Simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytifications If they are not proper, two complex algebraic varieties can be nonisomorphic yet have isomorphic analytifications.  I've heard informal examples (often involving moduli spaces), but am not sure of the references.  

What are the simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytificaitons?

By "simplest", I mean by one of the following measures.


*

*(best)  An example whose proof is as elementary as possible, and ideally short.  This of course requires proof that the complex algebraic varieties are nonisomorphic, and that the analytifications are isomorphic.

*A known example that is simple to state, but may have a complicated proof.  (Ideally there should be a reference.)

*An expected, folklore, or conjectured example.  
 A: I believe the following is an elementary example: Let $X$ be an affine smooth curve of geometric genus at least one. Let $L$ be a non-trivial algebraic line bundle
on $X$ (easy to produce such things). Then $L$ is analytically trivial because $X$ is a Stein space ($H^1(X, O_X)=0$) with trivial integral $H^2$. Hence, there is an
analytic isomorphism $L\simeq X\times A^1$. We see that there is no algebraic isomorphism
by noting:

Suppose there is an isomorphism $$f: L\simeq X\times A^1$$
of algebraic varieties. Then $L$ and $X\times A^1$ are isomorphic as algebraic line bundles.

Proof: For any fiber $L_y$ of $L$, if we consider the composite $$p\circ f: L_y\rightarrow X\times A^1 \stackrel{p}{\rightarrow} X$$
of $f$ with the projection, it must be constant, since $X$ is not rational. Hence, we have
$$f(L_y)\subset z\times A^1$$ for some point $z$. Since the map $L_y\rightarrow A^1$ thus obtained is injective,
it must be of the form $ax+b$ for non-zero $a$, that is, $f$ induces an isomorphism
$$L_y\simeq z\times A^1.$$ Also by injectivity, we see that $y\neq y'$ implies
$f(L_y)=z\times A^1$ and $f(L_{y'})=z'\times A^1$ for $z\neq z'$.
Now let $s:X\rightarrow L$ be the zero section. Then $$\phi=p\circ f\circ s: X\rightarrow X$$ is an injective map, and hence,
an automorphism. Thus, $$(\phi^{-1}\times 1)\circ f:L\rightarrow X\times A^1 \stackrel{\phi^{-1}\times 1}{\rightarrow} X\times A^1$$ is a map preserving the fibers of the projections to $X$. Now let $$q:X\times A^1 \rightarrow A^1$$
be the other projection, and $$h=q\circ (\phi^{-1}\times 1)\circ f\circ s.$$ So $h(y)$ is the image in $A^1$ under $(\phi^{-1}\times 1)\circ f$  of the origin of
$L_y$. We use this function to get a fiber-preserving isomorphism $g: X\times A^1\simeq X\times A^1$ that sends
$(y,\lambda )$ to $(y, \lambda-h(y))$. So finally, $$g\circ (\phi^{-1}\times 1)\circ f: L\simeq X\times A^1$$ is a fiber-preserving isomorphism of
varieties  that furthermore preserves the origins of each fiber.  It must then be fiber-wise
an isomorphism of vector spaces. Thus, it is an isomorphism of line bundles.

Added: The argument above can be easily modified to show that if $X$ is an irrational smooth curve and $L$ and $M$ are line bundles on $X$, then any isomorphism of algebraic varieties $$f:L\simeq M$$ is of the form
$$f=T_s\circ \tilde{\phi} \circ g$$
where $$\tilde{\phi}:\phi^*M\rightarrow M$$ is the base-change map for an automorphism $\phi$ of $X$, $$g:L\simeq \phi^*M$$ is an isomorphism of line bundles, and $$T_s:M\rightarrow M$$ is translation by a section $s:X\rightarrow M$ of $M$.
Since the algebraic automorphism group of an affine irrational curve is finite, we see, by varying $L$, that for  $X$ as above, there is in fact a 
continuum of distinct algebraic structures 
on the analytic space $X\times A^1$.
A: Dear Ravi,
maybe the simplest example is one by Serre: the holomorphic Stein surface $\mathbb C^\ast\times  \mathbb C^\ast $ underlies two non-isomorphic smooth complex algebraic varieties.
1) $\mathbb G_m \times \mathbb G_m$
2) An open subset $U\subset L$ of a $\mathbb P^1$-bundle $L$ on an an elliptic (complete!) curve $E$, obtained by deleting a section $S$  of said bundle: $U=L\setminus S$. That variety $U$ is not affine and has a huge Picard group, namely that of the elliptic curve $E$ :
$$Pic (U)=Pic (E) $$
So you can use two concepts to prove that  $U$ and $\mathbb G_m \times \mathbb G_m$ are not algebraically isomorphic: affineness and Picard. Actually you can use a third concept: just regular functions! Indeed $U$ has the strange property that its regular functions are constant:, just as if it were projective: $\Gamma(U, \mathcal  O_U)=\mathbb C$ . But it is far, far from projective since its analytification is Stein!
Details can be found in Hartshorne's Ample Subvarieties of Algebraic Varieties Chapter VI, §3,p.232 (Springer, LNM 156). A link to an earlier discussion is here .
Edit: I forgot to  say (but of course you know it better than I!)  that $\mathbb G_m \times \mathbb G_m$ has trivial Picard group: 
$$Pic(\mathbb G_m \times \mathbb G_m)=0$$
The way I see it is that $\mathbb G_m \times \mathbb G_m=Spec (A)$ where $A=S^{-1}\mathbb C[X,Y]$ with $S$ the multiplicative monoid consisting of the $X^iY^j$'s. So $A$ is a UFD (since it is a ring of fractions of a UFD) and its spectrum thus has trivial Picard group.
A slightly more geometric formulation is that we have a surjective group morphism $Pic(S) \to Pic(V) \to 0$ valid for every open subset $V\subset S$ of a locally factorial scheme $S$ [Hartshorne, Algebraic Geometry, page 133]. Apply to $S=\mathbb A^2$ which has trivial Picard group and to $V=\mathbb G_m \times \mathbb G_m$.
Second edit: Let us finally recall that the group of analytic line bundles on $\mathbb C^\ast\times  \mathbb C^\ast $ is $\mathbb Z$, more precisely that the first Chern class is an isomorphism 
$$c_1:Pic_{an}(\mathbb C^\ast\times  \mathbb C^\ast)\ =H^1(\mathbb C^\ast\times  \mathbb C^\ast,\mathcal O^\ast)\stackrel {\sim}{\to} H^2(\mathbb C^\ast\times  \mathbb C^\ast,\mathbb Z)=\mathbb Z $$
This follows as usual from the long cohomology exact sequence associated to the exponential exact sequence $0\to\mathbb Z\to \mathcal O \to \mathcal O^\ast \to 0$  and from the vanishing of the cohomology groups of the coherent sheaf $\mathcal O$ due to Steinness of $ \mathbb C^\ast\times  \mathbb C^\ast$, namely: $H^1(\mathbb C^\ast\times  \mathbb C^\ast,\mathcal O)=H^2(\mathbb C^\ast\times  \mathbb C^\ast,\mathcal O)=0$ 
A: If you look at local rings, it is easy to construct such examples. For example, if $X,Y$ are smooth of same dimension, then for any point $x\in X,y\in Y$, the completions of $O_{X,x}, O_{Y,y}$ are isomorphic, but of course the algebraic local rings are not necessarily isomorphic (for example, if $X,Y$ are not birational, then even their fraction fields are not isomorphic.) 
A: Let $E$ be an elliptic curve. The moduli space $M_E$ of line bundles with a connection on $E$ is an $\mathbb{A}^1$ bundle over $Pic^0(E) \cong E$. In particular, $E$ can be recoved from $M_E$ as the Albanese variety (of any compactification). As an analytic variety $M_E \cong(\mathbb{C}^{\times})^2$ since a complex line bundle with connection is exactly the same as a character of the fundamental group (which is $\mathbb{Z}^2$ in this case). So for all $E$ the analytifications are isomorphic whereas the moduli spaces are not isomorphic as algebraic varieties.
A similar construction works for higher genus curves as well and also for higher rank vector bundles. I believe this observation is originally due to Serre, but I do not know the precise reference.
A: By one of those coincidences so common in mathematics, two days after I asked this question, something unrelated I was thinking about led me to this wonderful paper of Hanspeter Kraft, which points out (among many other things) that $\mathbb{C}^3$ has other algebraic structures, for example the hypersurface in $\mathbb{A}^4$ given by $x + x^2 y + z^3 + t^2 = 0$ (Peter Russell showed it was analytically $\mathbb{C}^3$, and Makar-Limanov showed that it is not algebraically isomorphic to $\mathbb{A}^3$).
See also Ilya Nikokoshev's great question here (which also links to Kraft's paper)
A: I've been asking around a little, and nobody seems to be able to tell whether or not Russell's hypersurface is analytically $\mathbb C^3$. Adrien Dubouloz shows in this article that the Makar-Limanov invariant of its product with $\mathbb C$ is trivial. Maybe that helps.
