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Looks like it is easier to find example of commutative rings rather than non commutative rings. Prabably the easiest examples of the former are $\mathbb{Z}$ and $\mathbb{Z}_n$. We can find elaborations on these two commutative rings in various literatures including here and here. These are quite simple and easy to comprehend.

However, the examples on simple (non commutative kind) are not that easy. One example is found here and it has been mentioned as "one of the simplest examples of a non-commutative ring". This is also on the easier side.

EDIT: The following two are also commutative rings.

Other examples are given in this enumeration. But as you can see, examples like Gaussian integers or Eisenstein integers are difficult for starters to comprehend.

Do you think you can give one or two simple examples on non-commutative rings, based on every day numbers?

If it is that difficult, perhaps some insight comments why this is difficult would be welcome.

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What's so difficult about the Gaussian integers? If you know about the complex numbers and the concept of a subring, then they are a completely natural example. –  kahen May 30 '13 at 6:12
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How are the Gaussian integers not commutative? –  user79202 May 30 '13 at 6:13
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Both the examples @MMA listed as "difficult for starters" are, in fact, noted to be commutative in the articles he linked to. –  AWertheim May 30 '13 at 6:15
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5 Answers

up vote 2 down vote accepted

How about the ring of quaternions? This is an example of a non commutative ring and I don't think it's super difficult to wrap your head around. Understanding the ring of quaternions is also in fact very useful in understanding representations of rotations and groups of isometries for polyhedra.

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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.

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An alternate example to show that endomorphism rings are often not commutative (which doesn't depend on manufacturing a not commutative ring ahead of time): if $M$ is an $n>1$ dimensional vector space, then its endomorphism ring of linear transformations is a square matrix ring, which is definitely not commutative for $n>1$. –  rschwieb May 30 '13 at 19:46
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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!

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What about the ring of matrices? It is non-commutative and perhaps as easy to understand as the real numbers. You may also consider the subset of invertible matrices.

You can consider the abelian group of matrices and define the multiplication as

$ X*Y := XY - YX $.

Then you have a non-commutative ring. This is in fact the Lie algebra of all matrices.

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I believe the op was looking rings other than the ring of matrices, which he has listed in his original post. –  AWertheim May 30 '13 at 6:23
    
Yes, I mentioned about the ring of matrices in my question. –  MMA May 30 '13 at 6:23
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Then quaternions are the next best, as mentioned below. –  Vishal May 30 '13 at 6:25
    
You can look at the infinite version of matrices. The ring of all bounded operators on a Banach space. However, you would probably say that it is difficult. –  Vishal May 30 '13 at 6:26
    
@AWertheim: this isn't actually the ring of matrices though. It's the same set of matrices, but the multiplication is defined differently -- so it's a different ring. –  Ehsaan Nov 23 '13 at 22:25
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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.

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This ring is also known as the Weyl algebra, for reference. –  Ryan Reich May 30 '13 at 18:47
    
@Ryan Yes, that's probably the most common of various names that have been given to said algebra of differential operators. –  Key Ideas May 30 '13 at 18:50
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