Suppose $d\mid m$. Show that $\mathbb{Z}/d\mathbb{Z}$ is not an injective $\mathbb{Z}/m\mathbb{Z}$-module when $\exists p$ prime with $p\mid d$ and $p\mid \frac{m}{d}$.

$\textbf{My attempt:}$

We have that $(a+m\mathbb{Z})(b+d\mathbb{Z}) \mapsto ab + d\mathbb{Z}$ defines $\mathbb{Z}/d\mathbb{Z}$ as a $\mathbb{Z}/m\mathbb{Z}$-module as usual.

Using Baer's criterion I want to show that there exists an ideal of $\mathbb{Z}/m\mathbb{Z},$ say $J$, and an $R$-module map $f:J \to \mathbb{Z}/d\mathbb{Z}$ which $\textbf{cannot}$ be extended to a map $f_*:\mathbb{Z}/m\mathbb{Z} \to \mathbb{Z}/d\mathbb{Z}$.

Take the ideal $\langle \frac{m}{d} \rangle$ in $\mathbb{Z}/m\mathbb{Z}$ and map $$f:\left\langle \frac{m}{d} \right\rangle \xrightarrow{\simeq} \mathbb{Z}/d\mathbb{Z}$$ $$ a\frac{m}{d} + m\mathbb{Z} \mapsto a + d\mathbb{Z}$$

Then this is a module isomorphism.

If we could extend this to a map $f_*$ then as $f$ is surjective we'd get the following short exact sequence: $$ 0 \rightarrow ker(f_*) \xrightarrow{i} \mathbb{Z}/m\mathbb{Z} \xrightarrow{f_*} \mathbb{Z}/d\mathbb{Z} \rightarrow 0$$

This sequence splits because $f$ was an isomorphism (so $f_*\circ f^{-1} = f\circ f^{-1} = id$ witnesses the splitting) [$\textbf{is this logic valid?}$].

So $\mathbb{Z}/m\mathbb{Z}=ker(f_*)\oplus\mathbb{Z}/d\mathbb{Z}$ and $ker(f_*)$ as a submodule of $\mathbb{Z}/m\mathbb{Z}$ is of the form $\mathbb{Z}/e\mathbb{Z}$ for some $e.$ [$\textbf{Is this true?}$]
So that $\mathbb{Z}/m\mathbb{Z}=\mathbb{Z}/e\mathbb{Z}\oplus\mathbb{Z}/d\mathbb{Z}$. I want to show that this is the desired contradiction.

Now I want to use the condition $p\mid d$ and $p\mid\frac{m}{d}$ to show that $\mathbb{Z}/m\mathbb{Z} \ne \mathbb{Z}/e\mathbb{Z}\oplus\mathbb{Z}/d\mathbb{Z}$ but I'm not sure how. Tried using orders of elements but got a little muddled...

If I can do this (and I think it can be done), is this approach all ok, or have I made a mistake?

  • 2
    $\begingroup$ $\Bbb Z/e\Bbb Z\oplus\Bbb Z/d\Bbb Z$ has $p^2$ elements with $px=0$. $\endgroup$ – Lord Shark the Unknown May 10 '18 at 16:43
  • $\begingroup$ I think an easier approach is to find an exact sequence $0\to\Bbb Z/d\Bbb Z\to M\to N\to0$ which doesn't split. $\endgroup$ – Lord Shark the Unknown May 10 '18 at 16:44
  • $\begingroup$ I see, and that is equivalent to not being injective? (Just met this concept today). Is this approach valid even if slightly heavy handed? $\endgroup$ – SEWillB May 10 '18 at 16:46
  • $\begingroup$ See also this question. $\endgroup$ – Arnaud D. Oct 18 '18 at 12:05

I think the logic of your argument is correct. The first of your questioned logic is a straightforward computation, which you have written in your post. To elaborate the second of your question (the "Is this true?" part), note that $\ker(f_*)$ is a subgroup of the cyclic group $\mathbb Z/m\mathbb Z$ and it is also a cyclic group. Taking Lagrange theorem into account we'll see $\ker(f_*)=\mathbb Z/e\mathbb Z$ and $e\mid m$. Since $\mathbb Z/m\mathbb Z\simeq\mathbb Z/e\mathbb Z\oplus \mathbb Z/d\mathbb Z$, the cardinality forces $e=\frac{m}{d}$. Note that $(d,\frac{m}{d})>1$ and thus the direct sum is not a cyclic group, a contradiction.

If you are allowed to use more homological-algebraic arguments, there is another approach to this problem. Firstly if we observe that $\mathbb Z/m\mathbb Z$ is quasi-Frobenius whenever $m>0$, then this problem is reduced to an equivalent form: if $d\mid m$ and $(d,\frac{m}{d})>1$, then $\mathbb Z/d\mathbb Z$ is not a projective $\mathbb Z/m\mathbb Z$-module. If $m$ is square-free, the problem is verified trivially. If $m$ is not square-free, it can be shown that for any decomposition $m=rs$, $$\operatorname{proj. dim}_{\mathbb Z/m\mathbb Z}(\mathbb Z/r\mathbb Z)=\operatorname{proj. dim}_{\mathbb Z/m\mathbb Z}(\mathbb Z/s\mathbb Z)-1$$ whenever $(r,s)>1$. Then the problem is clear: if $\mathbb Z/d\mathbb Z$ is an injective (namely projective) $\mathbb Z/m\mathbb Z$-module, then we have $$0=\operatorname{proj. dim}_{\mathbb Z/m\mathbb Z}(\mathbb Z/d\mathbb Z)=\operatorname{proj. dim}_{\mathbb Z/m\mathbb Z}(\mathbb Z/\frac{m}{d}\mathbb Z)-1$$ and $$\operatorname{proj. dim}_{\mathbb Z/m\mathbb Z}(\mathbb Z/\frac{m}{d}\mathbb Z)=\operatorname{proj. dim}_{\mathbb Z/m\mathbb Z}(\mathbb Z/d\mathbb Z)-1=-1,$$ which leads to an obvious contradiction $\operatorname{proj. dim}_{\mathbb Z/m\mathbb Z}(\mathbb Z/\frac{m}{d}\mathbb Z)=1=-1$.

  • 1
    $\begingroup$ Thanks so much for this; currently working through a homological algebra course so hopefully will meet these concepts soon! $\endgroup$ – SEWillB May 11 '18 at 12:02

maybe we can calculate the injective:

First,we can show a fact: if $R$ is PID,then for any non-zero ideal $I$,$R/I$ is self-injective.and in this case $R/I$ is injective cogenerator of $R/I$ and every finite generated module canbe embedded into $\coprod_{i=1}^{s}R_i$ where $R_i=R$.

Second,if $m=p^lp_1^{l_1}…p_k^{l_k}$,where $p,p_1,…p_k$ are coprime ,then $\mathbb Z_m\cong \mathbb{Z}_{p^l}\coprod \mathbb Z_{p_1^{l_1}}\coprod …\coprod \mathbb Z_{p_k^{l_k}}$.

Remark that all $\mathbb Z_{p^l},…\mathbb Z_{p_k^{l_k}}$ are indecomposable since they have maximal submodule.

Hence,the summand given above are all indecomposable injective module.so all finite generated injective is finite direct sum of these indecomposable injective.

Since $p\mid d$ and $p\mid m/d$ we know it can't be injective.


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