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I'm reading David Eisenbud, Commutative Algebra with a View Toward Algebraic Geometry. At page 13, Chapter $0$, he says: "... if $e_1,\ldots,e_n$ is a complete set of orthogonal idempotents in a commutative ring, then $R=Re_1\times\cdots\times Re_n$ is a direct product decomposition."

How can be proved this?

Federica Maggioni
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2 Answers2

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You have $1_{R} = \sum_{i=1}^{n} e_{i}$ and $e_{i}e_{j} = \delta_{ij}e_{i}$ for each $i,j.$ From here on in, it is just a matter of writing down consequences of that.

Geoff Robinson
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Define a map $f:R\to Re_1\times\cdots\times Re_n$ by $f(x)=(xe_1,\dots,xe_n)$ and prove that $f$ is a surjective ring homomorphism with $\ker f=(0)$. (Note that $Re_i$ are unitary commutative rings and $f$ is a homomorphism of unitary rings.)

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    ok ok, the preimage of $(x_1e_1,\ldots,x_ne_n)$ is $x:=x_1e_1+\ldots+x_ne_n$ since $x\cdot e_i=x_i$ and $x_i=x_ie_i$, because $e_i$ is easily checked to be the unit of $Re_i$. – Federica Maggioni Apr 21 '13 at 15:23
  • How do you show the kernel is zero!? I seem to get stuck because there could be zero divisors. Could you please explain the reasoning there? – User0112358 May 27 '15 at 01:22
  • @User0112358 By assumption, the $e_i$ form a complete set of orthogonal idempotents, meaning that $e_1 + \cdots + e_n = 1$. So if $x$ is in the kernel, then $0 = xe_1 = \cdots = xe_n$, so $0 = x(e_1 + \cdots + e_n) = x$. – Timothy Chow Jul 13 '23 at 04:07