If $langle A,B rangle =langle C,D rangle$, then $langle A-lambda,B-mu rangle = $?
$begingroup$
Let $A,B,C,D in mathbb{C}[x,y]$, with $deg(A),deg(B),deg(C),deg(D) geq 1$.
Assume that the ideal generated by $A$ and $B$ equals the ideal
generated by $C$ and $D$, namely,
$langle A,B rangle =langle C,D rangle$.
Let $lambda,mu in mathbb{C}$.
$langle A-lambda,B-mu rangle = $ ? in terms of $C$ and $D$.
What we know is that $A=F_1C+G_1D$ and $B=F_2C+G_2D$, for some $F_1,G_1,F_2,G_2 in mathbb{C}[x,y]$.
Therefore, $A-lambda=F_1C+G_1D-lambda$ and $B-mu=F_2C+G_2D-mu$, but I guess that this does not tell much..
I do not mind to concentrate on the special case where $C in mathbb{C}[x]$ is separable (= has no multiple roots) and $D=y$, namely:
If $langle A,B rangle =langle f(x),y rangle$ with $f$ separable, then $langle A-lambda,B-mu rangle = $ ? in terms of $f(x)$ and $y$.
Remarks for the special case:
(1) It seems that $langle f(x),y rangle$ with $f$ separable is a radical ideal.
(2) $A=F_1f(x)+G_1y$ and $B=F_2f(x)+G_2y$, for some $F_1,G_1,F_2,G_2 in mathbb{C}[x,y]$.
Take $y=0$ and get: $A(x,0)=F_1(x,0)f(x)$ and $B(x,0)=F_2(x,0)f(x)$.
(This shows that $f(x)$ divides $gcd(A(x,0),B(x,0))$).
Then we can write:
$A=epsilon y+A(x,0)= epsilon y+F_1(x,0)f(x)=epsilon y +uf(x)$
and
$B=delta y+B(x,0)= delta y+F_2(x,0)f(x)=delta y +vf(x)$,
where $epsilon,delta in mathbb{C}[x,y]$ and $u=F_1(x,0),v=F_2(x,0) in mathbb{C}[x]$.
Perhaps similar arguments to the ones presented in the answer to this question (but one has to be careful, since there we had $(y,x-c)$ and here we have $(y,f(x))$) show that $epsilon v- delta u in mathbb{C}$. Therefore, $deg_y(epsilon)=deg_y(delta)$, which is quite a restrictive condition.
Thank you very much!
polynomials commutative-algebra ideals
asked Dec 16 '18 at 21:22
user237522user237522
2,1671617
$endgroup$
add a comment |
$begingroup$
Let $A,B,C,D in mathbb{C}[x,y]$, with $deg(A),deg(B),deg(C),deg(D) geq 1$.
Assume that the ideal generated by $A$ and $B$ equals the ideal
generated by $C$ and $D$, namely,
$langle A,B rangle =langle C,D rangle$.
Let $lambda,mu in mathbb{C}$.
$langle A-lambda,B-mu rangle = $ ? in terms of $C$ and $D$.
What we know is that $A=F_1C+G_1D$ and $B=F_2C+G_2D$, for some $F_1,G_1,F_2,G_2 in mathbb{C}[x,y]$.
Therefore, $A-lambda=F_1C+G_1D-lambda$ and $B-mu=F_2C+G_2D-mu$, but I guess that this does not tell much..
I do not mind to concentrate on the special case where $C in mathbb{C}[x]$ is separable (= has no multiple roots) and $D=y$, namely:
If $langle A,B rangle =langle f(x),y rangle$ with $f$ separable, then $langle A-lambda,B-mu rangle = $ ? in terms of $f(x)$ and $y$.
Remarks for the special case:
(1) It seems that $langle f(x),y rangle$ with $f$ separable is a radical ideal.
(2) $A=F_1f(x)+G_1y$ and $B=F_2f(x)+G_2y$, for some $F_1,G_1,F_2,G_2 in mathbb{C}[x,y]$.
Take $y=0$ and get: $A(x,0)=F_1(x,0)f(x)$ and $B(x,0)=F_2(x,0)f(x)$.
(This shows that $f(x)$ divides $gcd(A(x,0),B(x,0))$).
Then we can write:
$A=epsilon y+A(x,0)= epsilon y+F_1(x,0)f(x)=epsilon y +uf(x)$
and
$B=delta y+B(x,0)= delta y+F_2(x,0)f(x)=delta y +vf(x)$,
where $epsilon,delta in mathbb{C}[x,y]$ and $u=F_1(x,0),v=F_2(x,0) in mathbb{C}[x]$.
Perhaps similar arguments to the ones presented in the answer to this question (but one has to be careful, since there we had $(y,x-c)$ and here we have $(y,f(x))$) show that $epsilon v- delta u in mathbb{C}$. Therefore, $deg_y(epsilon)=deg_y(delta)$, which is quite a restrictive condition.
Thank you very much!
polynomials commutative-algebra ideals
asked Dec 16 '18 at 21:22
user237522user237522
2,1671617
$endgroup$
$begingroup$
My instantaneous thought is that you could use the fact that if you make the vector connecting the point to each vertex, then the three pairs of them will make angles summing to 2π if it's inside the triange, & <2π if it's outside ... & you could probably reduce that criterion to a relation of dotproducts without explicitly computing the angles. ¶ Or that the sum of the three areas will be greater than the area of the triangle if it's outside.
$endgroup$
– AmbretteOrrisey
Dec 17 '18 at 10:54
$begingroup$
@AmbretteOrrisey, thanks for the comment. Please, I did not understand it, could you explain?
$endgroup$
– user237522
Dec 17 '18 at 20:09
add a comment |
$begingroup$
Let $A,B,C,D in mathbb{C}[x,y]$, with $deg(A),deg(B),deg(C),deg(D) geq 1$.
Assume that the ideal generated by $A$ and $B$ equals the ideal
generated by $C$ and $D$, namely,
$langle A,B rangle =langle C,D rangle$.
Let $lambda,mu in mathbb{C}$.
$langle A-lambda,B-mu rangle = $ ? in terms of $C$ and $D$.
What we know is that $A=F_1C+G_1D$ and $B=F_2C+G_2D$, for some $F_1,G_1,F_2,G_2 in mathbb{C}[x,y]$.
Therefore, $A-lambda=F_1C+G_1D-lambda$ and $B-mu=F_2C+G_2D-mu$, but I guess that this does not tell much..
I do not mind to concentrate on the special case where $C in mathbb{C}[x]$ is separable (= has no multiple roots) and $D=y$, namely:
If $langle A,B rangle =langle f(x),y rangle$ with $f$ separable, then $langle A-lambda,B-mu rangle = $ ? in terms of $f(x)$ and $y$.
Remarks for the special case:
(1) It seems that $langle f(x),y rangle$ with $f$ separable is a radical ideal.
(2) $A=F_1f(x)+G_1y$ and $B=F_2f(x)+G_2y$, for some $F_1,G_1,F_2,G_2 in mathbb{C}[x,y]$.
Take $y=0$ and get: $A(x,0)=F_1(x,0)f(x)$ and $B(x,0)=F_2(x,0)f(x)$.
(This shows that $f(x)$ divides $gcd(A(x,0),B(x,0))$).
Then we can write:
$A=epsilon y+A(x,0)= epsilon y+F_1(x,0)f(x)=epsilon y +uf(x)$
and
$B=delta y+B(x,0)= delta y+F_2(x,0)f(x)=delta y +vf(x)$,
where $epsilon,delta in mathbb{C}[x,y]$ and $u=F_1(x,0),v=F_2(x,0) in mathbb{C}[x]$.
Perhaps similar arguments to the ones presented in the answer to this question (but one has to be careful, since there we had $(y,x-c)$ and here we have $(y,f(x))$) show that $epsilon v- delta u in mathbb{C}$. Therefore, $deg_y(epsilon)=deg_y(delta)$, which is quite a restrictive condition.
Thank you very much!
polynomials commutative-algebra ideals
asked Dec 16 '18 at 21:22
user237522user237522
2,1671617
$endgroup$
Let $A,B,C,D in mathbb{C}[x,y]$, with $deg(A),deg(B),deg(C),deg(D) geq 1$.
Assume that the ideal generated by $A$ and $B$ equals the ideal
generated by $C$ and $D$, namely,
$langle A,B rangle =langle C,D rangle$.
Let $lambda,mu in mathbb{C}$.
$langle A-lambda,B-mu rangle = $ ? in terms of $C$ and $D$.
What we know is that $A=F_1C+G_1D$ and $B=F_2C+G_2D$, for some $F_1,G_1,F_2,G_2 in mathbb{C}[x,y]$.
Therefore, $A-lambda=F_1C+G_1D-lambda$ and $B-mu=F_2C+G_2D-mu$, but I guess that this does not tell much..
I do not mind to concentrate on the special case where $C in mathbb{C}[x]$ is separable (= has no multiple roots) and $D=y$, namely:
If $langle A,B rangle =langle f(x),y rangle$ with $f$ separable, then $langle A-lambda,B-mu rangle = $ ? in terms of $f(x)$ and $y$.
Remarks for the special case:
(1) It seems that $langle f(x),y rangle$ with $f$ separable is a radical ideal.
(2) $A=F_1f(x)+G_1y$ and $B=F_2f(x)+G_2y$, for some $F_1,G_1,F_2,G_2 in mathbb{C}[x,y]$.
Take $y=0$ and get: $A(x,0)=F_1(x,0)f(x)$ and $B(x,0)=F_2(x,0)f(x)$.
(This shows that $f(x)$ divides $gcd(A(x,0),B(x,0))$).
Then we can write:
$A=epsilon y+A(x,0)= epsilon y+F_1(x,0)f(x)=epsilon y +uf(x)$
and
$B=delta y+B(x,0)= delta y+F_2(x,0)f(x)=delta y +vf(x)$,
where $epsilon,delta in mathbb{C}[x,y]$ and $u=F_1(x,0),v=F_2(x,0) in mathbb{C}[x]$.
Perhaps similar arguments to the ones presented in the answer to this question (but one has to be careful, since there we had $(y,x-c)$ and here we have $(y,f(x))$) show that $epsilon v- delta u in mathbb{C}$. Therefore, $deg_y(epsilon)=deg_y(delta)$, which is quite a restrictive condition.
Thank you very much!
polynomials commutative-algebra ideals
polynomials commutative-algebra ideals
asked Dec 16 '18 at 21:22
user237522user237522
2,1671617
asked Dec 16 '18 at 21:22
user237522user237522
2,1671617
edited Dec 17 '18 at 10:42
user237522
asked Dec 16 '18 at 21:22
user237522user237522
2,1671617
asked Dec 16 '18 at 21:22
user237522user237522
2,1671617
asked Dec 16 '18 at 21:22
user237522user237522
2,1671617
2,1671617
$begingroup$
My instantaneous thought is that you could use the fact that if you make the vector connecting the point to each vertex, then the three pairs of them will make angles summing to 2π if it's inside the triange, & <2π if it's outside ... & you could probably reduce that criterion to a relation of dotproducts without explicitly computing the angles. ¶ Or that the sum of the three areas will be greater than the area of the triangle if it's outside.
$endgroup$
– AmbretteOrrisey
Dec 17 '18 at 10:54
$begingroup$
@AmbretteOrrisey, thanks for the comment. Please, I did not understand it, could you explain?
$endgroup$
– user237522
Dec 17 '18 at 20:09
add a comment |
$begingroup$
My instantaneous thought is that you could use the fact that if you make the vector connecting the point to each vertex, then the three pairs of them will make angles summing to 2π if it's inside the triange, & <2π if it's outside ... & you could probably reduce that criterion to a relation of dotproducts without explicitly computing the angles. ¶ Or that the sum of the three areas will be greater than the area of the triangle if it's outside.
$endgroup$
– AmbretteOrrisey
Dec 17 '18 at 10:54
$begingroup$
@AmbretteOrrisey, thanks for the comment. Please, I did not understand it, could you explain?
$endgroup$
– user237522
Dec 17 '18 at 20:09
$begingroup$
My instantaneous thought is that you could use the fact that if you make the vector connecting the point to each vertex, then the three pairs of them will make angles summing to 2π if it's inside the triange, & <2π if it's outside ... & you could probably reduce that criterion to a relation of dotproducts without explicitly computing the angles. ¶ Or that the sum of the three areas will be greater than the area of the triangle if it's outside.
$endgroup$
– AmbretteOrrisey
Dec 17 '18 at 10:54
$begingroup$
My instantaneous thought is that you could use the fact that if you make the vector connecting the point to each vertex, then the three pairs of them will make angles summing to 2π if it's inside the triange, & <2π if it's outside ... & you could probably reduce that criterion to a relation of dotproducts without explicitly computing the angles. ¶ Or that the sum of the three areas will be greater than the area of the triangle if it's outside.
$endgroup$
– AmbretteOrrisey
Dec 17 '18 at 10:54
$begingroup$
@AmbretteOrrisey, thanks for the comment. Please, I did not understand it, could you explain?
$endgroup$
– user237522
Dec 17 '18 at 20:09
$begingroup$
@AmbretteOrrisey, thanks for the comment. Please, I did not understand it, could you explain?
$endgroup$
– user237522
Dec 17 '18 at 20:09
add a comment |
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$begingroup$
My instantaneous thought is that you could use the fact that if you make the vector connecting the point to each vertex, then the three pairs of them will make angles summing to 2π if it's inside the triange, & <2π if it's outside ... & you could probably reduce that criterion to a relation of dotproducts without explicitly computing the angles. ¶ Or that the sum of the three areas will be greater than the area of the triangle if it's outside.
$endgroup$
– AmbretteOrrisey
Dec 17 '18 at 10:54
$begingroup$
@AmbretteOrrisey, thanks for the comment. Please, I did not understand it, could you explain?
$endgroup$
– user237522
Dec 17 '18 at 20:09