Kurs:Vektorbündel und Abschlussoperationen (MSRI 2012)/Abschnitt 1

Systems of linear equations

We start with some linear algebra. Let ${}K$ be a field. We consider a system of linear homogeneous equations over ${}K$,

${}f_{11}t_{1}+\cdots +f_{1n}t_{n}=0\,,$

${}f_{21}t_{1}+\cdots +f_{2n}t_{n}=0\,,$

$\vdots$

${}f_{m1}t_{1}+\cdots +f_{mn}t_{n}=0\,,$

where the ${}f_{ij}$ are elements in ${}K$. The solution set to this system of homogeneous equations is a vector space ${}V$ over ${}K$ (a subvector space of $K^{n}$), its dimension is ${}n-\operatorname {rk} (A)$, where

${}A=(f_{ij})_{ij}\,$

is the matrix given by these elements. Additional elements  ${}f_{1},\ldots ,f_{m}\in K$  give rise to the system of inhomogeneous linear equations,

${}f_{11}t_{1}+\cdots +f_{1n}t_{n}=f_{1}\,,$

${}f_{21}t_{1}+\cdots +f_{2n}t_{n}=f_{2}\,,$

$\vdots$

${}f_{m1}t_{1}+\cdots +f_{mn}t_{n}=f_{m}\,.$

The solution set ${}T$ of this inhomogeneous system may be empty, but nevertheless it is tightly related to the solution space of the homogeneous system. First of all, there exists an action

$V\times T\longrightarrow T,\,(v,t)\longmapsto v+t,$

because the sum of a solution of the homogeneous system and a solution of the inhomogeneous system is again a solution of the inhomogeneous system. This action is a group action of the group ${}(V,+,0)$ on the set ${}T$. Moreover, if we fix one solution  ${}t_{0}\in T$ 

(supposing that at least one solution exists), then there exists a bijection

$V\longrightarrow T,\,v\longmapsto v+t_{0}.$

This means that the group ${}V$ acts simply transitive on ${}T$, and so ${}T$ can be identified with the vector space ${}V$, however not in a canonical way.

Suppose now that ${}X$ is a geometric object (a topological space, a manifold, a variety, a scheme, the spectrum of a ring) and that instead of elements in the field ${}K$ we have functions

$f_{ij}\colon X\longrightarrow K$

on ${}X$ (which are continuous, or differentiable, or algebraic). We form the matrix of functions  ${}A={\left(f_{ij}\right)}_{ij}$,  which yields for every point  ${}P\in X$  a matrix ${}A(P)$ over ${}K$. Then we get from these data the space

${}V={\left\{(P;t_{1},\ldots ,t_{n})\mid A(P){\begin{pmatrix}t_{1}\\\vdots \\t_{n}\end{pmatrix}}=0\right\}}\subseteq X\times K^{n}\,$

together with the projection to ${}X$. For a fixed point  ${}P\in X$,  the fiber ${}V_{P}$ of ${}V$ over ${}P$ is the solution space to the corresponding system of homogeneous linear equations given by inserting ${}P$ into ${}f_{ij}$. In particular, all fibers of the map

$V\longrightarrow X,$

are vector spaces (maybe of non-constant dimension). These vector space structures yield an addition

$V\times _{X}V\longrightarrow V,\,(P;s_{1},\ldots ,s_{n};t_{1},\ldots ,t_{n})\longmapsto (P;s_{1}+t_{1},\ldots ,s_{n}+t_{n}),$

(only points in the same fiber can be added). The mapping

$X\longrightarrow V,\,P\longmapsto (P;0,\ldots ,0),$

is called the zero-section .

Suppose now that additional functions

$f_{1},\ldots ,f_{m}\colon X\longrightarrow K$

are given. Then we can form the set

${}T={\left\{(P;t_{1},\ldots ,t_{n})\mid A(P){\begin{pmatrix}t_{1}\\\vdots \\t_{n}\end{pmatrix}}={\begin{pmatrix}f_{1}(P)\\\vdots \\f_{n}(P)\end{pmatrix}}\right\}}\subseteq X\times K^{n}\,$

with the projection to ${}X$. Again, every fiber ${}T_{P}$ of ${}T$ over a point  ${}P\in X$  is the solution set to the system of inhomogeneous linear equations which arises by inserting ${}P$ into ${}f_{ij}$ and ${}f_{i}$. The actions of the fibers ${}V_{P}$ on ${}T_{P}$ (coming from linear algebra) extend to an action

$V\times _{X}T\longrightarrow T,\,(P;t_{1},\ldots ,t_{n};s_{1},\ldots ,s_{n})\longmapsto (P;t_{1}+s_{1},\ldots ,t_{n}+s_{n}).$

Also, if a (continuous, differentiable, algebraic) map

$s\colon X\longrightarrow T$

with  ${}s(P)\in T_{P}$  exists, then we can construct a (continuous, differentiable, algebraic) isomorphism between ${}V$ and ${}T$. However, different from the situation in linear algebra (which corresponds to the situation where ${}X$ is just one point), such a section does rarely exist.

These objects ${}T$ have new and sometimes difficult global properties which we try to understand in these lectures. We will work mainly in an algebraic setting and restrict to the situation where just one equation

${}f_{1}T_{1}+\cdots +f_{n}T_{n}=f\,$

is given. Then in the homogeneous case (${}f=0$) the fibers are vector spaces of dimension ${}n-1$ or ${}n$, and the later holds exactly for the points  ${}P\in X$  where  ${}f_{1}(P)=\ldots =f_{n}(P)=0$.  In the inhomogeneous case the fibers are either empty or of dimension ${}n-1$ or ${}n$. We give some typical examples.

Example

We consider the line  ${}X={\mathbb {A} }_{K}^{1}$  (or ${}X=K,\mathbb {R} ,{\mathbb {C} }$ etc.) with the (identical) function ${}x$. For  ${}f_{1}=x$  and  ${}f=0$,  i.e. for the homogeneous equation  ${}xt=0$,  the geometric object ${}V$ consists of a horizontal line (corresponding to the zero-solution) and a vertical line over  ${}x=0$.  So all fibers except one are zero-dimensional vector spaces. For the inhomogeneous equation  ${}xt=1$,  ${}T$ is a hyperbola, and all fibers are zero-dimensional with the exception that the fiber over  ${}x=0$  is empty.

For the homogeneous equation  ${}0t=0$,  ${}V$ is just the affine cylinder over the base line. For the inhomogeneous equation  ${}0t=x$,  ${}T$ consists of one vertical line, almost all fibers are empty.

Example

Let ${}X$ denote a plane (like $K^{2},\mathbb {R} ^{2},{\mathbb {A} }_{K}^{2}$) with coordinate functions ${}x$ and ${}y$. We consider an inhomogeneous linear equation of type

${}x^{a}t_{1}+y^{b}t_{2}=x^{c}y^{d}\,.$

The fiber of the solution set ${}T$ over a point ${}\neq (0,0)$ is one-dimensional, whereas the fiber over ${}(0,0)$ has dimension two (for ${}a,b,c,d\geq 1$). Many properties of ${}T$ depend on these four exponents.

In (most of) these example we can observe the following behavior. On an open subset, the dimension of the fibers is constant and equals ${}n-1$, whereas the fiber over some special points degenerates to an ${}n$-dimensional solution set (or becomes empty).

Forcing algebras

We describe now the algebraic setting of systems of linear equations depending on a base space. For a commutative ring ${}R$, its spectrum  ${}X=\operatorname {Spec} {\left(R\right)}$  is a topological space on which the ring elements can be considered as functions. The value of  ${}f\in R$  at a prime ideal  ${}P\in \operatorname {Spec} {\left(R\right)}$  is just the image of ${}f$ under the ring homomorphism ${}R\rightarrow R/P\rightarrow \kappa (P)=Q(R/P)$. In this interpretation, a ring element is a function with values in different fields. Suppose that ${}R$ contains a field ${}K$. Then an element  ${}f\in R$  gives rise to the ring homomorphism

$K[Y]\longrightarrow R,\,Y\longmapsto f,$

which gives rise to a scheme morphism

$\operatorname {Spec} {\left(R\right)}\longrightarrow \operatorname {Spec} {\left(K[Y]\right)}\cong {\mathbb {A} }_{K}^{1}.$

This is another way to consider ${}f$ as a function on ${}\operatorname {Spec} {\left(R\right)}$ with values in the affine line.

The following construction appeared first in the work of Hochster in the context of solid closure.

Definition

Let ${}R$ be a commutative ring and let ${}f_{1},\ldots ,f_{n}$ and ${}f$ be elements in ${}R$. Then the ${}R$-algebra

$R[T_{1},\ldots ,T_{n}]/{\left(f_{1}T_{1}+\cdots +f_{n}T_{n}-f\right)}$

is called the forcing algebra of these elements (or these data).

The forcing algebra ${}B$ forces ${}f$ to lie inside the extended ideal ${}{\left(f_{1},\ldots ,f_{n}\right)}B$ (hence the name). For every ${}R$-algebra ${}S$ such that  ${}f\in {\left(f_{1},\ldots ,f_{n}\right)}S$  there exists a (non unique) ring homomorphism ${}B\rightarrow S$ by sending ${}T_{i}$ to the coefficient  ${}s_{i}\in S$  in an expression  ${}f=s_{1}f_{1}+\cdots +s_{n}f_{n}$. 

The forcing algebra induces the spectrum morphism

$\operatorname {Spec} {\left(B\right)}\longrightarrow \operatorname {Spec} {\left(R\right)}.$

Over a point  ${}P\in X=\operatorname {Spec} {\left(R\right)}$,  the fiber of this morphism is given by

$\operatorname {Spec} {\left(B\otimes _{R}\kappa (P)\right)},$

and we can write

${}B\otimes _{R}\kappa (P)=\kappa (P)[T_{1},\ldots ,T_{n}]/{\left(f_{1}(P)T_{1}+\cdots +f_{n}(P)T_{n}-f(P)\right)}\,,$

where ${}f_{i}(P)$ means the evaluation of the ${}f_{i}$ in the residue class field. Hence the ${}\kappa (P)$-points in the fiber are exactly the solutions to the inhomogeneous linear equation  ${}f_{1}(P)T_{1}+\cdots +f_{n}(P)T_{n}=f(P)$.  In particular, all the fibers are (empty or) affine spaces.

Forcing algebras and closure operations

Let ${}R$ denote a commutative ring and let  ${}I={\left(f_{1},\ldots ,f_{n}\right)}$  be an ideal. Let  ${}f\in R$  and let

${}B=R[T_{1},\ldots ,T_{n}]/{\left(f_{1}T_{1}+\cdots +f_{n}T_{n}-f\right)}\,$

be the corresponding forcing algebra and

$\varphi \colon \operatorname {Spec} {\left(B\right)}\longrightarrow \operatorname {Spec} {\left(R\right)}$

the corresponding spectrum morphism. How are properties of ${}\varphi$ (or of the ${}R$-algebra ${}B$) related to certain ideal closure operations?

We start with some examples. The element ${}f$ belongs to the ideal ${}I$ if and only if we can write  ${}f=r_{1}f_{1}+\cdots +r_{n}f_{n}$  with  ${}r_{i}\in R$.  By the universal property of the forcing algebra this means that there exists an ${}R$-algebra-homomorphism

$B\longrightarrow R,$

hence  ${}f\in I$  holds if and only if ${}\varphi$ admits a scheme section. This is also equivalent to

$R\longrightarrow B$

admitting an ${}R$-module section or ${}B$ being a pure ${}R$-algebra (so for forcing algebras properties might be equivalent which are not equivalent for arbitrary algebras).

The radical of an ideal

Now we look at the radical of the ideal ${}I$,

${}\operatorname {rad} {\left(I\right)}={\left\{f\in R\mid f^{k}\in I{\text{ for some }}k\right\}}\,.$

The importance of the radical comes mainly from Hilbert's Nullstellensatz, saying that for algebras of finite type over an algebraically closed field there is a natural bijection between radical ideals and closed algebraic zero-sets. So geometrically one can see from an ideal only its radical. As this is quite a coarse closure operation we should expect that this corresponds to a quite coarse property of the morphism ${}\varphi$ as well. Indeed, it is true that  ${}f\in \operatorname {rad} {\left(I\right)}$  if and only if ${}\varphi$ is surjective. This is true since the radical of an ideal is the intersection of all prime ideals in which it is contained. Hence an element ${}f$ belongs to the radical if and only if for all residue class homomorphisms

$\theta \colon R\longrightarrow \kappa ({\mathfrak {p}})$

where ${}I$ is sent to ${}0$, also ${}f$ is sent to ${}0$. But this means for the forcing equation that whenever the equation degenerates to ${}0$, then also the inhomogeneous part becomes zero, and so there will always be a solution to the inhomogeneous equation.

Exercise: Define the radical of a submodule inside a module.

Integral closure of an ideal

Another closure operation is integral closure . It is defined by

${}{\overline {I}}={\left\{f\in R\mid f^{k}+a_{1}f^{k-1}+\cdots +a_{k-1}f+a_{k}=0{\text{ for some }}k{\text{ and }}a_{i}\in I^{i}\right\}}\,.$

This notion is important for describing the normalization of the blow up of the ideal ${}I$. Another characterization (assume that ${}R$ is noetherian) is that there exists a  ${}z\in R$,  not contained in any minimal prime ideal of ${}R$, such that  ${}zf^{n}\in I^{n}$  holds for all ${}n$. Another equivalent property - the valuative criterion - is that for all ring homomorphisms

$\theta \colon R\longrightarrow D$

to a discrete valuation domain ${}D$ the containment  ${}\theta (f)\in \theta (I)D$  holds.

The characterization of the integral closure in terms of forcing algebras requires some notions from topology. A continuous map

$\varphi \colon X\longrightarrow Y$

between topological spaces ${}X$ and ${}Y$ is called a submersion , if it is surjective and if ${}Y$ carries the image topology (quotient topology) under this map. This means that a subset  ${}W\subseteq Y$  is open if and only if its preimage ${}\varphi ^{-1}(W)$ is open. Since the spectrum of a ring endowed with the Zarisiki topology is a topological space, this notion can be applied to the spectrum morphism of a ring homomorphism. With this notion we can state that  ${}f\in {\bar {I}}$  if and only if the forcing morphism

$\varphi \colon \operatorname {Spec} {\left(B\right)}\longrightarrow \operatorname {Spec} {\left(R\right)}$

is a universal submersion (universal means here that for any ring change ${}R\rightarrow R'$ to a noetherian ring ${}R'$, the resulting homomorphism ${}R'\rightarrow B'$ still has this property). The relation between these two notions stems from the fact that also for universal submersions there exists a criterion in terms of discrete valuation domains: A morphism of finite type between two affine noetherian schemes is a universal submersion if and only if the base change to any discrete valuation domain yields a submersion. For a morphism

$Z\longrightarrow \operatorname {Spec} {\left(D\right)}$

(${}D$ a discrete valuation domain) to be a submersion means that above the only chain of prime ideals in ${}\operatorname {Spec} {\left(D\right)}$, namely  ${}(0)\subset {\mathfrak {m}}_{D}$,  there exists a chain of prime ideals  ${}{\mathfrak {p}}'\subseteq {\mathfrak {q}}'$  in ${}Z$ lying over this chain. This pair-lifting property holds for a universal submersion

$\operatorname {Spec} {\left(S\right)}\longrightarrow \operatorname {Spec} {\left(R\right)}$

for any pair of prime ideals  ${}{\mathfrak {p}}\subseteq {\mathfrak {q}}$  in ${}\operatorname {Spec} {\left(R\right)}$. This property is stronger than lying over (which means surjective) but weaker than the going-down or the going-up property (in the presence of surjectivity).

If we are dealing only with algebras of finite type over the complex numbers ${}{\mathbb {C} }$, then we may also consider the corresponding complex spaces with their natural topology induced from the euclidean topology of ${}{\mathbb {C} }^{n}$. Then universal submersive with respect to the Zariski topology is the same as submersive in the complex topology (the target space needs to be normal).

Example

Let ${}K$ be a field and consider  ${}R=K[X]$.  Since this is a principal ideal domain, the only interesting forcing algebras (if we are only interested in the local behavior around ${}(X)$) are of the form ${}K[X,T]/{\left(X^{n}T-X^{m}\right)}$. For  ${}m\geq n$  this ${}K[X]$-algebra admits a section (corresponding to the fact that  ${}X^{m}\in {\left(X^{n}\right)}$ ), and if  ${}n\geq 1$  there exists an affine line over the maximal ideal ${}(X)$. So now assume  ${}m<n$.  If  ${}m\geq 0$  then we have a hyperbola mapping to an affine line, with the fiber over ${}(X)$ being empty, corresponding to the fact that ${}1$ does not belong to the radical of ${}{\left(X^{n}\right)}$ for  ${}n\geq 1$.  So assume finally  ${}1\leq m<n$.  Then ${}X^{m}$ belongs to the radical of ${}{\left(X^{n}\right)}$, but not to its integral closure (which is the identical closure on a one-dimensional regular ring). We can write the forcing equation as  ${}X^{n}T-X^{m}=X^{m}{\left(X^{n-m}T-1\right)}$.  So the spectrum of the forcing algebra consists of a (thickend) line over ${}(X)$ and of a hyperbola. The forcing morphism is surjective, but it is not a submersion. For example, the preimage of  ${}V(X)=\{(X)\}$  is a connected component hence open, but this single point is not open.

Example

Let ${}K$ be a field and let  ${}R=K[X,Y]$  be the polynomial ring in two variables. We consider the ideal  ${}I={\left(X^{2},Y\right)}$  and the element ${}X$. This element belongs to the radical of this ideal, hence the forcing morphism

$\operatorname {Spec} {\left(K[X,Y,T_{1},T_{2}]/{\left(X^{2}T_{1}+YT_{2}+X\right)}\right)}\longrightarrow \operatorname {Spec} {\left(K[X,Y]\right)}$

is surjective. We claim that it is not a submersion. For this we look at the reduction modulo ${}Y$. In  ${}K[X,Y]/(Y)\cong K[X]$  the ideal ${}I$ becomes ${}{\left(X^{2}\right)}$ which does not contain ${}X$. Hence by the valuative criterion for integral closure, ${}X$ does not belong to the integral closure of the ideal. One can also say that the chain  ${}V(X,Y)\subset V(Y)$  in the affine plane does not have a lift (as a chain) to the spectrum of the forcing algebra.

For the ideal

${}I={\left(X^{2},Y^{2}\right)}\,$

and the element ${}XY$ the situation looks different. Let

$\theta \colon K[X,Y]\longrightarrow D$

be a ring homomorphism to a discrete valuation domain ${}D$. If ${}X$ or ${}Y$ is mapped to ${}0$, then also ${}XY$ is mapped to ${}0$ and hence belongs to the extended ideal. So assume that ${}\theta (X)=u\pi ^{r}$ and ${}\theta (Y)=v\pi ^{s}$, where ${}\pi$ is a local parameter of ${}D$ and ${}u$ and ${}v$ are units. Then  ${}\theta (XY)=uv\pi ^{r+s}$  and the exponent is at least the minimum of ${}2r$ and ${}2s$, hence

${}\theta (XY)\in {\left(\pi ^{2r},\pi ^{2s}\right)}={\left(\theta {\left(X^{2}\right)},\theta {\left(Y^{2}\right)}\right)}D\,.$

So ${}XY$ belongs to the integral closure of ${}{\left(X^{2},Y^{2}\right)}$ and the forcing morphism

$\operatorname {Spec} {\left(K[X,Y,T_{1},T_{2}]/{\left(X^{2}T_{1}+Y^{2}T_{2}+XY\right)}\right)}\longrightarrow \operatorname {Spec} {\left(K[X,Y]\right)}$

is a universal submersion.

Continuous closure

Suppose now that  ${}R={\mathbb {C} }[X_{1},\ldots ,X_{k}]$.  Then every polynomial  ${}f\in R$  can be considered as a continuous function

$f\colon {\mathbb {C} }^{k}\longrightarrow {\mathbb {C} },\,{\left(x_{1},\ldots ,x_{k}\right)}\longmapsto f{\left(x_{1},\ldots ,x_{k}\right)},$

in the complex topology. If  ${}I={\left(f_{1},\ldots ,f_{n}\right)}$  is an ideal and  ${}f\in R$  is an element, we say that ${}f$ belongs to the continuous closure of ${}I$, if there exist continuous functions

$g_{1},\ldots ,g_{n}\colon {\mathbb {C} }^{k}\longrightarrow {\mathbb {C} }$

such that

${}f=\sum _{i=1}^{n}g_{i}f_{i}\,$

(as an identity of functions). The same definition works for ${}{\mathbb {C} }$-algebras of finite type.

It is not at all clear at once that there may exist polynomials  ${}f\notin I$  but inside the continuous closure of ${}I$. For ${}{\mathbb {C} }[X]$ it is easy to show that the continuous closure is (like the integral closure) just the ideal itself. We also remark that when we would only allow holomorphic functions ${}g_{1},\ldots ,g_{n}$ then we could not get something larger. However, with continuous functions

$g_{1},g_{2}\colon {\mathbb {C} }^{2}\longrightarrow {\mathbb {C} }$

we can for example write

${}X^{2}Y^{2}=g_{1}X^{3}+g_{2}Y^{3}\,.$

Continuous closure is always inside the integral closure and hence also inside the radical. The element ${}XY$ does not belong to the continuous closure of  ${}I=(X^{2},Y^{2})$,  though it belongs to the integral closure of ${}I$. In terms of forcing algebras, an element ${}f$ belongs to the continuous closure if and only if the complex forcing mapping

$\varphi _{\mathbb {C} }\colon \operatorname {Spec} {\left(B\right)}_{\mathbb {C} }\longrightarrow \operatorname {Spec} {\left(R\right)}_{\mathbb {C} }$

(between the corresponding complex spaces) admits a continuous section.