Q:

Prove that $R(+,.)$ is a division ring but I disproved it

QUESTION: 
Let $R=\left[\begin{matrix}\alpha & \beta \\ \bar\beta &
 \bar\alpha\end{matrix}\right]\in \mathbf{M_2(\mathbb{C})} $ where 
  $\bar\alpha,\bar\beta$ denote the conjugates of $\alpha, \beta$ 
  respectively. Prove that $R$ is a division ring but not field under 
  the usual matrix addition and multiplication.

MY ATTEMPT: 
I am comfortable with what I have to do and what I have to prove. I have successfully proved that it is not a field as the matrix multiplication is not commutative. But instead of proving that it is a division ring, I have disproved it.
We know that, 

A division ring, also called a skew field, is a ring in which division
  is possible. Specifically, it is a nonzero ring in which every
  nonzero element a has a multiplicative inverse, i.e., an element $x$
  with $a·x = x·a = 1$. Stated differently, a ring is a division ring if
  and only if the group of units equals the set of all nonzero elements.

Now the condition in bold is what I have shown not to hold. Actually the inverse of a matrix exists iff the matrix is not singular.
But $\left|\begin{matrix}\alpha & \beta \\ \bar\beta &
 \bar\alpha\end{matrix}\right|=|\alpha|^2-|\beta|^2$ which can be $0$ if $|\alpha|=|\beta|$ which holds for infinitely many $\alpha,\beta \in \mathbb{C}$.
So $R(+,.)$ is not a division ring since it contains an infinite number of non-invertible matrices.
Am I right or wrong? Please help.

A:

You are right about $$R=\left\{\begin{bmatrix}\alpha & \beta \\ \bar\beta &
 \bar\alpha\end{bmatrix}\mid  \alpha,\beta\in\mathbb{C}\right\} $$, especially since $e=\frac12\begin{bmatrix}1&1\\ 1&1\end{bmatrix}$ would create zero divisors : $e(1-e)=0$.
But presumably what you're reading is about 
$$R=\left\{\begin{bmatrix}\alpha & \beta \\ -\bar\beta &
 \bar\alpha\end{bmatrix}\mid  \alpha,\beta\in\mathbb{C}\right\} $$
which is isomorphic to Hamilton's quaternions, and is a division ring, of course.

