# finitely generated right R-module which is not cyclic

Let $A$ be a finitely generated right $R$-module which is not cyclic. Prove that there exist $B \le A_R$ maximal with respect to the property that $A/B$ is not cyclic.

since $A$ is finitely generated, then every proper submodule of $A$ is contained in a maximal submodule of $A$. (because of Zorn's Lemma) so there is a maxsimal submodule of $A$ ..

no i have not tried . thank you but if Module $A$ is not Cylic, is $A/B$ is not cyclic ? why ?

• What have you come up with so far? What approach would you like to try? – rschwieb Mar 31 '14 at 17:31
• Edited sentence is much clearer: thank you! – rschwieb Mar 31 '14 at 17:53
• @user104847: you should answer to comments as comment, not in question section.please edit! – user 1 Mar 31 '14 at 18:47

If we let the set of submodules $\mathcal{S}=\{B\subseteq A\mid A/B \text{ not cyclic } \}$ be partially ordered by inclusion, we should try to see if Zorn's lemma applies, so that we can extract maximal elements.
First of all, the statement "$A$ is not cyclic" just means that $\mathcal{S}$ contains at least $\{0\}$, so this poset is nonempty. Score one for Zorn's lemma!
Now to see if chains in $\mathcal{S}$ have upper bounds in $\mathcal{S}$.
Consider an increasing chain of indexed submodules $B_i$. If $\cup_i B_i\in \mathcal{S}$, then we are done, because Zorn's lemma will tell us there is a maximal element in the poset.
Suppose to the contrary that $\cup_i B_i\notin \mathcal{S}$, and so $\cup_i B_i$ is cyclic and generated by an element, call it $x+\cup_i B_i$. Let's also label the generators $g_1,\ldots g_n$ for $A$. We now know we can find $r_i$ such that $xr_i\equiv g_i$ mod $\cup_iB_i$. This means that $xr_i-g_i\in \cup_iB_i$ for $1\leq i\leq n$. Now just pick an $j$ sufficiently large such that $xr_i-g_i\in B_j$ for $1\leq i\leq n$. (This is possible since there are only finitely many generators, and the $B_i$ are linearly ordered.)
Now you can take it from here. Prove that this implies $A/B_j$ is cyclic, which is a contradiction. Thus the assumption that $\cup_iB_i\notin \mathcal{S}$ was incorrect, and Zorn's lemma applies.