How to calculate Colon Ideals in GCD Domains in particula in UFD Domains

Theorem:

Let RR be a GCD domain, I⊴RI \trianglelefteq R and S⊆RS \subseteq R. Moreover let I=<iα>α∈JI=<i_{\alpha}>_{\alpha \in J} and S={sβ}β∈KS=\{s_{\beta}\}_{\beta \in K} for some index sets I,KI,K. Then the following relation holds:

⋂β∈K<iαgcd(iα,sβ)>α∈J⊆I:S\begin{equation} \bigcap_{\beta \in K} \Big<\frac{i_{\alpha}}{gcd(i_{\alpha},s_{\beta})}\Big>_{\alpha \in J} \subseteq I:S \end{equation}

proof: We can take <S><S> instead SS as I:<S>=I:SI:<S>=I:S. This enables us to eliminate maybe a few of sβs_{\beta}'s and take only {sβ}\{s_{\beta}\} that generates <S><S>. As, I:S=I:<S>=I:<{sβ}>=I:{sβ}β∈KI:S=I:<S>=I:<\{s_{\beta}\}>=I:\{s_{\beta}\}_{\beta \in K} which is sometimes useful to reduce computation.
Coming back to proof, take
x∈⋂β∈K<iαgcd(iα,sβ)>α∈Jx\in \bigcap_{\beta \in K} \Big<\frac{i_{\alpha}}{gcd(i_{\alpha},s_{\beta})}\Big>_{\alpha \in J}
⇒x∈<iαgcd(iα,sβ)>α∈J,∀β∈K\Rightarrow x\in \Big<\frac{i_{\alpha}}{gcd(i_{\alpha},s_{\beta})}\Big>_{\alpha \in J}, \forall \beta \in K

⇒x=∑α∈J′⊆JJ′ is finiterαiαgcd(iα,sβ),∀β∈K\Rightarrow x=\sum_{\underset{J' \text{ is finite}}{\alpha \in J' \subseteq J}}r_{\alpha}\frac{i_{\alpha}}{gcd(i_{\alpha},s_{\beta})}, \forall \beta \in K
⇒xsβ=∑α∈J′⊆JJ′ is finiterαiαsβgcd(iα,sβ),∀β∈K\Rightarrow xs_{\beta}=\sum_{\underset{J' \text{ is finite}}{\alpha \in J' \subseteq J}}r_{\alpha}\frac{i_{\alpha}s_{\beta}}{gcd(i_{\alpha},s_{\beta})}, \forall \beta \in K
⇒xsβ=∑α∈J′⊆JJ′ is finiterαiαtα,β,∀β∈K\Rightarrow xs_{\beta}=\sum_{\underset{J' \text{ is finite}}{\alpha \in J' \subseteq J}}r_{\alpha}i_{\alpha}t_{\alpha, \beta}, \forall \beta \in K
where, tα,β=sβgcd(iα,sβ)t_{\alpha,\beta}=\frac{s_{\beta}}{gcd(i_{\alpha},s_{\beta})} and hence xsβ∈Ixs_{\beta}\in I. This means x∈I:{sβ}=I:Sx\in I:\{s_{\beta}\}=I:S.

Corollary:

To find the equality in (1) we have to check is there any element outside ⋂β∈K<iαgcd(iα,sβ)>α∈J\bigcap_{\beta \in K} \Big<\frac{i_{\alpha}}{gcd(i_{\alpha},s_{\beta})}\Big>_{\alpha \in J}. To find element outside this ideal we can look at non-zero element in R⋂β∈K<iαgcd(iα,sβ)>α∈J\frac{R}{\bigcap_{\beta \in K} \Big<\frac{i_{\alpha}}{gcd(i_{\alpha},s_{\beta})}\Big>_{\alpha \in J}}. For a non-zero element x+⋂β∈K<iαgcd(iα,sβ)>α∈Jx+\bigcap_{\beta \in K} \Big<\frac{i_{\alpha}}{gcd(i_{\alpha},s_{\beta})}\Big>_{\alpha \in J}, it is sufficient to check wheter x∈I:Sx \in I:S as corresponding nonzero elements in RR will be x+ix+i for all i∈⋂β∈K<iαgcd(iα,sβ)>α∈Ji\in \bigcap_{\beta \in K} \Big<\frac{i_{\alpha}}{gcd(i_{\alpha},s_{\beta})}\Big>_{\alpha \in J}. But if x+i∈I:Sx+i \in I:S so is xx.
Now if we have such xx then we have another ideal <x,⋂β∈K<iαgcd(iα,sβ)>α∈J>⊆I:S\Big<x,\bigcap_{\beta \in K} \Big<\frac{i_{\alpha}}{gcd(i_{\alpha},s_{\beta})}\Big>_{\alpha \in J}\Big> \subseteq I:S and repeat the above process. Else, we get the equality.

Example:

Let R=F[x,y]R=F[x,y], I=<xy2,x2y>I=<xy^2,x^2y> and S={y2}S=\{y^2\}.
By equation (1) we have,
<xy2gcd(xy2,y2),x2ygcd(x2y,y2)>=<x,x2>=<x>⊆I:S\Big<\frac{xy^2}{gcd(xy^2,y^2)}, \frac{x^2y}{gcd(x^2y,y^2)}\Big>=<x,x^2>=<x>\subseteq I:S

Now, find non-zero element in F[x,y]<x>≅F[y]\frac{F[x,y]}{<x>}\cong F[y]. Any non zero element of F[y]F[y] is of the form f(y)f(y) and taking pre-image under th natural isomorphism (ϕ(f(x,y)+<x>)=f(x,y)(mod x)\phi(f(x,y)+<x>)=f(x,y)(mod \text{ } x))

So, ϕ−1(f(y))=f(y)+<x>\phi^{-1}(f(y))=f(y)+<x>.
And hence check whether f(y)∈I:Sf(y) \in I:S?
f(y)y2∉If(y)y^2 \notin I as elements in II must have atleast 1 degree in indeterminate xx. Hence f(y)=0f(y)=0 and hence
I:S=<x>I:S=<x>