Let $R$ be a commutative ring. Two idempotents $e$ and $f$ are called **orthogonal** if $ef = 0$. The archetypal example is $(0,1)$ and $(1,0)$ in a product ring $R\times S$.

Let $e$ and $f$ be orthogonal idempotents. Then the ideal $(e,f)$ is equal to the ideal $(e + f)$. To see, this first note that $(e + f)\subseteq (e,f)$. On the other hand:

$$(1-e)(e + f) = e + f – e – ef = f$$

Therefore $f \in (e + f)$. Switching $e$ and $f$ in this calculation shows that $e\in (e + f)$. Using the fact that $e + f$ is also an idempotent, we see that by induction, if $e_1,\dots,e_n$ are pairwise orthogonal idempotents, then the ideal $(e_1,\dots,e_n)$ is generated by the single element $e_1 + \dots e_n$.

Now suppose $e$ and $f$ are idempotents that are not necessarily orthogonal. Then $(e,f)$ is *still* a principal ideal. To see this, consider the element $e – ef$. The calculation

$$(e – ef)^2 = e – 2ef + ef = e – ef$$

shows that $e – ef$ is an idempotent. Furthermore, $(e,f) = (e – ef,f)$ and $e-ef$ and $f$ are orthogonal idempotents. By what we discussed in the previous paragraph, $(e,f) = (e-ef,f)$ is generated by $e – ef + f$.

Everything we did assumed $R$ was commutative. But what if we foray into the land of noncommutative rings? Is it still true that a left-ideal generated by finitely many idempotents is also generated by a single idempotent? Any ideas?

## 2 Comments

It might not help for the general problem, but the answer to your final question is true for C*-algebras, where "finitely generated" has the same meaning as in "purely algebraic" theory, i.e. we are not taking closed ideals.

See Lemma 2.1 in http://arxiv.org/abs/1402.4411

Thanks for the reference!