Concerning the ring of continuous functions on $mathbb{R}$
$begingroup$
It is not difficult to check that the set of continuous functions from $mathbb{R}$ to $mathbb{R}$ is a ring (an $mathbb{R}$-algebra),
and similarly (if I am not wrong), the set of continuous functions from $[a,b]$ to $mathbb{R}$ is a ring (an $mathbb{R}$-algebra), where $a,b in mathbb{R}$, $a < b$,
denote it by $R_{[a,b]}$.
Let $R_{[0,1]}[X,Y]$ be the polynomial ring in $X$ and $Y$ over $R_{[0,1]}$.
Denote, for all $t in [0,1]$: $f(t)=sqrt{t}$, $g(t)=t$, $h(t)=1$, and let $A=fX+gY$, $W=hX+fY$
We have, $operatorname{Jac}(A,W):=A_xW_y-A_yW_x=f^2-gh=t-t=0$.
Here, $W notin R_{[0,1]}[A]$, since $W=hX+fY=frac{1}{f}(fX+gY)=frac{1}{f}A notin R[A]$ (since $frac{1}{f}(t)=frac{1}{sqrt{t}}$ is not defined for $0$).
However, $A$ does not have a Jacobian mate in $R_{[0,1]}[X,Y]$, since for every $B in R_{[0,1]}[X,Y]$:
$operatorname{Jac}(A,B)=fB_y-gB_x=fB_y-f^2B_x=f(B_y-fB_x) notin R_{[0,1]}^{times}$.
First question: Does there exist $A,B,W in R_{[0,1]}[X,Y]$ satisfting the following 3 conditions:
(1) $operatorname{Jac}(A,W)=0$.
(2) $W notin R_{[0,1]}[A]$.
(3) $operatorname{Jac}(A,B) in R_{[0,1]}^{times}$.
Perhaps Weierstrass function may help? Perhaps trigonometric functions may help?
It seems plausible that $A$ may not be linear, namely, $A$ will have total degree $>1$.
Second question: What if we consider complex (analytic) functions instead of real continuous functions?
Third question, general case: Replace those rings of functions by other rings.
The general case has one plausible easy answer, thanks to Warning 1.1.17, adjusted to my question: Let $R:=frac{mathbb{Z}[t]}{(2t)}$, $R[X,Y]$, $A=X-bar{t}X^2$, $B=Y$, $W=X$.
We have:
(1) $operatorname{Jac}(A,W)=(1-2bar{t}X)0-01=10-01=0$.
(2) $W=X notin R[X-bar{t}X^2]$ from considerations of degrees.
(3) $operatorname{Jac}(A,B)=1$.
Therefore, in the general case we should add a fourth condition:
(4) $R$ is a $mathbb{Q}$-algebra.
Any hints and comments are welcome!
real-analysis complex-analysis commutative-algebra jacobian real-algebraic-geometry
edited Dec 8 '18 at 21:26
user26857
39.4k124183
asked Dec 8 '18 at 19:19
user237522user237522
2,1691617
$endgroup$
add a comment |
$begingroup$
It is not difficult to check that the set of continuous functions from $mathbb{R}$ to $mathbb{R}$ is a ring (an $mathbb{R}$-algebra),
and similarly (if I am not wrong), the set of continuous functions from $[a,b]$ to $mathbb{R}$ is a ring (an $mathbb{R}$-algebra), where $a,b in mathbb{R}$, $a < b$,
denote it by $R_{[a,b]}$.
Let $R_{[0,1]}[X,Y]$ be the polynomial ring in $X$ and $Y$ over $R_{[0,1]}$.
Denote, for all $t in [0,1]$: $f(t)=sqrt{t}$, $g(t)=t$, $h(t)=1$, and let $A=fX+gY$, $W=hX+fY$
We have, $operatorname{Jac}(A,W):=A_xW_y-A_yW_x=f^2-gh=t-t=0$.
Here, $W notin R_{[0,1]}[A]$, since $W=hX+fY=frac{1}{f}(fX+gY)=frac{1}{f}A notin R[A]$ (since $frac{1}{f}(t)=frac{1}{sqrt{t}}$ is not defined for $0$).
However, $A$ does not have a Jacobian mate in $R_{[0,1]}[X,Y]$, since for every $B in R_{[0,1]}[X,Y]$:
$operatorname{Jac}(A,B)=fB_y-gB_x=fB_y-f^2B_x=f(B_y-fB_x) notin R_{[0,1]}^{times}$.
First question: Does there exist $A,B,W in R_{[0,1]}[X,Y]$ satisfting the following 3 conditions:
(1) $operatorname{Jac}(A,W)=0$.
(2) $W notin R_{[0,1]}[A]$.
(3) $operatorname{Jac}(A,B) in R_{[0,1]}^{times}$.
Perhaps Weierstrass function may help? Perhaps trigonometric functions may help?
It seems plausible that $A$ may not be linear, namely, $A$ will have total degree $>1$.
Second question: What if we consider complex (analytic) functions instead of real continuous functions?
Third question, general case: Replace those rings of functions by other rings.
The general case has one plausible easy answer, thanks to Warning 1.1.17, adjusted to my question: Let $R:=frac{mathbb{Z}[t]}{(2t)}$, $R[X,Y]$, $A=X-bar{t}X^2$, $B=Y$, $W=X$.
We have:
(1) $operatorname{Jac}(A,W)=(1-2bar{t}X)0-01=10-01=0$.
(2) $W=X notin R[X-bar{t}X^2]$ from considerations of degrees.
(3) $operatorname{Jac}(A,B)=1$.
Therefore, in the general case we should add a fourth condition:
(4) $R$ is a $mathbb{Q}$-algebra.
Any hints and comments are welcome!
real-analysis complex-analysis commutative-algebra jacobian real-algebraic-geometry
edited Dec 8 '18 at 21:26
user26857
39.4k124183
asked Dec 8 '18 at 19:19
user237522user237522
2,1691617
$endgroup$
add a comment |
$begingroup$
It is not difficult to check that the set of continuous functions from $mathbb{R}$ to $mathbb{R}$ is a ring (an $mathbb{R}$-algebra),
and similarly (if I am not wrong), the set of continuous functions from $[a,b]$ to $mathbb{R}$ is a ring (an $mathbb{R}$-algebra), where $a,b in mathbb{R}$, $a < b$,
denote it by $R_{[a,b]}$.
Let $R_{[0,1]}[X,Y]$ be the polynomial ring in $X$ and $Y$ over $R_{[0,1]}$.
Denote, for all $t in [0,1]$: $f(t)=sqrt{t}$, $g(t)=t$, $h(t)=1$, and let $A=fX+gY$, $W=hX+fY$
We have, $operatorname{Jac}(A,W):=A_xW_y-A_yW_x=f^2-gh=t-t=0$.
Here, $W notin R_{[0,1]}[A]$, since $W=hX+fY=frac{1}{f}(fX+gY)=frac{1}{f}A notin R[A]$ (since $frac{1}{f}(t)=frac{1}{sqrt{t}}$ is not defined for $0$).
However, $A$ does not have a Jacobian mate in $R_{[0,1]}[X,Y]$, since for every $B in R_{[0,1]}[X,Y]$:
$operatorname{Jac}(A,B)=fB_y-gB_x=fB_y-f^2B_x=f(B_y-fB_x) notin R_{[0,1]}^{times}$.
First question: Does there exist $A,B,W in R_{[0,1]}[X,Y]$ satisfting the following 3 conditions:
(1) $operatorname{Jac}(A,W)=0$.
(2) $W notin R_{[0,1]}[A]$.
(3) $operatorname{Jac}(A,B) in R_{[0,1]}^{times}$.
Perhaps Weierstrass function may help? Perhaps trigonometric functions may help?
It seems plausible that $A$ may not be linear, namely, $A$ will have total degree $>1$.
Second question: What if we consider complex (analytic) functions instead of real continuous functions?
Third question, general case: Replace those rings of functions by other rings.
The general case has one plausible easy answer, thanks to Warning 1.1.17, adjusted to my question: Let $R:=frac{mathbb{Z}[t]}{(2t)}$, $R[X,Y]$, $A=X-bar{t}X^2$, $B=Y$, $W=X$.
We have:
(1) $operatorname{Jac}(A,W)=(1-2bar{t}X)0-01=10-01=0$.
(2) $W=X notin R[X-bar{t}X^2]$ from considerations of degrees.
(3) $operatorname{Jac}(A,B)=1$.
Therefore, in the general case we should add a fourth condition:
(4) $R$ is a $mathbb{Q}$-algebra.
Any hints and comments are welcome!
real-analysis complex-analysis commutative-algebra jacobian real-algebraic-geometry
edited Dec 8 '18 at 21:26
user26857
39.4k124183
asked Dec 8 '18 at 19:19
user237522user237522
2,1691617
$endgroup$
It is not difficult to check that the set of continuous functions from $mathbb{R}$ to $mathbb{R}$ is a ring (an $mathbb{R}$-algebra),
and similarly (if I am not wrong), the set of continuous functions from $[a,b]$ to $mathbb{R}$ is a ring (an $mathbb{R}$-algebra), where $a,b in mathbb{R}$, $a < b$,
denote it by $R_{[a,b]}$.
Let $R_{[0,1]}[X,Y]$ be the polynomial ring in $X$ and $Y$ over $R_{[0,1]}$.
Denote, for all $t in [0,1]$: $f(t)=sqrt{t}$, $g(t)=t$, $h(t)=1$, and let $A=fX+gY$, $W=hX+fY$
We have, $operatorname{Jac}(A,W):=A_xW_y-A_yW_x=f^2-gh=t-t=0$.
Here, $W notin R_{[0,1]}[A]$, since $W=hX+fY=frac{1}{f}(fX+gY)=frac{1}{f}A notin R[A]$ (since $frac{1}{f}(t)=frac{1}{sqrt{t}}$ is not defined for $0$).
However, $A$ does not have a Jacobian mate in $R_{[0,1]}[X,Y]$, since for every $B in R_{[0,1]}[X,Y]$:
$operatorname{Jac}(A,B)=fB_y-gB_x=fB_y-f^2B_x=f(B_y-fB_x) notin R_{[0,1]}^{times}$.
First question: Does there exist $A,B,W in R_{[0,1]}[X,Y]$ satisfting the following 3 conditions:
(1) $operatorname{Jac}(A,W)=0$.
(2) $W notin R_{[0,1]}[A]$.
(3) $operatorname{Jac}(A,B) in R_{[0,1]}^{times}$.
Perhaps Weierstrass function may help? Perhaps trigonometric functions may help?
It seems plausible that $A$ may not be linear, namely, $A$ will have total degree $>1$.
Second question: What if we consider complex (analytic) functions instead of real continuous functions?
Third question, general case: Replace those rings of functions by other rings.
The general case has one plausible easy answer, thanks to Warning 1.1.17, adjusted to my question: Let $R:=frac{mathbb{Z}[t]}{(2t)}$, $R[X,Y]$, $A=X-bar{t}X^2$, $B=Y$, $W=X$.
We have:
(1) $operatorname{Jac}(A,W)=(1-2bar{t}X)0-01=10-01=0$.
(2) $W=X notin R[X-bar{t}X^2]$ from considerations of degrees.
(3) $operatorname{Jac}(A,B)=1$.
Therefore, in the general case we should add a fourth condition:
(4) $R$ is a $mathbb{Q}$-algebra.
Any hints and comments are welcome!
real-analysis complex-analysis commutative-algebra jacobian real-algebraic-geometry
real-analysis complex-analysis commutative-algebra jacobian real-algebraic-geometry
edited Dec 8 '18 at 21:26
user26857
39.4k124183
asked Dec 8 '18 at 19:19
user237522user237522
2,1691617
edited Dec 8 '18 at 21:26
user26857
39.4k124183
asked Dec 8 '18 at 19:19
user237522user237522
2,1691617
edited Dec 8 '18 at 21:26
user26857
39.4k124183
edited Dec 8 '18 at 21:26
user26857
39.4k124183
edited Dec 8 '18 at 21:26
user26857
39.4k124183
39.4k124183
asked Dec 8 '18 at 19:19
user237522user237522
2,1691617
asked Dec 8 '18 at 19:19
user237522user237522
2,1691617
asked Dec 8 '18 at 19:19
user237522user237522
2,1691617
2,1691617
add a comment |
add a comment |
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