## 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.

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