In a graded ring $S=\oplus_{k=0}^{\infty}S_k$, denote $m=\oplus_{k=1}^{\infty}S_k$, call an ideal $I$ to be saturated if $I=\cup_{n=1}^{\infty}(I\colon m^n)$. Is the sum of two saturated ideals still saturated?
$$\cup_{n=0}^\infty((I+J)\colon m^n)=\cup_{n=0}^\infty(I\colon m^n)+\cup_{n=0}^\infty(J\colon m^n)$$
Is there a counterexample?
Edit: The question arises when I try to understand global complete intersection, Hartshorne Ex 8.4(a). We define a scheme of codimension $r$ to be complete intersection if the saturated homogeneous ideal us generated by $r$ elements. It is claimed that a subscheme is complete intersection iff it is the scheme intersection of $r$ hypersurfaces $H_i$.
I am not clear how to show
(1) A hypersurface is a complete intersection. That is for a subscheme of codimension $1$, its saturated homogeneous ideal is generated by one element.
(2) Intersection of hypersurfaces is complete intersection. That is the saturated ideal of the sum of saturated principal ideals $(f_i)$ (assuming (1)) is saturated.
As the question arises here, so would it be better to have an counterexample with $I,J$ saturated, the sum ideal $I+J$ define a nonempty subsceme of $\mathbb{P}^n$?
No. I don't think that is correct.
Let $S = k[x,y]$, where $k$ is a field. Let $I = (x)$ and $J = (y)$. Then both $I,J$ are saturated. But $I+J = (x,y)$ is not.