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Let be $C$ a commutative ring such that $1+1=0$ and $0\neq 1$. I want to prove:

There exist rings $C_i$ such that $C\cong C_1\times \cdots\times C_n$ if and only if $C$ has a sub-ring isomorphic to $\mathbb{F}_2^n$

Some ideas:

  • If two rings are isomorphic then has the same characteristic.
  • If a Ring has characteristic $n$ then it has a sub-ring isomorphic to $\mathbb{Z}_n.$
  • A boolean ring is generated by copies of $\mathbb{Z}_2.$

Can help me how order these ideas?

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    $\begingroup$ What are all the $C_i$ supposed to be? $\endgroup$
    – Randall
    Feb 10, 2022 at 17:10
  • $\begingroup$ Your $\iff$ statement has no conditions on the left side. What is the question? $\endgroup$ Feb 10, 2022 at 17:12
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    $\begingroup$ But what about them? $C=\mathbb{Z} \times \mathbb{Z}$ is a commutative ring (that is a direct product) with no subrings iso to $\mathbb{F}_2$ or direct copies thereof. What's the question? $\endgroup$
    – Randall
    Feb 10, 2022 at 17:13
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    $\begingroup$ The idempontents aren’t a subring. But the ideals generated by individual idempotents is a subring. $\endgroup$ Feb 10, 2022 at 17:14
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    $\begingroup$ Your title is incomplete. That is a risk of trying to put the whole question in the title. Ask the question in the body of the question, even when it fits in the title. The title has another purpose - to let readers decide quickly if they can help. The body should have the complete details. $\endgroup$ Feb 10, 2022 at 17:20

1 Answer 1

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If $e_1,\dots,e_n$ are the standard idempotents of $\mathbb F_2^n,$ then $e_i\neq 0,1$ are idempotents in $C,$ and $e_ie_j=0$ when $i\neq j.$ Let $e_{n+1}=1-(e_1+\cdots+e_n).$

Show $e_{n+1}$ is an idempotent and $e_ie_{n+1}=0$ for $i<n+1,$ and $1=e_1+\cdots+e_{n+1}.$

Then let $C_i=e_iC.$ Show the $C_i$ are rings. Show $C\cong C_1\times \cdots\times C_{n+1}.$ Then $$C\cong C_1\times \cdots\times C_{n-1}\times\left(C_{n}\times C_{n+1}\right)$$

Note, while it is possible that $e_{n+1}$ might be $0,$ the ring $C_n\times C_{n+1}$ is not a zero ring.

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  • $\begingroup$ Grateful for your proof's sketch. $\endgroup$ Feb 10, 2022 at 17:52

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