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I'm trying to prove that all quaternion algebras over $\mathbb{F}_p$, for $p$ an odd prime, are isomorphic to $M_2(\mathbb{F}_p)$.

When $p = 1 \pmod 4$, $-1$ is a square in the field and we can construct an obvious sort of "basis" of matrices for $M_2(\mathbb{F}_p)$ that maps into $\{1,i,j,k\}$ (and preserves multiplicative relations).

When $p = 3 \pmod 4$, however, I'm not totally sure how to construct such matrices. The total number of squares in the field, $(p+1)/2$ seems like it might be relevant ... but it's not connecting for me. Some guidance would be appreciated.

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  • $\begingroup$ What happens with the isomorphic image of singular matrices? $\endgroup$ – Marc Bogaerts Nov 18 '16 at 11:40
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First, some notation. For $a,b \in F^\times$, denote by $\left(\frac{a,b}{F}\right)$ the quaternion algebra over the field $F$ defined by $$ i^2 = a, \qquad j^2 = b, \qquad ji = -ij \, . $$ Given a quaternion algebra as above, we can form the "field extension" $K = F[i] = \frac{F[x]}{(x^2 - a)}$. I write "field extension" in quotes because if $a$ is a square in $F$, then $K$ is not a field: if $a = c^2$, then $$ K = \frac{F[x]}{(x^2 - a)} = \frac{F[x]}{(x^2 - c^2)} \cong \frac{F[x]}{(x - c)} \times \frac{F[x]}{(x + c)} \cong F \times F. $$ If $K$ is a field, we have the field norm \begin{align*} N_{K/F}: K &\to F\\ c_1 + c_2 i &\mapsto (c_1 + c_2 i)(c_1 - c_2 i) = {c_1}^2 - {c_2}^2 i^2 = {c_1}^2 - a{c_2}^2 \, . \end{align*}

Theorem 5.4.4 (p. 63) of this forthcoming book by John Voight states multiple criteria for a quaternion algebra to be split. The one I will use is the following:

Proposition: Let $B=\left(\frac{a,b}{F}\right)$ be a quaternion algebra over a field $F$ with $\operatorname{char}(F) \neq 2$. Then $$ B \cong M_2(F) \iff b \in N_{K/F}(K^\times) $$ where $K = F[i]$. (If $K$ is not a field, we take $N_{K/F}(K^\times) = F^\times$.)

For extensions of finite fields the norm map is surjective (see here or here, for example), which implies, together with the above theorem, that every quaternion algebra $B$ over a finite field is split.

We can find this isomorphism to $M_2(\mathbb{F}_p)$ more explicitly as follows. For any $a \in \mathbb{F}_p^{\times}$, the quaternion algebra $B_0 = \left(\frac{a,1}{\mathbb{F}_p}\right)$ is split via the isomorphism \begin{align*} i &\mapsto \begin{pmatrix} 0 & a\\ 1 & 0 \end{pmatrix}\\ j &\mapsto \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \, . \end{align*} Given $a,b \in \mathbb{F}_p^{\times}$, we will show that $B = \left(\frac{a,b}{\mathbb{F}_p}\right)$ is isomorphic to $B_0$.

Since the norm map $N = N_{K/\mathbb{F}_p}$ is surjective, then there exists $\alpha = c_1 + c_2 i \in K$ such that $N(\alpha) = \frac{1}{b}$. Observe that $$ j \alpha = j (c_1 + c_2 i) = c_1 j + c_2 ji = c_1j - c_2 ij = (c_1 - c_2 i)j = \overline{\alpha} j \, . $$ Letting $J = \alpha j$, then $1, i, J, i J$ is a basis for $B$, and \begin{align*} J^2 = (\alpha j)^2 = \alpha j \alpha j = \alpha \overline{\alpha} j^2 = N(\alpha) b = \frac{1}{b} b = 1 \end{align*} Thus $B \cong \left(\frac{a,1}{\mathbb{F}_p}\right)$.

See also Exercise 3.13 on p. 39, which outlines a different proof of the result using group theoretic techniques.

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A technologica way to do this: the Brauer group of a finite field is trivial, so all central simple algebras are matricial.

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