# Identity of a ring as two different sums of idempotents

Let $R$ be any ring with identity $1_R$.

Prove that if there exist idempotents $e_1,..., e_n, e'_1,...,e'_n \in R$ such that

$$1_R = e_1 +...+ e_n = e'_1+... e'_n$$ then the following conditions are equivalent:

(i) $Re_i \simeq_R Re'_i$ for all $1 \leq i \leq n$

(ii) There exists an invertible element $a \in R^\times$ such that $e'_i = ae_ia^{-1}$ for all $1 \leq i \leq n$

My attempt:

$\bullet$ $(ii) \Rightarrow (i)$

Define a ring homomoprhism $$\psi : Re'_i \longrightarrow Re_i$$

such that $\psi(x) = xa$.

It is well defined and we check that its image lies in $Re'_i$ :

$$\psi(x) = \psi(xe'_i) = xe'_ia=x(aa^{-1})e'_ia=xae_i \in Re_i$$

To prove surjectivity just note that for any $r \in R$, we have $$re_i= \psi(ra^{-1}e'_i)$$

Injectivity follows because $a$ is a unit, and therefore $xa=0 \Longleftrightarrow x=0$, thus Ker $\psi = \{0\}$

$\bullet$ $(i) \Rightarrow (ii)$

For each $1 \leq i \leq n$ let $\phi_i$ be an isomoprhism between $Re_i$ and $Re'_i$.

Define $\phi \in$ End($R$) such that $\phi(x) = \phi_i(x)$ whenever $x \in Re_i$.

Now is the part I get stuck, I feel I am heading in the right direction and I have tried to play with the images of $e_i$ but I can't seem to get close to proving what I want.

• Where did you find this exercise? I normally put more conditions on idempotents, so I'm probably going to use stuff I shouldn't when I try this. Also, is $\phi$ well defined? And I would try to play with the images of $1$, since this is the only invertible element of $R$ that you know. – Mathematician 42 May 5 '16 at 23:18
• If you know that $e_ie_j=0$ for $i\ne j$ and $e'_ie'_j=0$ for $i\ne j$, then $a=\sum_i\phi_i^{-1}(e_i')$ and $a^{-1}=\sum_i \phi_i(e_i)$ will do, where $\phi_i:Re_i\to Re_i'$ are the given isomorphisms. – san May 6 '16 at 2:34
• It was one exercise my prof gave to us. I don't know which book. – proofromthebook May 6 '16 at 21:53
• For (ii) ⇒ (i): I cannot see any reason that $Re_i$ to be a ring. If you are working with noncommutative rings then it is just a left $R$-module, although the same argument works. – Orat May 11 '16 at 18:57

If you know that $e_ie_j=0$ for $i\ne j$ and $e′_ie′_j=0$ for $i\ne j$ (a condition forgotten to mention by the OP), then $a=\sum_i \phi_i^{-1}(e′_i)$ and $a^{−1}=\sum_i \phi_i(e_i)$ will do the job, where $\phi_i:R e_i\to Re′_i$ are the given left $R$-module isomorphisms.

• It's really kind of miraculous how the module isomorphisms produce an inner automorphism – rschwieb May 12 '16 at 3:22

A simple example to show that this is not in general true without the orthogonality condition:

Let $R$ be a ring of characteristic $2$ with two conjugate but non-equal idempotents $e$ and $f$, so that $Re$ and $Rf$ are isomorphic as left $R$-modules. For example, take $R=M_2(\mathbb{F}_2)$, $e=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $f=\begin{pmatrix}0&0\\0&1\end{pmatrix}$.

Then $$1=1+e+e+f+f=1+e+f+e+f.$$

As it is already commented by Mathematician 42 and san, this statement is usually stated with an orthogonality hypothesis (you can find, for example, Exercise 7 in here). So it is likely that your professor forget to mention about it.

You (and perhaps your professor) should be careful that the definition of an idempotent decomposition $e = e_1 + \dotsb + e_n$ requires not only $e_i^2 = e_i$ but also $e_ie_j = \delta_{ij}e_j$ (op. cit.), which I also had some trouble.

• Yes Orat, I should have added that, my professor told us that afterwards – proofromthebook May 11 '16 at 19:59