Let I be an ideal in a commutative ring with 1 A. Let $x,y \in A$. When does the formula $(I,xy)=(I,x) \cap (I,y)$ hold?
One may prove this is true when $x \neq y$ are irreducibles and $(x,y) \cap I= 0$, but this is very restrictive.
Is there any more general lemma? I need this as a tool for ideal decomposition in polynomial rings $\mathbb{K}[x_{1},...,x_{n}]$.
First, let me point out that in your condition you have to additionally specify $x\neq y$, otherwise we have $(I, x^2) = (I,x)$ which is not correct in general.
If one of $x,y$ is in $I$ (say $x\in I$), the statement is equivalent to $I = I\cap (I, y)$, which is always true.
The inclusion $(I, xy)\subset (I, x)\cap (I, y)$ is clear. We're interested in the other inclusion. We have the following inclusion:
$$(I,x)\cdot (I,y)= (I + (x))\cdot (I+(y)) = I^2 + I\cdot (x) + I \cdot (y) + (xy) \subset I + (xy) = (I,xy)$$
So, a more general sufficient condition would be that $(I,x)\cap (I,y) = (I,x)\cdot(I,y)$, or equivalently $(I,x)\cap (I,y) \subset(I,x)\cdot(I,y)$, which is the case if $(I,x)$ and $(I,y)$ are coprime (see below). This is the case, for example, if $(x)$ and $(y)$ are coprime, or $I$ and $(x)$ are coprime, or $I$ and $(y)$ are coprime, because:
$$(I, x) + (I, y) = I + (x) + (y)$$
It's only a sufficient condition, of course.
Note we have the following equation of ideals $\mathfrak{a},\mathfrak{b},\mathfrak{c}\subset A$:
$$(\mathfrak{a}+\mathfrak{b})(\mathfrak{a}\cap\mathfrak{b}) = \mathfrak{a}(\mathfrak{a}\cap \mathfrak{b})+(\mathfrak{a}\cap\mathfrak{b})\mathfrak{b} \subset \mathfrak{a}\mathfrak{b} \subset \mathfrak{a}\cap \mathfrak{b}$$
That means if $\mathfrak{a}$ and $\mathfrak{b}$ are coprime, i.e. $\mathfrak{a}+\mathfrak{b}=A$, we have $\mathfrak{a}\cap\mathfrak{b} = \mathfrak{a}\mathfrak{b}$.