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 $x,y$...
    $endgroup$
    – Math_QED
    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 $x,y$...
    $endgroup$
    – Math_QED
    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

57238




57238








  • 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 $x,y$...
    $endgroup$
    – Math_QED
    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 $x,y$...
    $endgroup$
    – Math_QED
    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 $x,y$...
$endgroup$
– Math_QED
Dec 20 '18 at 18:14






$begingroup$
This is evident $(x,y)$ is notation for ideal generated by $x,y$...
$endgroup$
– Math_QED
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

votes


















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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3047748%2fideal-of-kx-y-need-for-some-precisions%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









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

166k10171396




166k10171396












  • $begingroup$
    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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3047748%2fideal-of-kx-y-need-for-some-precisions%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Aardman Animations

Image
Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com
Read more

Are they similar matrix

Image
0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19
Read more

“minimization” problem in Euclidean space related to orthonormal basis

Image
3 $begingroup$ I have stumbled upon an exercice for second year undegraduate student majoring in economics which I find quite demanding. I have an idea for the solution, but it seems awfully complicated, and I am wondering if there was a simpler way to solve it. This goes as follows: Let $(E, langle, rangle)$ be a euclidean space of dimension $n geq 2$ . 1) For any $x in E$ , show that $||x|| = sqrt{n}$ if and only if there exists $(e_1, dots, e_n)$ an orthonormal basis of $E$ such that $x =e_1+ ldots + e_n$ . 2) For any $(x,y) in E$ , show that $||x|| = sqrt{n}$ , $||y|| = sqrt{frac{n(n+1)(2n+1)}{6}}$ and $langle x,y rangle = frac{n(n+1)}{2}$ if and only if there eixsts $(e_1, ldots, e_n)$ an orthonormal basis of $E$ such that $x = e_1 + ldots + e_n$ and $y = e_1 + 2e_2 + ldots + ne_n$ The first q
Read more