Factoting a polyonimal using minimal polynomial. [duplicate]
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Why does the minimal polynomial of α divide all polynomials for which α is a root?
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Let's say that $p$ is the minimal polynomial of $a$ over a field. Then is it true that if $P$ is any other polynomial which has $a$ among its roots there exists some polynomial $g$ so that $P(a) =p(a) cdot g(a) $?
abstract-algebra polynomials
asked Dec 23 '18 at 9:25
JustAnAmateurJustAnAmateur
1096
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Dec 23 '18 at 18:18
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This question already has an answer here:
Why does the minimal polynomial of α divide all polynomials for which α is a root?
3 answers
Let's say that $p$ is the minimal polynomial of $a$ over a field. Then is it true that if $P$ is any other polynomial which has $a$ among its roots there exists some polynomial $g$ so that $P(a) =p(a) cdot g(a) $?
abstract-algebra polynomials
asked Dec 23 '18 at 9:25
JustAnAmateurJustAnAmateur
1096
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marked as duplicate by Bill Dubuque abstract-algebra
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Dec 23 '18 at 18:18
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$begingroup$
This question already has an answer here:
Why does the minimal polynomial of α divide all polynomials for which α is a root?
3 answers
Let's say that $p$ is the minimal polynomial of $a$ over a field. Then is it true that if $P$ is any other polynomial which has $a$ among its roots there exists some polynomial $g$ so that $P(a) =p(a) cdot g(a) $?
abstract-algebra polynomials
asked Dec 23 '18 at 9:25
JustAnAmateurJustAnAmateur
1096
$endgroup$
This question already has an answer here:
Why does the minimal polynomial of α divide all polynomials for which α is a root?
3 answers
Let's say that $p$ is the minimal polynomial of $a$ over a field. Then is it true that if $P$ is any other polynomial which has $a$ among its roots there exists some polynomial $g$ so that $P(a) =p(a) cdot g(a) $?
This question already has an answer here:
Why does the minimal polynomial of α divide all polynomials for which α is a root?
3 answers
abstract-algebra polynomials
abstract-algebra polynomials
asked Dec 23 '18 at 9:25
JustAnAmateurJustAnAmateur
1096
asked Dec 23 '18 at 9:25
JustAnAmateurJustAnAmateur
1096
asked Dec 23 '18 at 9:25
JustAnAmateurJustAnAmateur
1096
asked Dec 23 '18 at 9:25
JustAnAmateurJustAnAmateur
1096
asked Dec 23 '18 at 9:25
JustAnAmateurJustAnAmateur
1096
1096
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Dec 23 '18 at 18:18
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1 Answer
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Yes.
Use polynomial division to write $P(X)=q(X)p(X)+r(X)$ with $deg r<deg p$ and note that $r(a)=0$. Conclude that $requiv 0$.
answered Dec 23 '18 at 9:28
Hagen von EitzenHagen von Eitzen
2843
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1 Answer
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1 Answer
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$begingroup$
Yes.
Use polynomial division to write $P(X)=q(X)p(X)+r(X)$ with $deg r<deg p$ and note that $r(a)=0$. Conclude that $requiv 0$.
answered Dec 23 '18 at 9:28
Hagen von EitzenHagen von Eitzen
2843
$endgroup$
add a comment |
$begingroup$
Yes.
Use polynomial division to write $P(X)=q(X)p(X)+r(X)$ with $deg r<deg p$ and note that $r(a)=0$. Conclude that $requiv 0$.
answered Dec 23 '18 at 9:28
Hagen von EitzenHagen von Eitzen
2843
$endgroup$
add a comment |
$begingroup$
Yes.
Use polynomial division to write $P(X)=q(X)p(X)+r(X)$ with $deg r<deg p$ and note that $r(a)=0$. Conclude that $requiv 0$.
answered Dec 23 '18 at 9:28
Hagen von EitzenHagen von Eitzen
2843
$endgroup$
Yes.
Use polynomial division to write $P(X)=q(X)p(X)+r(X)$ with $deg r<deg p$ and note that $r(a)=0$. Conclude that $requiv 0$.
answered Dec 23 '18 at 9:28
Hagen von EitzenHagen von Eitzen
2843
answered Dec 23 '18 at 9:28
Hagen von EitzenHagen von Eitzen
2843
answered Dec 23 '18 at 9:28
Hagen von EitzenHagen von Eitzen
2843
answered Dec 23 '18 at 9:28
Hagen von EitzenHagen von Eitzen
2843
2843
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