## Two facts about annihilators

Let $R$ be a ring with 1, let $M$ be a left $R$-module, let $N \subseteq M$ be a submodule, and let $I \subseteq R$ be a right ideal.

1. Prove that $N \subseteq \mathsf{Ann}_M(\mathsf{Ann}_R(N))$. Show by example that the inclusion may be strict.
2. Prove that $I \subseteq \mathsf{Ann}_R(\mathsf{Ann}_M(I))$. Show by example that the inclusion may be strict.

1. Let $n \in N$, and let $a \in \mathsf{Ann}_R(N)$. Certainly then $a \cdot n = 0$. Thus $N \subseteq \mathsf{Ann}_M(\mathsf{Ann}_R(N))$.

Now let $R = \mathbb{Z}$, let $M = \mathbb{Z}/(2) \times \mathbb{Z}/(6)$, and let $N = 0 \times (3)/(6)$. We claim that $\mathsf{Ann}_R(N) = (2)$. To see this, suppose $r \in R$ annihilates $N$. Then in particular, $r \cdot (\overline{0}, \overline{3}) = (\overline{0}, \overline{3r}) = 0$. Then $3r \equiv 0$ mod 6, so that $r \equiv 0$ mod 2. Now let $(\overline{0}, \overline{3k}) \in N$ and note that $2 \cdot (\overline{0}, \overline{3k}) = (\overline{0}, \overline{6k}) = 0$. So $\mathsf{Ann}_R(N) = (2)$. Now we claim that $\mathsf{Ann}_M(\mathsf{Ann}_R(N)) = \mathbb{Z}/(2) \times (3)/(6)$. To see this, suppose $(\overline{a}, \overline{b}) \in \mathsf{Ann}_M(\mathsf{Ann}_R(N))$. Then in particular, $2 \cdot (\overline{a}, \overline{b}) = (\overline{2a}, \overline{2b}) = 0$, and we have $2a \equiv 0$ mod 2 and $2b \equiv 0$ mod 6. Then $b \equiv 0$ mod 3 and we have no restrictions on $a$; thus $(\overline{a}, \overline{b}) \in \mathbb{Z}/(2) \times (3)/(6)$. Suppose now that $(\overline{a}, \overline{3b}) \in \mathbb{Z}/(2) \times (3)/(6)$ and let $2t \in (2)$. Then $2t \cdot (\overline{a}, \overline{3b}) = (\overline{2ta}, \overline{6tb}) = 0$, as desired. Note that $N \subsetneq \mathsf{Ann}_M(\mathsf{Ann}_R(N))$.

2. Let $a \in I$ and let $m \in \mathsf{Ann}_M(I)$. Certainly then $a \cdot m = 0$, so that in fact $I \subseteq \mathsf{Ann}_R(\mathsf{Ann}_M(I))$.

Now let $R = \mathbb{Z}$, let $M = \mathbb{Z}/(2) \times \mathbb{Z}/(6)$, and let $I = (4)$. We claim that $\mathsf{Ann}_M(I) = \mathbb{Z}/(2) \times (3)/(6)$. To see this, suppose $(\overline{a}, \overline{b})$ annihilates $(4)$. Then in particular, $4 \cdot (\overline{a}, \overline{b}) = (\overline{4a}, \overline{4b}) = 0$. Then $4b \equiv 0$ mod 6, and we have $b \equiv 0$ mod 3. hence $(\overline{a}, \overline{b}) \in \mathbb{Z}/(2) \times (3)/(6)$. Suppose now that $(\overline{a}, \overline{3b}) \in \mathbb{Z}/(2) \times (3)/(6)$ and let $4t \in (4)$. Then $4t \cdot (\overline{a}, \overline{3b}) = (\overline{4ta}, \overline{12tb}) = 0$ as desired. Thus $\mathsf{Ann}_M(I) = \mathbb{Z}/(2) \times (3)/(6)$. Next we claim that $\mathsf{Ann}_R(\mathsf{Ann}_M(I)) = (2)$. To see this, suppose $k \in \mathbb{Z}$ annihilates $\mathbb{Z}/(2) \times (3)/(6)$. In particular, $k \cdot (\overline{0}, \overline{3}) = (\overline{0}, \overline{3k}) = 0$. So $3k \equiv 0$ mod 6, and thus $k \equiv 0$ mod 2. Suppose now that $(\overline{a}, \overline{3b}) \in \mathbb{Z}/(2) \times (3)/(6)$. Since $2 \cdot (\overline{a}, \overline{3b}) = 0$, we have $(2) = \mathsf{Ann}_R(\mathsf{Ann}_M((4)))$. Note then that $(4) \subsetneq \mathsf{Ann}_R(\mathsf{Ann}_M((4)))$.