'Solving' $y=(AF+BG)y+(Au+Bv)(x-alpha)(x-beta)$
$begingroup$
Let $A,B,F,G in mathbb{C}[x,y]$, $u,v in mathbb{C}[x]$, $alpha,beta in mathbb{C}$, $alpha neq beta$.
Assume that $u$ and $v$ have no common zero,
and that ${alpha,beta}$ are not zeros of $u$ or $v$.
(I am not sure if I wish to assume that $F$ and $G$ are relatively prime or not).
Further assume that the following equation holds: $y=(AF+BG)y+(Au+Bv)(x-alpha)(x-beta)$.
What can be said about $A,B,F,G,u,v$? I do not know exactly what exactly I am looking for; I wish to obtain 'simpler' (= of a specific form/ with less variables/ a connection between) $A,B,F,G,u,v$.
Perhaps this equation leads to some contradiction?
Taking $y=0$, we get that: $(A(x,0)u+B(x,0)v)(x-alpha)(x-beta)=0$.
This shows that, for every $gamma notin {alpha,beta}$:
$A(gamma,0)u(gamma)+B(gamma,0)v(gamma)=0$.
In other words, the polynomial $A(x,0)u(x)+B(x,0)v(x)$ has infinitely many zeros,
which is impossible unless it is the zero polynomial; is it true that then $A(x,0)=v(x)$ and $B(x,0)=-u(x)$?
Taking $x=alpha$, we get that:
$y=(A(alpha,y)F(alpha,y)+B(alpha,y)G(alpha,y))y$.
Then,
$A(alpha,y)F(alpha,y)+B(alpha,y)G(alpha,y)=1$.
Similarly, taking $x=beta$ we get that:
$A(beta,y)F(beta,y)+B(beta,y)G(beta,y)=1$.
By this question,
$A=a_ny^n+cdots+a_1y+ v tilde{v}=tilde{A}y+vtilde{v}$
and
$B=b_my^m+cdots+b_1y- u tilde{v}=tilde{B}y-utilde{v}$,
for some $tilde{A},tilde{B} in k[x,y]$.
Therefore, $y=((tilde{A}y+vtilde{v})F+(tilde{B}y-utilde{v})G)y+((tilde{A}y+vtilde{v})u+(tilde{B}y-utilde{v})v)(x-alpha)(x-beta)$
becomes
$y=((tilde{A}y+vtilde{v})F+(tilde{B}y-utilde{v})G)y+(tilde{A}yu+tilde{B}yv)(x-alpha)(x-beta)$.
Then, $1=((tilde{A}y+vtilde{v})F+(tilde{B}y-utilde{v})G)+(tilde{A}u+tilde{B}v)(x-alpha)(x-beta)$.
I think that more can be said, and actually $AF+BG=Cy+D$, where $C,D in k[x]$
(by differentiating with respect to $y$); I will add the details later.
(I really apologize if this equation seems a little peculiar, it is just that I have encountered it in a calculation I did, and wish to understand it better).
Any comments are welcome!
algebraic-geometry polynomials commutative-algebra
asked Jan 5 at 22:12
user237522user237522
2,1801617
$endgroup$
add a comment |
$begingroup$
Let $A,B,F,G in mathbb{C}[x,y]$, $u,v in mathbb{C}[x]$, $alpha,beta in mathbb{C}$, $alpha neq beta$.
Assume that $u$ and $v$ have no common zero,
and that ${alpha,beta}$ are not zeros of $u$ or $v$.
(I am not sure if I wish to assume that $F$ and $G$ are relatively prime or not).
Further assume that the following equation holds: $y=(AF+BG)y+(Au+Bv)(x-alpha)(x-beta)$.
What can be said about $A,B,F,G,u,v$? I do not know exactly what exactly I am looking for; I wish to obtain 'simpler' (= of a specific form/ with less variables/ a connection between) $A,B,F,G,u,v$.
Perhaps this equation leads to some contradiction?
Taking $y=0$, we get that: $(A(x,0)u+B(x,0)v)(x-alpha)(x-beta)=0$.
This shows that, for every $gamma notin {alpha,beta}$:
$A(gamma,0)u(gamma)+B(gamma,0)v(gamma)=0$.
In other words, the polynomial $A(x,0)u(x)+B(x,0)v(x)$ has infinitely many zeros,
which is impossible unless it is the zero polynomial; is it true that then $A(x,0)=v(x)$ and $B(x,0)=-u(x)$?
Taking $x=alpha$, we get that:
$y=(A(alpha,y)F(alpha,y)+B(alpha,y)G(alpha,y))y$.
Then,
$A(alpha,y)F(alpha,y)+B(alpha,y)G(alpha,y)=1$.
Similarly, taking $x=beta$ we get that:
$A(beta,y)F(beta,y)+B(beta,y)G(beta,y)=1$.
By this question,
$A=a_ny^n+cdots+a_1y+ v tilde{v}=tilde{A}y+vtilde{v}$
and
$B=b_my^m+cdots+b_1y- u tilde{v}=tilde{B}y-utilde{v}$,
for some $tilde{A},tilde{B} in k[x,y]$.
Therefore, $y=((tilde{A}y+vtilde{v})F+(tilde{B}y-utilde{v})G)y+((tilde{A}y+vtilde{v})u+(tilde{B}y-utilde{v})v)(x-alpha)(x-beta)$
becomes
$y=((tilde{A}y+vtilde{v})F+(tilde{B}y-utilde{v})G)y+(tilde{A}yu+tilde{B}yv)(x-alpha)(x-beta)$.
Then, $1=((tilde{A}y+vtilde{v})F+(tilde{B}y-utilde{v})G)+(tilde{A}u+tilde{B}v)(x-alpha)(x-beta)$.
I think that more can be said, and actually $AF+BG=Cy+D$, where $C,D in k[x]$
(by differentiating with respect to $y$); I will add the details later.
(I really apologize if this equation seems a little peculiar, it is just that I have encountered it in a calculation I did, and wish to understand it better).
Any comments are welcome!
algebraic-geometry polynomials commutative-algebra
asked Jan 5 at 22:12
user237522user237522
2,1801617
$endgroup$
add a comment |
$begingroup$
Let $A,B,F,G in mathbb{C}[x,y]$, $u,v in mathbb{C}[x]$, $alpha,beta in mathbb{C}$, $alpha neq beta$.
Assume that $u$ and $v$ have no common zero,
and that ${alpha,beta}$ are not zeros of $u$ or $v$.
(I am not sure if I wish to assume that $F$ and $G$ are relatively prime or not).
Further assume that the following equation holds: $y=(AF+BG)y+(Au+Bv)(x-alpha)(x-beta)$.
What can be said about $A,B,F,G,u,v$? I do not know exactly what exactly I am looking for; I wish to obtain 'simpler' (= of a specific form/ with less variables/ a connection between) $A,B,F,G,u,v$.
Perhaps this equation leads to some contradiction?
Taking $y=0$, we get that: $(A(x,0)u+B(x,0)v)(x-alpha)(x-beta)=0$.
This shows that, for every $gamma notin {alpha,beta}$:
$A(gamma,0)u(gamma)+B(gamma,0)v(gamma)=0$.
In other words, the polynomial $A(x,0)u(x)+B(x,0)v(x)$ has infinitely many zeros,
which is impossible unless it is the zero polynomial; is it true that then $A(x,0)=v(x)$ and $B(x,0)=-u(x)$?
Taking $x=alpha$, we get that:
$y=(A(alpha,y)F(alpha,y)+B(alpha,y)G(alpha,y))y$.
Then,
$A(alpha,y)F(alpha,y)+B(alpha,y)G(alpha,y)=1$.
Similarly, taking $x=beta$ we get that:
$A(beta,y)F(beta,y)+B(beta,y)G(beta,y)=1$.
By this question,
$A=a_ny^n+cdots+a_1y+ v tilde{v}=tilde{A}y+vtilde{v}$
and
$B=b_my^m+cdots+b_1y- u tilde{v}=tilde{B}y-utilde{v}$,
for some $tilde{A},tilde{B} in k[x,y]$.
Therefore, $y=((tilde{A}y+vtilde{v})F+(tilde{B}y-utilde{v})G)y+((tilde{A}y+vtilde{v})u+(tilde{B}y-utilde{v})v)(x-alpha)(x-beta)$
becomes
$y=((tilde{A}y+vtilde{v})F+(tilde{B}y-utilde{v})G)y+(tilde{A}yu+tilde{B}yv)(x-alpha)(x-beta)$.
Then, $1=((tilde{A}y+vtilde{v})F+(tilde{B}y-utilde{v})G)+(tilde{A}u+tilde{B}v)(x-alpha)(x-beta)$.
I think that more can be said, and actually $AF+BG=Cy+D$, where $C,D in k[x]$
(by differentiating with respect to $y$); I will add the details later.
(I really apologize if this equation seems a little peculiar, it is just that I have encountered it in a calculation I did, and wish to understand it better).
Any comments are welcome!
algebraic-geometry polynomials commutative-algebra
asked Jan 5 at 22:12
user237522user237522
2,1801617
$endgroup$
Let $A,B,F,G in mathbb{C}[x,y]$, $u,v in mathbb{C}[x]$, $alpha,beta in mathbb{C}$, $alpha neq beta$.
Assume that $u$ and $v$ have no common zero,
and that ${alpha,beta}$ are not zeros of $u$ or $v$.
(I am not sure if I wish to assume that $F$ and $G$ are relatively prime or not).
Further assume that the following equation holds: $y=(AF+BG)y+(Au+Bv)(x-alpha)(x-beta)$.
What can be said about $A,B,F,G,u,v$? I do not know exactly what exactly I am looking for; I wish to obtain 'simpler' (= of a specific form/ with less variables/ a connection between) $A,B,F,G,u,v$.
Perhaps this equation leads to some contradiction?
Taking $y=0$, we get that: $(A(x,0)u+B(x,0)v)(x-alpha)(x-beta)=0$.
This shows that, for every $gamma notin {alpha,beta}$:
$A(gamma,0)u(gamma)+B(gamma,0)v(gamma)=0$.
In other words, the polynomial $A(x,0)u(x)+B(x,0)v(x)$ has infinitely many zeros,
which is impossible unless it is the zero polynomial; is it true that then $A(x,0)=v(x)$ and $B(x,0)=-u(x)$?
Taking $x=alpha$, we get that:
$y=(A(alpha,y)F(alpha,y)+B(alpha,y)G(alpha,y))y$.
Then,
$A(alpha,y)F(alpha,y)+B(alpha,y)G(alpha,y)=1$.
Similarly, taking $x=beta$ we get that:
$A(beta,y)F(beta,y)+B(beta,y)G(beta,y)=1$.
By this question,
$A=a_ny^n+cdots+a_1y+ v tilde{v}=tilde{A}y+vtilde{v}$
and
$B=b_my^m+cdots+b_1y- u tilde{v}=tilde{B}y-utilde{v}$,
for some $tilde{A},tilde{B} in k[x,y]$.
Therefore, $y=((tilde{A}y+vtilde{v})F+(tilde{B}y-utilde{v})G)y+((tilde{A}y+vtilde{v})u+(tilde{B}y-utilde{v})v)(x-alpha)(x-beta)$
becomes
$y=((tilde{A}y+vtilde{v})F+(tilde{B}y-utilde{v})G)y+(tilde{A}yu+tilde{B}yv)(x-alpha)(x-beta)$.
Then, $1=((tilde{A}y+vtilde{v})F+(tilde{B}y-utilde{v})G)+(tilde{A}u+tilde{B}v)(x-alpha)(x-beta)$.
I think that more can be said, and actually $AF+BG=Cy+D$, where $C,D in k[x]$
(by differentiating with respect to $y$); I will add the details later.
(I really apologize if this equation seems a little peculiar, it is just that I have encountered it in a calculation I did, and wish to understand it better).
Any comments are welcome!
algebraic-geometry polynomials commutative-algebra
algebraic-geometry polynomials commutative-algebra
asked Jan 5 at 22:12
user237522user237522
2,1801617
asked Jan 5 at 22:12
user237522user237522
2,1801617
edited Jan 7 at 4:59
user237522
asked Jan 5 at 22:12
user237522user237522
2,1801617
asked Jan 5 at 22:12
user237522user237522
2,1801617
asked Jan 5 at 22:12
user237522user237522
2,1801617
2,1801617
add a comment |
add a comment |
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