A fact about the annihilators of torsion modules over a PID

Let R be a principal ideal domain, let B be a torsion (left, unital) R-module, and let p \in R be prime. Prove that if pb = 0 for some nonzero b \in B, then \mathsf{Ann}(B) \subseteq (p).

[Note: I’m not sure why D&F assume that B is torsion. This proof doesn’t use that fact as far as I can tell. Read with caution.]

Recall that \mathsf{Ann}(B) \subseteq R is an ideal (by this previous exercise). Since R is a principal ideal domain, we have \mathsf{Ann}(B) = (t) for some t \in R.

Consider the cyclic submodule (b)_R generated by b. Again, \mathsf{Ann}((b)_R) \subseteq R is an ideal, so \mathsf{Ann}((b)_R) = (s) for some s. By our assumption, p \in (s), so that p = sr for some r \in R. Since p is prime, either p|r or p|s. If p|r, then we have r = pq for some q \in R. Now pqs = p, so that qs = 1; in particular, s is a unit, and so \mathsf{Ann}((b)_R) = R. But then 0 = 1 \cdot b = b, a contradiction. So p|s, and thus (s) \subseteq (p).

Since \mathsf{Ann}(B) \subseteq \mathsf{Ann}((b)_R), we have (t) \subseteq (s), and so (t) \subseteq (p), as desired.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: