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 .

• 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! Oct 5, 2019 at 17:08
• @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 $r, a \in R .$ Oct 5, 2019 at 17:27

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.)

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.

• 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 . Oct 4, 2019 at 12:12
• Ok. By the way, I pretty sure my original solution was wrong, so I made an edit Oct 4, 2019 at 20:22
• Okay, Sir, I will try to see your proof in later, after studying all these things, Thank You !!! Oct 5, 2019 at 1:39
• You can use \Bbb Q \langle x_1, x_2 \rangle to render $\Bbb Q \langle x_1, x_2 \rangle$ with angle brackets. Oct 5, 2019 at 4:33
• @Jonathan Dunay,Sir, then shall I unaccept this answer ??? Oct 5, 2019 at 5:04