687: Over Field, n-Degree Polynomial Has at Most n Roots

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description/proof of that over field, n-degree polynomial has at most n roots

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Topics

About: ring

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

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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.

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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.

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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.

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Main Body

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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:
$0<deg(p(x))$
$⟹$
$|S|\le deg(p(x))$, where $|S|$ 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 $0<deg(p(x))$.

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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 $p(x)$, $S$, $|S|\le deg(p(x))$.

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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 $q_1(x)$. $p(s_2)=(s_2-s_1)q_1(s_2)=0$. $s_2-s_1\ne 0$, 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 $deg(p(x))<k$, the right hand side would be larger-than-$deg(p(x))$-degree, a contradiction. So, $k\le deg(p(x))$.

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References

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