Polynomring in zwei Variablen/Monomiale Beispiele/Integraler Abschluss und Submersion/Beispiel/en

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

$\mathrm{Spec}(K[X,Y,T_1,T_2]/(X^2{T}_1+Y{T}_2+X))⟶\mathrm{Spec}(K[X,Y])$

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

$I=(X^2,Y^2)$

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

$\theta :K[X,Y]⟶D$

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

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

So $XY$ belongs to the integral closure of $(X^2,Y^2)$ and the forcing morphism

$\mathrm{Spec}(K[X,Y,T_1,T_2]/(X^2{T}_1+Y^2{T}_2+XY))⟶\mathrm{Spec}(K[X,Y])$

is a universal submersion.