If $A$ is a finitely generated $k$-algebra (with identity), for a field $k$, I am interested in whether or not a cyclic $A$-module can have a non-cyclic submodule. I think the answer is yes, and this is the progress I have made, but I'm not super sure if my example is valid or not.

It is useful to consider the following construction. For $M$ a $A$-module, and $v \in M$, define the annihilator $Ann(v) = \{ a \in A \mid a\cdot m =0\}$. Then one can show that $Ann(v)$ is a left ideal of $A$.

Then one can show that $M$ is cyclic if and only if $M \cong A / I$, for some left ideal $I$ of $A$. Further, I think, we can choose $I$ such that $I = Ann(v)$ for some $v \in M$. That is to say I think that an $A$-module $M$ is cyclic if and only if $M \cong A/Ann(v)$ for some $v\in M$. (Included an attempted proof of this at the bottom)

This lead me to the following construction

Let $k$ be a field, $A = k[X,Y]$, $I = \left<X,Y\right>_{A}$, and $F = A / I^{2}$. Then $F$ is cyclic since it is a quotient of $A$ by the left ideal $I^{2}$. Then consider $J = \left<X,Y\right> = \{ aX + bY \mid a,b\in k, X^{2} = Y^{2} = XY = 0 \} \subset F$.

Then for $a,b\in k$ not both zero, we see $Ann(aX+bY) = \left<X,Y\right> = I$, but $A/I \cong k$, so certainly $J \not\cong A / Ann(j)$ for any $j \in J$, and hence is not cyclic.

Would anyone be able to verify this example?

$\textbf{Claim: }$ Let $A$ be a $k$-algebra (with identity), then an $A$-module, $M$ say, is cyclic if and only if $M \cong A /ann(v)$ for some $v \in M$.

$\textbf{Proof: }$ Suppose first that $M$ is cyclic, then $M = \left<v\right>_{A}$ for some $v\in M$. Then define the map $\phi : A \rightarrow M$ such that $a \mapsto a\cdot v$. Then $\phi$ is an $A$-module homomorphism with $\text{Im}(\phi) = M$, and so $M \cong A / \text{Ker} (\phi)$. But $\text{Ker}(\phi) = \{ a \in A \mid a\cdot v = 0\} = Ann(v)$, and so $M \cong A / Ann(v)$.

Suppose now that $M \cong A / Ann(v)$ for some $v\in M$, if $M = \{0\}$ then in particular it is cyclic, so assume now that $M \neq \{0\}$. Let $\operatorname{\pi}$ be the quotient map, and let $1_{A}$ be the identity in A. Let $\operatorname{\pi} : A \rightarrow M$ be the quotient map. Let $u = \operatorname{\pi} (1_{A})$, then since $M \neq \{0\}$, $u \neq 0$, and then $M = \left<u\right>$ and so in particular is cyclic.

$\textbf{Remark 1: }$ The backwards direction of the above claim does not use any special properties of the annihilator, it is in fact just a special case of that fact that $A/I$ is cyclic for any left ideal $I$ of $A$.

$\textbf{Remark 2: }$ It is certainly true that a submodule of a cyclic $A$-module is finitely generated. This can be seen since $A$ is a finitely generated $k$-algebra, and so in particular is Noetherian. Then any cyclic $A$-module is a quotient of a Noetherian algebra, and so is Noetherian, and hence any sub-module of a cyclic $A$-module is finitely generated.

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    $\begingroup$ What about just $I=(x,y)$ in $R=k[x,y]$? That is the most commonly used example of a non PID, and it is certainly a 2 generated submodule of a cyclic submodule. $\endgroup$ – rschwieb May 20 '18 at 15:06
  • $\begingroup$ @rschwieb Thanks! That's a far simpler example. I overlooked that any ideal of a ring with a unit is a submodule of the cyclic submodule $(1)$. Thanks! $\endgroup$ – Adam Higgins May 20 '18 at 15:13

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