definition of root of polynomial in polynomials ring over commutative ring
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of polynomials ring over commutative ring.
Target Context
- The reader will have a definition of root of polynomial in polynomials ring over commutative ring.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( R\): \(\in \{\text{ the commutative rings }\}\)
\( R [x]\): \(= \text{ the polynomials ring over } R\)
\( p (x)\): \(\in R [x]\)
\(*r\): \(\in R\)
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Conditions:
\(\exists q (x) \in R [x] (p (x) = (x - r) q (x))\)
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2: Note
We have not defined 'root' as \(p (r) = 0\).
We need to distinguish what are implied when \(R\) is a field and when \(R\) is not necessarily a field.
When \(R\) is a field, this definition equals \(p (r) = 0\): if \(p (x) = (x - r) q (x)\), \(p (r) = (r - r) q (r) = 0 q (r) = 0\); if \(p (r) = 0\), \(p (x) = (p - r) q (x)\), by the proposition that for the polynomials ring over any field and any nonconstant polynomial, if and only if the evaluation of the polynomial at a field element is 0, the polynomial can be factorized with x - the element: when \(p (x)\) is any constant, \(p (x) = 0\) and \(p (x) = (x - r) 0\) anyway.
But when \(R\) is not necessarily a field, if \(p (x) = (x - r) q (x)\), \(p (r) = (r - r) q (r) = 0 q (r) = 0\); but the proposition that for the polynomials ring over any field and any nonconstant polynomial, if and only if the evaluation of the polynomial at a field element is 0, the polynomial can be factorized with x - the element cannot be applied because it uses the fact that \(R [x]\) is a Euclidean domain, which (the fact) has been proved based on the requirement that \(R\) is a field.
So, in general, this definition guarantees that \(p (r) = 0\), but we have not proved that \(p (r) = 0\) guarantees this definition.
For a \(p (x)\), there may be no root and there may be some multiple roots.
It is crucial to be aware that what \(R [x]\) we are thinking in: \(p (x) = x^2 - 2\) has no root with \(p (x)\) regarded in \(\mathbb{Q} [x]\), but it has the roots, \(\sqrt{2}, - \sqrt{2}\), with \(p (x)\) regarded in \(\mathbb{R} [x]\).
