Let $(R, \mathfrak m)$ be an Artinian local ring with every non-maximal ideal being principal. Then is $R$ a principal ideal ring ?
closed as off-topic by user26857, José Carlos Santos, Xander Henderson, Saad, B. Mehta Jun 10 '18 at 1:45
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Take $R=k[x,y]/(x,y)^2$. Then $(x,y)$ is a $2$-dimensional $k$-vector space and clearly not principal. Let $I \subseteq R$ with $I \neq R$. Then $I \subseteq (x,y)$ with the induce vector space structure. So, for $I \neq (x,y)$ the $k$-dimension of $I$ is at most $1$, showing that $I$ is principal.