Description of $\mathbb Z/p^2\mathbb Z$ as an extension of $\mathbb Z/p\mathbb Z$ The ring $\mathbb Z/p^2\mathbb Z$ has $\mathbb Z/p\mathbb Z$ as an ideal, with a quotient ring again isomorphic to $\mathbb Z/p\mathbb Z$. Is there a way to describe the structure of $\mathbb Z/p^2\mathbb Z$ based on this observation?
I tried writing elements of $\mathbb Z/p^2\mathbb Z$ as $kp+i$ with $0\leq k, i\leq p-1$. This representation is unique, and it's not difficult to get expressions for the representation of a sum and product of two elements expressed with that representation. But the problem is that we don't have $kp+i\equiv k'+pi'\mod p^2$ if and only if $k\equiv k'\mod p$ and $i\equiv i'\mod p$, so I can't see how to actually use this to describe the structure  of $\mathbb Z/p^2\mathbb Z$.
 A: The ring $A=\mathbb Z/p^2\mathbb Z$ has an ideal $I=(p)$ which is nilpotent and, in fact, has square zero. The quotient $A/I$ is isomorphic to $\mathbb Z/p\mathbb Z$ and we therefore have what is sometimes called an infinitessimal extension of rings $$0\to I\to A\to A/I\to 0$$ There is a whole theory dealing with such extensions, and they are classified using MacLane cohomology (the original reference is [S. MacLane. Homologie des anneaux et des modules. Coll. Topologie Algébrique (1956), pp. 55–80 Louvain] 
A: Forgive me for writing $\mathbb{Z}_n$ to mean $\mathbb{Z} / n \mathbb{Z}$ below.
The funny thing about the cyclic rings, namely $\mathbb{Z}$ and the rings $\mathbb{Z}_n, n \in \mathbb{N}$, is that their multiplicative structure is completely determined by their additive structure (together with the data of which element is $1$). In fact each of these rings is canonically isomorphic to the endomorphism ring of its underlying abelian group. More generally, if $R$ is a commutative ring and $R/I$ a quotient of it, then $R/I$ as an $R$-algebra is canonically isomorphic to the endomorphism $R$-algebra of $R/I$ as an $R$-module. 
So everything you need to know about $\mathbb{Z}_{p^2}$ as a ring is contained in the data of $\mathbb{Z}_{p^2}$ as an abelian group, and so you can use the language of group extensions and cohomology. In fact abelian extensions of abelian groups are completely described by the Ext functor, and the fact that the extension
$$0 \to \mathbb{Z}_p \to \mathbb{Z}_{p^2} \to \mathbb{Z}_p \to 0$$
does not split reflects the fact that the group $\text{Ext}^1(\mathbb{Z}_p, \mathbb{Z}_p)$ is nonzero (in fact it's $\mathbb{Z}_p$ again). 
A: The structure of the ring $\mathbb Z/p^2\mathbb Z$ is the same as the structure of the group $\mathbb Z/p^2\mathbb Z$ and both can be extracted from the canonical projection $\mathbb Z \to \mathbb Z/p^2\mathbb Z$ which has kernel $p^2\mathbb Z$.
As a group, $\mathbb Z/p^2\mathbb Z$ is cyclic and cannot be written as the product of two smaller groups.
