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I have troubles in showing this fact. Let $R$ be a ring, and assume that $e_1,\dots,e_n$ are central idempotents elements of $R$, i.e. idempotents contained in the center of $R$. Assume moreover that the following holds: $$\sum_{i=1}^n e_i=1.$$

Then prove that $$R=e_1R\times e_2R\times\cdots\times e_nR,$$

that is, the internal direct sum as an abelian group, in which the multiplication is defined componentwise.

Thanks in advance.

EDIT in the comments below the crucial points has been touched. Is it inessential that we assume $e_ie_j=0\quad \forall i\neq j$ or may this stronger hypothesis be recovered from the ones we are given in the text? (P.S. the problem lies in an handout my teacher gave to me and i've copied word by word)

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  • $\begingroup$ Consider the homomorphism: $a \in R \mapsto (ae_1,ae_2,...,ae_n)$ $\endgroup$ – francis-jamet Dec 6 '11 at 18:27
  • $\begingroup$ yes but what about proving that the sum is direct? $\endgroup$ – uforoboa Dec 6 '11 at 18:28
  • $\begingroup$ ok... can you please post a brief solution to my question? so then i may accept the answer and thank you :) $\endgroup$ – uforoboa Dec 6 '11 at 18:50
  • $\begingroup$ If $R= (\mathbb{Z}/ 2\mathbb{Z})^3, e_1=(1,0,0), e_2=(0,1,0), e_3=(0,0,1), e_4=e_5=(1,1,1)$, it seems false. Don't you have the condition $e_i e_j=0$ if $i \neq j$ ? $\endgroup$ – francis-jamet Dec 6 '11 at 18:58
  • $\begingroup$ you get exactly the point. To be honest the condition is not cited in the text but to me it sounded essential.. Perhaps there is the fact that they are central elements in $R$ so that condition may be recovered from the hypothesis... $\endgroup$ – uforoboa Dec 6 '11 at 19:05
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There is a counterexample.

If $R= \mathbb{Z} /2 \mathbb{Z}$, and $e_1=e_2=e_3=\bar{1}$.

$e_1,e_2$ and $e_3$ are central idempotents.

$e_1+e_2+e_3=\bar{1}$.

$e_i R=R$ for all $i \in \{1,2,3\}$.

But $R \neq R \times R \times R$, because $R$ has 2 elements, and $R \times R \times R$ has $8$ elements.

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