On $\bigcap_{I \lhd R} (I+\text{ann}_R I)$ in Artinian Gorenstein local ring

Question

Let $(R, \mathfrak m,k)$ be an Artinian Gorenstein local ring. Hence, $\text{ann}_R (\text{ann}_R I)=I$ for every ideal $I$ of $R$. Moreover, $k \cong \text{ann}_R \mathfrak m\subseteq I$ for every non-zero ideal $I$ of $R$.

My question is: Is it true that $$\bigcap_{I \lhd R} (I+\text{ann}_R I)\subseteq \text{ann}_R \mathfrak m ?$$

Answer 1

No. It occurred to me that a uniserial ring might be a good candidate, which led to the following.

It's easy to check that in $F[x]/(x^4)$, we have

$$\bigcap_{I\lhd R} (I+\mathrm{ann}_R(I))=(x^2+(x^4))$$

But $\mathrm{ann}((x+(x^4))=(x^3+(x^4))$.