Zero Divisors of the Quaternion Algebra

Introduction

This article studies the zero divisors of the quaternion algebra $\mathbb{H}$. It follows Norm and Invertibility in the Quaternion Algebra, above this article in this category, which established the criterion for invertibility and the three-way classification of the elements of $\mathbb{H}$, and it opens the element theory of the block: the zero divisors of the complexified block, of the idempotent and the nilpotent type, are Quaternionic Norm Regularity and Degeneracy in the Biquaternion Algebra, §The Zero Divisors, in the neighbouring Biquaternion category, and the vocabulary of pure and non-pure is used throughout. The goal here is to characterise the zero divisors of the quaternion product, to name the two families the complexified theory separates, and to prove that over $\mathbb{R}$ both are empty.

A divisor is relative to a product. The definition is stated once for a general product in Algebras: A General Introduction; the unqualified "zero divisor" means "zero divisor of the quaternion product", the multiplication of the quaternion algebra.

Every claim is proved or stated as a definition, no physics is invoked and no examples are given. The algebra $\mathbb{H}$, its natural conjugation and its remarkable subspaces are assumed from Introduction to the Quaternion Algebra and Introduction to the Remarkable Subspaces of the Quaternions, and the invertibility criterion and the quaternion norm from Norm and Invertibility in the General Plain Algebra. The annihilator of an element, the two multiplication operators and their rank are Annihilators of the Quaternion Algebra, below this article in this category.

Conventions

Throughout this article the quaternion basis is written $e_t = 1, \vec e_x, \vec e_y, \vec e_z$, with $\vec e_k^2 = -e_t$ and $\vec e_x\vec e_y = \vec e_z$, $\vec e_y\vec e_z = \vec e_x$, $\vec e_z\vec e_x = \vec e_y$. A general quaternion is

$$ \tilde q = q_t e_t + q_x \vec e_x + q_y \vec e_y + q_z \vec e_z = q_t e_t + \vec q, \qquad \vec q = q_x \vec e_x + q_y \vec e_y + q_z \vec e_z, \qquad q_\mu \in \mathbb{R}, $$

with scalar part $q_t = \mathrm{Sc}(\tilde q)$ and vector part $\vec q$. The natural conjugation is $\tilde q^{\natural} = q_t e_t - \vec q$. Over $\mathbb{R}$ the bar is the identity and the star coincides with the natural conjugation, so the chapter has exactly two general products, the plain one $\tilde q_1\tilde q_2$ and the quaternionic one $\tilde q_1^{\natural}\tilde q_2$; the plain product is the one of this article. A product of the space is written $\star$ when the statement is one about an arbitrary product and only the fact that it is a product is used, as in §Definition.

The plain form is $\langle\tilde q_1,\tilde q_2\rangle = \mathrm{Sc}(\tilde q_1\tilde q_2) = q_{1t}q_{2t} - (\vec q_1,\vec q_2)$, indefinite of signature $(1,3)$; the quaternion form is $\langle\tilde q_1,\tilde q_2\rangle_{\natural} = \mathrm{Sc}(\tilde q_1^{\natural}\tilde q_2) = q_{1t}q_{2t} + (\vec q_1,\vec q_2)$, positive definite. The quaternion norm is $\langle\tilde q\rangle_{\natural} = \tilde q\tilde q^{\natural} = \tilde q^{\natural}\tilde q = q_t^2 + |\vec q|^2$, a positive real vanishing only at $\tilde q = 0$, and it is multiplicative, $\langle\tilde q_1\tilde q_2\rangle_{\natural} = \langle\tilde q_1\rangle_{\natural}\langle\tilde q_2\rangle_{\natural}$, by Norm and Invertibility in the Quaternion Algebra, above this article in this category. The centre is the scalar line $\mathbb{R}e_t$; the vector subspace is $\mathrm{Vect}(\mathbb{H}) = \{\vec q\}$, of dimension three.

Definition

Definition. Let $\star$ be a product on $\mathbb{H}$ and let $\tilde q_1\in\mathbb{H}$. The element $\tilde q_1$ is a left zero divisor of $\star$ when $\tilde q_1\neq0$ and there is a nonzero $\tilde q_2$ with $\tilde q_1\star\tilde q_2 = 0$; it is a right zero divisor of $\star$ when $\tilde q_1\neq0$ and there is a nonzero $\tilde q_2$ with $\tilde q_2\star\tilde q_1 = 0$; and it is a zero divisor of $\star$ when it is one or the other. An element that is both a left and a right zero divisor of $\star$ is a two-sided zero divisor of $\star$.

The requirement that both $\tilde q_1$ and $\tilde q_2$ be nonzero is essential. In particular, the element $\tilde q_1 = 0$ is not a zero divisor of any product, even though $0\star\tilde q_2 = 0$ for any $\tilde q_2$.

On the products of the space. Substituting the product for $\star$ gives one definition to a product. On the quaternion product the two equations are $\tilde q_2\tilde q_3 = 0$ and $\tilde q_3\tilde q_2 = 0$; on $\tilde q_1^{\natural}\tilde q_2$ they are $\tilde q_2^{\natural}\tilde q_3 = 0$ and $\tilde q_3^{\natural}\tilde q_2 = 0$; and the cross product has every nonzero element as a zero divisor, since $\tilde q_2\wedge\tilde q_2 = 0$. Throughout the rest of this article, unqualified, the product is the quaternion product. A product that has a two-sided unit and no zero divisor is a division rule, and an algebra whose product is a division rule is a division algebra; the theorem of §The Criterion Is Never Satisfied Off the Origin shows that the plain product is one, so every nonzero element of the quaternion product is a unit.

The Criterion

The complexified block decides its zero divisors by the vanishing of $\tilde Q\tilde Q^{\natural}$, and the same criterion holds here, but its conclusion is reversed by the positivity of the norm.

Theorem (the criterion for the quaternion product). For the quaternion product, a nonzero quaternion $\tilde q$ is a zero divisor if and only if the product $\tilde q\tilde q^{\natural}$ vanishes:

$$ \tilde q\tilde q^{\natural} = 0. $$

Proof. Suppose $\tilde q_1 \neq 0$ and $\tilde q_1\tilde q_1^{\natural} = 0$. Since $\tilde q_1 \neq 0$ we also have $\tilde q_1^{\natural} \neq 0$. So $\tilde q_2 = \tilde q_1^{\natural}$ is a nonzero quaternion with $\tilde q_1\tilde q_2 = 0$, and $\tilde q_1$ is a zero divisor.

Conversely, suppose $\tilde q_1$ is a zero divisor: there is a nonzero $\tilde q_2$ with $\tilde q_1\tilde q_2 = 0$ or with $\tilde q_2\tilde q_1 = 0$. If $\tilde q_1\tilde q_1^{\natural} \neq 0$ then $\tilde q_1$ is invertible by the invertibility criterion. In the first case multiplying $\tilde q_1\tilde q_2 = 0$ on the left by $\tilde q_1^{-1}$ gives $\tilde q_2 = 0$; in the second case multiplying $\tilde q_2\tilde q_1 = 0$ on the right by $\tilde q_1^{-1}$ gives $\tilde q_2 = 0$. Either way a contradiction, so $\tilde q_1\tilde q_1^{\natural} = 0$.

The proof uses the quaternion product twice: the zero-divisor equation is the plain one, and the inverse is the inverse of the quaternion product, the only one of the two general products that has a two-sided unit. Neither ingredient is available for the quaternionic product, and that is why the criterion is a plain-product criterion.

The Criterion Is Never Satisfied Off the Origin

Over $\mathbb{R}$ the criterion has no solution away from the origin, because the product $\tilde q\tilde q^{\natural}$ is the quaternion norm.

Theorem (the zero-divisor set is empty). For the quaternion product, the only quaternion with $\tilde q\tilde q^{\natural} = 0$ is $\tilde q = 0$. Consequently $\mathbb{H}$ has no zero divisor: $\tilde q_1\tilde q_2 = 0$ forces $\tilde q_1 = 0$ or $\tilde q_2 = 0$.

Proof. By the quaternion norm, $\tilde q_1\tilde q_1^{\natural} = q_{1t}^2 + q_{1x}^2 + q_{1y}^2 + q_{1z}^2$, a sum of squares of real numbers. It vanishes only when every $q_{1\mu}$ vanishes, that is at $\tilde q_1 = 0$. Since a zero divisor is required to be nonzero, and every nonzero $\tilde q_1$ has $\tilde q_1\tilde q_1^{\natural} > 0$, the criterion admits no element. For the annihilating equation, if $\tilde q_2\tilde q_1 = 0$ with $\tilde q_1 \neq 0$, multiplying on the right by $\tilde q_1^{-1}$ gives $\tilde q_2 = 0$; the case $\tilde q_2 \neq 0$ is the same with the inverse on the left.

The Three-Way Classification

Combining the criterion for invertibility from the preceding article with the criterion for zero divisors, the elements of $\mathbb{H}$ are partitioned into three classes, and over $\mathbb{R}$ the middle class collapses.

Condition on $\tilde q\tilde q^{\natural}$	Condition on $\tilde q$	Conclusion
$\tilde q\tilde q^{\natural} \neq 0$	(automatically $\tilde q \neq 0$)	$\tilde q$ is invertible
$\tilde q\tilde q^{\natural} = 0$	$\tilde q = 0$	$\tilde q$ is the zero element
$\tilde q\tilde q^{\natural} = 0$	$\tilde q \neq 0$	$\tilde q$ is a zero divisor

The third row is empty: the second row exhausts the vanishing of $\tilde q\tilde q^{\natural}$. The classification therefore has the same three slots as the complexified block but one of them is empty, and every nonzero quaternion lies in the first row. This is the classification table of Norm and Invertibility in the Quaternion Algebra, above this article in this category, read against the criterion.

The Algebra Is a Division Algebra

By the vocabulary of Algebras: A General Introduction, a product with no zero divisor and a two-sided unit is a division rule, and an algebra whose product is a division rule is a division algebra. For the finite-dimensional algebra $\mathbb{H}$, the two clauses of the criterion agree: if every nonzero element is invertible then no nonzero element can annihilate another, and conversely if there are no zero divisors then every nonzero element has an inverse by the criterion above.

The quaternion algebra $\mathbb{H}$ contains no zero divisors, so it is a division algebra. This is the point at which the real block separates from the complexified one: the biquaternion algebra $\mathbb{B} = \mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}$ carries a genuine hypersurface of zero divisors, while $\mathbb{H}$ carries none. The reason is exactly the signature of the two norms. The biquaternion norm is the same sum of squares $\sum_\mu Q_\mu^2$, but computed in the complex coefficients, so it vanishes on a complex cone of real dimension six; over $\mathbb{R}$ the same expression is a sum of squares of real numbers and is positive definite, so the cone shrinks to the origin.

Theorem (Frobenius). Every finite-dimensional associative division algebra over $\mathbb{R}$ is isomorphic to $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$.

Proof. Quoted as standard. The dimension count is $1,2,4$, and $\mathbb{H}$ is the largest-dimensional associative one; the statement and its companions are The Specificities of the Quaternion Algebra, below this article in this category.

The Two Families of the Complexified Block

The complexified block splits its zero divisors into two families according to the value of the scalar part $Q_t$. A biquaternion is pure when $Q_t = 0$ and non-pure when $Q_t \neq 0$, and the criterion reads differently in the two cases. Writing $\tilde Q = Q_t e_t + \vec Q$ with $\vec Q$ a complex vector part,

$$ \tilde Q\tilde Q^{\natural} = Q_t^2 + Q_x^2 + Q_y^2 + Q_z^2 = Q_t^2 + (\vec Q,\vec Q), $$

so in the pure case the scalar contribution drops out and the condition is the isotropy of the complex vector part, while in the non-pure case the scalar contribution can cancel the vector one. The pure zero divisors are the nonzero nilpotents and the non-pure ones are the nonzero complex multiples of the idempotents.

Over $\mathbb{R}$ the two families inherit the split but lose their elements, and for the same reason in both cases: the criterion becomes the vanishing of a sum of squares of real coefficients.

Proposition (the pure family is empty). A pure quaternion $\vec q$ has $\vec q\vec q^{\natural} = |\vec q|^2 e_t$ and $\vec q^2 = -|\vec q|^2 e_t$, so

$$ \boxed{\; \vec q^2 = 0 \quad\Longleftrightarrow\quad |\vec q|^2 = q_x^2 + q_y^2 + q_z^2 = 0 \quad\Longleftrightarrow\quad \vec q = 0 \;} .$$

So there is no nonzero pure zero divisor, and the pure family of the complexified block has no real member.

Proof. For a pure quaternion the product with its conjugate is $\vec q\vec q^{\natural} = |\vec q|^2 e_t$ and the square is $\vec q^2 = -(\vec q,\vec q)e_t = -|\vec q|^2 e_t$, as the coordinate rule gives directly. A sum of three squares of real numbers vanishes only at $\vec q = 0$.

Proposition (the non-pure family is empty). A quaternion with $q_t \neq 0$ has

$$ \boxed{\; \tilde q\tilde q^{\natural} = q_t^2 + |\vec q|^2 \geq q_t^2 > 0 \;} ,$$

so it satisfies the criterion at no point. There is no non-pure zero divisor.

Proof. The scalar contribution $q_t^2$ is strictly positive and the vector contribution $|\vec q|^2$ is nonnegative; their sum cannot vanish when $q_t \neq 0$.

Corollary. Both families of the complexified block are empty over $\mathbb{R}$, and the zero-divisor set of the quaternion product is empty. The two families, read on the real algebra, are the classes of Idempotents of the Quaternion Algebra and Nilpotents of the Quaternion Algebra, below this article in this category: the idempotent class is the pair $\{0, e_t\}$, and the nilpotent class is empty, with square-zero set the single point $0$.

The Zero Divisor Set

The zero divisor set $\mathcal{Z} \subset \mathbb{H}$ is

$$ \boxed{\; \mathcal{Z} = \{\tilde q \in \mathbb{H} : \tilde q \neq 0, \; \tilde q\tilde q^{\natural} = 0\} = \varnothing \;} .$$

It is the complement of the invertible elements in the complement of the zero element, $\mathcal{Z} = \mathbb{H} \setminus (\{0\} \cup \mathbb{H}^\times)$, and here that complement is empty because $\mathbb{H}^\times = \mathbb{H}\setminus\{0\}$.

The complexified block reads the same formula as a complex hypersurface: the single complex equation $\tilde Q\tilde Q^{\natural} = 0$ in the four complex coefficients has complex dimension $3$ and real dimension $6$. Over $\mathbb{R}$ the equation is the quadratic form $q_t^2 + q_x^2 + q_y^2 + q_z^2 = 0$, whose real vanishing locus is the single point $0$; the origin is excluded from the zero-divisor set by the requirement that the element be nonzero, and what is left is nothing. The cone of the complexified block meets the real space only at the origin.

Remark (the cone is a point, and the point is excluded). The comparison is worth stating in one line. In $\mathbb{C}^4$ the equation cuts a three-complex-dimensional cone with the origin removed; in $\mathbb{R}^4$ the same equation cuts the origin alone, and the definition of the zero divisor removes it. The dimension drop is not a loss of rigour but the whole content of positive definiteness: the biquaternion norm can vanish on a nonzero element, the quaternion norm cannot.

The Zero Divisors and the Remarkable Subspaces

A zero divisor is an element on which $\tilde q\tilde q^{\natural}$ vanishes, so which subspaces contain zero divisors is the question of where the norm vanishes. Of the remarkable subspaces, the centre $\mathbb{R}e_t$ contains none: on the centre $\langle s e_t\rangle_{\natural} = s^2$ vanishes only at $s = 0$. The vector subspace $\mathrm{Vect}(\mathbb{H})$ contains none: on it $\langle \vec q\rangle_{\natural} = |\vec q|^2$ vanishes only at $\vec q = 0$. No other subspace is any different, the norm being positive definite on the whole space. So every remarkable subspace is free of zero divisors, and the only element of the space on which the criterion is satisfied is the origin, which is not a zero divisor by the definition of §Definition. The reading on the two subspaces is Introduction to the Remarkable Subspaces of the Quaternions, and the two restrictions of the norm are read in Norm and Invertibility in the Quaternion Algebra.

Summary

The zero divisors of the quaternion algebra are the nonzero elements on which $\tilde q\tilde q^{\natural}$ vanishes. Over $\mathbb{R}$ that product is the quaternion norm $q_t^2 + q_x^2 + q_y^2 + q_z^2$, a sum of squares of real numbers, so it vanishes only at the origin; the criterion admits no element, the zero divisor set is the empty set, and the algebra is a division algebra. The three-way classification of the elements keeps its invertible and zero slots and loses its zero-divisor slot, and every nonzero quaternion is a unit, $\tilde q^{-1} = \tilde q^{\natural}/\langle\tilde q\rangle_{\natural}$.

This is the opposite of the complexified block, whose zero divisors form the punctured complex cone $\{N = 0\}$ of complex dimension three and real dimension six, split into the pure nilpotent family and the non-pure idempotent family. The complexified cone is empty over $\mathbb{R}$ because the quaternion norm is positive definite, and the two families read on the real algebra are the classes of Idempotents of the Quaternion Algebra and Nilpotents of the Quaternion Algebra, below this article in this category: the idempotents are $0$ and $e_t$, and the nilpotent class is empty, its square-zero set being the single point $0$. Of the remarkable subspaces, the centre and the vector subspace contain none, and no nonzero element of either is a zero divisor.

Summary of Notation

Symbol	Meaning
$\mathbb{H}$	Quaternion algebra
$\tilde q = \sum_\mu q_\mu e_\mu$	General quaternion
$\tilde q^{\natural} = q_t e_t - \vec q$	Natural conjugation
$\langle\tilde q\rangle_{\natural} = \tilde q\tilde q^{\natural} = \sum_\mu q_\mu^2$	Quaternion norm, positive definite over $\mathbb{R}$
$\tilde q\tilde q^{\natural} = 0$	Algebraic criterion for a zero divisor
$\mathcal{Z} = \varnothing$	Zero divisor set
$\tilde\Pi^2 = \tilde\Pi$	Idempotent equation
$\tilde q^2 = 0$	Square-zero equation
$\mathbb{R}e_t$	The centre, the scalar line
$\mathrm{Vect}(\mathbb{H})$	The vector subspace