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I am not able to find out any example of non-commutative ring other than matrix ring which is easier to work on.please help

Martin Brandenburg
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ujjwal
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    Quaternions? They form a division ring, in fact: http://maths.ucd.ie/courses/mst2013/ch5.pdf – M. Vinay Jun 26 '14 at 17:39
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    You also have the free algebra of $n$ variables over a ring. It's like a polynomial ring, only you specifically prevent the variables from commuting. See my answer here for more details. – Arthur Jun 26 '14 at 17:41
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    Check out Lam's "A First Course in Noncommutative Rings". There are many examples in Section 1.1. –  Jun 26 '14 at 17:48
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    Group algebras with nonabelian groups. – anon Jun 26 '14 at 17:50

3 Answers3

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Let $G$ be a non-commutative group. Then the group ring $\mathbb{C}[G]$ is not a commutative ring.

Thomas
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the division ring of Quaternions is a non-commutative ring.

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Let $K$ be any field and let $\sigma$ be a nontrivial automorphism of $K$. Now define the set $$ R := K[x;\sigma] = \left\lbrace f(x) = \sum\limits_{i=0}^n a_ix^i \; \big| \; a_i \in K \right\rbrace$$ to be a ring with the additive operation in $R$ being the same as the additive operation of $K[x]$ (that is, $f(x) + g(x)$ is conventional polynomial addition as we know and love it) and define multiplication as follows: let $m,n \in \mathbb{Z}$. Then we define, for every $a,b \in K$, $$ (ax^m)(bx^n) = a\sigma^m(b)x^{m+n}.$$ Since $\sigma$ is a nontrivial automorphism it is easy to see that $R$ is a noncommutative ring. In fact, $R$ is a domain that is not integral, and we can, in some reasonable sense, say the degree to which $R$ is noncommutative is determined by the order of $\sigma$; if $\sigma$ is of order two, for instance, $R$ is noncommutative, but $R$ is sure trying its damndest to commute. If, however, $\sigma$ is of infinite order then $R$ is being a stubborn creature and will very, very rarely commute. As a point of interest, this ring is often called a Twisted Polynomial Ring.

Geoff
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