Let $X$ be a scheme. Assume we have two closed subschemes $Y\_1$, $Y\_2$ that cover $X$ set-theoretically. Are there closed subschemes $Y'\_1$, $Y'\_2$ with the same underlying sets, such that the natural map $Y'\_1\amalg Y'\_2\to X$ is schematically dominant? (Equivalently, we want the intersection of the ideal sheaves of $Y'\_1$, $Y'\_2$ to be zero).


This is true if $X$ is locally noetherian, as follows easily from the ''irredundant decomposition'' of EGA IV, (3,2.6) (a sheafification of primary decomposition).


But I cannot think of any other proof. Does anyone have a (non-noetherian) counterexample?


In the affine case, the problem translates as follows: Let $R$ be a ring with two ideals $I\_1$, $I\_2$ such that $I\_1\cap I\_2$ is contained in the radical. Find ideals $I'\_j$ ($j=1,2$)such that $\sqrt{I'\_j}=\sqrt{I\_j}$ and $I'\_1\cap I'\_2=\{0\}$.



Finite coverings by closed subschemes
Here's a counterexample.


$R=k[x,y\_1,y\_2,...]/((xy\_i)^{1+i}: i\ge1)$.


$I\_1=(x), I\_2=(y\_1,y\_2,...)$.


$I\_1,I\_2,I\_1 \cap I\_2$ are radical ideals and each element of $I\_1 \cap I\_2$ is nilpotent.


Some power of $x$, say $x^n$, is in $I'\_1$. Some power of $y\_n$, say $y\_n^m$, is in $I'\_2$. Then $x^ny\_n^m \ne 0$ is in $I'\_1 \cap I'\_2$.


