On a ring $R$ such that every subring of $R$ is an ideal . $\mathbf {The \ Problem \ is}:$ Give an example of a non-commutative ring $R$ (which may or may not contain the identity) such that every subring of $R$ is an ideal .
$\mathbf {My \ approach} :$ I found a proof of a problem that if a ring $R$ contains no divisors of $0$ and every subring of $R$ is an ideal, then $R$ is commutative . 
Again if $R$ has an identity and it satisfies the above stated property, then $R$ is either the "zero ring" $\{0\}$, $\mathbb Z$ or $\mathbb Z_n$ under the criterion that every subring of $R$ must contain the identity of $R .$
And, for any group $(R , +)$, if we define the multiplication operation such that $ab =0$ for all $a, b$ in $R$, then also the criterion would have been satisfied without the requirement of having an identity .
But, I tried some subrings of the matrix groups but failed .
 A: The Problem is:: Give an example of a noncommutative ring $R$
(which may or may not contain the identity) such that every
subring of $R$ is an ideal.

I will give an example which is a nonunital subring of
$M_4(\mathbb F_3)$.
Let
$$
A = \begin{bmatrix}
0&0&0&0\\
1&0&0&0\\
0&0&0&0\\
0&1&1&0\\
\end{bmatrix},
B = \begin{bmatrix}
0&0&0&0\\
0&0&0&0\\
1&0&0&0\\
0&-1&1&0\\
\end{bmatrix},\\
C = \begin{bmatrix}
0&0&0&0\\
0&0&0&0\\
0&0&0&0\\
1&0&0&0\\
\end{bmatrix}.
$$
These matrices were chosen so that they satisfy
$A^2=B^2=AB=-BA=C$ and $CX=0=XC$ for $X$
in the subring generated by $A, B, C$. 
Let $R$ be the subring of $M_4(\mathbb F_3)$
generated by $A, B$ and $C$.
Claim 1. $R$ is not commutative.
(Since $AB=C$, $BA=-C$, and $C\neq -C$.)
Claim 2. The ideal generated by $C$,
the subring generated by $C$, and the additive subgroup
generated by $C$ all coincide, and equal $\mathbb F_3\cdot C=\{0,\pm C\}$.
(This follows from the facts that (i) $CX=0=XC$ for $X\in R$,
and that (ii) $\mathbb F_3=\{0,\pm 1\}$ is a prime field.)
Claim 3. If $X\in R$ is nonzero, then the subring
generated by $X$ contains $C$.
[In fact, more is true: If $X\in R\setminus \{0\}$, then
$C$ is either a nonzero scalar multiple of $X$
or a nonzero scalar multiple of $X^2$. That is,
$C\in \{\pm X, \pm X^2\}$. This is enough
to prove that $C$ belongs to the subring generated by $X$.]
(Write $X = \alpha A + \beta B + \gamma C$ where
$\alpha, \beta,\gamma\in\mathbb F_3$ and not all are zero.
Compute from the relations that $X^2 = (\alpha^2+\beta^2)C$.
If $\alpha^2+\beta^2\neq 0$, then $C=(\alpha^2+\beta^2)^{-1}X^2$
and $C$ is a nonzero scalar multiple of $X^2$. 
Otherwise $\alpha^2+\beta^2=0$,
which forces $\alpha=\beta=0$.
[Here I am using that $\mathbb F_3$ contains no solution to $x^2+1=0$,
so  $\alpha^2+\beta^2=0$ implies $\alpha=\beta=0$.]
But if $\alpha=\beta=0$, we must have $\gamma\neq 0$, so
$C=\gamma^{-1} X$ is a nonzero scalar multiple of $X$.)
Claim 4. Every subring of $R$ is an ideal.
(Let $S$ be an arbitrary subring of $R$. If $S=\{0\}$, then $S$
is an ideal. If there exists $X\in S\setminus \{0\}$,
then, by Claim 3, $C$ belongs to the subring generated
by $X$, hence $C$ belongs to the larger subring $S$.
Now it follows from Claim 2 that the ideal $I_C$ generated by $C$ is contained
in $S$. It is easy to see that $RR=I_C$, so since $S\subseteq R$
we must have both $RS\subseteq RR=I_C\subseteq S$ and $SR\subseteq RR=I_C\subseteq S$.
This proves that $S$ is an ideal.)
A: Hint: I took a quotient ring of the free ring $\mathbb{Q}\langle x_1,x_2\rangle$ (See: noncommutative ring examples) and took a subring of that ring. My solution:

 To be specific, I took $\mathbb{Q}\langle x_1,x_2\rangle/(x_1x_2+x_2x_1,x_1^2-x_2^2,x_1^2-x_1x_2,x_1^3,x_2^3)$ and took the minimal subring containing both $\bar{x_1}$ and $\bar{x_2}$ (call this ring $R$). Notice that all elements of $R$ can be expressed uniquely as $a\bar{x_1}+b\bar{x_2}+c\bar{x_1}\bar{x_2}$. Observe that you could have constructed this by having elements of the ring be elements of the form $a\alpha+b\beta+c\gamma$ and declaring how multiplication works on $\alpha,\beta$ and $\gamma$. Verify that $R$ satisfies the desired conditions.

Edit: the solution above does not work if you are using coefficients in $\Bbb{Q}$; I believe that it does work if you work over something like $\Bbb{Z}_3$.
Edit: Upon considering, I don't believe that this works over the finite fields $\Bbb{Z}_p$ for prime $p$ either (including $\Bbb{Z}_3$). As a remark, it shouldn't work over $\Bbb{Z}$ either. I may have to try to think of another example.
