Let $R$ be a commutative ring with identity 1. The collection of ideals of $R$ is denoted by $S$ which forms a commutative semiring. $A(R)$ is the set of all ideals with nonzero annihilator i.e., $A(R)$ is an annihilating ideal that is an ideal $I$ of $R$ is said to be an annihilating ideal if there exists a nonzero ideal $J$ of $R$ such that $IJ = (0)$.
An ideal $I$ is said to be $K$-ideal if $x \in I, x+y \in I$, then $y\in I$.
We know that "in ring every ideal is $k$-ideal".
Therefore, I think that the ideals in semiring $S$ forms a $k$-ideal and the ideals in $A(R)$ forms $k$-ideal. Is my opinion is acceptable or not
I'll prefer to use $\mathcal I(R)$ for the semiring of ideals of $R$ and $\mathcal A(R)$ for the subset of annihilating ideals to try to keep the notation more readable.
Nope.
Let $R=\mathbb R[x]/(x^3)$ and $I_1$ be the ideal generated by $x$, $I_2$ be the ideal generated by $x^2$.
Then $A=\{0, I_1\}$ is a semiring ideal of $\mathcal I(R)$, and $I_1+I_2=I_1\in A$, but $I_2\notin A$.
Yes and the solution is trivial.
Suppose $I+J\in \mathcal A(R)$ for any ideals $I,J$. Then there exists a nonzero ideal $B$ such that $(I+J)B=\{0\}$. But then $\{0\}\subseteq JB\subseteq (I+J)B\subseteq \{0\}$. So both ideals are annihilator ideals. Assuming $I$ is an annihilator ideal never even comes into play. So not only does $\mathcal A(R)$ "additively absorb," it is even "additively saturated."