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We also suppose that the characteristic of a field is not $2.$

Definition 1. An algebra $B$ over $F$ is a quaternion algebra if there exist $i,j\in B$ such that $1,i,j,ij$ is an $F$-basis for $B$ and \begin{equation} i^2=a,j^2=b,\text{ and }ji=-ij \end{equation} for some $a,b$ in the multiplicative group $F^\times$ of units of $F$. We will denoted by $(a,b\mid F).$

Conversely, for $a,b\in F^\times$, $(a,b\mid F)$ exists. The ring ${\rm M}_2(F)$ of $2\times 2$-matrices with coefficients in $F$ is a quaternion algebra over $F$ with $F$-basis $$\left\{\begin{pmatrix} 1&0\\0&1 \end{pmatrix},\begin{pmatrix} 1&0\\0&-1 \end{pmatrix},\begin{pmatrix} 0&1\\1&0 \end{pmatrix},\begin{pmatrix} 1&0\\0&-1 \end{pmatrix}\begin{pmatrix} 0&1\\1&0 \end{pmatrix}\right\}.$$ There is an isomorphism $(1,1\mid F)\cong {\rm M}_2(F)$ of $F$-algebras induced by $$1\mapsto\begin{pmatrix} 1&0\\0&1 \end{pmatrix},i\mapsto\begin{pmatrix} 1&0\\0&-1 \end{pmatrix},j\mapsto\begin{pmatrix} 0&1\\1&0 \end{pmatrix},ij\mapsto\begin{pmatrix} 1&0\\0&-1 \end{pmatrix}\begin{pmatrix} 0&1\\1&0 \end{pmatrix}.$$ Like this, an isomorphism betwen quaternion algebras is a ring isomorphism that fixes the scalar term. Similarly, we also have $$1\mapsto\begin{pmatrix} 1&0\\0&1 \end{pmatrix},i\mapsto\begin{pmatrix} 0&1\\a&0 \end{pmatrix},j\mapsto\begin{pmatrix} 1&0\\0&-1 \end{pmatrix},ij\mapsto\begin{pmatrix} 0&-1\\a&0 \end{pmatrix}$$ is an isomorphism from any quaternion algebra $(a,1\mid F)$ to ${\rm M}_2(F).$ Similarly, we have $(1,b\mid F)\cong{\rm M}_2(F)$ via $$i\mapsto\begin{pmatrix} 1&0\\0&-1 \end{pmatrix},j\mapsto\begin{pmatrix} 0&b\\1&0 \end{pmatrix}.$$ Thus we also have shown $(1,1\mid F)\cong(a,1\mid F)\cong(1,b\mid F)\cong{\rm M}_2(F).$ If every element of $F$ is a square ($F$ is called quadratically closed), then $(a,b\mid F)\cong{\rm M}_2(F).$ Moreover, if $F$ is algebraically closed, then $(a,b\mid F)\cong{\rm M}_2(F).$

An natural question is "when does a quaternion algebra isomorphic to $M_2(F)$?". Morreover, I have a question when a quaternion algebra is not a division ring. We can classify all the cases not?

Tran Nam Son
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    IIRC $B$ is a division algbera if and only if the equation $$x_1^2-ax_2^2-bx_3^2+abx_4^2=0$$ has only the trivial solution $x_1=x_2=x_3=x_4=0$ in $F$. For example in the case of Hamilton's quaternions the equation reads $x_1^2+x_2^2+x_3^2+x_4^2=0$ with $x_i\in\Bbb{R}$. Otherwise $B$ is isomorphic to the matrix algebra. A caveat: the alternatives 1) $B\simeq M_2(F)$, 2) $B$ a division algebra are the only alternatives if $B$ is central simple. Alas, I don't remember whether central simplicity require something (like $ab\neq0$?). I could locate a source, but don't have time, sorry. – Jyrki Lahtonen Aug 01 '21 at 05:52
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    @JyrkiLahtonen: I think all quaternion algebras are central simple in char $\neq 2$, since they are cyclic algebras in this case. A great reference for the OP's question is "Central Simple Algebras and Galois Cohomology" Prop 1.1.7, which says TFAE: 1. $(a, b)$ is split; 2. $(a, b)$ is not a division algebra; 3. the norm map $N \colon (a, b) \to F$ has a nontrivial zero ($N$ is the equation Jyrki gives above); 4. The element $b$ is a norm from the field extension $F(\sqrt{a})/F$. – Alex Wertheim Aug 01 '21 at 09:53
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    Thanks @AlexWertheim. Appreciated. – Jyrki Lahtonen Aug 01 '21 at 10:54
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    See Voight's Quaternion Algebras, Main Theorem 5.4.4, p. 74 for many characterizations of splitting. – Viktor Vaughn Aug 02 '21 at 01:54

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