# Structure of Finite Commutative Rings

Is every finite commutative ring $A$ a direct product of finite algebras over $\mathbb Z/p^n$?

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You assume that $A$ has a unit? My first thought is that every finite commutative ring is Artinian. So by the structure theorem for Artinian rings may be written as a direct product of Local Artinian rings. So it suffices to prove the result for Local finite commutative rings. –  JSchlather Feb 17 '13 at 0:16
... and local algebras have only trivial idempotents, so that the characteristic has to be a prime power. –  Martin Brandenburg Feb 17 '13 at 0:36
yes, The ring is assumed to be unital. –  zacarias Feb 17 '13 at 13:37

A finite commutative ring is artinian, so it is a finite product of finite artinian local rings. Let $R$ be a finite commutative local ring and $\phi:\mathbb Z\to R$ the canonical ring homomorphism ($n\mapsto n1_R$). Then $\ker \phi=n\mathbb Z$, $n>1$. It follows that $\mathbb Z/n\mathbb Z$ is isomorphic to a subring of $R$. Since the idempotents of $R$ are trivial, the same holds for $\mathbb Z/n\mathbb Z$. This implies that $n$ is a power of a prime, and we are done. (More on finite local rings you can find here.)

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The classification of artinian rings is a total overkill here. –  Martin Brandenburg Feb 17 '13 at 4:06
@MartinBrandenburg If you think that the Chinese Remainder Theorem is an overkilling tool, then you are right! –  user26857 Feb 17 '13 at 9:43

Sure. A ring homomorphism $\prod_i R_i \to S$ corresponds to a decomposition $S =\prod_i S_i$ and a family of ring homomorphisms $R_i \to S_i$ (namely $S_j=S \overline{e_j}$ where $\overline{e_j}$ is the image of the canonical idempotent $e_j \in \prod_i R_i$). Thus, an algebra over a product is a product of algebras.

A finite ring has some characteristic $n \neq 0$, i.e. is an algebra over $\mathbb{Z}/n \cong \prod_p \mathbb{Z}/p^{v_p(n)}$.

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This is much simpler than what I had in mind. There's also some more of characterizing finite commutative rings here. –  JSchlather Feb 17 '13 at 0:35
@MartinBrandenburg: What exactly do you mean by "canonical idempotent"? –  Manos Feb 17 '13 at 16:24
The element $(0,\dotsc,1,\dotsc,0)$. –  Martin Brandenburg Sep 30 '13 at 22:18