description/proof of that over field, n-degree polynomial has at most n roots
Topics
About: ring
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of field.
- The reader knows a definition of polynomials ring over commutative ring.
- The reader admits the proposition that over any field, any polynomial and any nonzero polynomial divisor have the unique quotient and remainder.
- The reader admits the proposition that any field is an integral domain.
Target Context
- The reader will have a description and a proof of the proposition that over any field, any n-degree polynomial has at most n roots.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(F [x]\): \(= \text{ the polynomials ring over } F\)
\(p (x)\): \(\in F [x]\)
\(S\): \(= \{ \text{ the roots of } p (x)\}\), \(S \subseteq F\)
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Statements:
\(0 \lt deg (p (x))\)
\(\implies\)
\(\vert S \vert \le deg (p (x))\), where \(\vert S \vert\) denotes the cardinality of \(S\)
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When \(p (x) = 0\), is each \(s \in F\) called "root" of \(p (x)\)? Maybe. Anyway, in order to exclude that case, it has assumed that \(0 \lt deg (p (x))\).
2: Natural Language Description
For any field, \(F\), the polynomials ring over \(F\), \(F [x]\), any larger-than-\(0\)-degree polynomial, \(p (x) \in F [x]\), and the set of the roots of \(p (x)\), \(S\), \(\vert S \vert \le deg (p (x))\).
3: Proof
Whole Strategy: Step 1: for any 1st root, \(s_1 \in S\), factorize \(p (x)\) with \(x - s_1\) and the quotient; Step 2: for any 2nd root, \(s_2 \in S\), \(s_2\) is a root of the quotient, so, factorize the quotient with \(x - s_2\) and the new quotient, and so on; Step 3: see that the number of the roots cannot excess the degree of \(p (x)\).
Step 1:
Let us take any 1st (which does not mean any order of \(F\): just means that 1st-ly chosen) root, \(s_1 \in S\). \(p (x) = (x - s) q_1 (x)\), by Note for the proposition that over any field, any polynomial and any nonzero polynomial divisor have the unique quotient and remainder.
Step 2:
Let us take any 2nd root, \(s_2 \in S\). Let us see that \(s_2\) is a root of \(q_1 (x)\). \(p (s_2) = (s_2 - s_1) q_1 (s_2) = 0\). \(s_2 - s_1 \neq 0\), because otherwise, \(s_2 = s_1\). As \(F\) is an integral domain, by the proposition that any field is an integral domain, \(q_1 (s_2) = 0\). So, \(q_1 (x) = (x - s_2) q_2 (x)\), by Note for the proposition that over any field, any polynomial and any nonzero polynomial divisor have the unique quotient and remainder. So, \(p (x) = (x - s_1) (x - s_2) q_2 (x)\).
Likewise, as far as there are \(k\) roots, \(p (x) = (x - s_1) (x - s_2) (x - s_k) q_k (x)\).
Step 3:
If \(deg (p (x)) \lt k\), the right hand side would be larger-than-\(deg (p (x))\)-degree, a contradiction. So, \(k \le deg (p (x))\).
