$(R,m)$ local ring with finitely many ideals, $R/m$ is infinite, why $R$ is a PIR and uniserial? 
Let $(R,m)$ be a commutative local ring with finitely many ideals and the residue field $R/m$ is infinite. I want to show that $R$ is a PIR and also uniserial.

I know that $R$ is both notherian and artinian, so there exist a composition series for $R$, it means that there exist a natural number $n$ s.t. $m^n=0$.
But i don't know how to continue. (Also I'm not sure if what I have done is useful...)
 A: Let $(R, m, k)$ be the commutative local ring with finitely many ideals and $\#k = \infty$. We need the following claim.
Claim: $\dim_k m/m^2 \le 1$.
Let us see first that how the statements follows from $\dim_k m/m^2 = 1$ (if $\dim_k m/m^2 = 0$ then $m = 0$ by the Nakayama's lemma and nothing to prove). 


*

*$R$ is uniserial: Since $R$ is local, $R/m^2$ is uniserial by the claim. Hence $R$ is also uniserial.

*$R$ is PIR: Take an element $\pi \in m \setminus m^2$. By the claim, $m = (\pi) + m^2$. Hence we have $m = (\pi)$ by the Nakayama's lemma. From the above proof, all the ideals are of the form $m^k = (\pi^k)$ for some $k \ge 0$.
Proof of Claim: (The basic idea is already shown in the comment of Mohan.) Assume $\dim_k m/m^2 \ge 2$. Then we have two elements $x, y \in m$ such that $\{x + m^2, y + m^2\}$ is linearly independent in $m/m^2$. Consider ideals 
$$I_\lambda := \{\, rx + \lambda ry + z \mid r \in R,\ z \in m^2 \,\}$$ 
parametrized by $\lambda \in k$. Since $\#k = \infty$, this yields an infinite series of ideals. Contradiction.
 QED.
