Is there a non commutative ring with quotient ring is commutative?

Question

We have a result if $R$ is commutative, then so is any quotient ring $R/I$, for any ideal $I$. If we take the contrapositive, we see that if $R/I$ is non commutative, then $R$ is non commutative. Is there any noncommutative ring that has a commutative quotient ring? If so, give an example.

Answer 1

Consider a direct sum of a commutative ring $M$ and a non-commutative ring $N$, say $M \times N$. Then the quotient ring $(M \times N)/(\{0\} \times N)$ is isomorphic to $M$, which is commutative.

Answer 2

let $R$ be a noncommutative ring. Take $I=R$.

Answer 3

$R$ is an ideal of itself. $R/R\cong\{0\}$

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A non trivial example:

$U_n(\Bbb{R})=\{\begin{pmatrix}a&b\\0&c\end{pmatrix}:a,b,c\in\Bbb{R}\}$

$D_n(\Bbb{R})=\{\begin{pmatrix}a&0\\0&c\end{pmatrix}:a,c\in\Bbb{R}\}$

$U_n(\Bbb{R})$ is non commutative and $D_n(\Bbb{R})$ is commutative ring.

Define a map:

$\phi:U_n(\Bbb{R})\to D_n(\Bbb{R})$ by $$\phi(\begin{pmatrix}a&b\\0&c\end{pmatrix}) = \begin{pmatrix}a&0\\0&c\end{pmatrix} $$

Claim:

1. $\phi$ is a ring homomorphism from $U_n(\Bbb{R})$onto$ D_n(\Bbb{R})$

2. $\ker(\phi) =\{\begin{pmatrix}0&b\\0&0\end{pmatrix}:b\in\Bbb{R}\}$

3. $U_n(\Bbb{R})/\ker(\phi) \cong D_n(\Bbb{R})$