description/proof of that for polynomials ring over field and nonconstant polynomial, iff evaluation of polynomial at field element is 0, polynomial can be factorized with x - element
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of polynomials ring over commutative ring.
- The reader knows a definition of field.
- The reader admits the proposition that the polynomials ring over any field is a Euclidean domain.
Target Context
- The reader will have a description and a proof of 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.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(F [x]\): \(= \text{ the polynomials ring over } F\)
\(p (x)\): \(\in F [x]\)
\(\alpha\): \(\in F\)
//
Statements:
\(p (\alpha) = 0\)
\(\iff\)
\(\exists q (x) \in F [x] (p (x) = (x - \alpha) q (x))\)
//
2: Proof
Whole Strategy: use the fact that \(F [x]\) is a Euclidean domain; Step 1: suppose that \(p (\alpha) = 0\), express \(p (x)\) as \(= (x - \alpha) q (x) + r (x)\), and see that \(r (x) = 0\); Step 2: suppose that \(p (x) = (x - \alpha) q (x)\), and see that \(p (\alpha) = 0\).
Step 1:
Let us suppose that \(p (\alpha) = 0\).
\(F [x]\) is a Euclidean domain with the size function as taking the degree of the polynomial, by the proposition that the polynomials ring over any field is a Euclidean domain.
So, \(p (x) = (x - \alpha) q (x) + r (x)\) where \(r (x)\) is with degree smaller than 1, so, 0, which means that \(r (x) = r\), a constant.
\(0 = p (\alpha) = (\alpha - \alpha) q (\alpha) + r = 0 q (\alpha) + r = r\).
So, \(p (x) = (x - \alpha) q (x)\).
Step 2:
Let us suppose that \(p (x) = (x - \alpha) q (x)\).
\(p (\alpha) = (\alpha - \alpha) q (\alpha) = 0 q (\alpha) = 0\).
