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))).

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