ideal polynomial Let $R$ be a commutative ring and denote $A := R[x,y]$, $A_x$ le localization of $A$ by $\{ x^n \mid n \geq0\}$ and $A_y$ the same for $y$.
Let $I_x \subset A_x$, $I_y \subset A_y$  ideals s.t $I_x = I_y$ in $A_{xy}$, so $I_x$ and $I_y$ become equal in the localization by $xy$.
Does there exist a unique ideal $I$ of $A$ s.t. $I = I_x$ in $A_x$ and $I = I_y$ in $A_y$ ?
There is no uniqueness; see comment of Mohan. So the question is existence.
 A: Yes. See here for a solution to a generalization of this problem. In this specific case, let $\iota_x:A\rightarrow A_x$ and $\iota_y:A\rightarrow A_y$, and likewise $\tau_x:A_x\rightarrow A_{xy}$ and $\tau_y:A_y\rightarrow A_{xy}$, be the canonical localization maps. The equality condition that you express is then that $\tau_x(I_x)A_{xy}=\tau_y(I_y)A_{xy}$. Note also that $$\ker{\tau_x}=\{f\in A_x:\exists k\in\mathbb{N}\text{ such that }(y^k\big/1)f=0\big/1\in A_x\},$$ and that likewise $$\ker{\tau_y}=\{f\in A_y:\exists k\in\mathbb{N}\text{ such that }(x^k\big/1)f=0\big/1\in A_y\}$$
Now, let $I=\iota_x^{-1}(I_x)\cap\iota_y^{-1}(I_y)$. We claim that the image of $I$ in $A_x$ generates $I_x$. Thus let $J=\iota_x(I)A_x$ be the ideal of $A_x$ generated by the image of $I$. Because $I\subseteq\iota_x^{-1}(I_x)$, we have $\iota_x(I)\subseteq I_x$ and so certainly $J\subseteq I_x$. To show the other direction, let $a\big/x^m\in I_x$ be arbitrary, for some $a\in A$. Now, to show $a\big/x^m\in J$, it suffices to show $a\big/1\in J$, so we just need to find $b\in I$ and $n\in\mathbb{N}$ such that $a\big/1=b\big/x^n\in A_x$.
To do this, note that – because $a\big/1\in I_x$ – we have $a\big/1\in\tau_x(I_x)A_{xy}=\tau_y(I_y)A_{xy}$, so there is $f\in I_y$ and $l\in\mathbb{N}$ such that $a\big/1=f\big/x^l\in A_{xy}$. This means that $f-x^la\big/1\in\ker{\tau_y}$, and so there is some $k\in\mathbb{N}$ such that $(x^k\big/1)(f-x^la\big/1)=0\big/1\in A_y$. But this means $x^{k+l}a\big/1=(x^k\big/1)f\in I_y$, since $f\in I_y$, and so $x^{k+l}a\in\iota_y^{-1}(I_y)$. Also, because $a\big/1\in I_x$, we have $a\in\iota_x^{-1}(I_x)$, and so certainly $x^{k+l}a\in\iota_x^{-1}(I_x)$. Putting these two facts together gives $x^{k+l}a\in I$, and so letting $b=x^{k+l}a$ and $n=k+l$ gives $a\big/1=b\big/x^n\in J=\iota_x(I)A_x$ as desired.
Thus $\iota_x(I)A_x=I_x$, and by exactly the same argument we can show $\iota_y(I)A_y=I_y$, so we have proven existence. However, as Mohan points out, $I$ is certainly not unique, for the images of $I\cdot(x,y)<I$ in $A_x$ and $A_y$ will clearly generate the same ideals as the images of $I$. The problem is essentially that the ideal $(x, y)$ generated by $x$ and $y$ is not all of $A$; the answer I've linked to above shows how a related condition in the general case would allow for a proof of uniqueness.
