Existence of square root of a matrix Testing a method with the use of C.-H. theorem for  finding square roots of real $ 2 \times 2 $ matrices I have noticed that some matrices probably don't have  their square roots with  real and complex entries.   
An example the matrix $A= \begin{bmatrix}  0 & 1 \\ 0 & 0 \end{bmatrix}$.
However is it at all a proof that it is impossible  to extend somehow the field of entries in order to satisfy equation $B^2=A$ similar to the situation when many years ago solution of $a^2=-1$ seemed to be impossible to solve for real numbers hence imaginary numbers $i$ were introduced ?
Is it   possible to devise such numbers (...quaternions? octonions ? or others..) that $B^2=A$ would be however satisfied ?
Additionally, when we are sure in general case for $n \times n$ matrices that a square root exists  if we are free to vastly extend a field?

That follows is the continuation of the @Michael Burr 's post.We assume that the entries of $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ are in $R$, a non-commutative ring with unity and we want to solve $X^2=A$ where $X=\begin{pmatrix}p&q\\r&s\end{pmatrix}$. 
Necessarily $AX=XA$, that implies $p=s,r=0$. Finally, $X=\begin{pmatrix}p&q\\0&p\end{pmatrix}$ where $p^2=0,pq+qp=1$.
The simplest example: if $R=M_2(K)$ (where $K$ is a field), then a particular square root (there are many other solutions) is obtained for 
$p=\begin{pmatrix}0&1\\0&0\end{pmatrix},q=\begin{pmatrix}x&y\\1&-x\end{pmatrix}$.
