Erzwingende Algebra/Abschlussoperation für Ideale/Verschiedene Beispiele/en/Textabschnitt

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.

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.