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Let $K$ be an algebraic closed field, $f$ is an irreducible polynomial in $K[X,Y]$,and $f(0,0)=0$. Denote $A(V(f))$ as the coordinate ring $K[X,Y]/(f)$.

Now I don't konw how to show that $A(V(f))/(X,Y)^n A(V(f))$ is a local ring.

Can any one give me some hints?

Thank you very much.

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    Maximal ideals of your quotient correspond to maximal ideals $\mathfrak{m} \subset K[X, Y]$ containing $(X, Y)^n$ and $f$. Maximal ideals are prime -- can you use that to show that $X, Y \in \mathfrak{m}$?2011-12-04

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Put $A=k[X,Y] \;$ . Consider the ideals $I=(X,Y)^n \subset A \; $ and $J=(f(X,Y) )\subset A$.
The ring you are interested in is $(A/J)/(I.A/J)$ . The trick is to replace it by the isomorphic ring $R:=(A/I)/(J.A/I)$.

[This is permitted because taking the quotient of a ring $A$ successively by two ideals $I,J$ does not depend on the order in which you perform the quotients: think about this!]

But then everything becomes clear: $A/I=k[X,Y]/(X,Y)^n \quad $ is local (see Dylan's comment) and so is any of its quotient, in particular our ring $R$.
(By the way: the irreducibility of $f$ is completely irrelevant.)