Ideal of K[x,y], need for some precisions.












2












$begingroup$


I was reading this topic to show that $(x-a,y-b)$ is an ideal of $K[x,y]$, where $K$ is a field. The answer suggests to show that $(x-a,y-b)$ is the kernel of the evaluation in $ev_{(a,b)} : K[x,y] rightarrow K$.
I understand that $ev_{(a,b)}(x-a)=0$ and $ev_{(a,b)}(y-b)=0$ so we have the inclusion $(x-a,y-b) subset Ker(ev_{(a,b)})$ but how to show the other inclusion?






ideals







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: use polynomial division
    $endgroup$
    – Václav Mordvinov
    Dec 20 '18 at 17:11










  • $begingroup$
    This is evident $(x,y)$ is notation for ideal generated by $x,y$...
    $endgroup$
    – user370967
    Dec 20 '18 at 18:14












  • $begingroup$
    @VáclavMordvinov but is there a polynômial division in K[x,y]?
    $endgroup$
    – roi_saumon
    Dec 21 '18 at 13:44
















2












$begingroup$


I was reading this topic to show that $(x-a,y-b)$ is an ideal of $K[x,y]$, where $K$ is a field. The answer suggests to show that $(x-a,y-b)$ is the kernel of the evaluation in $ev_{(a,b)} : K[x,y] rightarrow K$.
I understand that $ev_{(a,b)}(x-a)=0$ and $ev_{(a,b)}(y-b)=0$ so we have the inclusion $(x-a,y-b) subset Ker(ev_{(a,b)})$ but how to show the other inclusion?






ideals







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hint: use polynomial division
    $endgroup$
    – Václav Mordvinov
    Dec 20 '18 at 17:11










  • $begingroup$
    This is evident $(x,y)$ is notation for ideal generated by $x,y$...
    $endgroup$
    – user370967
    Dec 20 '18 at 18:14












  • $begingroup$
    @VáclavMordvinov but is there a polynômial division in K[x,y]?
    $endgroup$
    – roi_saumon
    Dec 21 '18 at 13:44














2












2








2





$begingroup$


I was reading this topic to show that $(x-a,y-b)$ is an ideal of $K[x,y]$, where $K$ is a field. The answer suggests to show that $(x-a,y-b)$ is the kernel of the evaluation in $ev_{(a,b)} : K[x,y] rightarrow K$.
I understand that $ev_{(a,b)}(x-a)=0$ and $ev_{(a,b)}(y-b)=0$ so we have the inclusion $(x-a,y-b) subset Ker(ev_{(a,b)})$ but how to show the other inclusion?






ideals







share|cite|improve this question









$endgroup$




I was reading this topic to show that $(x-a,y-b)$ is an ideal of $K[x,y]$, where $K$ is a field. The answer suggests to show that $(x-a,y-b)$ is the kernel of the evaluation in $ev_{(a,b)} : K[x,y] rightarrow K$.
I understand that $ev_{(a,b)}(x-a)=0$ and $ev_{(a,b)}(y-b)=0$ so we have the inclusion $(x-a,y-b) subset Ker(ev_{(a,b)})$ but how to show the other inclusion?






ideals




ideals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 20 '18 at 16:59









roi_saumonroi_saumon

63438




63438








  • 1




    $begingroup$
    Hint: use polynomial division
    $endgroup$
    – Václav Mordvinov
    Dec 20 '18 at 17:11










  • $begingroup$
    This is evident $(x,y)$ is notation for ideal generated by $x,y$...
    $endgroup$
    – user370967
    Dec 20 '18 at 18:14












  • $begingroup$
    @VáclavMordvinov but is there a polynômial division in K[x,y]?
    $endgroup$
    – roi_saumon
    Dec 21 '18 at 13:44














  • 1




    $begingroup$
    Hint: use polynomial division
    $endgroup$
    – Václav Mordvinov
    Dec 20 '18 at 17:11










  • $begingroup$
    This is evident $(x,y)$ is notation for ideal generated by $x,y$...
    $endgroup$
    – user370967
    Dec 20 '18 at 18:14












  • $begingroup$
    @VáclavMordvinov but is there a polynômial division in K[x,y]?
    $endgroup$
    – roi_saumon
    Dec 21 '18 at 13:44








1




1




$begingroup$
Hint: use polynomial division
$endgroup$
– Václav Mordvinov
Dec 20 '18 at 17:11




$begingroup$
Hint: use polynomial division
$endgroup$
– Václav Mordvinov
Dec 20 '18 at 17:11












$begingroup$
This is evident $(x,y)$ is notation for ideal generated by $x,y$...
$endgroup$
– user370967
Dec 20 '18 at 18:14






$begingroup$
This is evident $(x,y)$ is notation for ideal generated by $x,y$...
$endgroup$
– user370967
Dec 20 '18 at 18:14














$begingroup$
@VáclavMordvinov but is there a polynômial division in K[x,y]?
$endgroup$
– roi_saumon
Dec 21 '18 at 13:44




$begingroup$
@VáclavMordvinov but is there a polynômial division in K[x,y]?
$endgroup$
– roi_saumon
Dec 21 '18 at 13:44










1 Answer
1






active

oldest

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0












$begingroup$

Hint:
Write
$ev_{(a,b)} = ev_{(0,0)} circ phi $, where $phi$ is the automorphism of $K[x,y]$ given by $phi(f(x,y))=f(x+a,y+b)$.



Then
$$
ker ev_{(a,b)} = ev_{(a,b)}^{-1}(0) = phi^{-1}(ev_{(0,0)}^{-1}(0)) = phi^{-1}(ker ev_{(0,0)})
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I am not sure I see how to make use of the hint.
    $endgroup$
    – roi_saumon
    Dec 21 '18 at 13:46










  • $begingroup$
    @roi_saumon, see my edited answer
    $endgroup$
    – lhf
    Dec 21 '18 at 15:36












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1 Answer
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1 Answer
1






active

oldest

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active

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active

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0












$begingroup$

Hint:
Write
$ev_{(a,b)} = ev_{(0,0)} circ phi $, where $phi$ is the automorphism of $K[x,y]$ given by $phi(f(x,y))=f(x+a,y+b)$.



Then
$$
ker ev_{(a,b)} = ev_{(a,b)}^{-1}(0) = phi^{-1}(ev_{(0,0)}^{-1}(0)) = phi^{-1}(ker ev_{(0,0)})
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I am not sure I see how to make use of the hint.
    $endgroup$
    – roi_saumon
    Dec 21 '18 at 13:46










  • $begingroup$
    @roi_saumon, see my edited answer
    $endgroup$
    – lhf
    Dec 21 '18 at 15:36
















0












$begingroup$

Hint:
Write
$ev_{(a,b)} = ev_{(0,0)} circ phi $, where $phi$ is the automorphism of $K[x,y]$ given by $phi(f(x,y))=f(x+a,y+b)$.



Then
$$
ker ev_{(a,b)} = ev_{(a,b)}^{-1}(0) = phi^{-1}(ev_{(0,0)}^{-1}(0)) = phi^{-1}(ker ev_{(0,0)})
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I am not sure I see how to make use of the hint.
    $endgroup$
    – roi_saumon
    Dec 21 '18 at 13:46










  • $begingroup$
    @roi_saumon, see my edited answer
    $endgroup$
    – lhf
    Dec 21 '18 at 15:36














0












0








0





$begingroup$

Hint:
Write
$ev_{(a,b)} = ev_{(0,0)} circ phi $, where $phi$ is the automorphism of $K[x,y]$ given by $phi(f(x,y))=f(x+a,y+b)$.



Then
$$
ker ev_{(a,b)} = ev_{(a,b)}^{-1}(0) = phi^{-1}(ev_{(0,0)}^{-1}(0)) = phi^{-1}(ker ev_{(0,0)})
$$






share|cite|improve this answer











$endgroup$



Hint:
Write
$ev_{(a,b)} = ev_{(0,0)} circ phi $, where $phi$ is the automorphism of $K[x,y]$ given by $phi(f(x,y))=f(x+a,y+b)$.



Then
$$
ker ev_{(a,b)} = ev_{(a,b)}^{-1}(0) = phi^{-1}(ev_{(0,0)}^{-1}(0)) = phi^{-1}(ker ev_{(0,0)})
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 21 '18 at 15:36

























answered Dec 20 '18 at 17:52









lhflhf

167k11172404




167k11172404












  • $begingroup$
    I am not sure I see how to make use of the hint.
    $endgroup$
    – roi_saumon
    Dec 21 '18 at 13:46










  • $begingroup$
    @roi_saumon, see my edited answer
    $endgroup$
    – lhf
    Dec 21 '18 at 15:36


















  • $begingroup$
    I am not sure I see how to make use of the hint.
    $endgroup$
    – roi_saumon
    Dec 21 '18 at 13:46










  • $begingroup$
    @roi_saumon, see my edited answer
    $endgroup$
    – lhf
    Dec 21 '18 at 15:36
















$begingroup$
I am not sure I see how to make use of the hint.
$endgroup$
– roi_saumon
Dec 21 '18 at 13:46




$begingroup$
I am not sure I see how to make use of the hint.
$endgroup$
– roi_saumon
Dec 21 '18 at 13:46












$begingroup$
@roi_saumon, see my edited answer
$endgroup$
– lhf
Dec 21 '18 at 15:36




$begingroup$
@roi_saumon, see my edited answer
$endgroup$
– lhf
Dec 21 '18 at 15:36


















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