Let $R$ be a (commutative or non-commutative) unital ring, and let $e \in R$ be an idempotent. It is well known (and relatively easy to prove) that $eR = \mathfrak i_1 \oplus \mathfrak i_2$, for some right ideals $\mathfrak i_1, \mathfrak i_2 \subseteq R$, if and only if there exist orthogonal idempotents $e_1, e_2 \in R$ with $e = e_1 + e_2$ such that $\mathfrak i_1 = e_1 R$ and $\mathfrak i_2 = e_2R$.
Question. Do you know of any explicit reference to this (elementary but very useful) little fact?
I was pretty sure I would have found it mentioned in [1], but I can't see it there.
Edit (2021-07-02). In a comment under this post, @rschwieb wrote:
[…] Regarding the "easy to prove" part: it's obvious $\mathfrak i_1=fR$ for some idempotent $f$, but how do you cook up $e_1$ from $e$ and $f$? I tried a few things I know about generating idemopotents, but the orthogonality with $e-e_1$ wasn't working out.
Assume first that $eR = \mathfrak i_1 \oplus \mathfrak i_2$, where $e \in R$ is an idempotent and $\mathfrak i_1, \mathfrak i_2$ are right ideals of $R$. Then
$$
R = eR \oplus (1_R – e)R = \mathfrak i_1 \oplus \mathfrak i_2 \oplus (1_R – e)R,
$$
so we conclude from, say, [1, Ch. III, Sect. 7, Proposition 2] that there is a Peirce basis $(e_1, e_2, e_3)$ of $R$ such that $ \mathfrak i_1=e_1R$, $ \mathfrak i_2=e_2 R$, and $(1_R-e)R=e_3R$. In particular, this yields $eR=e_1R \oplus e_2R$, showing that $e = e_1a + e_2b$ for some $a, b \in R$; moreover, $1_R-e = e_3c$ for a certain $c\in R$. As a result, we have $1_R = e_1a + e_2b + e_3c$, and hence $e_1 = e_1 1_R = e_1a$ and $e_2 = e_2 1_R = e_2b$. It follows $e = e_1 + e_2$, and so we are done with the "only if" part.
Now suppose that $e = e_1 + e_2$ for some orthogonal idempotents $e_1, e_2 \in R$, and let $\mathfrak i_1 := e_1 R$ and $\mathfrak i_2 := e_2 R$. Since $e_1 = e e_1$ and $e_2 = e e_2$, it is clear that $$
eR = (e_1 + e_2)R \subseteq e_1 R + e_2 R \subseteq e R + eR = eR,
$$
that is, $eR = e_1R + e_2 R$. On the other hand, if $e_1 a = e_2 b$ for some $a, b \in R$, then $0_R = e_1 e_2 b = e_1^2 a = e_1 a$. So, $e_1R$ and $e_2R$ intersect trivially, implying that $eR = e_1R \oplus e_2R$ (as wished).
Bibliography.
- [1] N. Jacobson, Structure of Rings, Amer. Math. Soc. Colloq. Publ. 37, Amer. Math. Soc., 1964.
Best Answer
[Sorry for answering my own question, but I've just found the answer.]
If $e = 0_R$, the conclusion is obvious. Otherwise, the result is a well-known characterization of primitive idempotents; for an explicit reference, see Proposition (21.8) in