## Idempotents and Decomposition

Let R be a general ring, not necessarily commutative. An element xR is said to be idempotent if x2x.

Note

An endomorphism f of an R-module M (i.e. $f\in \text{End}_R M$) is an idempotent if and only if f is a projection, i.e. M = ker(f) ⊕ im(f) and fMM projects onto im(f). Indeed ⇐ is obvious, and conversely if f is idempotent, we have:

• Every mM is just f(m) + (m – f(m)). The first term is in im(f); the second term lies in ker(f) since f(m – f(m)) = f(m) – f2(m) = f(m) – f(m) = 0. So M = ker(f) + im(f).
• Any element of ker(f) ∩ im(f) can be written as f(m) such that f(f(m)) = 0. But this means f(m) = f2(m) = 0, so ker(f) ∩ im(f) = 0.

Throughout this article, we shall focus on idempotents which commute, i.e. effe. A set of idempotents {ei} is said to be orthogonal if eiej = 0 for all ij. The following are easy to prove.

1. The sum of two orthogonal idempotents is also an idempotent.
2. If e is any idempotent, then e and 1-e are orthogonal idempotents.

The key result is the following.

Theorem. Let R be any ring. There is a 1-1 correspondence between the following:

• a decomposition $R = I_1 \oplus \ldots \oplus I_n$ as a direct sum of left ideals, and
• orthogonal idempotents $e_1, \ldots, e_n$ such that $e_1 + \ldots e_n = 1.$

Proof

The correspondence is given as follows: for any $R = \oplus I_i$, write $1 = \sum_i x_i$ with $x_i \in I_i.$ Then each xi is idempotent since $x_i = x_i\cdot 1 = \sum_{j=1}^n x_i x_j$ and each $x_i x_j \in I_j$ since Ij is a left module. Since R is the direct sum of Ij‘s we have $x_i x_j = 0$ for all ji and $x_i = x_i^2$ for all i. Hence the $\{x_i\}$ are orthogonal idempotents.

Conversely given {ei}, let us define Ii := Rei which is a left ideal. Note that $R = \sum_i I_i$ since 1 lies in the RHS. On the other hand, an element of $Re_1 \cap (Re_2 + \ldots + Re_n)$ can be written as $r_1 e_1 = r_2 e_2 + \ldots + r_n e_n.$ Then $r_1 e_1 = r_1 e_1^2 = r_2 e_2 e_1 + \ldots + r_n e_n e_1 = 0$ since the ei‘s are orthogonal. Similarly, $Re_i \cap (Re_1 + \ldots + \hat{Re_i} + \ldots + Re_n) = 0$ for all i and thus $R = I_1 \oplus \ldots \oplus I_n.$

It remains to show that the two constructions are mutually inverse.

• Start with $R = \oplus I_i$ and write $1 = \sum_i e_i$ with $e_i \in I_i.$ We need to show $I_i = Re_i.$
• Since $e_i \in I_i$ we have $I_i \supseteq Re_i.$
• Conversely, if $x\in I_i$ write $x = x\cdot 1 = \sum_j xe_j$ where we have $xe_j \in Re_j \subseteq I_j$ from what we just proved. Since $x\in I_i$ we see that $xe_j = 0$ for all ji and so $x = xe_i \in Re_i.$

Finally, start with orthogonal ei‘s summing to 1 and let $I_i := Re_i.$ Clearly $1=\sum_i e_i$ with $e_i\in I_i$ and this is the unique representation. ♦

Note that under the above correspondence,

• Ii = 0 if and only if ei =0;
• any idempotent eR gives orthogonal {e, 1-e}, and so RRe ⊕ R(1-e).

## Indecomposable Left Ideals

Now suppose R is an artinian ring (and hence noetherian by the Hopkins–Levitzki theorem). The Krull-Schmidt theorem says that R is a direct sum of indecomposable projective modules Ii. Such modules correspond to primitive idempotents.

Definition. A non-zero idempotent e is said to be primitive if it cannot be written as a sum of non-zero orthogonal idempotents $e = f_1 + f_2.$

Proposition. If e is an idempotent, then Re is indecomposable if and only if e is primitive.

Proof

Indeed, if $e = f_1 + f_2$ then $Re = Rf_1 \oplus Rf_2$ is the direct sum of two non-zero left modules.

• First, Rf1 ∩ Rf2 = 0: indeed, if $xf_1 = yf_2$ then $xf_1 = xf_1^2 = yf_2 f_1 = 0$ so we do have a direct sum on the RHS.
• Clearly $R(f_1 + f_2) \subseteq Rf_1 \oplus Rf_2.$
• Finally for $xf_1 + yf_2$ in the RHS, we have $xf_1 + yf_2 = (xf_1 + yf_2)(f_1 + f_2).$

Conversely, if ReI ⊕ J, express ex+y and we have a sum of two orthogonal idempotents (proof left to the reader). ♦

In writing R as a direct sum of indecomposable projective modules, the terms are unique up to isomorphism and permutation, but this does not mean the corresponding idempotents are unique. Specifically, we can have $R \cong \oplus I_i'$ where $I_i' \cong I_i$ as R-modules but they are distinct left ideals of R.

## Examples

Suppose R is the ring of 2×2 upper-triangular matrices with real entries. We have the following decomposition:

$R = \left\{\begin{pmatrix} * & * \\ 0 & *\end{pmatrix}\right\} = \left\{ \begin{pmatrix} * & 0 \\ 0 & 0 \end{pmatrix}\right\} \oplus \left\{ \begin{pmatrix} 0 & * \\ 0 & * \end{pmatrix}\right\} = \left \{\begin{pmatrix} a & a \\ 0 & 0 \end{pmatrix} \right\} \oplus \left\{ \begin{pmatrix} 0 & * \\ 0 & * \end{pmatrix} \right\}.$

Correspondingly, we have the (orthogonal) idempotents:

$I = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & -1 \\ 0 & 1\end{pmatrix}.$

For an example in group algebras, let K be a field whose characteristic is not 2 or 3. Then $K[S_3]$ is semisimple and we have the following orthogonal idempotents:

\begin{aligned} 1 = &\small\frac 1 6 (1 + (1,2) + (1,3) + (2,3) + (1,2,3) + (1,3,2)) + \\ &\frac 1 6 (1 - (1,2) - (1,3) - (2,3) + (1,2,3) + (1,3,2)) + \\ &\frac 1 3 (1 + (1,2) - (1,3) - (1,3,2)) + \\ &\frac 1 3 (1 - (1,2) + (1,3) - (1,2,3)). \end{aligned}

## Central Idempotents

Recall that the centre of a ring, denoted Z(R), is the set of all xR, such that xyyx for all yR. This is a commutative subring of R. The centre of the group algebra is easy to describe.

Lemma. If R is commutative, then Z(R[G]) is the free R-module with basis:

$e_C :=(\sum_{g\in C} g) \in R[G]$ where C ⊆ G is a conjugancy class.

For example if GS3, then Z(R[G]) is the free R-module with basis e, (1,2)+(2,3)+(1,2) and (1,2,3)+(1,3,2).

Proof

Let $\alpha = \sum_{x\in G} c_x x$ where $c_x \in R$; this lies in the centre iff gα = αg for all gG. Multiplying gives

$g\alpha = \sum_{x\in G} c_x gx, \quad \alpha g = \sum_{x\in G} c_x xg.$

It follows that α lies in Z(R[G]) iff for all gx in G$c_{g^{-1}x} = c_{xg^{-1}}$ or equivalently for all gx, $c_{g^{-1}xg} = c_x.$ ♦

In particular, this means if R is a complete discrete valuation ring with uniformizer π, residue field k and field of fractions K, then Z(k[G]) = Z(R[G])/πZ(R[G]) and $Z(K[G]) = K\otimes_R Z(R[G]).$

Next, we look at idempotents.

Definition. An idempotent e of ring R is said to be central if it lies in Z(R).

Correspondingly we have the following

Theorem. There is a bijection between:

• an isomorphism $R \cong R_1 \times \ldots \times R_n$ as a product of rings;
• a decomposition $R = I_1 \oplus \ldots \oplus I_n$ as a direct sum of (two-sided) ideals;
• an expression $1 =e_1 + \ldots + e_n$ as a sum of orthogonal central idempotents.

Proof

First we prove the correspondence between the second and third sets.

By the earlier correspondence $R = \oplus I_i$ gives us $1 = \sum_i e_i$ where $e_i \in I_i$ is a collection of n orthogonal idempotents. Let us show that $e_i$ commutes with all x in R: indeed, $xe_i, e_i x\in I_i$ since $I_i$ is a two-sided ideal. Now $x = x\cdot 1 = \sum_i xe_i$ and $x = 1\cdot x = \sum_i e_i x$ and since $R = \oplus I_i$ is a direct sum, matching components gives us $xe_i = e_i x.$

Conversely, suppose $1 = \sum_i e_i$ where $\{e_i\}$ are orthogonal central idempotents. The prior correspondence gives $I_i := Re_i,$ which is a two-sided ideal since $e_i$ commutes with everything.

The correspondence between the first and second collections is left as an exercise. ♦

As before, let us consider the case where the decomposition is maximal.

Definition. Let e be a non-zero central idempotent (in Z(R)). We say e is centrally primitive if we cannot write e as a sum of two non-zero orthogonal central idempotents.

Note

If R is artinian, then we can write $R = \oplus I_i$ as a finite direct sum of ideals, where each $I_i$ cannot be decomposed further as a direct sum of two non-zero ideals. This corresponds to writing $1 = \sum_i e_i$ as a sum of orthogonal central idempotents which are centrally primitive.

A central idempotent e can be centrally primitive without being primitive, i.e. e can be written as a sum of two non-zero orthogonal idempotents, but neither of these is central. We will see an explicit example later.

Unlike the case of general idempotents, we have:

Proposition. For any ring R, if

$1 = e_1 + \ldots + e_r = f_1 + \ldots + f_s$

where each of $\{e_1, \ldots, e_r\}$ and $\{f_1, \ldots, f_s\}$ is a set of centrally primitive central idempotents which are orthogonal, then r=s and there is a permutation σ of {1,…,r} such that $e_i = f_{\sigma(i)}$ for all i.

Proof

First note that $e_1, \ldots, e_r$ are distinct: indeed if e is orthogonal to itself, then $0 = e\cdot e = e.$ Same goes for $f_1, \ldots, f_s.$

Next, consider $f_j = 1\cdot f_j = \sum_i e_i f_j.$ Since $f_j$ is centrally primitive and $e_i f_1, e_i f_2, \ldots, e_i f_s$ are orthogonal central idempotent:

$(e_i f_j) \cdot (e_i f_k) = e_i^2 f_j f_k = e_i f_j f_k = \begin{cases} e_i f_j, &\text{ if } j=k,\\ 0, &\text{ otherwise,}\end{cases}$

it follows that we must have $f_j = e_i f_j$ for some unique i and all remaining terms are zero. Likewise, for this i, we have $e_i = e_i f_k$ for some unique k. So we have $f_j = e_i f_j = e_i f_j f_k$ and since $f_j \ne 0,$ we must have j=k and so $e_i = f_j.$ Since $e_1, \ldots, e_r$ are distinct, as are $f_1, \dots, f_s,$ the result follows. ♦

Example

Let us find all central idempotents of $\mathbf{Q}[S_3].$ Note that its centre is spanned by 1, a := (1,2)+(1,3)+(2,3) and b := (1,2,3)+(1,3,2). These satisfy

$a^2 = 3 + 3b, \quad b^2 = b + 2, \quad ab = ba = 2a.$

Now we can write:

$1 = \frac 1 3 (2-b) + \frac 1 6 (1-a+b) + \frac 1 6 (1+a+b)$

which is a sum of orthogonal central idempotents, which gives an isomorphism $Z(\mathbf{Q}[S_3]) \cong \mathbf{Q} \times \mathbf{Q} \times \mathbf{Q}$ of rings. Since this decomposition is clearly maximal, the above three idempotents are all centrally primitive. Note that the first term is centrally primitive but not primitive since

$\frac 1 3(2-b) = \frac 1 3 (1 + (1,2) - (1,3) - (1,3,2)) + \frac 1 3 (1 - (1,2) + (1,3) - (1,2,3)).$

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