# Replacing two idempotents with one

Posted by Jason Polak on 27. December 2017 · 2 comments · Categories: commutative-algebra · Tags:

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?

# Projective Principal Ideals, Idempotent Annihilators

Given an idempotent $e$ in a ring $R$, the right ideal $eR$ is projective as a right $R$-module. In fact, $eR + (1-e)R$ is actually a direct sum decomposition of $R$ as a right $R$-module. An easy nontrivial example is $\Z\oplus\Z$ with $e = (1,0)$.

Fix an $a\in R$. If $aR$ is a projective right $R$-module, however, that doesn't mean that $a$ is an idempotent. In fact $aR$ is projective whenever $a$ is a nonzerodivisor, and in this case $aR$ is just isomorphic to $R$ itself as a right $R$-module.

So how do idempotents come into play in general? It turns out we have to look at annihilators! The right annihilator of $e$ is the right ideal $(1-e)R$. Indeed, $e(1-e) = 0$. And, if $er = 0$, then $(1 – e)r = r$, so anything that annihilates $e$ is a multiple of $(1-e)$. So we see that the annihilator of $eR$ is $(1-e)R$.

What about in general? It turns out that if $aR$ is projective, the right annihilator of $a$ must be of the form $eR$ for an idempotent $e$. Indeed, if $aR$ is projective, then the map $R\to aR$ given by $r\mapsto ar$ has a splitting $\varphi:aR\to R$. I'll leave it as an exercise to show that the right annihilator of $a$ is $(1 – \varphi(a))R$, and that $1 – \varphi(a)$ is in fact an idempotent.

Conversely, if the right annihilator of an $a\in R$ is of the form $eR$ for some idempotent, then multiplication by $1-e$ gives the splitting of the natural map $R\to aR$, so $aR$ must be projective.