Why isn't $\mathbb C$ the center of $\mathbb H$ (the ring of real quaternions)? I don't get what is wrong with my proof below that $Z(\mathbb H) = \mathbb C$.

If $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in M_2(\mathbb C)$ and $A$ commutes with every $x = \begin{pmatrix} \alpha & \beta \\ -\bar \beta & \bar \alpha \end{pmatrix}$ in $\mathbb H$, then:
$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}  \begin{pmatrix} \alpha & \beta \\ -\bar \beta & \bar \alpha \end{pmatrix} =  \begin{pmatrix} \alpha & \beta \\ -\bar \beta & \bar \alpha \end{pmatrix}  \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
for all $\alpha, \beta \in \mathbb C$, i.e.
$$\begin{pmatrix} a \alpha - b\bar \beta & a\beta + b \bar \alpha \\ c\alpha - d\bar \beta & c\beta + d\bar \alpha \end{pmatrix} = \begin{pmatrix} \alpha a + \beta c & \alpha b + \beta d \\ - \bar \beta a + \bar \alpha c & - \bar \beta b + \bar \alpha d \end{pmatrix}$$
for all $\alpha, \beta \in \mathbb C$.
Plugging in $\beta = 1$ then $\beta = i$, we get from comparing the first entries that $b = c = 0$. This gives $a=d$ and nothing more. Hence $A = \text{diag}(a,a)$, $a\in \mathbb C$. Conversely, any such matrix evidently commutes with all the elements of $\mathbb H$. So
$$Z(\mathbb H) = \left\{ \text{diag}(a,a): a \in \mathbb C\right \} \equiv \mathbb C$$

 A: What you're missing is that $\operatorname{diag}(a,a)$ is not in $\mathbb H$ at all unless $a$ is real.
A: You want to find the matrices in $\mathbb{H}$ that commute with all matrices in $\mathbb{H}$. So you should start with
$$\newcommand\quat[2]{\begin{pmatrix} #1 & #2 \\ -\overline{#2} & \overline{#1} \end{pmatrix}}
\quat{a}{b}\quat{\alpha}{\beta}=
\quat{\alpha}{\beta}\quat{a}{b}
$$
Hence
$$
\quat{a\alpha-b\bar{\beta}}{a\beta+b\bar{\alpha}}
=
\quat{\alpha a-\beta\bar{b}}{\alpha b+\beta\bar{a}}
$$
that, of course, reduces to
$$
\begin{cases}
a\alpha-b\bar{\beta}=\alpha a-\beta\bar{b} \\[6px]
a\beta+b\bar{\alpha}=\alpha b+\beta\bar{a}
\end{cases}
$$
Plugging in $\beta=1$, we get
\begin{cases}
a\alpha-b=\alpha a-\bar{b} \\[6px]
a+b\bar{\alpha}=\alpha b+\bar{a}
\end{cases}
With $\alpha=0$, we obtain $b=\bar{b}$ and $a=\bar{a}$. Plugging in $\beta=i$ we get $bi=-i\bar{b}$, so $b=-\bar{b}$. Hence $b=0$.
It's known that a matrix of the form
$$
\begin{pmatrix} a & 0 \\ 0 & a\end{pmatrix}
$$
with real $a$ (that is $a=\bar{a}$) commutes with every element of $\mathbb{H}$, actually it commutes with every matrix in $M_2(\mathbb{C})$.
