Split-Complex Ideals and Peirce Decomposition

Introduction

This article treats the ideal theory of the split-complex algebra $\mathbb{D}$ and the Peirce decomposition attached to its idempotents. It is the two-dimensional counterpart of Biquaternion Ideals and Peirce Decomposition, and the contrast is complete: the biquaternion algebra $\mathbb{B}\cong M_2(\mathbb{C})$ is simple, with no nontrivial two-sided ideals, while $\mathbb{D}$ is a commutative product $\mathbb{R}\oplus\mathbb{R}$ with two nontrivial two-sided ideals and a finite Boolean lattice of them.

The article records the definition of an ideal in a commutative algebra, determines the two-sided ideals, exhibits the two minimal ideals $\mathbb{R}\Pi_\pm$, writes the Peirce decomposition of $\mathbb{D}$ with respect to an idempotent, computes the lattice of ideals, and identifies the radical. The failure of simplicity is stated with its reason, and compared throughout with the simplicity of $\mathbb{B}$ over $\mathbb{C}$.

Placement. The article is third in the Algebra group, after Split-Complex Algebra and Split-Complex Idempotents and Projections, and before Split-Complex Zero Divisors and Worked Examples in the Split-Complex Algebra. It uses the idempotents of the second article; the norm and the invertibility criterion belong to Split-Complex Norm and Invertibility in the Topology group.

Conventions. The algebra is $\mathbb{D} = \mathbb{R}[x]/(x^2-1)$, with basis $1$, $j$, $j^2 = +1$, general element $A = a+ja'$, conjugate $\bar A = a-ja'$, idempotents $\Pi_\pm = \tfrac12(1\pm j)$ and idempotent coordinates $A = A_+\Pi_1 + A_-\Pi_2$. The base is the field $\mathbb{R}$; the algebra is commutative, so no distinction between left, right and two-sided ideals arises for the ambient ring.

Ideals in an Algebra

Definition. A subset $I \subseteq \mathbb{D}$ is an ideal if

$$ I \text{ is an additive subgroup}, \qquad B \in I \implies AB \in I \ \text{and} \ BA \in I. $$

Since $\mathbb{D}$ is commutative, the two conditions coincide, and every ideal is two-sided; there is no separate theory of left and right ideals. The ideals $\{0\}$ and $\mathbb{D}$ are the trivial ideals; an ideal is proper if it is neither, and maximal if it is proper and contained in no larger proper ideal, minimal if it is nonzero and contains no smaller nonzero ideal.

A principal ideal is one generated by a single element, written $(A) = A\mathbb{D} = \mathbb{D}A$. The lattice of all ideals, ordered by inclusion, is denoted $\mathrm{Id}(\mathbb{D})$, with meet $I\cap J$ and join $I+J$.

Since every ideal is closed under multiplication by $\Pi_1$ and by $\Pi_2$, and $A = A\Pi_1 + A\Pi_2$, every ideal decomposes along the idempotent splitting; this is the mechanism behind the computation below.

The Two-Sided Ideals

Theorem. The ideals of $\mathbb{D}$ are exactly the four sets

$$ \{0\}, \qquad \mathbb{R}\Pi_1 = \mathbb{D}\Pi_1, \qquad \mathbb{R}\Pi_2 = \mathbb{D}\Pi_2, \qquad \mathbb{D}. $$

Proof. Each of the three nontrivial sets listed is an ideal: $\mathbb{R}\Pi_1$ is closed under addition, and $A(\lambda \Pi_1) = \lambda A \Pi_1 = \lambda A_+ \Pi_1 \in \mathbb{R}\Pi_1$ for every $A$, so it is an ideal; the same argument gives $\mathbb{R}\Pi_2$; and $\mathbb{D}$ is trivially an ideal. Conversely, let $I \neq \{0\}$ be an ideal and choose $A \in I$ nonzero. If both idempotent coordinates of $A$ are nonzero then $A$ is a unit by Split-Complex Algebra, so $1 = A^{-1}A \in I$ and $I = \mathbb{D}$. Otherwise one coordinate of $A$ vanishes, and $A$ is a nonzero multiple of $\Pi_1$ or of $\Pi_2$ by the idempotent classification; say $A = \lambda \Pi_1$ with $\lambda \neq 0$, so that $\Pi_1 = \lambda^{-1}A \in I$. Now take any $B \in I$. If there is such a $B$ with $B_- \neq 0$, then $B\Pi_2 = B_-\Pi_2 \in I$, so $\Pi_2 = B_-^{-1}B\Pi_2 \in I$, and then $1 = \Pi_1 + \Pi_2 \in I$, giving $I = \mathbb{D}$. If on the other hand every $B \in I$ has $B_- = 0$, then $I \subseteq \mathbb{R}\Pi_1$; since $I \neq 0$ and $\Pi_1 \in I$, we get $I = \mathbb{R}\Pi_1$. The case $A = \lambda \Pi_2$ is the same with the roles of $+$ and $-$ exchanged and gives $I = \mathbb{R}\Pi_2$ or $I = \mathbb{D}$. So the list is complete.

Corollary (failure of simplicity). $\mathbb{D}$ is not simple: it has exactly two nontrivial proper ideals, both minimal and both maximal. The algebra is the direct product of the two fields

$$ \mathbb{D} \cong \mathbb{R}\Pi_1 \times \mathbb{R}\Pi_2, \qquad \mathbb{R}\Pi_1 \cong \mathbb{R}, \quad \mathbb{R}\Pi_2 \cong \mathbb{R}, $$

and each of the two ideals is a field in its own right.

The reason. The algebra is commutative and generated by an element $j$ satisfying the reducible polynomial $x^2 - 1 = (x-1)(x+1)$; the generator therefore has two distinct eigenvalues $+1$ and $-1$ already in the base field, and the algebra splits as the product of the corresponding eigencomponents. In the biquaternion algebra the situation is different: $\mathbb{B}\cong M_2(\mathbb{C})$ has no nontrivial two-sided ideal at all, because its complex dimension $4$ is a square and $M_2(\mathbb{C})$ is a full matrix algebra over a field, hence simple. Simplicity of $\mathbb{B}$ and non-simplicity of $\mathbb{D}$ are thus the two ends of the same phenomenon: the split-complex algebra is the commutative, reducible, low-dimensional member of the family, and the biquaternion algebra is its simple matrix quotient in four dimensions.

Artinian, Semisimple, Length Two

Since $\mathbb{D}$ is finite-dimensional over the field $\mathbb{R}$, it is Artinian as a ring: every descending chain of ideals stabilises. It is also Noetherian for the same reason.

Proposition. $\mathbb{D}$ is semisimple in the sense of having zero Jacobson radical, and it is the product of the two simple algebras $\mathbb{R}\Pi_1$ and $\mathbb{R}\Pi_2$.

Proof. The Jacobson radical of a product of fields is zero, since a field has no nonzero nilpotent elements and is its own maximal ideal; and $\mathbb{D}\cong\mathbb{R}\times\mathbb{R}$ via the idempotent decomposition.

Proposition (length). As a module over itself, $\mathbb{D}$ has length two, with composition series

$$ \{0\} \subset \mathbb{R}\Pi_1 \subset \mathbb{D}, \qquad \mathbb{R}\Pi_1 \cong \mathbb{R}, \qquad \mathbb{D}/\mathbb{R}\Pi_1 \cong \mathbb{R}\Pi_2 \cong \mathbb{R}. $$

Proof. The inclusion is proper at each step, $\mathbb{R}\Pi_1$ is simple as a module because it is a field summand, and the quotient $\mathbb{D}/\mathbb{R}\Pi_1$ is isomorphic to $\mathbb{R}\Pi_2$ through $A \mapsto A\Pi_2$, again simple. No shorter or longer chain is possible because the dimension is $2$ and every nonzero proper ideal is one-dimensional.

So $\mathbb{D}$ has the same module length as a two-dimensional vector space, in contrast with $\mathbb{B}$, whose regular module has length two with simple constituents of complex dimension two. Both algebras are semisimple of length two; they differ in that the constituents of $\mathbb{D}$ are one-dimensional real, in keeping with its smaller dimension.

Idempotents and the Peirce Decomposition

Let $e$ be an idempotent of $\mathbb{D}$ and write $f = 1-e$ for its complement. Multiplication by $e$ and by $f$ splits the algebra into four additive subgroups,

$$ \mathbb{D} = \mathbb{D}_{11}\oplus\mathbb{D}_{12}\oplus\mathbb{D}_{21}\oplus\mathbb{D}_{22}, $$

where

$$ \mathbb{D}_{11} = e\mathbb{D}e, \qquad \mathbb{D}_{12} = e\mathbb{D}f, \qquad \mathbb{D}_{21} = f\mathbb{D}e, \qquad \mathbb{D}_{22} = f\mathbb{D}f. $$

This is the Peirce decomposition of $\mathbb{D}$ with respect to $e$. It is the refinement of the two-term decomposition $\mathbb{D} = \mathbb{D}e\oplus\mathbb{D}f$: the summand $\mathbb{D}e$ is $\mathbb{D}_{11}\oplus\mathbb{D}_{21}$ and the summand $\mathbb{D}f$ is $\mathbb{D}_{12}\oplus\mathbb{D}_{22}$.

Take $e = \Pi_1$, $f = \Pi_2$, the standard idempotent. Then

$$ \Pi_1\mathbb{D}\Pi_1 = \mathbb{R}\Pi_1, \qquad \Pi_2\mathbb{D}\Pi_2 = \mathbb{R}\Pi_2, \qquad \Pi_1\mathbb{D}\Pi_2 = 0, \qquad \Pi_2\mathbb{D}\Pi_1 = 0. $$

The two off-diagonal Peirce components vanish, because $\Pi_1\Pi_2 = 0$ annihilates every product alternating between the two idempotents, and the decomposition collapses to

$$ \mathbb{D} = \mathbb{D}_{11}\oplus\mathbb{D}_{22} = \mathbb{R}\Pi_1 \oplus \mathbb{R}\Pi_2, $$

of total dimension $1+1+0+0 = 2$. With respect to the idempotent $0$ or $1$ the Peirce decomposition is trivial: for $e = 1$ the only nonzero component is the whole algebra, and for $e = 0$ it is the zero ideal.

Definition. The Peirce components are the four blocks of the multiplication table of $\mathbb{D}$ under the splitting $1 = \Pi_1 + \Pi_2$; the vanishing of the off-diagonal blocks is the statement that the splitting is a direct product decomposition.

The Idempotents as Matrix Units

In the idempotent basis $\{\Pi_1, \Pi_2\}$ the standard idempotents are the two diagonal matrix units

$$ \Pi_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \qquad \Pi_2 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, $$

acting on the column $\begin{pmatrix} A_+ \\ A_- \end{pmatrix}$ of idempotent coordinates, and they satisfy the matrix-unit relations

$$ \Pi_1^2 = \Pi_1, \qquad \Pi_2^2 = \Pi_2, \qquad \Pi_1\Pi_2 = \Pi_2\Pi_1 = 0, \qquad \Pi_1 + \Pi_2 = I. $$

These are the defining relations of the diagonal matrix units of a $2\times 2$ matrix algebra. Because $\mathbb{D}$ is commutative, the two off-diagonal units $E_{12}$ and $E_{21}$ are absent, and the algebra is not a full matrix algebra; the four Peirce components are the four blocks of this basis, and the vanishing of the off-diagonal blocks is exactly the statement that the matrix picture is diagonal. This is the structural contrast with $\mathbb{B}\cong M_2(\mathbb{C})$, whose idempotents are general rank-one projections with off-diagonal units present.

The Lattice of Ideals

The four ideals of $\mathbb{D}$ are $\{0\}$, $\mathbb{R}\Pi_1$, $\mathbb{R}\Pi_2$, $\mathbb{D}$, with the inclusions

$$ \{0\} \subset \mathbb{R}\Pi_1 \subset \mathbb{D}, \qquad \{0\} \subset \mathbb{R}\Pi_2 \subset \mathbb{D}, $$

and no inclusion between $\mathbb{R}\Pi_1$ and $\mathbb{R}\Pi_2$ (they are independent, and their intersection is $\{0\}$).

meet	value	join	value
$\mathbb{R}\Pi_1 \cap \mathbb{R}\Pi_2$	$\{0\}$	$\mathbb{R}\Pi_1 + \mathbb{R}\Pi_2$	$\mathbb{D}$
$\mathbb{R}\Pi_1 \cap \mathbb{D}$	$\mathbb{R}\Pi_1$	$\mathbb{R}\Pi_1 + \mathbb{D}$	$\mathbb{D}$
$\{0\} \cap I$	$\{0\}$	$\{0\} + I$	$I$

Theorem. The lattice $\mathrm{Id}(\mathbb{D})$ is a Boolean lattice with four elements, isomorphic to the power set of a two-point set:

$$ \mathrm{Id}(\mathbb{D}) \cong \mathcal{P}(\{+,-\}), \qquad \{0\} \leftrightarrow \varnothing, \quad \mathbb{R}\Pi_1 \leftrightarrow \{+\}, \quad \mathbb{R}\Pi_2 \leftrightarrow \{-\}, \quad \mathbb{D} \leftrightarrow \{+,-\}. $$

Proof. The map is the ideal generated by the indicator subset under the isomorphism $\mathbb{D}\cong\mathbb{R}\oplus\mathbb{R}$: the ideal of $\mathbb{R}\oplus\mathbb{R}$ generated by the indicator vector of a subset $S$ is the set of pairs supported on $S$, and these four sets are exactly the ideals. Under this identification the complement of the ideal associated with $S$ is the ideal associated with the complementary subset $S^{\mathrm{c}}$, meets are intersections and joins are unions, so the lattice is the power-set lattice $\mathcal{P}(\{+,-\})$, which is Boolean.

The lattice is finite, distributive and complemented. This is a further contrast with $\mathbb{B}$, whose two-sided ideal lattice is the two-element chain $\{0\}\subset\mathbb{B}$ (simplicity), while its one-sided ideal lattice is a projective line over $\mathbb{C}$; in $\mathbb{D}$ the left, right and two-sided ideals coincide, so there is a single finite lattice, and it is the Boolean one above.

The Radical

Proposition. The Jacobson radical, the nilradical and the prime radical of $\mathbb{D}$ all vanish.

Proof. The nilradical is the set of nilpotent elements. If $A^n = 0$, write $A = A_+\Pi_1 + A_-\Pi_2$; then $A^n = A_+^n\Pi_1 + A_-^n\Pi_2 = 0$, so $A_+^n = 0$ and $A_-^n = 0$ in $\mathbb{R}$, whence $A_+ = A_- = 0$ and $A = 0$. Hence the nilradical is $\{0\}$. The Jacobson radical of the product of fields $\mathbb{R}\times\mathbb{R}$ is zero, since the maximal ideals are the two coordinate ideals and their intersection is zero.

So $\mathbb{D}$ is reduced (no nonzero nilpotent) and semiprime. This is the opposite of the dual-number algebra $\mathbb{D}' = \mathbb{R}[\varepsilon]/(\varepsilon^2)$, whose maximal ideal is nonzero and nilpotent, and it is the reason $\mathbb{D}$ is much closer in behaviour to a product of fields than to a local ring.

The Real Structure and Comparison with $\mathbb{B}$

The algebra $\mathbb{D}$ is defined over $\mathbb{R}$, and its two minimal ideals are defined over $\mathbb{R}$: each is the real span of the corresponding idempotent, and the unique non-trivial automorphism of $\mathbb{D}$, the conjugation $\bar{\cdot}$, exchanges them,

$$ \overline{\Pi_1} = \Pi_2, \qquad \overline{\Pi_2} = \Pi_1, $$

while fixing neither. So the pair of ideals is a single Galois orbit under the conjugation, and the algebra carries no proper $\mathbb{R}$-subalgebra splitting it into one ideal only. In the biquaternion algebra the analogous real structure is the complex conjugation fixing the quaternion subspace; here it is the exchange of the two components.

feature	$\mathbb{B} \cong M_2(\mathbb{C})$	$\mathbb{D} \cong \mathbb{R}\oplus\mathbb{R}$
simple?	yes	no
two-sided ideals	$\{0\}, \mathbb{B}$	$\{0\}, \mathbb{R}\Pi_1, \mathbb{R}\Pi_2, \mathbb{D}$
lattice	two-element chain	four-element Boolean lattice
one-sided ideals	projective line over $\mathbb{C}$	the same four ideals
radical	$0$	$0$
length of the regular module	$2$	$2$
reason for (non-)simplicity	full matrix algebra over a field	product of two copies of $\mathbb{R}$

The failure of simplicity in $\mathbb{D}$ is therefore not a defect but the algebraic expression of its defining decomposition: the two minimal ideals are the two summands of $\mathbb{D}\cong\mathbb{R}\oplus\mathbb{R}$, and the four-element Boolean lattice is the lattice of subsets of that two-point spectrum.

Summary

An ideal of the commutative algebra $\mathbb{D}$ is a two-sided ideal, and there are exactly four: $\{0\}$, the two minimal ones $\mathbb{R}\Pi_1$ and $\mathbb{R}\Pi_2$, and $\mathbb{D}$. The two minimal ideals are the two summands of the direct product $\mathbb{D}\cong\mathbb{R}\times\mathbb{R}$, and the lattice of ideals is the four-element Boolean lattice of subsets of $\{+,-\}$. The algebra is Artinian, semisimple and reduced, of module length two, with zero Jacobson radical and no nonzero nilpotent.

With respect to an idempotent $e$, the Peirce decomposition splits $\mathbb{D}$ into the four blocks $e\mathbb{D}e$, $e\mathbb{D}f$, $f\mathbb{D}e$, $f\mathbb{D}f$; for the standard idempotent $\Pi_1$ the two off-diagonal blocks vanish and the decomposition collapses to $\mathbb{D} = \mathbb{R}\Pi_1\oplus\mathbb{R}\Pi_2$. The idempotents $\Pi_\pm$ are central and orthogonal, and no element connects the two components, because the algebra is commutative. This is the failure of simplicity of $\mathbb{D}$, against the simplicity of the biquaternion algebra $\mathbb{B}\cong M_2(\mathbb{C})$ over $\mathbb{C}$; the reason is that $\mathbb{D}$ is generated by an element with two distinct eigenvalues in its base field, so it splits as a product of two copies of $\mathbb{R}$, while $\mathbb{B}$ is a full matrix algebra and has no nontrivial two-sided ideal.

Summary of Notation

Symbol	Meaning
$\mathbb{D}$	Split complex algebra
$A = a + j a'$	General split complex number
$\Pi_\pm = \tfrac12(1\pm j)$	Standard idempotents
$I$, $(A) = A\mathbb{D}$	Ideal, principal ideal
$\mathrm{Id}(\mathbb{D})$	Lattice of ideals
$\mathbb{R}\Pi_1, \mathbb{R}\Pi_2$	The two minimal (and maximal proper) ideals
$\mathbb{D}_{11} = \Pi_1\mathbb{D}\Pi_1$	Peirce component, $= \mathbb{R}\Pi_1$
$\mathbb{D}_{22} = \Pi_2\mathbb{D}\Pi_2$	Peirce component, $= \mathbb{R}\Pi_2$
$\mathbb{D}_{12}, \mathbb{D}_{21}$	Off-diagonal Peirce components, both $0$
$f = 1-e$	Complement of an idempotent
$\operatorname{rad}(\mathbb{D})$	Jacobson radical of $\mathbb{D}$, $= 0$; the nilradical and prime radical vanish as well