Ciro Santilli @cirosantilli 34

Unlike over non-commutative rings, polynomials do look like proper polynomials over commutative ring.

In particular, Hilbert's tenth problem is about polynomials over the integers, which is a commutative ring, and therefore brings mindshare to this definition.

The usual definition of a polynomial is over a field as shown at polynomial over a field.

However, there is nothing in the immediate definition that prevents us from having a ring instead, i.e. a field but without the commutative property and inverse elements.

The only thing is that then we would need to differentiate between different orderings of the terms of multivariate polynomial, e.g. the following would all be potentially different terms:while for a field they would all go into a single term:so when considering a polynomial over a ring we end up with a lot more more possible terms.

If the ring is a commutative ring however, polynomials do look like proper polynomials: Section "Polynomial over a commutative ring".

A Ring can be seen as a generalization of a field where:

Addition however has to be commutative and have inverses, i.e. it is an Abelian group.

The simplest example of a ring which is not a full fledged field and with commutative multiplication are the integers. Notably, no inverses exist except for the identity itself and -1. E.g. the inverse of 2 would be 1/2 which is not in the set. More specifically, the integers are a commutative ring.

A polynomial ring is another example with the same properties as the integers.

The simplest non-commutative, non-division is is the set of all 2x2 matrices of real numbers:Note that $GL(n)$ is not a ring because you can by addition reach the zero matrix.