# Are complements of non-isomorphic affine hypersurfaces necessarily non-isomorphic?

Fix $n$ and let $H_1$ and $H_2$ be two hypersurfaces in ${\bf A}^n$ (not necessarily smooth or irreducible, but we'll assume reduced). If the complements $U_i = {\bf A}^n \setminus H_i$ are isomorphic as schemes, does this imply that the $H_i$ are isomorphic?

From asking other people, the answer is yes via Euler characteristic arguments when the $H_i$ are smooth, but I have in mind singular examples.

As for the ground field, the polynomials I have in mind are defined over ${\bf Z}$, so choose whatever field you like. Probably I'd prefer an answer for the complex numbers though.

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Your question appears as the 'Complement problem' in Hanspeter Kraft's article Challenging problems on affine n-space and it seems from that article that the problem is still open for $H_1, H_2$ irreducible. Even the case of $n=2$ is non-trivial. In this case, the article mentions a result of Kraft and Vust saying that the curves $H_1$ and $H_2$ in $A^2$ are isomorphic if their genus is $\ge 1$.

In the case where $H_1$ and $H_2$ are reducible however, I suspect that there are many counterexamples. I.e., take $H_1=Z(xy(x-y))$ and $H_2=Z(xy(y-1))$, the complements of these curves are isomorphic.

EDIT: A better example, due to Vust:

Take $C_1=Z(y^2-x^3)$ and $C_2=Z(y)$ and let $f:A^2\to A^2$ be the map $f(x,y)=(xy,y)$. Then $f^{-1}(C_1\cup C_2) = C_2\cup C$ where $C$ is the curve $Z(x^3y-1)$, which is disjoint from $C_2$. It follows that $f$ induces an isomorphism $$A^2\setminus C_1\cup C_2\simeq A^2\setminus C\cup C_2.$$

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This is unfortunate, but good to know. Thanks. – Steven Sam Dec 7 '10 at 3:31

The paper of Jeremy Blanc, "The correspondence between a plane curve and its complement" (Crelle 2009) shows that two irreducible plane curves can have isomorphic complements in P^2 without being isomorphic. I would look at his argument and see if you can arrange for the affine complements to be isomorphic as well (if indeed you really care about the distinction between A^n and P^n here.)

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