# Finite coverings by closed subschemes

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 $$I'_1\cap I'_2=\{0\}$$.

$$R=k[x,y_1,y_2,...]/((xy_i)^{1+i}: i\ge1)$$.
$$I_1=(x), I_2=(y_1,y_2,...)$$.
$$I_1,I_2,I_1 \cap I_2$$ 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 $$I'_1 \cap I'_2$$.