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).

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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). ♦

warningIn 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}

blue-lin

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 Gc_{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.

warningA 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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