Suppose that $B,C \subset A^{n} \subset P^{n}$ be an affine variety (irreducible closed set in Zariski closure) which are isomorphic, where $A^{n}$ and $P^{n}$ denotes an affine $n$-space and projective $n$-space over algebraically closed field. Then, are $\overline{B}$ and $\overline{C}$, projective closures in $P^{n},$ isomorphic as a projective variety? Can we deduce exact map from $B \cong C$ between $\overline{B}$ and $\overline{C}?$ Honestly, I don't know how to find a map between $\overline{B}\setminus B$ to $\overline{C} \setminus C$.
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It is possible that $B,C$ are isomorphic but their closures are not. Pick coordinates $x,y$ on $\Bbb A^2$, and let $B=V(y)\subset \Bbb A^2$ and $C=V(y-x^3)\subset \Bbb A^2$. These are both isomorphic to $\Bbb A^1$, as they both have coordinate algebras $k[x]$. Their projective closures are $V(y)\subset \Bbb P^2$ and $V(yz^2-x^3)\subset \Bbb P^2$. This first variety is a copy of $\Bbb P^1$ and thus smooth, while the second is singular at $[0:1:0]$, thus they cannot be isomorphic.
KReiser
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That is a brilliant example! Thank you very much! – user124697 Apr 27 '20 at 19:07