Recently I am working with exercises in The block theory of finite group algebras by Markus Linckelmann. I have a question about uniserial module. In Exercise 1.10.24, It says that
Let $A$ be a $k$-algebra and $M$ a uniserial $A$-module. Show that $M$ is finitely generated and isomorphic to a quotient of the regular $A$-module $A$.
In this book $k$ represents a unital commutative ring and $A$ is always a finite dimension $k$-algebra that is finitely generated as a $k$-module.
I want to construct a surjective map from $A$ to $M$ just as the proof of showing simple modules are regular, but I can't find such a map. How to construct this isomorphism? and do uniserial modules always have such a property?
As noted in the comments, not all uniserial modules need be finitely generated, even under the conditions in the OP. Let $A=k=\mathbb{Z}$, let $p$ be a prime, and let $M$ be the uniserial module $\mathbb{Z}_{p^\infty}$. Certainly $M$ is not finitely generated.
There are positive results. In fact, a nonzero uniserial module $M$ is cyclic if and only if it has a maximal submodule. To see this, suppose $N$ is a maximal submodule and $m\in M\setminus N$. Since $Rm$ is not contained in $N$, it must properly contain $N$, whence $M=Rm$. Since every finitely generated nonzero module has maximal submodules, this shows that every finitely generated uniserial module is cyclic.
Let $J$ be the Jacobson radical of $A$ and suppose $J$ is nilpotent and $A/J$ is semisimple. (Rings $A$ with these properties are called semiprimary. Artinian rings are semiprimary.) If $M$ is a nonzero uniserial module, the nilpotency of $J$ implies $JM\subsetneq M$. As $M/JM$ is a nonzero semisimple module, it must have a maximal submodule. Pulling this back, we see $M$ has a maximal submodule and thus $M$ is cyclic.
If $k$ is a commutative Artinian ring and $A$ is a $k$-algebra that is finitely generated as a $k$-module, then $A$ is Artinian. Thus we get a positive answer to the OP: every uniserial $A$-module will be cyclic in this case.