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How does one prove that ring of Hamilton Quaternions with coefficients coming from the field $\mathbb{Z}/p\mathbb{Z}$ is not a divison ring.

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Chandru, you seem to have aged rapidly overnight :-) –  Robin Chapman Aug 13 '10 at 16:59
@Robin Chapman: I am laughing like hell..thats a nice one. –  anonymous Aug 13 '10 at 17:03
No sledgehammer needed. There is a solution in Z/pZ to the equation x^2 + y^2 + 1 = 0, so in the quaternions over Z/pZ the quaterion xi + yj + k has norm 0 without being 0. Thus it is a (nonzero) zero divisor. (As for why x^2 + y^2 + 1 = 0 in Z/pZ has a solution, write the equation as x^2 + 1 = -y^2 in Z/pZ and count the number of distinct values on both sides to see there is an overlap.) –  KCd Aug 13 '10 at 20:51

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up vote 3 down vote accepted

This answer builds on damiano's comment.

Because of Wedderburn's theorem, we know that there are no finite-dimensional non-commutative division algebras over $\mathbb{F}_p$. In particular, every quaternion algebra over $\mathbb{F}_p$ is split, i.e., isomorphic to the algebra $M_2(\mathbb{F}_p)$ of $2 \times 2$ matrices over $\mathbb{F}_p$. Can we see this directly by arguments particular to quaternion algebras?

Yes. For instance, a quaternion algebra is split iff its associated norm form

$N(x) = x \overline{x} = q(x_1,x_2,x_3,x_4)$

is isotropic -- i.e., there exists $(x_1,\ldots,x_4) \neq (0,\ldots,0)$ such that $q(x_1,\ldots,x_4) = 0$.

In the case of the Hamiltonian quaternion algebra,

$N(x) = x_1^2 + x_2^2 + x_3^2 + x_4^2$.

As Robin points out, one way to see that this form is isotropic over $\mathbb{F}_p$ is to realize that by Lagrange's Four Squares Theorem, $p$ is a sum of four integral squares, and then reduce modulo $p$.

This is overkill. The fact that any quadratic form in at least three variables over a finite field is isotropic is a special case of the (easy to prove) Chevalley-Warning theorem: see e.g.


Even this is more than is necessary: it follows from a simple counting argument that any nondegenerate quadratic form in at least two variables over $\mathbb{F}_p$ is universal, i.e., represents every nonzero element of the field. For the details, see e.g. p. 5 of


From this we get that every nondegenerate quadratic form in at least three variables over a finite field is isotropic, and in particular the norm form of any quaternion algebra over a finite field is split.

In fact the universality of binary quadratic forms over finite fields is the key to this result. Assuming the characteristic is not $2$, the quaternion algebra may be represented as $\langle a, b \rangle$. For a quaternion algebra $Q = \langle a, b \rangle$ over any field $F$ (of characteristic not $2$), a necessary and sufficient criterion for $Q$ to be split is that $b$ is a norm in the extension $F(\sqrt{a})/F$. Thus the universality of the norm form of $\mathbb{F}_p(\sqrt{a})/\mathbb{F}_p$ implies that all quaternion algebras over $\mathbb{F}_p$ are split.

For that matter, from the perspective of Galois cohomology, Wedderburn's theorem is not especially deep or difficult to prove. Using standard properties of cohomology of cyclic groups, it comes down to showing that for any finite degree extension $\mathbb{F}_{q^n}/\mathbb{F}_q$ of finite fields, the norm map is surjective, and this may also be proved by an elementary counting argument (or simply by a straightforward calculation).

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The sledgehammer is to invoke Wedderburn's theorem to the effect that every finite division ring is commutative.

For a slightly lighter sledgehammer invoke Legendre's four square theorem.

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Thanks for the hint. By the way i couldn't complete yesterdays problem. math.stackexchange.com/questions/2269/… I would like to have a detailed explanation as to how to prove that thing.Sorry, for talking this here, i felt that no one would look at that thread again. –  anonymous Aug 13 '10 at 16:57
Instead of Legendre's four square theorem, you may also use that every element of a finite field is a sum of two squares: this is an easy counting argument! –  damiano Aug 13 '10 at 17:21
Ivoking so many sledgehammer is going to shatter my head. Perhaps you will have to fix it. :) –  anonymous Aug 13 '10 at 17:21

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