# Examples of rings with idempotent elements

As a part of my studies in ring theory, I've encountered the concept of an idempotent element, i.e., an element $e$ such that $e^2=e$.

Are there some interesting examples of rings with idempotent elements?

As you realize, all rings have the idempotents $0$ and $1$, so the question is whether they have any others.

If a commutative ring has a non-trivial idempotent, then it is isomorphic to a product of two non-trivial rings. The same is true for a non-commutative ring, as long as the idempotent lies in its centre.

If $n$ is a positive integer which is not a prime power then $\mathbb{Z}/n\mathbb{Z}$ has nontrivial idempotents.

Matrix rings tend to have lots of idempotents, but not usually in their centres.

A group algebra $KG$ for a finite group $G$ over a characteristic zero field $K$ has the central idempotent $|G|^{-1}\sum_{g\in G}g$ and usually others.

Nontrivial idempotents are intimately connected to direct product decompositions. Generally, any idempotent $$\rm e^2 = e$$ yields a decomposition $$\rm\ R = e R + (1-e) R\$$ known as the Pierce Decomposition, and vice versa. This extends to any finite set of idempotents with sum 1 which are mutually orthogonal: $$\rm\ e_j\: e_k = 0\$$ if $$\rm\ j \ne k\$$. For example, the Chinese Remainder Theorem (CRT) for rings has this form.

Actually there are rings in which every element is idempotent. They are called boolean rings. Stone duality tells us that

$X \mapsto C_0(X,\mathbb{F}_2)$

is a duality between locally compact totally disconnected hausdorff spaces and boolean rings. Compactness here corresponds to unitality. For example, every finite boolean ring is isomorphic to $(\mathbb{F}_2)^n$ for some $n$.

Let $X$ be a disconnected space. Then the ring $C(X)$ of complex continuous functions has an idempotent.

In a sense, this is a motivating example, because if $R$ has a nontrivial idempotent, then $Spec(R)$ is disconnected. Since elements of $R$ can be viewed (loosely) as functions, we can construct an element of $R$ which is "one" on one half of $Spec(R)$ and "zero" on the other. (To make this precise, one thinks in terms of the sheaf of regular functions on the scheme $Spec(R)$.)

The endomorphism ring $\mathrm{End}(V)$ of a given vector space $V$ is riddled with idempotents. For instance, all projections $\pi=\pi_{L_1,L_2}$ of $V$ onto $L_1$ along $L_2$ where $V=L_1 \oplus L_2$ is a direct decompositon of $V$ are idempotents ($\pi(x)=\pi(x_1+x_2)=x_1$ where $x \in V$ and $x_k \in L_k,$ $k=1,2$).

Also, there are many hard/nice questions concering idempotents in group rings.

You can compute idempotents in any square matrix ring $M_n(K)$, $K$ a field, as follows: $A \in M_n(K)$ is idempotent if and only if $A^2 = A$. But this means that the polynomial $p(t) = t(t-1)$ annihilates $A$. So the minimal polynomial $m_A(t)$ of $A$ must divide $p(t)$. This leaves only four possibilities for $m_A(t)$:

1. $m_A(t) = 1$, in which case the unit matrix equals the zero matrix $I=0$ and $K = 0$.
2. $m_A(t) = t$, in which case $A = 0$.
3. $m_A(t) = t-1$, in which case $A=I$.
4. $m_A(t) = t(t-1)$ is the only non trivial one. It means that your matrix $A$ has only two eigenvalues: $0$ and $1$. For instance,

$A= \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \ .$

Since all of prime factors in the minimal polynomial have multiplicity $1$, all diagonalize, so it's fairly easy to get a picture of them: all are of the form $A = S^{-1}D S$, with $D$ a diagonal matrix with just $0$ and $1$ on the main diagonal.

Matrices of Olod's non-trivial projections belong to this case.