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Let $\mathbb{H} \subset M_2(\mathbb{C})$ be the set of matrices of the form:

$$A = \begin{pmatrix}z & -\bar{w}\\w & \bar{z}\end{pmatrix} \text{ where } z,w \in \mathbb{C}$$

$A^{-1} = \begin{pmatrix}z & -\bar{w}\\w & \bar{z}\end{pmatrix}^{-1} = \frac{1}{(\det A)} \begin{pmatrix}\bar{z} & \bar{w}\\-w & {z}\end{pmatrix} = \frac{1}{(\bar{z}z + w\bar{w})} \begin{pmatrix}\bar{z} & \bar{w}\\-w & z\end{pmatrix} = A^{-1}$

So, ${AA^{-1} = I \qquad\forall A \in \mathbb{H}^\ast = \mathbb{H}\setminus\{0\}}$

Hence, $\mathbb{H}$ is a division ring

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  • $\begingroup$ Is there a question?${}$ $\endgroup$ – José Carlos Santos Mar 17 '20 at 18:11
  • $\begingroup$ Does my work look ok? $\endgroup$ – user551155 Mar 17 '20 at 18:15
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    $\begingroup$ This is certainly the most relevant part (proving $\mathbb H$ to be a subring is comparatively easy). You might want to justify why $\frac{1}{z\overline z+w\overline w}\begin{pmatrix}\overline z&\overline w\\ -w&z\end{pmatrix}$ is in the form $\begin{pmatrix}a&-\overline b\\ b&\overline a\end{pmatrix}$. $\endgroup$ – user239203 Mar 17 '20 at 18:19
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    $\begingroup$ On second thought, proving that $AB\in \mathbb H$ for all $A,B\in\mathbb H$ is easy, but not for free. $\endgroup$ – user239203 Mar 17 '20 at 18:20
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It is correct, except that you should add to it that $z\overline z+w\overline w\in\mathbb R$. If $\lambda$ is an arbitrary complex number, it is not true that $\lambda\left(\begin{smallmatrix}z&-\overline w\\w&\overline z\end{smallmatrix}\right)\in\mathbb H$. And, instead of “$=A^{-1}$”, you should have written “$\in\mathbb H$”.

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