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Let $R$ be a (not necessarily) commutative ring and $e \in R$ some idempotent. Then the Pierce decomposition writes $$ eRe \oplus (1-e)Re \oplus eR(1-e) \oplus (1-e)R(1-e). $$ I tried to construct such an isomorphism, specifically consider the cartesian product $eRe \times eR(1-e) \times (1-e)Re \times (1-e)R(1-e)$, and the mapping $$ \varphi(r) = (ere, re - ere, er - ere, r - er - (re - ere)). $$ Then this is bijective. And also it is an isomorphism of $(R, +)$ and the additive group we get by componentwise addition on this cartesian product.

But the only case that multiplication makes any sense is when $e$ is central, i.e. commutes with all elements, otherwise I am not even able to define multiplication componentwise in the above case. But the Pierce decomposition, for example also here and in the above link, does not require the idempotents to be central. So I am unable how this decomposition should work out, and I am surprised by the above inconsistencies? So could anyone explain them?

Also for the preservation of the additive structure, and surjectivity and injectivity of $\varphi$ nowhere is it needed that we have $e^2 = e$.

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  • $\begingroup$ I'm not sure what you mean by "I am not even able to define multiplication componentwise". Even if $e$ is not central, $eRe$ and all the other components are closed under multiplication as subsets of $R$, so you can use that to define multiplication componentwise. Of course, the problem remains that $\varphi$ may not preserve multiplication. $\endgroup$ – Eric Wofsey Jan 14 '17 at 20:52
  • $\begingroup$ @EricWofsey Okay, yes surely the sets $eRe$ etc are subrings. So is there a way to give an isomorphism, i.e. are they isomorphic then? $\endgroup$ – StefanH Jan 14 '17 at 20:57
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The Peirce decomposition is only additive, not multiplicative. That is, $R$ is only isomorphic to $eRe \oplus (1-e)Re \oplus eR(1-e) \oplus (1-e)R(1-e)$ as an additive group, not as a ring (unless, as you observe, $e$ happens to be central).

If you want to get an isomorphism of rings, you have to instead put a different multiplication on the decomposition instead of the componentwise multiplication. The way to do this is to write the components as a $2\times 2$ matrix like $$\begin{pmatrix}eRe & eR(1-e) \\ (1-e)Re & (1-e)R(1-e)\end{pmatrix}$$ and use the usual matrix multiplication formula. That is, given $(a,b,c,d)\in eRe \oplus (1-e)Re \oplus eR(1-e) \oplus (1-e)R(1-e)$, you think of $(a,b,c,d)$ as the matrix $$\begin{pmatrix}a & b \\ c & d\end{pmatrix}.$$ To multiply two elements of $eRe \oplus (1-e)Re \oplus eR(1-e) \oplus (1-e)R(1-e)$, you then just multiply the corresponding matrices according to the usual formula for matrix multiplication.

(To provide some intuition for why this should work, consider the case that $R=M_2(k)$ and $e=\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}$. Then the four components of the Peirce decomposition just correspond to the four matrix entries of an element of $R$. For instance, $eRe$ is the set of matrices whose only nonzero entry is the top left entry, $eR(1-e)$ is the set of matrices whose only nonzero entry is the top right entry, and so on.)

The idempotence of $e$ is in fact needed in order to get the additive isomorphism, specifically to get that your map is surjective. For instance, if $R=\mathbb{Z}$ and $e=2$, then all the direct summands in the decomposition are isomorphic to $\mathbb{Z}$ as groups, so your map is a group-homomorphism $\mathbb{Z}\to\mathbb{Z}^4$, which cannot be surjective.

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