4

For every abelian groups $G$ let $\mathrm{d}G$ be its maximal divisible subgroup. Then $G \mapsto G/\mathrm{d}G$ is a right exact functor $\mathbf{Ab} \to \mathbf{Ab}$.

Let $$ 0 \to G \xrightarrow{i} H \xrightarrow{p} K \to 0 $$ be an exact sequence. I'm having trouble showing for the sequence $$ G/\mathrm{d}G \xrightarrow{i'} H/\mathrm{d}H \xrightarrow{p'} K/\mathrm{d}K \to 0 $$ (where $p' \colon h + \mathrm{d}H \mapsto p(h) + \mathrm{d}K$ ) that $\ker p' \subseteq \operatorname{im} i'$.

So let $h + \mathrm{d}H \in \ker p'$. Then $p(h) \in \mathrm{d}K$. I need to show that there is $h'$ in $\mathrm{d}H$ such that $p(h) = n p(h')$ for some integer $n$ in order to draw the conclusion.

Any help would be appreciated!

Jendrik Stelzner
  • 17,081
scsnm
  • 1,321

1 Answers1

2

Consider the short exact sequence: $$0\to\mathbb{Z}^{\mathbb{N}_{>0}}\stackrel i \to \mathbb{Z}^{\mathbb{N}_{>0}}\stackrel p\to \mathbb{Q}\to 0,$$ where $i(f_n)=(n+1)e_{n+1}-e_n$ and $p(e_n)=\frac1{n!}$ for $n\geq 1$.

Applying the functor results in the sequence: $$0\to\mathbb{Z}^{\mathbb{N}_{>0}}\stackrel i \to \mathbb{Z}^{\mathbb{N}_{>0}}\to 0\to 0,$$ which is not exact in the middle term.

Note $\mathbb{Z}^{\mathbb{N}_{>0}}$ denotes the direct sum of copies of $\mathbb{Z}$ indexed over $\mathbb{N}_{>0}$ (not the direct product).

tkf
  • 15,315
  • Nice example. More generally, given any exact sequence $F'\to F\to D\to 0$ with $F,F'$ reduced (e.g. free) and $D\neq 0$ is divisible, the sequence obtained by applying the functor in question will have the form $F'\to F\to 0\to 0,$ which is obviously not exact. – Ken Mar 12 '21 at 02:17