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.


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.

  • $\begingroup$ Yes, you are right. For example the matrix $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \in R$ has no inverse. $\endgroup$
    – user42761
    May 26 '16 at 13:45

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.


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