infinite ring with unity all whose proper subrings, which contains the unity of the ring, are commutative Let $R$ be a ring with unity (not necessarily commutative). Let us call a subset $S\subseteq R$ to be a subring of $R$ if it is a ring with the operations induced from $R$ and the unity of $R$ is inside $S$. 
My question is: If $R$ is an infinite ring with unity all whose proper subrings are commutative, then is $R$ commutative ? 
 A: Since we are not interested in rings without unity, I'll henceforth adopt the videogame notation "rng" = "ring", "ring" = "rng with unity" (with the intention of this being the last sentence where the word rng occurs).
The short answer is that there are such non-commutative rings, and one example is $$R=(\Bbb Z[X]\amalg \Bbb Z[Y])/(2X,2Y,YXY,XYX, X^2,Y^2)$$ i.e. the free ring over two variables quotiented by the relations $2X=2Y=0$, $X^2=Y^2=0$, $XYX=YXY=0$. A more concrete way to describe it is by endowing the $\Bbb Z$-module $$\Bbb Z\oplus \Bbb F_2X\oplus \Bbb F_2Y\oplus \Bbb F_2XY\oplus \Bbb F_2YX$$ with the product \begin{align}&(n_1+k_1X+h_1Y+r_1XY+s_1YX)(n_2+k_2X+h_2Y+r_2XY+s_2YX)=\\&=n_1n_2+(n_2k_1+k_2n_1)X+(n_2h_1+h_2n_1)Y+(k_1h_2+n_1r_2+n_2r_1)XY+\\&+(k_2h_1+n_1s_2+n_2s_1)YX\end{align}
Proving that the two are isomorphic is tantamount to checking that the latter is in fact a valid ring.
Where does this example come from? To fix a notation, call $\Bbb Z_R\langle S\rangle$ the least subring of $R$ that contains the set $S$. Two observations:


*

*A ring is a counterexample if and only if it is non-commutative and $R=\Bbb Z_R\langle a,b\rangle$ for all $a,b$ such that $ab\ne ba$.

*This is a bit more hand-wavy: we must not be allowed to say that $\Bbb Z_R\langle pa,pb,pab,pba\rangle$ is a proper non-commutative subring for any prime $p$.
The example comes from giving a minimal candidate for these euristics, and then hope that it satisfies (1)
Now, let's prove that this ring satisfies (1). Consider $a,b$ such that $ab-ba\ne 0$. Since $\Bbb Z$ is in the center of the ring, adding to $a$ and $b$ elements of $\Bbb Z$ does not change whether they commute or not, nor whether they generate the whole ring. Therefore we can assume that $n_a=n_b=0$.
It is now a matter of checking $256$ cases, but let's make it quicker: we are looking at elements in the form $G+Q$ with $G\in \Bbb F_2X\oplus\Bbb F_2Y$ and $Q\in\Bbb F_2 XY\oplus \Bbb F_2YX$. Notice that $$(G_a+Q_a)(G_b+Q_b)-(G_b+Q_b)(G_a+Q_a)=G_aG_b-G_bG_a$$ because the other products are $0$.
So we want to consider the ones whose "linear parts" don't commute. This is possible only if they are distinct and non-zero. Since the only possible options are $\{X,Y,X+Y\}$, there will be a pair $(c,d)$ out of $\{a,b,a-b\}$ such that $G_c=X$ and $G_d=Y$. Then, $cd=G_cG_d=XY$ and $dc=G_dG_c=YX$. All these are in $\Bbb Z_R\langle a,b\rangle$. So $\Bbb Z_R\langle a,b\rangle$ contains $XY$, $YX$ and some $c,d$ such that $G_c=X$ adn $G_d=Y$. Therefore, it contains $X$ and $Y$ as well. So it contains everything, QED.
