(Simple) Examples on Non-Commutative Rings

Solution 1:

Given an abelian group $M$, let $\operatorname{End}(M)$ denote the set of all homomorphisms $M \to M$ (i.e endomorphisms). This set becomes a ring under pointwise addition and composition.

To see that $\operatorname{End}(M)$ may not be a commutative ring, choose another noncommutative ring $R$ (you already know one). Left multiplication by elements of $R$ are endomorphisms of the underlying abelian group. Since there are elements $a, b \in R$ such that $ab \ne ba$, $\operatorname{End}(R, +)$ is noncommutative.

Solution 2:

Here's a slightly different example. If you are comfortable with doing algebra with polynomials, then I think you will find it easy to understand.

Take any finite group $G$ (say of order $n$). Now, you can formally make a set $\{\sum_{g\in G} \alpha_g g\mid \alpha_g\in \Bbb R\}$.

Since you know how to multiply the elements of $G$, you can extend the multiplication to this set by requiring linearity to hold. So, for example, if you have $a,b,c,d\in G$ and you want to multiply $(3a+2b)(c-5d)$, you would just distribute: $3ac-15ad+2bc-10bd$. Then $ac,ad,bc,bd$ would be multiplied to be elements of $G$, and you would wind back up in the set. Addition is just done by adding "like terms."

With these operations, the set is called the group ring $\Bbb R[G]$. In fact, $G$ doesn't have to be finite, but if $G$ is infinite then you need to only use finite sums in the set I gave above. Even more, $G$ doesn't have to be a group, it just needs to have an associatve multiplication defined on it, so it could be just a monoid or semigroup.

A monoid can informally be described by thinking of a group which does not require the existence of inverses. So, if $x$ is an indeterminate, then $S=\{1,x,x^2,x^3\dots\}$ is a monoid, because $x^ix^j=x^{i+j}$ is an associative multiplication. The monoid ring $\Bbb R[S]$ for this set is something you are really familiar with: it is usualy denoted $\Bbb R [x]$ and called "the ring of polynomials over $\Bbb R$"!

For another thing, you can use any ring you want besides $\Bbb R$! You could even use $\Bbb Z$ if you so chose, or whatever other ring you wanted.

I'm confident that if you stretch your intuition for multiplying polynomials by replacing the powers of $x$ with elements of a group (or monoid), you will quickly grasp what a group ring is.

Anyhow, let's get to the point. If you choose $G$ to be a group that is not Abelian (that is, there exists $g,h\in G$ such that $gh\neq hg$) then you know for sure $\Bbb R[G]$ is not abelian: because in the ring, $gh\neq hg$.

If you would like to experiment with the smallest group which isn't commutative, you'll have to begin with the symmetries of a triangle $S_3=\{1,\sigma,\sigma^2,\tau,\sigma\tau,\sigma^2\tau\}$. To review, the multiplication obeys the relations $\sigma^3=1=\tau^2$, and $\tau\sigma=\sigma^2\tau$.

Pick two elements $p,q$ of $\Bbb Z[S_3]$. Compute $\sigma p$ and $p\sigma$. Compute $p+q$ and $p-q$ and $pq$. Have fun!

Solution 3:

Fairly concrete examples of noncommutative rings arise in calculus when studying differential equations using operator algebra. For example, consider the ring of linear operators generated by the derivative $\, D = d/dx\,$ and the operation of multiplication by $\,x,\,$ i.e. $\, f\mapsto xf,\,$ where $\,(L+M)f = Lf + Mf\, $ and $\,(LM)(f) = L(Mf)),\,$ i.e. multiplication is composition of operators. This bivariate ring of differential polynomials $\,\Bbb R\langle x,D\rangle$ is noncommutative since $$ Dx = xD + 1\ \ \ {\rm i.e.}\ \ (Dx)(f) = D(xf) = x Df + f = (xD + 1) f$$

There is also a discrete analog for difference equations (recurrences), consisting of polynomials in the shift operator $\,Sf(n) = f(n\!+\!1),\,$ and multiplication by $\,n,\,$ i.e. $\, f\mapsto nf.\,$ Then

$$ S n = (n\!+\!1) S\ \ \ {\rm i.e.}\ \ \ (Sn)(f(n)) = S(nf(n)) = (n\!+\!1)f(n\!+\!1) = ((n\!+\!1)S)(f(n))$$

Both of these operator algebras prove useful because we can sometimes employ noncomutative analogs of familiar polynomial arithmetic, e.g. factoring operator polynomials in order to solve differential and difference equations, e.g. solving the Fibonacci recurrence by factoring it as $\,(S-\phi)(S-\bar \phi) f_n = 0,\,$ so $\,f_n = c\phi^n + d\bar \phi^n,\,$ or the noncommutative example here $$ (n\!-\!1)\ S^2\! - (3n\!-\!2)\ S + 2\,n\, =\, ((n\!-\!1)\, S - n)\ (S - 2)$$ Special cases go by various names, e.g the method of characteristic polynomials, or the classic Heaviside operator calculus, etc.