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Let $k$ be a field. If necessary, add assumptions on $k$ or just take $k=\mathbb{C}$.

It is easy to classify the ideals $I \subseteq k[x]$ such that $k[x]/I \to k[[x]]/(I)$ is an isomorphism, namely $I=(x^n)$ for some $n \in \mathbb{N}$.

Is it also possible to classify those ideals $I \subseteq k[x,y]$ such that $k[x,y]/I \to k[[x,y]]/(I)$ is an isomorphism? Clearly, $I=(x,y)^n$ is an example.

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Doesn't $I = (x^n, y^m)$ work ? In $k[[x]]$, the only ideals are of the form $I = (x^n)$. So maybe it is simpler to look at the problem backwards : Pick an ideal in $k[[x,y]]$, and intersect it with $k[x,y]$. –  Joel Cohen Nov 27 '12 at 11:42

1 Answer 1

up vote 5 down vote accepted

Let $I$ be such an ideal. If $I\not\subset (x,y)$, then $(I)=k[[x,y]]$, so $k[x,y]/I=0$ and $I=k[x,y]$.

Suppose now $I\subseteq (x,y)$. As $k[[x,y]]/I$ is a local ring, being quotient of a local ring, $k[x,y]/I$ is also local. So $V(I)$ is one point and is necessarily equal to $(0,0)$ as $(0,0)\in V(I)$. Therefore by Hilbert's Nullstellensatz, $\sqrt{I}=(x,y)$. Conversely, if $\sqrt{I}=(x,y)$, then $I$ contains a power of $(x,y)$, and $k[x,y]/I$ is Artian and noetherian, hence complete.

Conclusion, $k[x,y]/I\to k[[x,y]]/(I)$ is an isomorphism if and only if $\sqrt{I}\supseteq (x,y)$.

Edit 2: More generally, let $A$ be a (commutative) noetherian ring, $\hat{A}$ the completion of $A$ with respect to a maximal ideal $ \mathfrak m$, let $I$ be an ideal of $A$. Then the canonical homomorphism $A/I\to \hat{A}/I\hat{A}$ is an isomorphism if $\sqrt{I}\supseteq \mathfrak m$. The proof is exactly the same. On the other hand, the isomorphism implies that either $I=A$ or $\mathfrak m$ is the only maximal ideal containing $I$.

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Thanks a lot. When $R$ is an artinian ring with an ideal $J$, then the $J$-adic completion is $R/J^n$ for $n$ big enough. But in our case $R=k[x,y]/I$, $J=(x,y)$, we have $J^n = 0$ for $n$ big enough, so $R$ is complete. I hope this explains your conclusion "hence complete". Your proof works more generally with $k[[x_1,\dotsc,x_d]]$. At least if $k$ is algebraically closed, is this really necessary? –  Martin Brandenburg Nov 27 '12 at 19:45
    
One can say it is complete because the $(x,y)$-adic topology is discrete on the quotient. –  user18119 Nov 27 '12 at 20:49
    
How do you proof the more general statement? After all, we cannot apply Hilbert's Nullstellensatz. Perhaps we get $\mathfrak{m} \subseteq \mathrm{jac}(I)$ instead? –  Martin Brandenburg Nov 27 '12 at 21:13
    
You are right. So let's go back to $A=k[x_1,...,x_d]$. Nullstellensatz doesn't require $k$ to be algebraically closed in its form $\sqrt{I}=$ jacobson radical of $I$. –  user18119 Nov 27 '12 at 21:21
    
Yes, or in your general statement we just have to replace $\sqrt{I}$ by $\mathrm{jac}(I)$. For jacobson rings $A$ they coincide. –  Martin Brandenburg Nov 27 '12 at 21:31

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