I'm trying to show that every $\mathbb{Z}/(6)$ module is injective. My strategy is to use Baer's Criterion.

The only nontrivial ideals of $\mathbb{Z}/(6)$ are $(2)=(4)=\{0,2,4\}$ and $(3)=\{0,3\}$. Suppose $g:(2)\to Q$ is a morphism into any $\mathbb{Z}/(6)$ module $Q$. I try to extend it some $G$ on all of $\mathbb{Z}/(6)$. I know $G$ will be uniquely determined by $G(1)$. From $$ g(2)=G(2)=2G(1) $$ and $2g(4)=g(8)=g(2)=2G(1)$, I want to define $G(1)=g(4)$ and extend homomorphically. It seems to check out that $G$ is an extension of $g$. My concern is, is there some guarantee that just setting $G(1)=g(4)$ is ok? Will $G(1)$ be well-defined to extend to a module homomorphism on all of $\mathbb{Z}/(6)$?

  1. Every module over a field (i.e. vector space) is injective.

  2. If $C_1,C_2$ are categories with injective objects $I_1 \in C_1$, $I_2 \in C_2$, then $(I_1,I_2)$ is an injective object of $C_1 \times C_2$.

  3. If $R_1,R_2$ are rings, there is an equivalence of categories $\mathsf{Mod}(R_1 \times R_2) \simeq\mathsf{Mod}(R_1) \times \mathsf{Mod}(R_2)$.

  4. Chinese Remainder Theorem.

Combining these basic facts, we get that in $\mathsf{Mod}(\mathbb{Z}/6) \simeq \mathsf{Mod}(\mathbb{F}_2) \times \mathsf{Mod}(\mathbb{F}_3)$ every object is injective.

In order to make your proof work, use $\hom(\mathbb{Z}/n,A) \cong \{a \in A : n \cdot a = 0\}$, $f \mapsto f(1)$.

  • $\begingroup$ Thanks, I appreciate the category theoretic answer. Can you clarify how $\operatorname{Hom}(\mathbb{Z}/n,A)\cong\{a\in A:na=0\}$ implies my chosen homomorphism is suitable defined? $\endgroup$ – YN Chew Jan 21 '14 at 7:53
  • $\begingroup$ Can you try this yourself? $\endgroup$ – Martin Brandenburg Jan 21 '14 at 8:01
  • $\begingroup$ Is it simply that $6G(1)=6g(4)=g(24)=g(0)=0$? $\endgroup$ – YN Chew Jan 21 '14 at 8:07
  • $\begingroup$ Make sure that your proof isn't circular (you don't have $G$ yet). $\endgroup$ – Martin Brandenburg Jan 21 '14 at 9:20

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