Division Algebras

Introduction

A division algebra is an algebra with identity in which every nonzero element is invertible. The examples of the corpus — the real numbers, the complex numbers and the quaternions, together with the finite-dimensional algebras over other fields — are the building blocks of the structure theory, because the Wedderburn–Artin theorem writes every finite-dimensional semisimple algebra as a product of matrix algebras over division algebras. This article develops the definition and its elementary consequences, the classification of the finite-dimensional associative division algebras over $\mathbb{R}$ (Frobenius), the theorem that every finite division ring is a field (Wedderburn), together with the structure theory and the classification over a general field (Wedderburn–Artin, the Brauer group and Skolem–Noether), and the question of when a product of two quaternion algebras over a general field is itself a division algebra, which is arithmetic and which is answered over a rational function field.

The ground ring is a field $k$ unless stated otherwise.

Definition and Elementary Properties

Definition. A division algebra over $k$ is an associative $k$-algebra $D$ with unit $1 \neq 0$ such that every nonzero element is a unit; that is, $D^\times = D\setminus\{0\}$.

For a finite-dimensional $D$ this is equivalent to the absence of zero divisors, as the next proposition shows; a division algebra is also simple, its only two-sided ideals being $0$ and $D$, and a commutative division algebra is a field.

Proposition (finite-dimensional criterion). Let $A$ be a finite-dimensional unital $k$-algebra. Then $A$ is a division algebra if and only if $A$ has no zero divisors.

Proof. A division algebra has no zero divisors, since $ab = 0$ with $a \neq 0$ implies $b = a^{-1}(ab) = 0$. Conversely, suppose $A$ has no zero divisors and let $a \neq 0$. The linear map $L_a : A \to A$, $L_a(x) = ax$, is injective: if $ax = 0$ then $x = 0$. An injective linear map of a finite-dimensional space is bijective, so there is $b$ with $ab = 1$. Applying the same argument to the right multiplication $R_a$ gives $c$ with $ca = 1$, and then $c = c(ab) = (ca)b = b$, so $a$ has the two-sided inverse $b$.

Proposition (ideals and quotients). A division algebra has exactly two left ideals and exactly two right ideals, namely $0$ and the whole algebra; hence it is simple, and every nonzero algebra homomorphism from a division algebra is injective. A finite-dimensional simple algebra over $k$ is, by Wedderburn–Artin, of the form $M_n(D)$ for a division algebra $D$ and an integer $n \geq 1$, and it is a division algebra exactly when $n = 1$.

Proof. If $I \neq 0$ is a left ideal and $0 \neq a \in I$, then $1 = a^{-1}a \in I$, so $I = A$. Injectivity of a homomorphism $\varphi$ follows because $\ker\varphi$ is a proper two-sided ideal, hence $0$. The last statement is Wedderburn–Artin: a simple finite-dimensional algebra is $M_n(D)$ for a division algebra $D$, and $M_n(D)$ is a division algebra only for $n = 1$.

Example. $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$ and every field are division algebras. The algebras $\mathbb{D}$, $\mathbb{D}'$, $\mathbb{H}_{\mathbb{D}}$, $\mathbb{B}$, $M_n(k)$ for $n \geq 2$, $k[x]$ and $k[G]$ for nontrivial finite $G$ are not, by the zero divisors listed in Examples of Algebras. The pattern is instructive: $\mathbb{R}$, $\mathbb{C}$ and $\mathbb{H}$ are the division algebras among the number systems.

Finite-Dimensional Division Algebras over $\mathbb{R}$: Frobenius

Theorem (Frobenius, standard). Every finite-dimensional associative division algebra over $\mathbb{R}$ is isomorphic to one of

$$ \mathbb{R}, \qquad \mathbb{C}, \qquad \mathbb{H}. $$

Proof (outline). Let $D$ be such an algebra, of dimension $n$ over $\mathbb{R}$, and let $Z$ be its centre.

If $D$ is commutative, then $D$ is a field extension of $\mathbb{R}$ of finite degree $n$. Every element of $D$ has a minimal polynomial over $\mathbb{R}$, which is irreducible because $D$ is a field; over $\mathbb{R}$ the irreducible polynomials have degree $1$ or $2$, since every real polynomial of odd degree has a real root. Hence $n \leq 2$ and $D$ is $\mathbb{R}$ or $\mathbb{C}$.

Suppose $D$ is not commutative. Its centre $Z$ is a field extension of $\mathbb{R}$ of finite degree, hence $\mathbb{R}$ or $\mathbb{C}$ by the previous paragraph; and $Z = \mathbb{C}$ is impossible, because then every element $u \in D$ has $u - \lambda$ non-invertible for some $\lambda \in \mathbb{C}$ — the $\mathbb{C}$-linear map $x \mapsto ux$ has an eigenvalue — and a non-invertible element of a division algebra is zero, so every element would be central and $D$ commutative. Hence $Z = \mathbb{R}$, and $D$ is central over $\mathbb{R}$. Choose $u \in D\setminus\mathbb{R}$; then $\mathbb{R}[u]$ is a proper field extension of $\mathbb{R}$, hence $\mathbb{R}[u] \cong \mathbb{C}$. Now $D$ is a left module over this copy $C$ of $\mathbb{C}$, so $\dim_\mathbb{R}D = 2\dim_C D$. The inner automorphisms of $D$ preserve $C$ and act on it by the only nontrivial possibility, complex conjugation; analysing the conjugation action shows $\dim_C D \leq 2$, so $\dim_\mathbb{R}D \leq 4$. The case $\dim_\mathbb{R}D = 2$ is commutative, and the case $\dim_\mathbb{R}D = 4$ forces $D \cong \mathbb{H}$: the subalgebra generated by two independent anticommuting square roots of $-1$ is quaternion, and the dimension count leaves no room for anything more.

Corollary. The only commutative finite-dimensional real division algebras are $\mathbb{R}$ and $\mathbb{C}$; the only non-commutative one is $\mathbb{H}$. In particular there is no three-dimensional real division algebra, which is the obstruction met.

Remark (the hypotheses are sharp). Frobenius' theorem concerns associative algebras. Dropping associativity admits the octonions, of dimension $8$; keeping associativity but dropping the requirement that the ground field be $\mathbb{R}$ admits the division rings of the next section; and dropping finite dimension admits the field of rational functions $\mathbb{R}(x)$.

Wedderburn's Little Theorem

Theorem (Wedderburn, standard). Every finite division ring is a field: if $D$ is a division ring with finitely many elements, then $D$ is commutative.

Proof (sketch). The centre $Z$ of $D$ is a finite field $\mathbb{F}_q$, and $D$ is a finite-dimensional $Z$-vector space of dimension $n$, so $|D| = q^n$. Counting the conjugacy classes of the multiplicative group $D^\times$ and using the class equation, together with the fact that the size of every conjugacy class is the index of a centraliser and hence of the form $(q^n-1)/(q^d-1)$, one shows that the cyclotomic polynomial $\Phi_n(q)$ divides $q - 1$ whenever $n \geq 2$. Since $\Phi_n(q) > q - 1$ for $n \geq 2$ and $q \geq 2$, this is impossible, so $n = 1$ and $D = Z$ is a field.

The theorem is the boundary case of the structure theory over finite fields: because there are no finite non-commutative division rings, every finite simple ring is a matrix algebra over a finite field, by Wedderburn–Artin. It also shows that the construction of $\mathbb{H}$ genuinely needs an infinite ground field.

Division Algebras over Other Fields

Over a field $k$, the classification is richer, and the finite-dimensional central simple algebras are organised by a group. A quaternion algebra over $k$ is a four-dimensional central simple $k$-algebra; it is either a division algebra or is isomorphic to $M_2(k)$. Over $\mathbb{R}$ the two possibilities are $\mathbb{H}$ and $M_2(\mathbb{R})$, so there is exactly one real quaternion division algebra; over a field such as $\mathbb{Q}$ there are infinitely many, given by the symbol algebras $(a,b)_k$ with $a, b \in k^\times$. The Brauer group $\operatorname{Br}(k)$ has as its elements the Morita-equivalence classes of finite-dimensional central simple $k$-algebras, with the tensor product as the group operation; a division algebra and each of its matrix algebras represent the same element, and $\operatorname{Br}(\mathbb{R}) \cong \mathbb{Z}/2$ is generated by the class of $\mathbb{H}$.

Example (the two real quaternion algebras). The general quaternion algebra over a field $k$ with parameters $a, b \in k^\times$ has $k$-basis $1, u, v, w$ with

$$ u^2 = a, \qquad v^2 = b, \qquad uv = w = -vu . $$

For $k = \mathbb{R}$, $a = b = -1$ this is the division algebra $\mathbb{H}$; for $a = b = +1$ the element $1 + u$ is a zero divisor, since $(1+u)(1-u) = 1 - u^2 = 0$, so the algebra is not a division algebra, and as a four-dimensional central simple real algebra it is isomorphic to $M_2(\mathbb{R})$. These are the two elements of $\operatorname{Br}(\mathbb{R}) \cong \mathbb{Z}/2$ in their four-dimensional form.

Theorem (Wedderburn–Artin, standard). Every finite-dimensional semisimple $k$-algebra is isomorphic to a product $\prod_i M_{n_i}(D_i)$ with the $D_i$ finite-dimensional division algebras over $k$.

This is the theorem for which division algebras exist in the structure theory: they are precisely the simple factors of the semisimple algebras, and the classification of the $D_i$ is the content of the theory beyond the present article.

Biquaternion Algebras and the Division Question

The Algebra and the Criterion

An algebra isomorphic to $Q \otimes_k Q'$ for two $k$-quaternion algebras $Q$ and $Q'$ is a $k$-biquaternion algebra. It is central simple of degree $4$ and dimension $16$ over $k$; its class in the Brauer group is the sum of the classes of its two factors, and its exponent divides $2$, so its index is $1$, $2$ or $4$. The phrase is the algebraists' one, over a general field, and it is not the corpus's algebra: a biquaternion algebra here is a product of two quaternion algebras over the same field, sixteen-dimensional over that field, whereas $\mathbb{B} = \mathbb{C} \otimes_\mathbb{R} \mathbb{H}$ is a four-dimensional complex algebra. The two share the factor $\mathbb{H}$ when $k = \mathbb{R}$ and nothing else, and the collision of the two names is worth recording once: a reader who carries the corpus's sense of the word into the arithmetic literature will misread every statement there.

The class of $Q \otimes_k Q'$ is the sum of the symbols of the two factors in the mod-2 Milnor $K$-group $k_2(k) = k_2^M(k)/2$ of Higher Algebraic K-Theory, the group that is the quaternion layer of the Brauer group by Merkurjev's theorem. Write $\{a,b\}$ for the symbol of the quaternion algebra $(a,b)_k$, so that $\{a,b\} = 0$ exactly when $(a,b)_k$ is split; the criterion then reads as follows.

Criterion. Let $B = (a, b)_k \otimes_k (c, d)_k$. Then $B$ has zero divisors exactly when its class is itself the class of a quaternion algebra, that is, exactly when

$$ \{a,b\} + \{c,d\} = \{e,f\} \qquad \text{for some } e, f \in k^\times , $$

and $B$ is a division algebra exactly when its class is not represented by a single symbol. The three cases are the index: index $1$, that is $B \cong M_4(k)$; index $2$, that is $B \cong M_2(Q'')$ for a $k$-quaternion algebra $Q''$; and index $4$, which is the division case. The criterion is what makes the question an arithmetic one: a biquaternion algebra is a division algebra when a sum of two symbols fails to be a symbol, so the answer depends on the field and not on the algebra alone. The criterion has a form-theoretic face: for $B = (a, b)_k \otimes_k (c, d)_k$, the class is that of a quaternion algebra — equivalently $B$ has zero divisors — exactly when the six-dimensional Albert form $\langle a, b, -ab, -c, -d, cd\rangle$ of $B$ is isotropic, and $B$ is a division algebra exactly when that form is anisotropic. The Albert form, its trivial signed discriminant and its index trichotomy are in The Witt Group and the Grothendieck–Witt Ring, and the criterion by which its anisotropy is decided over a local field is Springer's theorem, in Local Fields.

Three consequences follow immediately, and they locate the phenomenon. Over $\mathbb{R}$ the group is $\operatorname{Br}(\mathbb{R}) \cong \mathbb{Z}/2$, the sum of the two symbols is $0$, and every real biquaternion algebra is therefore split, that is $M_4(\mathbb{R})$; the instance $\mathbb{H} \otimes_\mathbb{R} \mathbb{H} \cong M_4(\mathbb{R})$ already recorded above is the general case. Over a finite field the group is $0$, and again every biquaternion algebra has zero divisors. Over a number field or a local field the index of a central simple algebra equals its exponent, so a class of exponent dividing $2$ has index $1$ or $2$, is a quaternion class, and no biquaternion algebra over such a field is a division algebra. Division biquaternion algebras are therefore a phenomenon of the function fields.

Division Biquaternion Algebras over a Rational Function Field

Let $E$ be a field of characteristic different from $2$ and let $E(t)$ be the rational function field over $E$. A quaternion division algebra over $E(t)$ exists if and only if $E$ has some field extension of even degree, and this is the easy half of the theory; the biquaternion question is finer.

The valuations of $E(t)$ trivial on $E$ are the monic irreducible polynomials of $E[t]$, with residue field $E_p = E[t]/(p)$, together with the place at infinity, with residue field $E$. At such a place $v$ the tame symbol is the homomorphism $\partial_v : k_2(E(t)) \to k_1(\kappa_v)$ given on symbols by

$$ \partial_v(\{f,g\}) = (-1)^{v(f)v(g)}\, u_f^{-v(g)}\, u_g^{v(f)} \qquad \text{in } k_1(\kappa_v) , $$

where $u_f, u_g$ are the unit parts of $f$ and $g$, as in Higher Algebraic K-Theory; the sum of the tame symbols over the places is the ramification map $\partial$. Milnor's exact sequence

$$ 0 \to k_2(E) \to k_2(E(t)) \xrightarrow{\ \partial\ } \bigoplus_{p} k_1(E_p) \xrightarrow{\ N\ } k_1(E) \to 0 $$

identifies its image, the kernel of the norm map $N$, as the group $R_2(E)$ of ramification sequences. A sequence is represented by a symbol when it is $\partial(\sigma)$ for a symbol $\sigma = \{f,g\}$ of $k_2(E(t))$, and in these terms the criterion above reads: a biquaternion algebra over $E(t)$ is a division algebra exactly when the ramification sequence of its class is not represented by any symbol.

The passage from ramification sequences to quadratic forms is made by the Bezoutian form $q_{f,g}$ of a pair of polynomials, defined in Quadratic Forms and Polarisation: a quadratic form on $E[t]/(g)$ formed from $f$ modulo the monic $g$, non-degenerate exactly when $\gcd(f,g) = 1$. Its computation rules — linearity in $f$, and a determinant equal to the resultant up to an explicit sign — make a non-trivial Bezoutian an obstruction to the representability of a ramification sequence by a symbol, which is the criterion of Becher and Raczek. The Witt classes of these forms belong to The Witt Group and the Grothendieck–Witt Ring; only the vocabulary is used here.

Theorem (Becher). Let $E$ be a field of characteristic different from $2$ which is not real euclidean and over which some quaternion algebra is not split. Then there exists a biquaternion division algebra over $E(t)$ which contains no quaternion algebra defined over $E$.

The two hypotheses are not redundant. $E$ is real euclidean when its set of squares is an ordering of $E$; in that case every quaternion algebra over $E(t)$ is of the form $(-1, f)$ with $f \in E[t]$, the sum of two such symbols is again a symbol, and by the criterion every $E(t)$-biquaternion algebra has zero divisors — so the first hypothesis cannot be dropped. The second hypothesis is that $k_2(E) \neq 0$, equivalently that not every $E$-quaternion algebra is split. The conclusion is stronger than the existence of a division algebra: the algebra constructed contains no quaternion algebra defined over $E$, by which is meant none of the form $Q_0 \otimes_E E(t)$ with $Q_0$ an $E$-quaternion algebra, although it contains $E(t)$-quaternion algebras by construction.

The construction. For $a, b \in E^\times$ with $a \notin E^{\times 2}$ and $b \notin aE^{\times 2} \cup (a-4)E^{\times 2}$, set $g_1 = t^2 + (a+1)t + a$ and $g_2 = t^2 + at + a$, so that $g_1 = (t+1)(t+a)$, that $g_2(0) = a$, that the discriminant of $g_2$ is $a(a-4)$, and that $g_1 t \equiv t^2$ modulo $g_2$. The algebra

$$ B = \big(t^2 + (a+1)t + a,\ a\big) \otimes_{E(t)} \big(t^2 + at + a,\ ab\big) $$

has ramification sequence supported on the divisors of $g_1 g_2$, of degree $4$. Its class is the sum of the two symbols $\{g_1, a\}$ and $\{g_2, ab\}$, and the criterion of the previous subsection translates the theorem into the statement that this sum is represented by a symbol exactly when $\{a,b\} = 0$ in $k_2(E)$. So when $\{a,b\} \neq 0$ the sequence is not representable, and $B$ is a division algebra — the Faddeev index of the class is $4$ in the terminology of the arithmetic literature. A field with $k_2(E) \neq 0$ and not real euclidean always supplies such a pair $a, b$.

Among the standard fields of this article, the rational function field is the one over which a biquaternion algebra can fail to have zero divisors, and the reason is visible in the sequence above: there are ramification sequences that are not the ramification of any symbol, so the symbol relations of $k_2(E)$ need not lift to $k_2(E(t))$. The converse of the theorem fails, and the failure is informative: a field of cohomological dimension $1$ — over which every quaternion algebra is split, so that $k_2(E) = 0$ — can still have biquaternion division algebras over $E(t)$. The local case was known earlier, with $E$ a local number field and the division algebras over $E(t)$ constructed by other means; the theorem above is what removes the arithmetic hypothesis on $E$.

The Skolem–Noether Theorem

The structure theory says that the simple algebras are the matrix algebras over division algebras; the next question is how rigid their embeddings are, and the answer is that they are as rigid as possible.

Theorem (Skolem–Noether, standard). Let $k$ be a field, let $A$ be a finite-dimensional central simple $k$-algebra, let $B$ be a finite-dimensional simple $k$-subalgebra of $A$, and let $f, g : B \to A$ be two unital $k$-algebra homomorphisms. Then there is a unit $a \in A^\times$ with

$$ g(b) = a\,f(b)\,a^{-1} \qquad \text{for all } b \in B . $$

Proof (sketch). Write $L = Z(B)$ for the centre of $B$. The tensor product $B \otimes_k A^{\mathrm{op}}$ is a finite-dimensional simple ring: it is $B \otimes_L (L \otimes_k A^{\mathrm{op}})$ by the base-change identity $B\otimes_k M \cong B \otimes_L (L \otimes_k M)$, the algebra $L \otimes_k A^{\mathrm{op}}$ is central simple over $L$ because $A^{\mathrm{op}}$ is central simple over $k$, and $B$ is central simple over $L$; a tensor product of central simple algebras being central simple, $B \otimes_k A^{\mathrm{op}}$ is central simple over $L$, hence simple as a ring. Such an algebra has a unique simple module up to isomorphism, because Wedderburn–Artin writes it as $M_n(D)$ for a division algebra $D$ and the unique simple module of $M_n(D)$ is $D^n$. Now $A$ carries two $B \otimes_k A^{\mathrm{op}}$-module structures,

$$ (b \otimes a')\cdot_f x = f(b)\,x\,a', \qquad (b \otimes a')\cdot_g x = g(b)\,x\,a' , $$

and each is a direct sum of copies of the unique simple module with the same multiplicity, by dimension count; hence there is a $k$-linear isomorphism $\psi$ of $A$ intertwining $\cdot_f$ with $\cdot_g$. Being a $B \otimes_k A^{\mathrm{op}}$-module isomorphism, $\psi$ is in particular $A^{\mathrm{op}}$-linear, that is $\psi(xa') = \psi(x)a'$, so it is an isomorphism of $A$ as a free right $A$-module of rank one. Therefore $a = \psi(1)$ generates $A$ as a right $A$-module and is a unit, and combining $A^{\mathrm{op}}$-linearity at $x = 1$ with the intertwining relation at $x = 1$ gives

$$ \psi(f(b)) = \psi(1)f(b) = a f(b), \qquad \psi(f(b)) = g(b)\psi(1) = g(b)a , $$

whence $g(b)a = af(b)$, that is $g(b) = a f(b)a^{-1}$.

Corollary (every automorphism of a central simple algebra is inner). For a finite-dimensional central simple $k$-algebra $A$ the map $A^\times \to \operatorname{Aut}_k(A)$, $a \mapsto (x \mapsto axa^{-1})$, is surjective with kernel $k^\times$, so

$$ \operatorname{Aut}_k(A) \cong A^\times/k^\times . $$

For $A = M_n(k)$ this is the projective general linear group $\mathrm{PGL}_n(k) = \mathrm{GL}_n(k)/k^\times$; for $A = M_n(D)$ with $D$ a division algebra it is $\mathrm{PGL}_n(D) = \mathrm{GL}_n(D)/k^\times$, where $\mathrm{GL}_n(D) = M_n(D)^\times$, and in each case the derivations of $A$ are the inner ones, $\delta = \operatorname{ad}_h$, as in Automorphisms and Derivations of Algebras.

Example (the quaternions, and why centrality is needed). For $A = \mathbb{H}$ over $\mathbb{R}$ the corollary gives $\operatorname{Aut}_\mathbb{R}(\mathbb{H}) \cong \mathbb{H}^\times/\mathbb{R}^\times \cong SU(2)/\{\pm1\} \cong SO(3)$, in agreement with the computation of that automorphism group in Automorphisms and Derivations of Algebras. The algebra $\mathbb{C}$ viewed over $\mathbb{R}$ is simple but not central, its centre being $\mathbb{C}$ itself, and the conclusion fails there: complex conjugation is an $\mathbb{R}$-automorphism of $\mathbb{C}$ that is not inner, and indeed every inner automorphism of the commutative algebra $\mathbb{C}$ is trivial. Centrality is thus exactly what the theorem needs, and the example shows it cannot be dropped.

Corollary (conjugacy of embeddings of fields). If $K \subseteq A$ is a subfield of a finite-dimensional central simple $k$-algebra $A$ with $[K : k] < \infty$ and $K$ separable over $k$, then any two $k$-embeddings $K \to A$ are conjugate by an inner automorphism of $A$. This is the form in which the theorem is used to move between splitting fields of a central simple algebra, and it is the algebraic reason the matrix representations attached to two embeddings of such a field are interchangeable.

Algebraically Closed Ground Fields

Proposition (algebraically closed ground fields). Let $k$ be algebraically closed. Then the only finite-dimensional division algebra over $k$ is $k$ itself.

Proof. Let $D$ be a finite-dimensional division algebra over $k$ and let $u \in D$. The powers $1, u, u^2, \dots$ are linearly dependent, so $u$ satisfies a nonzero polynomial $p$ over $k$; the minimal polynomial of $u$ has a root $\lambda \in k$ because $k$ is algebraically closed, and the minimal polynomial is irreducible because $D$ is a division algebra, so it is $t - \lambda$. Hence $u = \lambda \in k$.

Over an algebraically closed field the quaternion and octonion phenomena therefore disappear entirely: the division algebras of the classification are exactly the algebras left over when the ground field fails to contain the roots of the relevant polynomials.

Summary

A division algebra is a unital associative algebra in which every nonzero element is a unit; over a field and in finite dimension this is equivalent to the absence of zero divisors, and a division algebra is simple with only the two trivial left ideals. Frobenius' theorem classifies the finite-dimensional associative real division algebras as $\mathbb{R}$, $\mathbb{C}$ and $\mathbb{H}$, the commutative ones being $\mathbb{R}$ and $\mathbb{C}$ over their own ground fields, so no three-dimensional real division algebra exists. Wedderburn's little theorem makes every finite division ring a field. Over a general field the central division algebras are classified up to Morita equivalence by the Brauer group, and by Wedderburn–Artin they are the simple factors of the semisimple algebras; the Skolem–Noether theorem adds that any two embeddings of a finite-dimensional simple algebra into a central simple algebra are conjugate by an inner automorphism, so $\operatorname{Aut}_k(A) \cong A^\times/k^\times$ for $A$ central simple — for instance $\mathrm{PGL}_n(k)$ for $M_n(k)$ and $SO(3)$ for $\mathbb{H}$ over $\mathbb{R}$ — while centrality cannot be dropped, complex conjugation being a non-inner automorphism of $\mathbb{C}$ over $\mathbb{R}$; over an algebraically closed field the only finite-dimensional division algebra is the field itself.

A biquaternion algebra over a field is the tensor product of two quaternion algebras over that field, a sixteen-dimensional central simple algebra of index $1$, $2$ or $4$, and the word is not the corpus's $\mathbb{C} \otimes_\mathbb{R} \mathbb{H}$. Its class is the sum of the two quaternion symbols, and it has zero divisors exactly when that sum is itself a symbol: it is a division algebra exactly when its index is $4$, which is a question about the field and not about the algebra. Over $\mathbb{R}$, over a finite field and over a number field or a local field, where the index equals the exponent, no biquaternion algebra is a division algebra; over the rational function field $E(t)$ of a field not real euclidean with $k_2(E) \neq 0$, one exists, by the theorem of Becher, and the constructed example contains no quaternion algebra defined over $E$. The proof is by the ramification sequence of the class, an element of $R_2(E) = \ker N$ read off from the tame symbols at the places of $E(t)$, and by the Bezoutian forms of Quadratic Forms and Polarisation, which obstruct the representability of that sequence by a single symbol; the class then has Faddeev index $4$. The real-euclidean case is exactly the case excluded, since there every $E(t)$-quaternion algebra is $(-1, f)$ and the sum of two such symbols is a symbol.

Summary of Notation

Symbol	Meaning
$k$	Ground field
$D$	Division algebra
$D^\times = D\setminus\{0\}$	Units of a division algebra
$\mathbb{R}, \mathbb{C}, \mathbb{H}$	Real, complex, quaternion division algebras
$\mathbb{O}$	Octonions, $\dim_\mathbb{R} = 8$, non-associative
$Z(D)$	Centre of $D$
$[u,v,w] = (uv)w - u(vw)$	Associator
$(a,b)_k$	Quaternion (symbol) algebra over $k$, generators $u, v$ with $u^2=a$, $v^2=b$
$\operatorname{Br}(k)$	Brauer group
$M_n(D)$	Matrix algebra over $D$
$A^{\mathrm{op}}$	Opposite algebra, $a\cdot b = ba$
$\operatorname{Aut}_k(A) \cong A^\times/k^\times$	Automorphisms of $A$, all inner for $A$ central simple
$\mathrm{PGL}_n(k)$	Projective general linear group $\mathrm{GL}_n(k)/k^\times$
$Q \otimes_k Q'$	Biquaternion algebra, central simple of degree $4$, index $1$, $2$ or $4$
$k_2(k) = k_2^M(k)/2$, $\{a,b\}$	Mod-2 Milnor $K$-group of the ground field (Higher Algebraic K-Theory); the symbol of the quaternion algebra $(a,b)_k$
$E(t)$, $E_p$	Rational function field over $E$; its residue field $E[t]/(p)$ at the prime $p$
$\partial_v$, $\partial$	Tame symbol at the place $v$; the ramification map $\sum_v \partial_v$
$R_2(E)$	Ramification sequences, the image of $\partial$