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

azimut
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zacarias
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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 00:16
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    ... and local algebras have only trivial idempotents, so that the characteristic has to be a prime power. – Martin Brandenburg Feb 17 '13 at 00:36
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    yes, The ring is assumed to be unital. – zacarias Feb 17 '13 at 13:37
  • Related: https://math.stackexchange.com/questions/1825661 – Watson Dec 11 '16 at 18:43

2 Answers2

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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 $\mathbb{Z}/n \cong \prod_p \mathbb{Z}/p^{v_p(n)}$.

Martin Brandenburg
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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.)

Community
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