# Forcing algebra/Closure operations for ideals/Examples/Section

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

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

be the corresponding forcing algebra and

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

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

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

${\displaystyle B\longrightarrow R,}$

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

${\displaystyle R\longrightarrow B}$

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

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

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

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 ${\displaystyle {}\varphi }$ as well. Indeed, it is true that ${\displaystyle {}f\in \operatorname {rad} (I)}$ if and only if ${\displaystyle {}\varphi }$ 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 ${\displaystyle {}f}$ belongs to the radical if and only if for all residue class homomorphisms

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

where ${\displaystyle {}I}$ is sent to ${\displaystyle {}0}$, also ${\displaystyle {}f}$ is sent to ${\displaystyle {}0}$. But this means for the forcing equation that whenever the equation degenerates to ${\displaystyle {}0}$, 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

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

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

${\displaystyle \theta \colon R\longrightarrow D}$

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

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

${\displaystyle \varphi \colon X\longrightarrow Y}$

between topological spaces ${\displaystyle {}X}$ and ${\displaystyle {}Y}$ is called a submersion, if it is surjective and if ${\displaystyle {}Y}$ carries the image topology (quotient topology) under this map. This means that a subset ${\displaystyle {}W\subseteq Y}$ is open if and only if its preimage ${\displaystyle {}\varphi ^{-1}(W)}$ 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 ${\displaystyle {}f\in {\bar {I}}}$ if and only if the forcing morphism

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

is a universal submersion (universal means here that for any ring change ${\displaystyle {}R\rightarrow R'}$ to a noetherian ring ${\displaystyle {}R'}$, the resulting homomorphism ${\displaystyle {}R'\rightarrow B'}$ 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

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

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

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

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

## Example

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

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

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

For the ideal

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

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

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

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

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

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

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

is a universal submersion.

Continuous closure

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

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

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

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

such that

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

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

It is not at all clear at once that there may exist polynomials ${\displaystyle {}f\notin I}$ but inside the continuous closure of ${\displaystyle {}I}$. For ${\displaystyle {}\mathbb {C} [X]}$ 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 ${\displaystyle {}g_{1},\ldots ,g_{n}}$ then we could not get something larger. However, with continuous functions

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

we can for example write

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

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

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

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