Idempotent

An idempotent $𝑒 \in 𝑅$ of a ring $𝑅$ is an element satisfying $(𝑒) 2 = 𝑒$ . #m/def/ring

Further terminology

$1$ and $0$ are idempotents, called trivial.
For any such $𝑒$ , we automatically have $(1 - 𝑒) 2 = 1 - 2 𝑒 + 𝑒 2 = 1 - 𝑒 .$ We call $1 - 𝑒$ the complementary idempotent of $𝑒$ .
Idempotents $𝑒, 𝑓$ are called orthogonal iff $𝑒 𝑓 = 0$ . Thus, in particular, $𝑒$ and $1 - 𝑒$ are orthogonal idempotents. As a result, nontrivial idempotents are zero-divisors.

Effect on modules

Let $𝑒 \in 𝑅$ be an idempotent. Then if $𝑓 : 𝑉 \to 𝑊$ is a (left) $𝑅$ -morphism, then $𝑓 (𝑒 𝑉) \subseteq 𝑒 𝑊$ . Thus left multiplication by $𝑒$ forms an endofunctor on $𝑅 𝖬 𝗈 𝖽$ , equipped with natural transformations from and to the identity endofunctor for projection and inclusion.

Moreover in $𝖠 𝖻$ we have a natural isomorphism

c|https://q.uiver.app/#q=WzAsNCxbMCwwLCJcXGxNb2R7Un0oUmUsVikiXSxbMCwyLCJcXGxNb2R7Un0oUmUsVykiXSxbMiwwLCJlViJdLFsyLDIsImVXIl0sWzAsMiwiXFxsYW1iZGFfViIsMCx7Im9mZnNldCI6LTF9XSxbMSwzLCJcXGxhbWJkYV9XIiwwLHsib2Zmc2V0IjotMX1dLFsyLDMsImYiXSxbMCwxLCJmXyoiLDJdLFsyLDAsIiIsMSx7Im9mZnNldCI6LTF9XSxbMywxLCIiLDAseyJvZmZzZXQiOi0xfV1d

where for $𝑔 \in 𝑅 𝖬 𝗈 𝖽 (𝑅 𝑒, 𝑉)$ we have $𝜆 𝑉 (𝑔) = 𝑔 (𝑒)$ .

Proof

The functoriality follows from $𝑓 (𝑒 𝑉) = 𝑒 𝑓 (𝑉) \subseteq 𝑒 𝑊$ , and the first diagram commutes by the fact $𝑉$ and $𝑊$ are left $𝑅$ -modules.

To see that $𝜆 𝑉$ has the correct codomain, note that for $𝑔 : 𝑅 𝑒 \to 𝑒 𝑉$ we have $𝑒 𝑔 (𝑒) = 𝑔 (𝑒 2) = 𝑔 (𝑒)$ . Clearly $𝜆 𝑉$ is a group homomorphism. Also, $𝜆 𝑉$ is a group monomorphism, because if $𝑔 (𝑒) = 𝑔' (𝑒)$ then $𝑔 (𝑟 𝑒) = 𝑟 𝑔 (𝑒) = 𝑟 𝑔' (𝑒) = 𝑔' (𝑟 𝑒)$ for any $𝑟 \in 𝑅$ . To see that it is a group epimorphism, let $𝑒 𝑣 \in 𝑒 𝑉$ . Then we can define $𝑔 : 𝑅 𝑒 \to 𝑉 : 𝑟 𝑒 \mapsto 𝑟 𝑒 𝑣$ , wherefore $𝜆 𝑉 (𝑔) = 𝑔 (𝑒) = 𝑒 2 𝑣 = 𝑒 𝑣$ .

Finally, naturality follows from the fact $𝑓 𝜆 𝑉 (𝑔) = 𝑓 𝑔 (𝑒) = 𝜆 𝑊 𝑓 * 𝑔$ .

The case when $𝑉 = 𝑅 𝑒$ gives a ring antiïsomorphism

𝜆 𝑅 𝑒 : E n d 𝑅 (𝑅 𝑒) 𝐨 𝐩 ≅ 𝑒 𝑅 𝑒

Proof

To see that $𝜆 𝑅 𝑒$ is a ring antiïsomorphism, let $𝑔, ℎ \in E n d 𝑅 (𝑅 𝑒)$ . Then

𝜆 𝑅 𝑒 (𝑔 ℎ) = 𝑔 ℎ (𝑒) = 𝑔 ℎ (𝑒 2) = 𝑔 (ℎ (𝑒) 𝑒) = ℎ (𝑒) 𝑔 (𝑒) = 𝜆 𝑅 𝑒 (ℎ) 𝜆 𝑅 𝑒 (𝑔)

as required.

Idempotent

Further terminology

Effect on modules

See also