Can a non-commutative ring contains identity? Can a non-commutative ring $R$ contains identity?
Suppose $R$ contains the identity element 1. Construct an ideal $Z(R) = \{a \in R \mid ra = ar\text{ for all }r \in R\}$. Since $1 \in Z(R)$, $R = Z(R)$. This implies that $R$ is commutative. Contradiction.
I can't see what it wrong with my proof. However, I don't think a non-commutative ring must not contain identity. Can anyone enlighten me please?
 A: Consider the ring $M_n(\mathbb{R})$ of all $n\times n$ matrices over $\mathbb{R}$. Clearly, this is a non-commutative ring but it has an identity, namely, the $\textbf{Identity Matrix}$
Actually you don't need to compute the center explicitly. For simplicity I will exhibit the case of $n=2$, the same can be generalized for larger values of $n$. 
Notice that since the Identity matrix is in $Z(M_2(\mathbb{R}))$, then for $Z(M_2(\mathbb{R}))$ to be an ideal we must have   $A \in Z(M_2(\mathbb{R}))$ for any matrix $A \in M_2(\mathbb{R})$. In particular, consider the matrix 
\begin{equation}A = \begin{bmatrix}
       0 &  0           \\[0.3em]
       1 & 0
     \end{bmatrix}.\end{equation} Then for the matrix $B = \begin{bmatrix}
       0 &  1           \\[0.3em]
       0 & 0
     \end{bmatrix}$ you can see that 
\begin{equation}AB = \begin{bmatrix}
       0 &  0           \\[0.3em]
       0 & 1
     \end{bmatrix} \neq BA = \begin{bmatrix}
       1 &  0           \\[0.3em]
       0 & 0
     \end{bmatrix},\end{equation} which shows that $A$ is not in $Z(M_2(\mathbb{R}))$ and hence $Z(M_2(\mathbb{R}))$ is not an ideal. 
$\textbf{Note:}$ For general $n$ pick matrices like we have picked $A$ and $B$ see that $Z(M_n(\mathbb{R}))$ is not an ideal in $M_2(\mathbb{R})$.
A: The ring of quaternions is non-commutative.
