Concerning $frac{mathbb{C}[x,y,z]}{langle x^2-1, yz rangle}$
$begingroup$
Let $R=frac{mathbb{C}[x,y,z]}{langle x^2-1, yz rangle}$.
For convenience, I will write $x,y,z in R$ instead of $bar{x},bar{y},bar{z}$.
In $R[X,Y]$, take $A=(x+iy)X+yY$, $B=yX+(x-iy)Y$, $C=zX$.
We have:
$operatorname{Jac}(A,B)=A_XB_Y-A_YB_X=(x+iy)(x-iy)-yy=x^2-i^2y^2-y^2=x^2+y^2-y^2=x^2=1$
$operatorname{Jac}(A,C)=A_XC_Y-A_YC_X=(x+iy)0-yz=-yz=0$.
Is it true that $C notin R[A]$?
I think/hope that the answer is yes (though I may be wrong):
Otherwise, $C in R[A]$ so there exists $h(t) in R[t]$ such that
$C=h(A)$. Since $deg(C)=deg(A)=1$, it follows that $h$ must be of degree $1$,
so $h(t)=lambda t + mu$ for some $lambda, mu in R$.
Therefore, $C=h(A)$ becomes $zX=lambda A + mu= lambda ((x+iy)X+yY)+ mu =lambda(x+iy)X+lambda yY+mu$.
Then we must have:
(i) $z=lambda(x+iy)$; (ii) $0=lambda y$; (iii) $0=mu$.
Now, (ii) implies (if I am not wrong) that $lambda=0$ or $lambda=z$.
But each of these two options contradicts (i):
The first option yields $z=lambda(x+iy)=0(x+iy)=0$,
and the second option yields $z=lambda(x+iy)=z(x+iy)$, so
$z(1-x-iy)=z-z(x+iy)=0$.
(I think that $z(1-x-iy)=z(1-x)-izy=z(1-x)-i0=z(1-x)$ is not zero).
Am I missing something trivial?
Thank you very much!
Motivation: Actually, it seems (but again I may be wrong) that the two-dimensional Jacobian Conjecture in $R[X,Y]$,
$R=frac{mathbb{C}[x,y,z]}{langle x^2-1, yz rangle}$, is false:
Just take $A=(x+iy)X+yY$, $B=yX+(x-iy)Y$.
We have seen above that $operatorname{Jac}(A,B)=1$.
It seems that $R[A,B] subsetneq R[X,Y]$, so $(X,Y) mapsto (A,B)$ is not an automorphism of $R[X,Y]$.
Moreover, if indeed $R[A,B] subsetneq R[X,Y]$, then Proposition 1.1.12 implies that the two-dimensional Jacobian Conjecture in $mathbb{C}[X,Y]$ is false..
It remains to show that $R[A,B] subsetneq R[X,Y]$. I am not sure yet how to prove this (if this is true..). Observe that $zA=z(x+iy)X+zyY=z(x+iy)X$
and $zB=zyX+z(x-iy)Y=z(x-iy)Y$.
It seems impossible to obtain $X$ and $Y$; am I right?
Thank you very much again!
polynomials complex-numbers commutative-algebra jacobian
asked Dec 14 '18 at 1:32
user237522user237522
2,1631617
$endgroup$
add a comment |
$begingroup$
Let $R=frac{mathbb{C}[x,y,z]}{langle x^2-1, yz rangle}$.
For convenience, I will write $x,y,z in R$ instead of $bar{x},bar{y},bar{z}$.
In $R[X,Y]$, take $A=(x+iy)X+yY$, $B=yX+(x-iy)Y$, $C=zX$.
We have:
$operatorname{Jac}(A,B)=A_XB_Y-A_YB_X=(x+iy)(x-iy)-yy=x^2-i^2y^2-y^2=x^2+y^2-y^2=x^2=1$
$operatorname{Jac}(A,C)=A_XC_Y-A_YC_X=(x+iy)0-yz=-yz=0$.
Is it true that $C notin R[A]$?
I think/hope that the answer is yes (though I may be wrong):
Otherwise, $C in R[A]$ so there exists $h(t) in R[t]$ such that
$C=h(A)$. Since $deg(C)=deg(A)=1$, it follows that $h$ must be of degree $1$,
so $h(t)=lambda t + mu$ for some $lambda, mu in R$.
Therefore, $C=h(A)$ becomes $zX=lambda A + mu= lambda ((x+iy)X+yY)+ mu =lambda(x+iy)X+lambda yY+mu$.
Then we must have:
(i) $z=lambda(x+iy)$; (ii) $0=lambda y$; (iii) $0=mu$.
Now, (ii) implies (if I am not wrong) that $lambda=0$ or $lambda=z$.
But each of these two options contradicts (i):
The first option yields $z=lambda(x+iy)=0(x+iy)=0$,
and the second option yields $z=lambda(x+iy)=z(x+iy)$, so
$z(1-x-iy)=z-z(x+iy)=0$.
(I think that $z(1-x-iy)=z(1-x)-izy=z(1-x)-i0=z(1-x)$ is not zero).
Am I missing something trivial?
Thank you very much!
Motivation: Actually, it seems (but again I may be wrong) that the two-dimensional Jacobian Conjecture in $R[X,Y]$,
$R=frac{mathbb{C}[x,y,z]}{langle x^2-1, yz rangle}$, is false:
Just take $A=(x+iy)X+yY$, $B=yX+(x-iy)Y$.
We have seen above that $operatorname{Jac}(A,B)=1$.
It seems that $R[A,B] subsetneq R[X,Y]$, so $(X,Y) mapsto (A,B)$ is not an automorphism of $R[X,Y]$.
Moreover, if indeed $R[A,B] subsetneq R[X,Y]$, then Proposition 1.1.12 implies that the two-dimensional Jacobian Conjecture in $mathbb{C}[X,Y]$ is false..
It remains to show that $R[A,B] subsetneq R[X,Y]$. I am not sure yet how to prove this (if this is true..). Observe that $zA=z(x+iy)X+zyY=z(x+iy)X$
and $zB=zyX+z(x-iy)Y=z(x-iy)Y$.
It seems impossible to obtain $X$ and $Y$; am I right?
Thank you very much again!
polynomials complex-numbers commutative-algebra jacobian
asked Dec 14 '18 at 1:32
user237522user237522
2,1631617
$endgroup$
$begingroup$
Unfortunately, I had an error and $C in R[A]$.. The correct argument is that (ii) implies that $lambda=z epsilon$, for an appropriate $epsilon in R$, for example. $epsilon=x-iy$. Then $lambda=z(x-iy)$ and we get that $lambda A=lambda((x+iy)X+yY)=lambda (x+iy)X+lambda yY=z(x-iy)(x+iy)X+z(x-iy)yY=z(x^2-i^2y^2)X+zy(x-iy)Y=z(x^2+y^2)X+0(x-iy)Y=z(1+y^2)X+0Y=(z+zy^2)X=(z+0)X=zX=C$
$endgroup$
– user237522
Dec 15 '18 at 19:16
$begingroup$
$(X,Y) mapsto (A,B)$ is an automorphism with inverse $(x,y) mapsto ((x-iy)X-yY,-yX+(x+iy)Y)$. Notice that, for example, $X$ is in the image of $(x,y) mapsto (A,B)$, since $X=(x-iy)A-yB$.
$endgroup$
– user237522
Dec 15 '18 at 19:42
add a comment |
$begingroup$
Let $R=frac{mathbb{C}[x,y,z]}{langle x^2-1, yz rangle}$.
For convenience, I will write $x,y,z in R$ instead of $bar{x},bar{y},bar{z}$.
In $R[X,Y]$, take $A=(x+iy)X+yY$, $B=yX+(x-iy)Y$, $C=zX$.
We have:
$operatorname{Jac}(A,B)=A_XB_Y-A_YB_X=(x+iy)(x-iy)-yy=x^2-i^2y^2-y^2=x^2+y^2-y^2=x^2=1$
$operatorname{Jac}(A,C)=A_XC_Y-A_YC_X=(x+iy)0-yz=-yz=0$.
Is it true that $C notin R[A]$?
I think/hope that the answer is yes (though I may be wrong):
Otherwise, $C in R[A]$ so there exists $h(t) in R[t]$ such that
$C=h(A)$. Since $deg(C)=deg(A)=1$, it follows that $h$ must be of degree $1$,
so $h(t)=lambda t + mu$ for some $lambda, mu in R$.
Therefore, $C=h(A)$ becomes $zX=lambda A + mu= lambda ((x+iy)X+yY)+ mu =lambda(x+iy)X+lambda yY+mu$.
Then we must have:
(i) $z=lambda(x+iy)$; (ii) $0=lambda y$; (iii) $0=mu$.
Now, (ii) implies (if I am not wrong) that $lambda=0$ or $lambda=z$.
But each of these two options contradicts (i):
The first option yields $z=lambda(x+iy)=0(x+iy)=0$,
and the second option yields $z=lambda(x+iy)=z(x+iy)$, so
$z(1-x-iy)=z-z(x+iy)=0$.
(I think that $z(1-x-iy)=z(1-x)-izy=z(1-x)-i0=z(1-x)$ is not zero).
Am I missing something trivial?
Thank you very much!
Motivation: Actually, it seems (but again I may be wrong) that the two-dimensional Jacobian Conjecture in $R[X,Y]$,
$R=frac{mathbb{C}[x,y,z]}{langle x^2-1, yz rangle}$, is false:
Just take $A=(x+iy)X+yY$, $B=yX+(x-iy)Y$.
We have seen above that $operatorname{Jac}(A,B)=1$.
It seems that $R[A,B] subsetneq R[X,Y]$, so $(X,Y) mapsto (A,B)$ is not an automorphism of $R[X,Y]$.
Moreover, if indeed $R[A,B] subsetneq R[X,Y]$, then Proposition 1.1.12 implies that the two-dimensional Jacobian Conjecture in $mathbb{C}[X,Y]$ is false..
It remains to show that $R[A,B] subsetneq R[X,Y]$. I am not sure yet how to prove this (if this is true..). Observe that $zA=z(x+iy)X+zyY=z(x+iy)X$
and $zB=zyX+z(x-iy)Y=z(x-iy)Y$.
It seems impossible to obtain $X$ and $Y$; am I right?
Thank you very much again!
polynomials complex-numbers commutative-algebra jacobian
asked Dec 14 '18 at 1:32
user237522user237522
2,1631617
$endgroup$
Let $R=frac{mathbb{C}[x,y,z]}{langle x^2-1, yz rangle}$.
For convenience, I will write $x,y,z in R$ instead of $bar{x},bar{y},bar{z}$.
In $R[X,Y]$, take $A=(x+iy)X+yY$, $B=yX+(x-iy)Y$, $C=zX$.
We have:
$operatorname{Jac}(A,B)=A_XB_Y-A_YB_X=(x+iy)(x-iy)-yy=x^2-i^2y^2-y^2=x^2+y^2-y^2=x^2=1$
$operatorname{Jac}(A,C)=A_XC_Y-A_YC_X=(x+iy)0-yz=-yz=0$.
Is it true that $C notin R[A]$?
I think/hope that the answer is yes (though I may be wrong):
Otherwise, $C in R[A]$ so there exists $h(t) in R[t]$ such that
$C=h(A)$. Since $deg(C)=deg(A)=1$, it follows that $h$ must be of degree $1$,
so $h(t)=lambda t + mu$ for some $lambda, mu in R$.
Therefore, $C=h(A)$ becomes $zX=lambda A + mu= lambda ((x+iy)X+yY)+ mu =lambda(x+iy)X+lambda yY+mu$.
Then we must have:
(i) $z=lambda(x+iy)$; (ii) $0=lambda y$; (iii) $0=mu$.
Now, (ii) implies (if I am not wrong) that $lambda=0$ or $lambda=z$.
But each of these two options contradicts (i):
The first option yields $z=lambda(x+iy)=0(x+iy)=0$,
and the second option yields $z=lambda(x+iy)=z(x+iy)$, so
$z(1-x-iy)=z-z(x+iy)=0$.
(I think that $z(1-x-iy)=z(1-x)-izy=z(1-x)-i0=z(1-x)$ is not zero).
Am I missing something trivial?
Thank you very much!
Motivation: Actually, it seems (but again I may be wrong) that the two-dimensional Jacobian Conjecture in $R[X,Y]$,
$R=frac{mathbb{C}[x,y,z]}{langle x^2-1, yz rangle}$, is false:
Just take $A=(x+iy)X+yY$, $B=yX+(x-iy)Y$.
We have seen above that $operatorname{Jac}(A,B)=1$.
It seems that $R[A,B] subsetneq R[X,Y]$, so $(X,Y) mapsto (A,B)$ is not an automorphism of $R[X,Y]$.
Moreover, if indeed $R[A,B] subsetneq R[X,Y]$, then Proposition 1.1.12 implies that the two-dimensional Jacobian Conjecture in $mathbb{C}[X,Y]$ is false..
It remains to show that $R[A,B] subsetneq R[X,Y]$. I am not sure yet how to prove this (if this is true..). Observe that $zA=z(x+iy)X+zyY=z(x+iy)X$
and $zB=zyX+z(x-iy)Y=z(x-iy)Y$.
It seems impossible to obtain $X$ and $Y$; am I right?
Thank you very much again!
polynomials complex-numbers commutative-algebra jacobian
polynomials complex-numbers commutative-algebra jacobian
asked Dec 14 '18 at 1:32
user237522user237522
2,1631617
asked Dec 14 '18 at 1:32
user237522user237522
2,1631617
edited Dec 14 '18 at 3:12
user237522
asked Dec 14 '18 at 1:32
user237522user237522
2,1631617
asked Dec 14 '18 at 1:32
user237522user237522
2,1631617
asked Dec 14 '18 at 1:32
user237522user237522
2,1631617
2,1631617
$begingroup$
Unfortunately, I had an error and $C in R[A]$.. The correct argument is that (ii) implies that $lambda=z epsilon$, for an appropriate $epsilon in R$, for example. $epsilon=x-iy$. Then $lambda=z(x-iy)$ and we get that $lambda A=lambda((x+iy)X+yY)=lambda (x+iy)X+lambda yY=z(x-iy)(x+iy)X+z(x-iy)yY=z(x^2-i^2y^2)X+zy(x-iy)Y=z(x^2+y^2)X+0(x-iy)Y=z(1+y^2)X+0Y=(z+zy^2)X=(z+0)X=zX=C$
$endgroup$
– user237522
Dec 15 '18 at 19:16
$begingroup$
$(X,Y) mapsto (A,B)$ is an automorphism with inverse $(x,y) mapsto ((x-iy)X-yY,-yX+(x+iy)Y)$. Notice that, for example, $X$ is in the image of $(x,y) mapsto (A,B)$, since $X=(x-iy)A-yB$.
$endgroup$
– user237522
Dec 15 '18 at 19:42
add a comment |
$begingroup$
Unfortunately, I had an error and $C in R[A]$.. The correct argument is that (ii) implies that $lambda=z epsilon$, for an appropriate $epsilon in R$, for example. $epsilon=x-iy$. Then $lambda=z(x-iy)$ and we get that $lambda A=lambda((x+iy)X+yY)=lambda (x+iy)X+lambda yY=z(x-iy)(x+iy)X+z(x-iy)yY=z(x^2-i^2y^2)X+zy(x-iy)Y=z(x^2+y^2)X+0(x-iy)Y=z(1+y^2)X+0Y=(z+zy^2)X=(z+0)X=zX=C$
$endgroup$
– user237522
Dec 15 '18 at 19:16
$begingroup$
$(X,Y) mapsto (A,B)$ is an automorphism with inverse $(x,y) mapsto ((x-iy)X-yY,-yX+(x+iy)Y)$. Notice that, for example, $X$ is in the image of $(x,y) mapsto (A,B)$, since $X=(x-iy)A-yB$.
$endgroup$
– user237522
Dec 15 '18 at 19:42
$begingroup$
Unfortunately, I had an error and $C in R[A]$.. The correct argument is that (ii) implies that $lambda=z epsilon$, for an appropriate $epsilon in R$, for example. $epsilon=x-iy$. Then $lambda=z(x-iy)$ and we get that $lambda A=lambda((x+iy)X+yY)=lambda (x+iy)X+lambda yY=z(x-iy)(x+iy)X+z(x-iy)yY=z(x^2-i^2y^2)X+zy(x-iy)Y=z(x^2+y^2)X+0(x-iy)Y=z(1+y^2)X+0Y=(z+zy^2)X=(z+0)X=zX=C$
$endgroup$
– user237522
Dec 15 '18 at 19:16
$begingroup$
Unfortunately, I had an error and $C in R[A]$.. The correct argument is that (ii) implies that $lambda=z epsilon$, for an appropriate $epsilon in R$, for example. $epsilon=x-iy$. Then $lambda=z(x-iy)$ and we get that $lambda A=lambda((x+iy)X+yY)=lambda (x+iy)X+lambda yY=z(x-iy)(x+iy)X+z(x-iy)yY=z(x^2-i^2y^2)X+zy(x-iy)Y=z(x^2+y^2)X+0(x-iy)Y=z(1+y^2)X+0Y=(z+zy^2)X=(z+0)X=zX=C$
$endgroup$
– user237522
Dec 15 '18 at 19:16
$begingroup$
$(X,Y) mapsto (A,B)$ is an automorphism with inverse $(x,y) mapsto ((x-iy)X-yY,-yX+(x+iy)Y)$. Notice that, for example, $X$ is in the image of $(x,y) mapsto (A,B)$, since $X=(x-iy)A-yB$.
$endgroup$
– user237522
Dec 15 '18 at 19:42
$begingroup$
$(X,Y) mapsto (A,B)$ is an automorphism with inverse $(x,y) mapsto ((x-iy)X-yY,-yX+(x+iy)Y)$. Notice that, for example, $X$ is in the image of $(x,y) mapsto (A,B)$, since $X=(x-iy)A-yB$.
$endgroup$
– user237522
Dec 15 '18 at 19:42
add a comment |
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$begingroup$
Unfortunately, I had an error and $C in R[A]$.. The correct argument is that (ii) implies that $lambda=z epsilon$, for an appropriate $epsilon in R$, for example. $epsilon=x-iy$. Then $lambda=z(x-iy)$ and we get that $lambda A=lambda((x+iy)X+yY)=lambda (x+iy)X+lambda yY=z(x-iy)(x+iy)X+z(x-iy)yY=z(x^2-i^2y^2)X+zy(x-iy)Y=z(x^2+y^2)X+0(x-iy)Y=z(1+y^2)X+0Y=(z+zy^2)X=(z+0)X=zX=C$
$endgroup$
– user237522
Dec 15 '18 at 19:16
$begingroup$
$(X,Y) mapsto (A,B)$ is an automorphism with inverse $(x,y) mapsto ((x-iy)X-yY,-yX+(x+iy)Y)$. Notice that, for example, $X$ is in the image of $(x,y) mapsto (A,B)$, since $X=(x-iy)A-yB$.
$endgroup$
– user237522
Dec 15 '18 at 19:42