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?