# 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.