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Unlike <Polynomial over a ring>[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.


Polynomial over a commutative ring

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The usual definition of a <polynomial> is over a <field (mathematics)> as shown at <polynomial over a field>.

However, there is nothing in the immediate definition that prevents us from having a <ring (mathematics)> instead, i.e. a <field (mathematics)> 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:
$$
2xxy + 2xyx + 2yxx +
x2xy + x2yx + y2xx +
xx2y + xy2x + yx2x +
xxy2 + xyx2 + yxx2
$$
while for a field they would all go into a single term:
$$
12x^2y
$$
so when considering a polynomial over a <ring (mathematics)> we end up with a lot more more possible terms.

If the <ring> is a <commutative ring> however, polynomials do look like proper polynomials: <Polynomial over a commutative ring>{full}.


Ciro Santilli @cirosantilli 37

 Incoming links: Polynomial over a ring