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 $\overline{C} \setminus C$.
Isomorphic affine variety induces isomorphism between its projective closure
algebraic-geometrygeneral-topology
Related Solutions
One can show that if the dimension of the affine variety is positive (equivalently, if it is infinite in the sense of our question) then the projective closure is strictly larger.
There are probably lots of ways to prove this, but one is by a version of Noether normalization. If you would like a detailed proof, let me know by a comment and I can give one.
(Another way to phrase this result is to say that a variety that is simultaneously affine and projective is necessarily finite. Scheme-theory mavens will recognize this as a special case of the more general statement that a morphism which is simultaneously affine and proper is finite.)
Added: Here is a sketch of a proof, as promised:
Let me begin with Noether normalization in a geometric form. Fix an afffine variety $V$ contained in $\mathbb A^n$, with projective closue $\overline{V}$. (Here and below I am always working over an algebraically closed field $k$.)
Assuming that $V$ is not all of $\mathbb A^n$, we see that $\overline{V}$ does not contain the hyperplane at infinity, and so we may choose a point $P$ lying in the hyperplane at infinity, but not lying in $\overline{V}$. We may also choose a different hyperplane $H$ (i.e. not the hyperplane at infinity) which doesn't contain $P$.
With $P$ and $H$ in hand, we may define the projection map $\pi: \mathbb P^n \setminus P \to H$, which maps any $Q \neq P$ to the intersection of the line $\ell$ joining $P$ and $Q$ with the hyperplane $H$. Restricting $\pi$ to $\overline{V}$, we obtain a map $V \to H$.
Now since $P$ is not contained in $\overline{V}$, none of the lines $\ell$ appearing in the projection map are contained in $V$, and so each of them meets $\overline{V}$ in only finitely many points. In particular, if $\overline{V}$ is infinite, so is its image under $\pi$.
Now by elimination theory, i.e. the fact that projective varieties are proper, we know that $\pi(\overline{V})$ is closed in $H$, i.e. is a projective variety in $H$, which is a projective space of dimension $n-1$.
Also, our choice of $P$ ensures that for any $Q \in \overline{V}$, the image $\pi(Q)$ lies at infinity if and only if $Q$ itself does. So $\pi(V)$ is an affine variety (in the affine space $H \cap \mathbb A^n$ of dimension $n-1$), and $\pi(\overline{V})$ is its projective closure.
What I have just done is prove Noether normalization, in a geometric form.
The result we want now follows immeidately, by induction on $n$. (Basically, the case when $V = \mathbb A^n$ is clear, and if $V$ is not all of $\mathbb A^n$, the preceding argument allows us to reduce the dimension of the ambient affine space by one.)
If the map from $X$ to the maxspec of $\mathcal O(X)$ is a bijection, then $X$ is indeed affine.
Here is an argument:
By assumption $X \to $ maxspec $\mathcal O(X)$ is bijective, thus quasi-finite, and so by (Grothendieck's form of) Zariski's main theorem, this map factors as an open embedding of $X$ into a variety that is finite over maxspec $\mathcal O(X)$. Any variety finite over an affine variety is again affine, and hence $X$ is an open subset of an affine variety, i.e. quasi-affine. So we are reduced to considering the case when $X$ is quasi-affine, which is well-known and straightforward.
(I'm not sure that the full strength of ZMT is needed, but it is a natural tool to exploit to get mileage out of the assumption of a morphism having finite fibres, which is what your bijectivity hypothesis gives.)
In fact, the argument shows something stronger: suppose that we just assume that the morphism $X \to $ maxspec $\mathcal O(X)$ has finite non-empty fibres, i.e. is quasi-finite and surjective. Then the same argument with ZMT shows that $X$ is quasi-affine. But it is standard that the map $X \to $ maxspec $\mathcal O(X)$ is an open immersion when $X$ is quasi-affine, and since by assumption it is surjecive, it is an isomorphism.
Note that if we omit one of the hypotheses of surjectivity or quasi-finiteness, we can find a non-affine $X$ satisfying the other hypothesis.
E.g. if $X = \mathbb A^2 \setminus \{0\}$ (the basic example of a quasi-affine, but non-affine, variety), then maxspec $\mathcal O(X) = \mathbb A^2$, and the open immersion $X \to \mathbb A^2$ is evidently not surjective.
E.g. if $X = \mathbb A^2$ blown up at $0$, then maxspec $\mathcal O(X) = \mathbb A^2$, and $X \to \mathbb A^2$ is surjective, but has an infinite fibre over $0$.
Caveat/correction: I should add the following caveat, namely that it is not always true, for a variety $X$ over a field $k$, that $\mathcal O(X)$ is finitely generated over $k$, in which case maxspec may not be such a good construction to apply, and the above argument may not go through. So in order to conclude that $X$ is affine, one should first insist that $\mathcal O(X)$ is finitely generated over $k$, and then that futhermore the natural map $X \to $ maxspec $\mathcal O(X)$ is quasi-finite and surjective.
(Of course, one could work more generally with arbitrary schemes and Spec rather than maxspec, but I haven't thought about this general setting: in particular, ZMT requires some finiteness hypotheses, and I haven't thought about what conditions might guarantee that the map $X \to $ Spec $\mathcal O(X)$ satisfies them.)
Incidentally, for an example of a quasi-projective variety with non-finitely generated ring of regular functions, see this note of Ravi Vakil's
Best Answer
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.