$\sum_{i \in I} \operatorname{ann}_l(\operatorname{ann}_r(S_i)) = \operatorname{ann}(\bigcap_{i \in I} \operatorname{ann}_r(S_i))$

Question

I'm trying to prove

$$\sum_{i \in I} \operatorname{ann}_l(\operatorname{ann}_r(S_i)) = \operatorname{ann}(\bigcap_{i \in I} \operatorname{ann}_r(S_i))$$

here $S_i \subseteq R$ for all $i \in I$ where $R$ is a ring and

$$\operatorname{ann}_l(S) = \{r \in R: r S = 0\}$$ $$\operatorname{ann}_r(S) = \{r \in R: Sr = 0\}$$

I was able to prove $\subseteq$ but I'm struggling at the other inclusion.

I know some basic identities like:

• $S$ is contained in the left annihilator of the right annihilator of $S$.
• The left annihilator of the right annihilator of the left annihilator of $S$ is the left annihilator of $S$.
• Intersection of left annihilators is the annihilator of the sum.

These may be useful in proving the result.

Thanks in advance for any help/suggestion!

Answer 1

The other inclusion is false in general.

For example, take the algebra $A:=k\langle x,y\rangle/(x^2,y^2)$, having $k$-basis the alternating products of $x$ and $y$.

Then $\mathrm{ann}_r(x)=xA$ and $\mathrm{ann}_r(y)=yA$, and their intersection is zero, so has annihilator all of $A$.

On the other hand, $\mathrm{ann}_l(xA)=Ax$ and $\mathrm{ann}_l(yA)=Ay$, and their sum is $Ax+Ay\neq A$.