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Divisibility of polynomials
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I am having trouble trying to prove the following statement:
Let $f = (f,f').g$ with $(f,f')not= 1$. Both $f$ and $g$ are polynomials with rational coefficients. Then, $(f,f')$ and $g$ share roots.
My idea goes like this:
I name $d = (f,f')$. Since $d not= 1$ then $f$ is not irreducible, so it may be written as $f(x) = p^2(x) , . a(x)$ for some polynomial $a$ where $p$ is an irreducible factor of $d$.
Then $f'= 2,p,p' , a + p^2 , a'$ = $p , (2 , p' , a + p , a')$. This means that $p|f'$.
On the other hand, $f' = d' , g + d , g'$, ant since $p$ divides both $d$ and $f$, it must divide $d' g$.
So I believe that I need to show that $(p,d') = 1$ because in that case I can assure $p$ divides $g$, so $g$ and $f$ will share roots, since $p$ will divide both of them; but I am missing this final step.
Any suggestion?
EDIT: I corrected my question. I originally wrote "$f$ and $g$ share roots" instead of "$(f,f')$ and $g$ share roots" which is what I was really looking for.
abstract-algebra divisibility irreducible-polynomials greatest-common-divisor
asked Jul 23 at 19:50
Javi
404
add a comment |Â
up vote
3
down vote
favorite
I am having trouble trying to prove the following statement:
Let $f = (f,f').g$ with $(f,f')not= 1$. Both $f$ and $g$ are polynomials with rational coefficients. Then, $(f,f')$ and $g$ share roots.
My idea goes like this:
I name $d = (f,f')$. Since $d not= 1$ then $f$ is not irreducible, so it may be written as $f(x) = p^2(x) , . a(x)$ for some polynomial $a$ where $p$ is an irreducible factor of $d$.
Then $f'= 2,p,p' , a + p^2 , a'$ = $p , (2 , p' , a + p , a')$. This means that $p|f'$.
On the other hand, $f' = d' , g + d , g'$, ant since $p$ divides both $d$ and $f$, it must divide $d' g$.
So I believe that I need to show that $(p,d') = 1$ because in that case I can assure $p$ divides $g$, so $g$ and $f$ will share roots, since $p$ will divide both of them; but I am missing this final step.
Any suggestion?
EDIT: I corrected my question. I originally wrote "$f$ and $g$ share roots" instead of "$(f,f')$ and $g$ share roots" which is what I was really looking for.
abstract-algebra divisibility irreducible-polynomials greatest-common-divisor
asked Jul 23 at 19:50
Javi
404
1
The degree of $(f,f')$ is not greater than the degree of $f'$, which is strictly smaller than the degree of $f$. Therefore, the degree of $g$ must be positive. From the equation $f=(f,f')cdot g$ you get that $g$ divides $f$. Therefore, all the roots of $g$ are also roots of $f$, and since $g$ has positive degree, there are some of such roots. The information that $(f,f')=1$ was not needed.
– user577471
Jul 23 at 19:54
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I am having trouble trying to prove the following statement:
Let $f = (f,f').g$ with $(f,f')not= 1$. Both $f$ and $g$ are polynomials with rational coefficients. Then, $(f,f')$ and $g$ share roots.
My idea goes like this:
I name $d = (f,f')$. Since $d not= 1$ then $f$ is not irreducible, so it may be written as $f(x) = p^2(x) , . a(x)$ for some polynomial $a$ where $p$ is an irreducible factor of $d$.
Then $f'= 2,p,p' , a + p^2 , a'$ = $p , (2 , p' , a + p , a')$. This means that $p|f'$.
On the other hand, $f' = d' , g + d , g'$, ant since $p$ divides both $d$ and $f$, it must divide $d' g$.
So I believe that I need to show that $(p,d') = 1$ because in that case I can assure $p$ divides $g$, so $g$ and $f$ will share roots, since $p$ will divide both of them; but I am missing this final step.
Any suggestion?
EDIT: I corrected my question. I originally wrote "$f$ and $g$ share roots" instead of "$(f,f')$ and $g$ share roots" which is what I was really looking for.
abstract-algebra divisibility irreducible-polynomials greatest-common-divisor
asked Jul 23 at 19:50
Javi
404
I am having trouble trying to prove the following statement:
Let $f = (f,f').g$ with $(f,f')not= 1$. Both $f$ and $g$ are polynomials with rational coefficients. Then, $(f,f')$ and $g$ share roots.
My idea goes like this:
I name $d = (f,f')$. Since $d not= 1$ then $f$ is not irreducible, so it may be written as $f(x) = p^2(x) , . a(x)$ for some polynomial $a$ where $p$ is an irreducible factor of $d$.
Then $f'= 2,p,p' , a + p^2 , a'$ = $p , (2 , p' , a + p , a')$. This means that $p|f'$.
On the other hand, $f' = d' , g + d , g'$, ant since $p$ divides both $d$ and $f$, it must divide $d' g$.
So I believe that I need to show that $(p,d') = 1$ because in that case I can assure $p$ divides $g$, so $g$ and $f$ will share roots, since $p$ will divide both of them; but I am missing this final step.
Any suggestion?
EDIT: I corrected my question. I originally wrote "$f$ and $g$ share roots" instead of "$(f,f')$ and $g$ share roots" which is what I was really looking for.
abstract-algebra divisibility irreducible-polynomials greatest-common-divisor
asked Jul 23 at 19:50
Javi
404
edited Jul 24 at 3:49
asked Jul 23 at 19:50
Javi
404
asked Jul 23 at 19:50
Javi
404
asked Jul 23 at 19:50
Javi
404
404
1
The degree of $(f,f')$ is not greater than the degree of $f'$, which is strictly smaller than the degree of $f$. Therefore, the degree of $g$ must be positive. From the equation $f=(f,f')cdot g$ you get that $g$ divides $f$. Therefore, all the roots of $g$ are also roots of $f$, and since $g$ has positive degree, there are some of such roots. The information that $(f,f')=1$ was not needed.
– user577471
Jul 23 at 19:54
add a comment |Â
1
The degree of $(f,f')$ is not greater than the degree of $f'$, which is strictly smaller than the degree of $f$. Therefore, the degree of $g$ must be positive. From the equation $f=(f,f')cdot g$ you get that $g$ divides $f$. Therefore, all the roots of $g$ are also roots of $f$, and since $g$ has positive degree, there are some of such roots. The information that $(f,f')=1$ was not needed.
– user577471
Jul 23 at 19:54
1
1
The degree of $(f,f')$ is not greater than the degree of $f'$, which is strictly smaller than the degree of $f$. Therefore, the degree of $g$ must be positive. From the equation $f=(f,f')cdot g$ you get that $g$ divides $f$. Therefore, all the roots of $g$ are also roots of $f$, and since $g$ has positive degree, there are some of such roots. The information that $(f,f')=1$ was not needed.
– user577471
Jul 23 at 19:54
The degree of $(f,f')$ is not greater than the degree of $f'$, which is strictly smaller than the degree of $f$. Therefore, the degree of $g$ must be positive. From the equation $f=(f,f')cdot g$ you get that $g$ divides $f$. Therefore, all the roots of $g$ are also roots of $f$, and since $g$ has positive degree, there are some of such roots. The information that $(f,f')=1$ was not needed.
– user577471
Jul 23 at 19:54
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
I think the conclusion that you are looking for is that all the roots of $f(x)$ are also roots of $g(x)$. Clearly any root of $g$ is also a root of $f$.
Take a particular root $a$ and let $f(x)=h(x)(x-a)^r$ where $(x-a)$ is not a factor of $h(x)$.
Then $f'(x)=r(x-a)^r-1h(x)+(x-a)^rh'(x)$ and the first of the terms is divisible by $(x-a)^r-1$ but not by $(x-a)^r$, whence $(f,f')$ is only divisible by $(x-a)^r-1$ and you must therefore have $(x-a)|g(x)$.
This is true for all the roots of $f$, and works equally if $r=1$, so it works certainly if there is at least one multiple root, which is the case in your question.
In your attempt you have isolated a squared factor. But it is easier to look at the maximum possible power, and to deal with different roots separately. As the subject develops the idea of a "valuation" comes up - and this naturally isolates the power belonging to a particular root - rather like taking prime factors one at a time when we are looking at integers. When we factorise integers we often identify the part which belongs to each prime.
answered Jul 23 at 21:07
Mark Bennet
76.4k773170
you are absolutely right: I made a mistake when asking my question. What I wanted to show is that (f,f') and g share roots. I will edit the question. Thank you for pointing out.
– Javi
Jul 24 at 3:45
add a comment |Â
up vote
1
down vote
Since $(f,f')ne 1$, $f$ and $f'$ share roots, say $a$: if the order of $f$ in $a$ is $n>1$, the order of $f'$ at $a$ is $n-1$, and the same for the order of $(f,f')$ at $a$. Hence the order of $g$ at $a$ must be $1$.
answered Jul 23 at 20:37
xarles
92059
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I think the conclusion that you are looking for is that all the roots of $f(x)$ are also roots of $g(x)$. Clearly any root of $g$ is also a root of $f$.
Take a particular root $a$ and let $f(x)=h(x)(x-a)^r$ where $(x-a)$ is not a factor of $h(x)$.
Then $f'(x)=r(x-a)^r-1h(x)+(x-a)^rh'(x)$ and the first of the terms is divisible by $(x-a)^r-1$ but not by $(x-a)^r$, whence $(f,f')$ is only divisible by $(x-a)^r-1$ and you must therefore have $(x-a)|g(x)$.
This is true for all the roots of $f$, and works equally if $r=1$, so it works certainly if there is at least one multiple root, which is the case in your question.
In your attempt you have isolated a squared factor. But it is easier to look at the maximum possible power, and to deal with different roots separately. As the subject develops the idea of a "valuation" comes up - and this naturally isolates the power belonging to a particular root - rather like taking prime factors one at a time when we are looking at integers. When we factorise integers we often identify the part which belongs to each prime.
answered Jul 23 at 21:07
Mark Bennet
76.4k773170
you are absolutely right: I made a mistake when asking my question. What I wanted to show is that (f,f') and g share roots. I will edit the question. Thank you for pointing out.
– Javi
Jul 24 at 3:45
add a comment |Â
up vote
1
down vote
accepted
I think the conclusion that you are looking for is that all the roots of $f(x)$ are also roots of $g(x)$. Clearly any root of $g$ is also a root of $f$.
Take a particular root $a$ and let $f(x)=h(x)(x-a)^r$ where $(x-a)$ is not a factor of $h(x)$.
Then $f'(x)=r(x-a)^r-1h(x)+(x-a)^rh'(x)$ and the first of the terms is divisible by $(x-a)^r-1$ but not by $(x-a)^r$, whence $(f,f')$ is only divisible by $(x-a)^r-1$ and you must therefore have $(x-a)|g(x)$.
This is true for all the roots of $f$, and works equally if $r=1$, so it works certainly if there is at least one multiple root, which is the case in your question.
In your attempt you have isolated a squared factor. But it is easier to look at the maximum possible power, and to deal with different roots separately. As the subject develops the idea of a "valuation" comes up - and this naturally isolates the power belonging to a particular root - rather like taking prime factors one at a time when we are looking at integers. When we factorise integers we often identify the part which belongs to each prime.
answered Jul 23 at 21:07
Mark Bennet
76.4k773170
you are absolutely right: I made a mistake when asking my question. What I wanted to show is that (f,f') and g share roots. I will edit the question. Thank you for pointing out.
– Javi
Jul 24 at 3:45
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I think the conclusion that you are looking for is that all the roots of $f(x)$ are also roots of $g(x)$. Clearly any root of $g$ is also a root of $f$.
Take a particular root $a$ and let $f(x)=h(x)(x-a)^r$ where $(x-a)$ is not a factor of $h(x)$.
Then $f'(x)=r(x-a)^r-1h(x)+(x-a)^rh'(x)$ and the first of the terms is divisible by $(x-a)^r-1$ but not by $(x-a)^r$, whence $(f,f')$ is only divisible by $(x-a)^r-1$ and you must therefore have $(x-a)|g(x)$.
This is true for all the roots of $f$, and works equally if $r=1$, so it works certainly if there is at least one multiple root, which is the case in your question.
In your attempt you have isolated a squared factor. But it is easier to look at the maximum possible power, and to deal with different roots separately. As the subject develops the idea of a "valuation" comes up - and this naturally isolates the power belonging to a particular root - rather like taking prime factors one at a time when we are looking at integers. When we factorise integers we often identify the part which belongs to each prime.
answered Jul 23 at 21:07
Mark Bennet
76.4k773170
I think the conclusion that you are looking for is that all the roots of $f(x)$ are also roots of $g(x)$. Clearly any root of $g$ is also a root of $f$.
Take a particular root $a$ and let $f(x)=h(x)(x-a)^r$ where $(x-a)$ is not a factor of $h(x)$.
Then $f'(x)=r(x-a)^r-1h(x)+(x-a)^rh'(x)$ and the first of the terms is divisible by $(x-a)^r-1$ but not by $(x-a)^r$, whence $(f,f')$ is only divisible by $(x-a)^r-1$ and you must therefore have $(x-a)|g(x)$.
This is true for all the roots of $f$, and works equally if $r=1$, so it works certainly if there is at least one multiple root, which is the case in your question.
In your attempt you have isolated a squared factor. But it is easier to look at the maximum possible power, and to deal with different roots separately. As the subject develops the idea of a "valuation" comes up - and this naturally isolates the power belonging to a particular root - rather like taking prime factors one at a time when we are looking at integers. When we factorise integers we often identify the part which belongs to each prime.
answered Jul 23 at 21:07
Mark Bennet
76.4k773170
edited Jul 23 at 21:14
answered Jul 23 at 21:07
Mark Bennet
76.4k773170
answered Jul 23 at 21:07
Mark Bennet
76.4k773170
answered Jul 23 at 21:07
Mark Bennet
76.4k773170
76.4k773170
you are absolutely right: I made a mistake when asking my question. What I wanted to show is that (f,f') and g share roots. I will edit the question. Thank you for pointing out.
– Javi
Jul 24 at 3:45
add a comment |Â
you are absolutely right: I made a mistake when asking my question. What I wanted to show is that (f,f') and g share roots. I will edit the question. Thank you for pointing out.
– Javi
Jul 24 at 3:45
you are absolutely right: I made a mistake when asking my question. What I wanted to show is that (f,f') and g share roots. I will edit the question. Thank you for pointing out.
– Javi
Jul 24 at 3:45
you are absolutely right: I made a mistake when asking my question. What I wanted to show is that (f,f') and g share roots. I will edit the question. Thank you for pointing out.
– Javi
Jul 24 at 3:45
add a comment |Â
up vote
1
down vote
Since $(f,f')ne 1$, $f$ and $f'$ share roots, say $a$: if the order of $f$ in $a$ is $n>1$, the order of $f'$ at $a$ is $n-1$, and the same for the order of $(f,f')$ at $a$. Hence the order of $g$ at $a$ must be $1$.
answered Jul 23 at 20:37
xarles
92059
add a comment |Â
up vote
1
down vote
Since $(f,f')ne 1$, $f$ and $f'$ share roots, say $a$: if the order of $f$ in $a$ is $n>1$, the order of $f'$ at $a$ is $n-1$, and the same for the order of $(f,f')$ at $a$. Hence the order of $g$ at $a$ must be $1$.
answered Jul 23 at 20:37
xarles
92059
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since $(f,f')ne 1$, $f$ and $f'$ share roots, say $a$: if the order of $f$ in $a$ is $n>1$, the order of $f'$ at $a$ is $n-1$, and the same for the order of $(f,f')$ at $a$. Hence the order of $g$ at $a$ must be $1$.
answered Jul 23 at 20:37
xarles
92059
Since $(f,f')ne 1$, $f$ and $f'$ share roots, say $a$: if the order of $f$ in $a$ is $n>1$, the order of $f'$ at $a$ is $n-1$, and the same for the order of $(f,f')$ at $a$. Hence the order of $g$ at $a$ must be $1$.
answered Jul 23 at 20:37
xarles
92059
answered Jul 23 at 20:37
xarles
92059
answered Jul 23 at 20:37
xarles
92059
answered Jul 23 at 20:37
xarles
92059
92059
add a comment |Â
add a comment |Â
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The degree of $(f,f')$ is not greater than the degree of $f'$, which is strictly smaller than the degree of $f$. Therefore, the degree of $g$ must be positive. From the equation $f=(f,f')cdot g$ you get that $g$ divides $f$. Therefore, all the roots of $g$ are also roots of $f$, and since $g$ has positive degree, there are some of such roots. The information that $(f,f')=1$ was not needed.
– user577471
Jul 23 at 19:54