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?
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1Hint: use polynomial division – Václav Mordvinov Dec 20 '18 at 17:11
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This is evident $(x,y)$ is notation for ideal generated by $x,y$... – Dec 20 '18 at 18:14
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@VáclavMordvinov but is there a polynômial division in K[x,y]? – roi_saumon Dec 21 '18 at 13:44
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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 $$ \ker ev_{(a,b)} = ev_{(a,b)}^{-1}(0) = \phi^{-1}(ev_{(0,0)}^{-1}(0)) = \phi^{-1}(\ker ev_{(0,0)}) $$
lhf
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