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$\mathbf {The \ Problem \ is}:$ 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 .

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    $\begingroup$ Engaging question, endorsed, +1!. Can you give a link/citing for a proof of the assertion made in the paragraph opening with "My approach"? Cheers! $\endgroup$ – Robert Lewis Oct 5 '19 at 17:08
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    $\begingroup$ @Robert Lewis, Sir, In a ring $R$, for any non zero element $a$, the centraliser of $a$, denoted by $C(a)$ is a subring of $R$, hence an ideal, and then for any $r \in R$, $(ar-ra)a = 0$ as $a \in C(a)$, then $ar = ra$ for all $r, a \in R .$ $\endgroup$ – Rabi Kumar Chakraborty Oct 5 '19 at 17:27
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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.)

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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.

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  • $\begingroup$ Sir, actually, I have just studied ring theory upto polynomial rings, irreducibility of polynomials etc, but I have seen the link you provided about free ring, I hope I shall study it after crossing these chapters thoroughly . $\endgroup$ – Rabi Kumar Chakraborty Oct 4 '19 at 12:12
  • $\begingroup$ Ok. By the way, I pretty sure my original solution was wrong, so I made an edit $\endgroup$ – Jonathan Dunay Oct 4 '19 at 20:22
  • $\begingroup$ Okay, Sir, I will try to see your proof in later, after studying all these things, Thank You !!! $\endgroup$ – Rabi Kumar Chakraborty Oct 5 '19 at 1:39
  • $\begingroup$ You can use \Bbb Q \langle x_1, x_2 \rangle to render $\Bbb Q \langle x_1, x_2 \rangle$ with angle brackets. $\endgroup$ – Travis Willse Oct 5 '19 at 4:33
  • $\begingroup$ @Jonathan Dunay,Sir, then shall I unaccept this answer ??? $\endgroup$ – Rabi Kumar Chakraborty Oct 5 '19 at 5:04

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