Over a PID which is not a field, no finitely generated module is injective

Let R be a principal ideal domain which is not a field. Prove that no finitely generated R-module is injective.

Let M be a finitely generated R-module. By Theorem 5 (FTFGMPID), we have M \cong_R R^t \oplus \mathsf{Tor}(M) for some t \in \mathbb{N}. By this previous exercise, M is injective if and only if R^t and \mathsf{Tor}(M) are injective. We claim that if M is injective, then M = 0.

We begin with a lemma.

Lemma: If M is a finitely generated torsion module over a principal ideal domain, then there exists a nonzero element r \in R such that rM = 0. Proof: By FTFGMPID, we have M \cong \bigoplus_i R/(p_i^{k_i}) for some primes p_i \in R and nonnegative natural numbers k_i. Now r = \prod_i p_i^{k_i} is nonzero and clearly annihilates M. \square

By Proposition 36 on page 396 of D&F, \mathsf{Tor}(M) (if nontrivial) cannot be injective, since we have rM = 0 for some r. So if M is injective, then \mathsf{Tor}(M) = 0.

Consider now R^t; if t \geq 1, it suffices to consider a single copy of R. If p \in R is not a unit, then pR = (p) \subsetneq R. So again by Proposition 36 on page 396, R is not injective, so that R^t is not injective if t \geq 1.

So if M is injective, then M = 0.

Conversely, the zero module is trivially injective since every short exact sequence 0 \rightarrow 0 \rightarrow M \rightarrow N \rightarrow 0 splits (trivially).

So over a principal ideal domain R which is not a field, no nontrivial module is injective.

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: