Concerning the ring of continuous functions on $mathbb{R}$












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$begingroup$


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







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 21:26









      user26857

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      39.4k124183










      asked Dec 8 '18 at 19:19









      user237522user237522

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