The Specificities of the Split-Complex Algebra

Introduction

The split-complex algebra $\mathbb{D} = \mathbb{R}[x]/(x^2-1)$ is one of the smallest structures of the family that is neither a field nor a division algebra, and this article collects the properties that single it out: it is the two-dimensional commutative associative unital $\mathbb{R}$-algebra generated by an element of square $+1$, it is reduced (it has no nonzero nilpotent; the only square-zero element is $0$) and yet not a field (it has zero divisors), its idempotents are exactly two nontrivial ones, and it is isomorphic to the product $\mathbb{R}\oplus\mathbb{R}$ of two copies of its base, hence semisimple and not simple, with the two idempotent lines as its only nontrivial ideals. It is the split analogue of the complex field, obtained from the same construction by changing the sign of the square.

The article records these properties, draws their consequences for the classes of element, and reads the degeneracy of the matrix models: the regular representation of $\mathbb{D}$ is a commutative algebra of $2\times2$ real matrices, all of them simultaneously diagonalisable, so the matrix picture carries no more than the diagonal algebra $\mathbb{R}\oplus\mathbb{R}$ already carries. The comparison closes the article: against $\mathbb{C}$, a field and simple, whose only idempotents are $0$ and $1$.

The Definition and the Base

Definition. The split-complex algebra is the quotient

$$ \mathbb{D} = \mathbb{R}[x]/(x^2-1), $$

the $\mathbb{R}$-algebra generated by one element $j$, the class of $x$, subject to the single relation $j^2 = 1$. Its elements are the expressions $Q = q+jq'$ with $q, q' \in \mathbb{R}$, and the multiplication is

$$ (q+jq')(c+jc') = (qc+q'c') + (qc'+q'c)j . $$

Proposition. $\mathbb{D}$ is a commutative associative unital $\mathbb{R}$-algebra, of dimension two over $\mathbb{R}$, with unit $1$ and commutative multiplication. Its commutativity and associativity are those of the field $\mathbb{R}$ read on the two coordinates.

Proof. The product displayed is obtained by expanding and using $j^2 = 1$; it is symmetric in $(q,q')$ and $(c,c')$, which is commutativity, and a direct expansion of the triple product confirms associativity. The unit is $1 = 1+0j$, and the two coordinates are added and multiplied independently, so no axiom beyond those of $\mathbb{R}$ is used.

The underlying real vector space is $\mathbb{D} \cong \mathbb{R}^2$ with basis $\{1, j\}$, and the algebra is therefore one of the two two-dimensional commutative $\mathbb{R}$-algebras that are not fields, the other being the dual-number algebra; the complex field, of the same dimension, is the field neighbour, and the three are the trio obtained by changing the defining polynomial, with $\mathbb{D}$ the split member.

The Sign of the Square and the Splitting

The defining polynomial $x^2-1$ factors over the base field,

$$ x^2 - 1 = (x-1)(x+1), $$

and this single fact governs the whole structure. In the language of the structures, the multiplication by $j$ is a real-linear map $J$ with $J^2=+\mathrm{id}$, where the complex structure of the field extension has $J^2=-\mathrm{id}$; the two structures are the two signs of the same square, and this is the flip that makes the Hermitian side of the split ring indefinite where the complex one is definite. For the complex field the defining polynomial is $x^2+1$, which is irreducible over $\mathbb{R}$, and the field stays a field; for the split-complex algebra the polynomial splits into two distinct linear factors already in $\mathbb{R}[x]$, so by the Chinese remainder theorem

$$ \boxed{\;\mathbb{D} = \mathbb{R}[x]/\big((x-1)(x+1)\big) \;\cong\; \mathbb{R}[x]/(x-1) \times \mathbb{R}[x]/(x+1) \;\cong\; \mathbb{R}\oplus\mathbb{R}\;}. $$

Proposition. The map

$$ \varphi : \mathbb{D} \longrightarrow \mathbb{R}\oplus\mathbb{R}, \qquad \varphi(q+jq') = (q+q',\, q-q') = (Q_+, Q_-), $$

is an isomorphism of $\mathbb{R}$-algebras onto the product ring with componentwise operations. Its inverse sends $(u,v)$ to $\tfrac{u+v}{2} + j\,\tfrac{u-v}{2}$.

Proof. The map is $\mathbb{R}$-linear and bijective because its matrix in the basis $\{1,j\}$ is $\left(\begin{smallmatrix}1&1\\1&-1\end{smallmatrix}\right)$, of determinant $-2 \neq 0$. It is multiplicative because $(q+q')(c+c') = (qc+q'c') + (qc'+q'c)$ is the first coordinate of the product and $(q-q')(c-c') = (qc+q'c') - (qc'+q'c)$ its second; so $\varphi(QB) = \varphi(Q)\varphi(B)$ with the componentwise product, and $\varphi(1) = (1,1)$.

The isomorphism is the ring property behind every specificity below. It transports the two field copies to the two idempotent lines $\mathbb{R}\Pi_+$ and $\mathbb{R}\Pi_-$, where

$$ \Pi_+ = \varphi^{-1}(1,0) = \tfrac12(1+j), \qquad \Pi_- = \varphi^{-1}(0,1) = \tfrac12(1-j), $$

and it reads the multiplication as $Q = Q_+\Pi_+ + Q_-\Pi_-$ with $Q_\pm = q \pm q'$.

The Idempotents and the Split Units

Theorem. The idempotents of $\mathbb{D}$ are exactly $0$, $1$, $\Pi_+$ and $\Pi_-$. The two nontrivial ones satisfy

$$ \Pi_+^2 = \Pi_+, \qquad \Pi_-^2 = \Pi_-, \qquad \Pi_+\Pi_- = 0, \qquad \Pi_+ + \Pi_- = 1 , $$

and they are central.

Proof. The image under $\varphi$ of an idempotent is an idempotent pair $(u,v)$ of $\mathbb{R}\oplus\mathbb{R}$, and an idempotent of a field is $0$ or $1$, so $u,v \in \{0,1\}$ and the four pairs give the four elements listed. The relations are read coordinatewise from the definition of $\varphi$, and centrality is commutativity.

So the algebra has exactly two nontrivial idempotents, orthogonal and complete. This is the property that a field can never have: a field has only the two trivial idempotents, because an idempotent $e$ satisfies $e(e-1) = 0$ and a field is an integral domain. The complex field therefore has no nontrivial idempotent, and its analogue of the split decomposition is the single field copy. The idempotents are also the primitive zero divisors, $\Pi_\pm^2 = \Pi_\pm$, and each annihilates the opposite line; the full description is Square Roots of One and the Idempotents of the Split-Complex Algebra, and Zero Divisors of the Split-Complex Algebra.

Corollary (the units). An element is invertible exactly when both idempotent coordinates are nonzero,

$$ \boxed{\;\mathbb{D}^\times = \{Q : Q_+ \neq 0,\; Q_- \neq 0\} \;\cong\; \mathbb{R}^\times \times \mathbb{R}^\times\;}, $$

and the inverse is componentwise, $Q^{-1} = Q_+^{-1}\Pi_+ + Q_-^{-1}\Pi_-$.

Proof. The element $Q$ is a unit iff $\varphi(Q) = (Q_+,Q_-)$ is a unit of $\mathbb{R}\oplus\mathbb{R}$, and a pair is a unit iff both of its coordinates are nonzero.

The group of units is thus a direct product of two copies of the multiplicative group of the base. In the field $\mathbb{C}$ the unit group is $\mathbb{C}^\times$, connected; in $\mathbb{D}$ it is $\mathbb{R}^\times\times\mathbb{R}^\times$, which has four connected components, indexed by the signs of $Q_+$ and $Q_-$. The norm, the criterion $\langle Q\rangle_{*} \neq 0$ and the components are Norm and Invertibility in the General Split-Complex Algebra, where the group of units is treated as a Lie group.

Reduced, Not a Field

Two multiplicative properties sit side by side and pull in opposite directions: $\mathbb{D}$ has no nonzero nilpotent, and yet not every nonzero element is a unit.

Theorem (reduced). There is no nonzero nilpotent in $\mathbb{D}$; the only square-zero element is $0$, and the nilradical vanishes.

Proof. Under $\varphi$ a power becomes the pair of powers, $\varphi(Q^n) = (Q_+^n, Q_-^n)$, so $Q^n = 0$ iff $Q_+^n = 0$ and $Q_-^n = 0$, which forces $Q_+ = Q_- = 0$ over the field $\mathbb{R}$; hence $Q = 0$.

Theorem (not a field). $\mathbb{D}$ is not an integral domain and not a field: it has zero divisors, $\Pi_+\Pi_- = 0$ with both factors nonzero, and the nonzero non-units are exactly its zero divisors.

Proof. The product $\Pi_+\Pi_- = 0$ is the displayed relation, with $\Pi_+,\Pi_- \neq 0$, so $\mathbb{D}$ has zero divisors. In a commutative ring every nonzero non-unit of a finite-dimensional algebra over a field is a zero divisor, since multiplication by it is a singular $\mathbb{R}$-linear map with nonzero kernel; and conversely a zero divisor is not a unit. The criterion in idempotent coordinates is that $Q$ is a non-unit iff $Q_+Q_- = 0$, that is iff it lies on the null cone $\mathcal{N} = \mathbb{R}\Pi_+\cup\mathbb{R}\Pi_-$, which is Zero Divisors of the Split-Complex Algebra.

So the non-units of $\mathbb{D}$ are precisely the null cone, origin included: the two lines through the origin. The pair of theorems is the whole distance between the split-complex algebra and a field: reducedness says that the failure of invertibility is never caused by nilpotence, and the zero divisors say that it is caused instead by the splitting of the algebra into two field copies, each of which can be switched off independently.

Semisimple, Not Simple

Theorem (not simple). The only two-sided ideals of $\mathbb{D}$ are $\{0\}$, $\mathbb{R}\Pi_+$, $\mathbb{R}\Pi_-$ and $\mathbb{D}$. The two nontrivial proper ideals are minimal and maximal, and they are the two idempotent lines.

Proof. The classification is Ideals and Peirce Decomposition of the Split-Complex Algebra: each of $\mathbb{R}\Pi_+$ and $\mathbb{R}\Pi_-$ is an ideal because $Q(\lambda\Pi_+) = \lambda Q_+\Pi_+$, and any nonzero ideal containing an element with both coordinates nonzero contains a unit and is the whole ring, while an ideal supported on one coordinate is contained in the corresponding line. The proof is by the criterion of invertibility above.

Theorem (semisimple). $\mathbb{D}$ is Artinian and semisimple; its Jacobson radical, nilradical and prime radical all vanish.

Proof. The ring is finite-dimensional over the field $\mathbb{R}$, hence Artinian. By the isomorphism $\mathbb{D}\cong\mathbb{R}\times\mathbb{R}$ it is a product of two fields, whose Jacobson radical is the product of the radicals of the factors, that is $0$, and a product of fields has no nonzero nilpotent.

The two theorems are the exact statement that $\mathbb{D}$ is the product of two fields and not a field: the failure of simplicity is measured by the two minimal ideals, which are the two summands of the product decomposition.

Consequences for the Classes of Element

The specificities just proved are the reason the classes of element of $\mathbb{D}$ differ from those of the field and the matrix algebra. The classes are the subject of The Classes of Element of the Split-Complex Algebra; here only the consequences are named.

The zero divisors are non-empty. A field has none; $\mathbb{D}$ has the null cone $\mathbb{R}^\times\Pi_+\cup\mathbb{R}^\times\Pi_-$. The reason is the reducibility of $x^2-1$: the splitting into two field copies makes it possible for an element to be supported on only one of them, and such an element annihilates the other copy.
The idempotents are non-trivial. A field has only $0$ and $1$; $\mathbb{D}$ has the two additional ones $\Pi_\pm$. The reason is the same splitting, each summand of the product supplying its own identity.
The nilpotents and the square-zero elements stay trivial. A product of fields is reduced, so there is no nonzero nilpotent and the only square-zero element of the plain product is $0$, exactly as in a field. In this respect $\mathbb{D}$ is not more complicated than $\mathbb{C}$: it is the zero divisors and the idempotents, not the nilpotents, that the splitting creates.
The square roots of the central values change in number. The value $+1$ gains the two non-scalar roots $\pm j$, because $j$ is a square root of $1$ in a ring where the generator squares to $+1$; the value $-1$ loses its roots, because a split square has non-negative real part. The two effects are collected in The Classes of Element of the Split-Complex Algebra.

The pattern is that the positively defined classes (idempotents, zero divisors, roots of $+1$) grow, while the negatively defined ones (nilpotents, roots of $-1$) shrink or vanish. The split-complex algebra is the case in which no off-diagonal matrix units are present because the algebra is commutative.

The Degeneracy of the Matrix Models

The regular representation of $\mathbb{D}$ is faithful on the underlying real space of dimension two and lands in the $2\times2$ real matrices,

$$ \mathsf{M}_2 : \mathbb{D} \longrightarrow M_2(\mathbb{R}), \qquad \mathsf{M}_2(q+jq') = \begin{pmatrix} q & q' \\ q' & q \end{pmatrix}, $$

the matrix of multiplication by $Q = q+jq'$ in the basis $\{1,j\}$: the first column is the coordinates of $Q \cdot 1 = Q$ and the second those of $Q \cdot j = q'+jq$. These are the matrices of hyperbolic type, symmetric with equal diagonal entries, and their determinant and trace are

$$ \det\mathsf{M}_2(Q) = q^2 - q'^2 = \langle Q\rangle_{*}, \qquad \operatorname{tr}\mathsf{M}_2(Q) = 2q , $$

the determinant and the trace of the regular element, the determinant being the norm $\langle Q\rangle_{*}$ of The Two Forms and the Two Algebraic Norms of the Split-Complexoid Space.

Proposition (degeneracy). The image $\mathsf{M}_2(\mathbb{D})$ is a commutative two-dimensional subalgebra of $M_2(\mathbb{R})$ equal to its own commutant, and a single change of basis diagonalises all of its elements at once:

$$ \boxed{\;\mathsf{M}_2(Q) \;\sim\; \begin{pmatrix} Q_+ & 0 \\ 0 & Q_- \end{pmatrix}, \qquad Q_\pm = q\pm q'\;}. $$

Proof. The image is commutative because $\mathbb{D}$ is, and an $\mathbb{R}$-linear map $\mathsf{M}_2(B)$ commuting with every $\mathsf{M}_2(Q)$ commutes in particular with $\mathsf{M}_2(j) = \left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$ and with $\mathsf{M}_2(1) = I$, which forces $B$ to have the same shape, so the commutant is the image. The conjugating basis is the idempotent basis $\{\Pi_+,\Pi_-\}$: in it the matrices are diagonal with entries $Q_\pm$, since $Q\Pi_+ = Q_+\Pi_+$ and $Q\Pi_- = Q_-\Pi_-$.

Corollary. The matrix model $\mathsf{M}_2$ is the algebra $\mathbb{R}\oplus\mathbb{R}$ in disguise: in the idempotent basis it is the diagonal subalgebra of $M_2(\mathbb{R})$, its centre is itself, and the image carries no idempotent other than the four diagonal projections and no nilpotent other than $0$.

The degeneracy is the matrix form of commutativity: the split-complex model is a commutative subalgebra that diagonalises completely. The $4\times4$ real model of $\mathbb{D}$, the left regular representation on $\mathbb{D}\oplus\mathbb{D}$, is the same algebra with the two diagonal entries repeated, so it adds nothing: it extends the $2\times2$ model by two dimensions that carry no new invariant. The models are The Matrix Element Representations of the Split-Complexoid, and the element reading of the regular matrix is Split-Complex Regular Element Representation.

Summary

The split-complex algebra $\mathbb{D} = \mathbb{R}[x]/(x^2-1)$ is the two-dimensional commutative associative unital $\mathbb{R}$-algebra generated by an element of square $+1$. Its defining polynomial splits, and by the Chinese remainder theorem it is the product of two copies of its base, $\mathbb{D}\cong\mathbb{R}\oplus\mathbb{R}$, the isomorphism being $Q\mapsto(Q_+,Q_-)$ with $Q_\pm = q\pm q'$.

Four properties single it out. It is reduced: it has no nonzero nilpotent, and $0$ is its only square-zero element. It is not a field: its nonzero non-units are its zero divisors, the null cone $\mathbb{R}\Pi_+\cup\mathbb{R}\Pi_-$. It has exactly two nontrivial idempotents $\Pi_\pm$, central, orthogonal and complete, which a field cannot have. And it is semisimple and not simple: its only two-sided ideals are $\{0\}$, the two lines $\mathbb{R}\Pi_\pm$ and $\mathbb{D}$, and its radical vanishes.

The consequences for the classes of element are that the positively defined classes grow and the negatively defined ones do not: the zero divisors are present and the idempotents are non-trivial because of the splitting, while the nilpotents and the square-zero elements of the plain product remain the single element $0$, and the roots of $+1$ gain the non-scalar pair $\pm j$ while the roots of $-1$ vanish. The regular representation degenerates to the commutative subalgebra $\left\{\begin{pmatrix}q&q'\\q'&q\end{pmatrix}\right\}$ of $M_2(\mathbb{R})$, isomorphic to the diagonal subalgebra, with determinant $q^2-q'^2$ and trace $2q$: a commutative algebra equal to its own commutant, and the $4\times4$ model adds nothing. Against the complex field, a field and simple with no idempotent but the trivial ones, the split-complex algebra is the product-of-fields case, the minimal one in which the zero divisors and the non-trivial idempotents of the family appear.

Summary of Notation

Symbol	Meaning
$\mathbb{D} = \mathbb{R}[x]/(x^2-1)$	Split-complex algebra
$Q = q + jq'$	General split-complex number
$\varphi(Q) = (q+q', q-q')$	Isomorphism $\mathbb{D}\cong\mathbb{R}\oplus\mathbb{R}$
$Q_\pm = q\pm q'$	Idempotent coordinates
$\Pi_\pm = \tfrac12(1\pm j)$	The two nontrivial idempotents
$\mathbb{R}\Pi_+$, $\mathbb{R}\Pi_-$	The two idempotent lines, the only nontrivial ideals
$\mathbb{D}^\times = \{Q_+ \neq 0, Q_- \neq 0\}$	Group of units, $\cong\mathbb{R}^\times\times\mathbb{R}^\times$
$\mathcal{N} = \{Q_+Q_- = 0\}$	Null cone, the set of nonzero non-units
$\mathsf{M}_2(Q) = \begin{pmatrix} q & q' \\ q' & q \end{pmatrix}$	Regular matrix of $Q$, of determinant $\langle Q\rangle_{*}$ and trace $2q$
$\dim_\mathbb{R}\mathbb{D} = 2$	Dimension of the algebra over its base