Mixing
'Solutions' to a specific equation in $K[t]$
$begingroup$
Let $K$ be a field of characteristic zero. Let $f,g,h in K[t]$ be three separable polynomials, with no common zeros.
Denote $deg(f)=a,deg(g)=b,deg(h)=c$, $a geq 2, b geq 1, c geq 1$,
and denote the roots of $f,g,h$ ,respectively, by $a_i,b_j,c_k$, $1 leq i leq a$, $1 leq j leq b$, $1 leq k leq c$.
Let $u,v,w in K[t]$, with $deg(u) < a, deg(v) < a$.
If $ug+vh=wf$, then is there something interesting to say about $f,g,h,u,v,w$ and their degrees?
The only thing I have managed to obtain thus far is that, in the special case where $w=0$, we have: $ug+vh=0$, so $ug=-vh$, and since $g$ and $h$ are relatively prime, it follows that $g$ divides $v$ and $h$ divides $u$.
Hence, $v=ghat{v}$ and $u=hhat{u}$, and then $hhat{u}g+ghat{v}h=0$, so $hat{u}=-hat{v}$, so $v=ghat{v}$ and $u=-hhat{v}$, which shows that $gcd(u,v)=hat{v}$.I do not know what is the answer I am looking for, but I hope that it would be 'nice'.
I prefer not to assume that $w=0$.
Any hints and comments are welcome!
polynomials commutative-algebra
asked Jan 9 at 1:14
user237522user237522
2,1691617
$endgroup$
add a comment |
$begingroup$
Let $K$ be a field of characteristic zero. Let $f,g,h in K[t]$ be three separable polynomials, with no common zeros.
Denote $deg(f)=a,deg(g)=b,deg(h)=c$, $a geq 2, b geq 1, c geq 1$,
and denote the roots of $f,g,h$ ,respectively, by $a_i,b_j,c_k$, $1 leq i leq a$, $1 leq j leq b$, $1 leq k leq c$.
Let $u,v,w in K[t]$, with $deg(u) < a, deg(v) < a$.
If $ug+vh=wf$, then is there something interesting to say about $f,g,h,u,v,w$ and their degrees?
The only thing I have managed to obtain thus far is that, in the special case where $w=0$, we have: $ug+vh=0$, so $ug=-vh$, and since $g$ and $h$ are relatively prime, it follows that $g$ divides $v$ and $h$ divides $u$.
Hence, $v=ghat{v}$ and $u=hhat{u}$, and then $hhat{u}g+ghat{v}h=0$, so $hat{u}=-hat{v}$, so $v=ghat{v}$ and $u=-hhat{v}$, which shows that $gcd(u,v)=hat{v}$.I do not know what is the answer I am looking for, but I hope that it would be 'nice'.
I prefer not to assume that $w=0$.
Any hints and comments are welcome!
polynomials commutative-algebra
asked Jan 9 at 1:14
user237522user237522
2,1691617
$endgroup$
$begingroup$
By division algorithm, there is always $u,v$ with degrees less than $a$ and $ug+vh=wf$. What else would you like to say?
$endgroup$
– Mohan
Jan 9 at 1:36
$begingroup$
Thank you for your comment. I am not sure what else I would like to say.. (I will try to think about it more and will update my question, in case I will have something interesting to say/ask).
$endgroup$
– user237522
Jan 9 at 1:51
add a comment |
$begingroup$
Let $K$ be a field of characteristic zero. Let $f,g,h in K[t]$ be three separable polynomials, with no common zeros.
Denote $deg(f)=a,deg(g)=b,deg(h)=c$, $a geq 2, b geq 1, c geq 1$,
and denote the roots of $f,g,h$ ,respectively, by $a_i,b_j,c_k$, $1 leq i leq a$, $1 leq j leq b$, $1 leq k leq c$.
Let $u,v,w in K[t]$, with $deg(u) < a, deg(v) < a$.
If $ug+vh=wf$, then is there something interesting to say about $f,g,h,u,v,w$ and their degrees?
The only thing I have managed to obtain thus far is that, in the special case where $w=0$, we have: $ug+vh=0$, so $ug=-vh$, and since $g$ and $h$ are relatively prime, it follows that $g$ divides $v$ and $h$ divides $u$.
Hence, $v=ghat{v}$ and $u=hhat{u}$, and then $hhat{u}g+ghat{v}h=0$, so $hat{u}=-hat{v}$, so $v=ghat{v}$ and $u=-hhat{v}$, which shows that $gcd(u,v)=hat{v}$.I do not know what is the answer I am looking for, but I hope that it would be 'nice'.
I prefer not to assume that $w=0$.
Any hints and comments are welcome!
polynomials commutative-algebra
asked Jan 9 at 1:14
user237522user237522
2,1691617
$endgroup$
Let $K$ be a field of characteristic zero. Let $f,g,h in K[t]$ be three separable polynomials, with no common zeros.
Denote $deg(f)=a,deg(g)=b,deg(h)=c$, $a geq 2, b geq 1, c geq 1$,
and denote the roots of $f,g,h$ ,respectively, by $a_i,b_j,c_k$, $1 leq i leq a$, $1 leq j leq b$, $1 leq k leq c$.
Let $u,v,w in K[t]$, with $deg(u) < a, deg(v) < a$.
If $ug+vh=wf$, then is there something interesting to say about $f,g,h,u,v,w$ and their degrees?
The only thing I have managed to obtain thus far is that, in the special case where $w=0$, we have: $ug+vh=0$, so $ug=-vh$, and since $g$ and $h$ are relatively prime, it follows that $g$ divides $v$ and $h$ divides $u$.
Hence, $v=ghat{v}$ and $u=hhat{u}$, and then $hhat{u}g+ghat{v}h=0$, so $hat{u}=-hat{v}$, so $v=ghat{v}$ and $u=-hhat{v}$, which shows that $gcd(u,v)=hat{v}$.I do not know what is the answer I am looking for, but I hope that it would be 'nice'.
I prefer not to assume that $w=0$.
Any hints and comments are welcome!
polynomials commutative-algebra
polynomials commutative-algebra
asked Jan 9 at 1:14
user237522user237522
2,1691617
asked Jan 9 at 1:14
user237522user237522
2,1691617
asked Jan 9 at 1:14
user237522user237522
2,1691617
asked Jan 9 at 1:14
user237522user237522
2,1691617
asked Jan 9 at 1:14
user237522user237522
2,1691617
2,1691617
$begingroup$
By division algorithm, there is always $u,v$ with degrees less than $a$ and $ug+vh=wf$. What else would you like to say?
$endgroup$
– Mohan
Jan 9 at 1:36
$begingroup$
Thank you for your comment. I am not sure what else I would like to say.. (I will try to think about it more and will update my question, in case I will have something interesting to say/ask).
$endgroup$
– user237522
Jan 9 at 1:51
add a comment |
$begingroup$
By division algorithm, there is always $u,v$ with degrees less than $a$ and $ug+vh=wf$. What else would you like to say?
$endgroup$
– Mohan
Jan 9 at 1:36
$begingroup$
Thank you for your comment. I am not sure what else I would like to say.. (I will try to think about it more and will update my question, in case I will have something interesting to say/ask).
$endgroup$
– user237522
Jan 9 at 1:51
$begingroup$
By division algorithm, there is always $u,v$ with degrees less than $a$ and $ug+vh=wf$. What else would you like to say?
$endgroup$
– Mohan
Jan 9 at 1:36
$begingroup$
By division algorithm, there is always $u,v$ with degrees less than $a$ and $ug+vh=wf$. What else would you like to say?
$endgroup$
– Mohan
Jan 9 at 1:36
$begingroup$
Thank you for your comment. I am not sure what else I would like to say.. (I will try to think about it more and will update my question, in case I will have something interesting to say/ask).
$endgroup$
– user237522
Jan 9 at 1:51
$begingroup$
Thank you for your comment. I am not sure what else I would like to say.. (I will try to think about it more and will update my question, in case I will have something interesting to say/ask).
$endgroup$
– user237522
Jan 9 at 1:51
add a comment |
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$begingroup$
By division algorithm, there is always $u,v$ with degrees less than $a$ and $ug+vh=wf$. What else would you like to say?
$endgroup$
– Mohan
Jan 9 at 1:36
$begingroup$
Thank you for your comment. I am not sure what else I would like to say.. (I will try to think about it more and will update my question, in case I will have something interesting to say/ask).
$endgroup$
– user237522
Jan 9 at 1:51