f(x) has a repeated root if and only if . $+ neq $
$begingroup$
I want to prove that f(x) has a repeated root if and only if $<f(x)>+<f'(x)> neq <1>$.
I managed to prove that a repeated root implies $<f(x)>+<f'(x)> neq <1>$, and I know how to prove that if $gcd(f(x),f'(x))neq 1$ then f(x) has repeated root.
So it is left to show that $<f(x)>+<f'(x)> neq <1>$ implies $gcd(f(x),f'(x))neq 1$, but I have no clue how to prove it.
I'll appreciate any lead.
abstract-algebra field-theory
asked Dec 13 '18 at 14:18
ron653ron653
85
$endgroup$
add a comment |
$begingroup$
I want to prove that f(x) has a repeated root if and only if $<f(x)>+<f'(x)> neq <1>$.
I managed to prove that a repeated root implies $<f(x)>+<f'(x)> neq <1>$, and I know how to prove that if $gcd(f(x),f'(x))neq 1$ then f(x) has repeated root.
So it is left to show that $<f(x)>+<f'(x)> neq <1>$ implies $gcd(f(x),f'(x))neq 1$, but I have no clue how to prove it.
I'll appreciate any lead.
abstract-algebra field-theory
asked Dec 13 '18 at 14:18
ron653ron653
85
$endgroup$
add a comment |
$begingroup$
I want to prove that f(x) has a repeated root if and only if $<f(x)>+<f'(x)> neq <1>$.
I managed to prove that a repeated root implies $<f(x)>+<f'(x)> neq <1>$, and I know how to prove that if $gcd(f(x),f'(x))neq 1$ then f(x) has repeated root.
So it is left to show that $<f(x)>+<f'(x)> neq <1>$ implies $gcd(f(x),f'(x))neq 1$, but I have no clue how to prove it.
I'll appreciate any lead.
abstract-algebra field-theory
asked Dec 13 '18 at 14:18
ron653ron653
85
$endgroup$
I want to prove that f(x) has a repeated root if and only if $<f(x)>+<f'(x)> neq <1>$.
I managed to prove that a repeated root implies $<f(x)>+<f'(x)> neq <1>$, and I know how to prove that if $gcd(f(x),f'(x))neq 1$ then f(x) has repeated root.
So it is left to show that $<f(x)>+<f'(x)> neq <1>$ implies $gcd(f(x),f'(x))neq 1$, but I have no clue how to prove it.
I'll appreciate any lead.
abstract-algebra field-theory
abstract-algebra field-theory
asked Dec 13 '18 at 14:18
ron653ron653
85
asked Dec 13 '18 at 14:18
ron653ron653
85
edited Dec 13 '18 at 14:22
ron653
asked Dec 13 '18 at 14:18
ron653ron653
85
asked Dec 13 '18 at 14:18
ron653ron653
85
asked Dec 13 '18 at 14:18
ron653ron653
85
85
add a comment |
add a comment |
1 Answer
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$begingroup$
Okay, the gcd of two polynomials is (by definition) a monic polynomial. So if $gcd(f,f')ne 1$, it must be a monic polynomial of degree $geq 1$. This polynomial is both, a divisor of $f$ and a divisor of $f'$.
Conversely, if $langle frangle +langle f'rangle ne K[x]$, then $langle frangle +langle f'rangle$ is a proper ideal of $K[x]$ and so an ideal of the form $langle grangle$, since $K[x]$ is a principle ideal domain.
Thus $af + bf'=g$ for some $a,b$. Hence, $g$ is a common divisor.
answered Dec 13 '18 at 14:21
WuestenfuxWuestenfux
4,6491413
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$begingroup$
Okay, the gcd of two polynomials is (by definition) a monic polynomial. So if $gcd(f,f')ne 1$, it must be a monic polynomial of degree $geq 1$. This polynomial is both, a divisor of $f$ and a divisor of $f'$.
Conversely, if $langle frangle +langle f'rangle ne K[x]$, then $langle frangle +langle f'rangle$ is a proper ideal of $K[x]$ and so an ideal of the form $langle grangle$, since $K[x]$ is a principle ideal domain.
Thus $af + bf'=g$ for some $a,b$. Hence, $g$ is a common divisor.
answered Dec 13 '18 at 14:21
WuestenfuxWuestenfux
4,6491413
$endgroup$
add a comment |
$begingroup$
Okay, the gcd of two polynomials is (by definition) a monic polynomial. So if $gcd(f,f')ne 1$, it must be a monic polynomial of degree $geq 1$. This polynomial is both, a divisor of $f$ and a divisor of $f'$.
Conversely, if $langle frangle +langle f'rangle ne K[x]$, then $langle frangle +langle f'rangle$ is a proper ideal of $K[x]$ and so an ideal of the form $langle grangle$, since $K[x]$ is a principle ideal domain.
Thus $af + bf'=g$ for some $a,b$. Hence, $g$ is a common divisor.
answered Dec 13 '18 at 14:21
WuestenfuxWuestenfux
4,6491413
$endgroup$
add a comment |
$begingroup$
Okay, the gcd of two polynomials is (by definition) a monic polynomial. So if $gcd(f,f')ne 1$, it must be a monic polynomial of degree $geq 1$. This polynomial is both, a divisor of $f$ and a divisor of $f'$.
Conversely, if $langle frangle +langle f'rangle ne K[x]$, then $langle frangle +langle f'rangle$ is a proper ideal of $K[x]$ and so an ideal of the form $langle grangle$, since $K[x]$ is a principle ideal domain.
Thus $af + bf'=g$ for some $a,b$. Hence, $g$ is a common divisor.
answered Dec 13 '18 at 14:21
WuestenfuxWuestenfux
4,6491413
$endgroup$
Okay, the gcd of two polynomials is (by definition) a monic polynomial. So if $gcd(f,f')ne 1$, it must be a monic polynomial of degree $geq 1$. This polynomial is both, a divisor of $f$ and a divisor of $f'$.
Conversely, if $langle frangle +langle f'rangle ne K[x]$, then $langle frangle +langle f'rangle$ is a proper ideal of $K[x]$ and so an ideal of the form $langle grangle$, since $K[x]$ is a principle ideal domain.
Thus $af + bf'=g$ for some $a,b$. Hence, $g$ is a common divisor.
answered Dec 13 '18 at 14:21
WuestenfuxWuestenfux
4,6491413
edited Dec 13 '18 at 14:35
answered Dec 13 '18 at 14:21
WuestenfuxWuestenfux
4,6491413
answered Dec 13 '18 at 14:21
WuestenfuxWuestenfux
4,6491413
answered Dec 13 '18 at 14:21
WuestenfuxWuestenfux
4,6491413
4,6491413
add a comment |
add a comment |
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