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Like any commutative ring $\mathbb{Z}/p^m\mathbb{Z}$ is an algebra over itself. I feel that there is also an structure of a $\mathbb{Z}/p^{m+1}\mathbb{Z}$ Algebra on $\mathbb{Z}/p^m\mathbb{Z}$ but I fail to show this formally. Can anybody help?

Secondly, is there any easy resolution of the latter algebra?

Thank you

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    You might want to think about the simplest case, say $\mathbb{Z}/4\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$. – 2017-01-17
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    What do you mean by a "resolution"? – 2017-01-17
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    @hardmath: My Intuition would be to multiply representatives but I fail to Show that this is well-defined. – 2017-01-17
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    @EricWofsey: An exact sequence of $\mathbb{Z}/p^{m+1}\mathbb{Z}$-algebras augmenting on $\mathbb{Z}/p^{m}\mathbb{Z}$. – 2017-01-17
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    What do you mean by an exact sequence of algebras? Do you just mean you want a free resolution of $\mathbb{Z}/p^m\mathbb{Z}$ as a $\mathbb{Z}/p^{m+1}\mathbb{Z}$-module? – 2017-01-17
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    Oh yes, a free resolution of $\mathbb{Z}/p^{m+1}\mathbb{Z}$-modules. – 2017-01-17
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    I don't know what your definition of "$S$ has the structure of an $R$-algebra" is, but mine (at least for commutative rings) is that we have chosen a ring homomorphism $R \rightarrow S$. Now what ring homomorphisms $\mathbb{Z}/p^{m+1} \rightarrow \mathbb{Z}/p^{m}$ do you see? – 2017-01-17
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    I think of a bilinear map $S\times S\to S$ but I know the other approach. You send $z+ p^{m+1}Z$ to $z+ p^{m}Z$? – 2017-01-17
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    Ok, i got it now. Considering the respective groups there are only two possible homomorphisms, one of which is trivial. Thanks a lot for this. I would still be glad if someone can help me with the free resolution of $\mathbb{Z}/p^{m+1}\mathbb{Z}$-modules! – 2017-01-17

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