Example of unital non-commutative ring with $(ab)^2=(ba)^2$ for all $a,b$

I'm trying to exhibit a unital, non-commutative ring $R$ such that $(ab)^2=(ba)^2$ for all $a,b\in R$.

This is an exercise out of Herstein's Topics in Algebra. In the previous exercise, I showed that if $R$ was unital, had the same property regarding squares, but also had the property that $2a=0$ implied $a=0$ for any $a\in R$, then $R$ was commutative.

Therefore I've tried several things with the quaternions, the $2\times 2$ matrices, the $3\times 3$ matrices, and even block matrices, all over $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z}$. I can't seem to come up with anything, though, and I'd appreciate a hint.

Solution 1:

I like your idea of trying matrices over $\mathbb{Z}/2\mathbb{Z}$. However, you don't want to take all matrices, as you have two many idempotents, $x$ such that $x^2 = x$.

For instance $a = E_{11} = \left[\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right]$ and $b=E_{11}+E_{12} = \left[\begin{smallmatrix}1&1\\0&0\end{smallmatrix}\right]$. In general $E_{ij} \cdot E_{jk} = E_{ik}$ and if $j\neq k$ then $E_{ij} \cdot E_{k\ell} = 0$. Hence $ab=b$ but $ba=a$; of course $a^2 = a$ and $b^2=b$.

So the hint is to take a subring of the matrix ring that is not commutative but that avoids having too many idempotents. Idempotents are like partial 1s and partial 0s. What we want is for our elements to either be “1” or “0” not both. In more standard language, to avoid idempotents we make sure every element is either a unit ($x$ such that $x^{-1}$ exists) or nilpotent ($x$ such that $x^n=0$).

The very first idea along these lines doesn't quite work, but is a good idea: we take a basis with one element a unit (the identity matrix $E_{11}+E_{22}$) and the other element nilpotent ($x=E_{12}$). It is not too hard to show that this vector space is closed under multiplication, so we do get a ring. And in this ring $(ab)^2 = (ba)^2$ for all $a,b$. However, this ring consists of only $0,1,x,1+x$ and it is not hard to check it is commutative; it is $\mathbb{Z}[x]/(2,x^2)$.

You can fix the commutativity problem without too much trouble though:

Just use larger matrices. Let $R$ be the vector space with basis $1=E_{11}+E_{22}+E_{33}$, $x=E_{12}$, $y=E_{23}$, and $z=E_{13}$ over the field $\mathbb{Z}/2\mathbb{Z}$. Since $xy=z$, but $$xx=yy=zz=yx=xz=zx=yz=zy=0$$ multiplication is very easy. For $$A=\begin{bmatrix} a & b & d \\ 0 & a & c \\ 0 & 0 & a \end{bmatrix}, \qquad B = \begin{bmatrix} e & f & h \\ 0 & e & g \\ 0 & 0 & e \end{bmatrix}$$ we get $$(AB)^2 - (BA)^2 = \begin{bmatrix} 0 & 0 & 2ae(bh-df) \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ which is $0$ as long as $2=0$.

Solution 2:

Rather than construct a single example, one could try to prove enough about such rings that examples become clear.

Definition: Define $t:R^2 \to R : (a,b) \mapsto (ab)^2 - (ba)^2$. Call a ring “$t$-good” if the image of $t$ is 0.

Example: If $R$ is a basic, local $k$-algebra with $J^3=0$, and $k$ of characteristic 2, then $R$ is $t$-good.

The rest of this answer tries to make that example seem like a logical consequence of the definition. However, at several points I had to make assumptions that simplified matters, but that might not be true in general. These are marked “(not WLOG)” since WLOG is normally reserved for “without loss of generality” rather than “with loss of generality.”

Proposition: An idempotent of a $t$-good ring is central.

Proof: If a ring is $t$-good then $t(a,b) - t(1-a,b) = ab^2 - b^2a$ must be 0 as well, so the square of every element is in the center. In particular, if $e$ is an idempotent of a $t$-good ring, then it is central (that is $e=e^2$ implies $ae=ea$ for all $a \in R$ as long as $R$ is $t$-good). $\square$

Lemma: A central idempotent is the same as a direct product factorization of the ring.

Proof: Let $f=1-e$ and notice that we get a ring direct product $R = Re \times Rf$. $Re$ is a ring not just a left ideal since $(re)(se) = rse^2 = (rs)e$ using that $e$ is a central idempotent. Similarly $f=(1-e)^2 = 1-2e+e^2 = 1-2e+e = 1-e=f$ so $f$ is another central idempotent, so $Rf$ is another ring. If $x \in Re \cap Rf$ then $x=re=s(1-e)$ so $xe = re^2 = re = x$ on the one hand, but $xe =s(1-e)e = s(e-e^2) = s(e-e) = 0$ on the other, so $x=0$ and $Re \cap Rf = 0$ so we get a direct product. Every element of $R$ is in this direct product, because $r = r1 = r(e + (1-e)) = r(e+f) = re+rf \in Re \times Rf$. $\square$

We can basically ignore direct products however, since:

Proposition: A direct product of rings is $t$-good iff every factor is $t$-good.

Proof: This is simply because the $i$th coordinate of $t(\mathbf{a},\mathbf{b})$ is just $t$ applied to the $i$th coordinates of $\mathbf{a}$ and $\mathbf{b}$. $\square$

Assumption: (WLOG) $R$ is not a direct product of two proper subrings.

We call such a ring “connected”, though some call it directly indecomposable as a ring.

Proposition: A connected $t$-good ring has no idempotents.

Assumption: (not WLOG) $R$ is a finite dimensional $k$-algebra.

Proposition: In any non-commutative $t$-good ring, $2$ is not a unit. Thus if a $k$-algebra is $t$-good, then $k$ has characteristic 2.

Proof: Now $t(1-a,1-b)-t(1-a,b)-t(a,1-b)+t(a,b) = 2(ab-ba)$, so if 2 is a unit, then $R$ is commutative. We are specifically interested in non-commutative examples, so we cannot have $2$ be a unit. $\square$

Proposition: A connected finite-dimensional $k$-algebra that is $t$-good must be local.

Proof: A finite dimensional algebra is local iff it has no non-trivial idempotents. Since we assume the algebra is connected, it has no central idempotents. Since we assume the algebra is $t$-good, it has no non-central idempotents. $\square$

Assumption: (not WLOG) $R/J$ is separable over $k$.

Assumption: (not WLOG) $K=R/J$ is commutative.

For instance, if $k$ is finite both of these are automatic.

Lemma: $R$ is a $K$-algebra and as $K$-vector space is isomorphic to $K \oplus J$ where $K$ is a central subfield of $R$ and $J$ is a nilpotent ideal.

Proof: This is the Wedderburn-Malçev theorem. $\square$

Assumption: (WLOG) $R/J \cong k$, that is, $R$ is a basic, local $k$-algebra.

Proposition: In a $t$-good ring, $x(ab+ba)=(ab+ba)x$.

Proof: $t(1-x,a-b) + t(x,a) + t(x,b) - t(1-x,a)-t(1-x,b)-t(x,a-b) = xab+xba-abx-bax. \square$

What I would like to say is that in fact $x(ab) = (ab)x$ when $a,b \in J$.

Assumption: (not WLOG) Assume $J^2$ is central.

Proposition: Every basic, local, finite-dimensional $k$-algebra whose Jacobson radical squared is central is $t$-good.

Proof: We calculate $(a+j)(b+k) = ab + bj + ak + jk$ versus $(b+k)(a+j) = ab+bj+ak+kj$ which differ only in the $jk$ versus $kj$. $((a+j)(b+k))^2 - ((b+k)(a+j))^2$ therefore mostly cancels. The only parts that can remain are those involving $jk$ or $kj$. Note however, that $J^2$ is central, so that all that remains $2ab(jk-kj) + 2bj(jk-kj) + 2ak(jk-kj)$. If $J^3=0$ this simplifies further to $2ab(jk-kj)$.

Note that if $J^3=0$, then $J^2$ is automatically central, since it commutes with the central subfield $K$ and of course also with $J$ since $J(J^2) = (J^2)J = 0$.