Higher annihilators of an ideal

Let R be a ring with 1, let M be a left R-module, and let I be an ideal of R. Prove that \bigcup_{k \in \mathbb{N}^+} \mathsf{Ann}_M(I^k) is a submodule of M.

We claim that for each k \geq 1, \mathsf{Ann}_M(I^k) \subseteq \mathsf{Ann}_M(I^{k+1}). To see this, let m \in \mathsf{Ann}_M(I^k) and let a \in I^{k+1}. Now by definition, a is a finite sum of k+1-fold products of elements in I. Note that every k-fold product of elements in I is annihilated my M, so that in fact a \cdot m = 0. Thus m \in \mathsf{Ann}_M(I^{k+1}).

So \bigcup_{k \in \mathbb{N}^+} \mathsf{Ann}_M(I^k) is the union of a chain of submodules. By this previous exercise, it is a submodule of M.

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  • Yun  On November 13, 2011 at 11:50 pm

    Perhaps I’m misreading this, but should you be taking a union instead of an intersection of the annihilators?

    • nbloomf  On November 14, 2011 at 8:07 am

      You’re right. This isn’t the first time I’ve mixed up ‘cap’ and ‘cup’…


  • versus  On November 16, 2011 at 2:55 pm

    What do you mean by fold product?

    • nbloomf  On November 17, 2011 at 10:42 am

      A k-fold product is just a product with k factors. For example, if I is an ideal, then the ideal product I^2 consists of all possible finite sums of 2-fold products of elements from I.

  • Gobi Ree  On February 16, 2012 at 7:23 pm

    Typo in the problem: \cap should be \cup

    • nbloomf  On March 17, 2012 at 2:54 pm


      Looking back, it seems I’ve fixed that typo before. :/

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