]] [[Noetherian ring|Noetherian]] [[local ring|local]] [[ring]] of [[Krull dimension]] $1$ $\implies$ [[DVR]]. $\mathcal{O}_{X, \xi}$ is regular because $\text{dim }\mathcal{O}_{X, \xi}=1$ and we have assumed that the [[scheme]] $X$ is [[regular scheme|regular]] in codimension $1$. [^3]: To reiterate: $\xi$ is the generic point of $Y$, *not* of $X$! [^4]: Here we employ that fact that the function field $K(X)$ is also the [[field of fractions]] of $\mathcal{O}_{X, \xi}$. Indeed, $\mathcal{O}_{X ,\xi} \cong A_{\mathfrak{p}}$ for some [[affine scheme|open affine neighborhood]] $U=\text{Spec }A$ of $\xi$ and some $\mathfrak{p} \in \text{Spec }A$, and $\text{Frac }A=\text{Frac }A_{\mathfrak{p}}$ (just 'adding more denominators'). $\text{Frac }A=K(X)$ per the discussion in [[generic point of an integral scheme]]. [^5]: Recall that $Y$ is [[integral scheme|integral]], hence [[irreducible scheme|irreducible]], and also assumed [[closed set|closed]] as a subset of $X$, so the proposition [[computing codimension with generic point]] applies. [^6]: Here we employ that fact that $K(X)$ is the [[field of fractions]] of $\mathcal{O}_{X, \xi}$. Indeed, $\mathcal{O}_{X ,\xi} \cong A_{\mathfrak{p}}$ for some [[affine scheme|open affine neighborhood]] $U=\text{Spec }A$ of $\xi$ and some $\mathfrak{p} \in \text{Spec }A$, and $\text{Frac }A=\text{Frac }A_{\mathfrak{p}}$ (just 'adding more denominators'). $\text{Frac }A=K(X)$ per the discussion in [[generic point of an integral scheme]]. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` part when using the convention $\nu(0)=\infty$. ---- #### References > [!backlink] > {CODE_BLOCK_PLACEHOLDER} > [!frontlink] > {CODE_BLOCK_PLACEHOLDER} if assuming $\nu(0)=\infty$. ---- #### References > [!backlink] > {CODE_BLOCK_PLACEHOLDER} > [!frontlink] > {CODE_BLOCK_PLACEHOLDER} part when using the convention $\nu(0)=\infty$. ---- #### References > [!backlink] > {CODE_BLOCK_PLACEHOLDER} > [!frontlink] > {CODE_BLOCK_PLACEHOLDER}