6 pure typography : suppressed "\" and added ":"

Dear Ravi,
maybe the simplest example is \ one by Serre. The : 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: $H^1(\mathbb C^\ast\times \mathbb C^\ast,\mathcal O)=H^2(\mathbb C^\ast\times \mathbb C^\ast,\mathcal O)=0$

5 Added Second edit. displayed all calculations of Pic

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.The 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: $H^1(\mathbb C^\ast\times \mathbb C^\ast,\mathcal O)=H^2(\mathbb C^\ast\times \mathbb C^\ast,\mathcal O)=0$

4 Added reference To Hartshorne's Lecture Notes. Added constancy of regular functions on U

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 $E$ .

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.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 Hartshorne, Algebraic Geometry, page 133]. Apply to $S=\mathbb A^2$ which has trivial Picard group and to $V=\mathbb G_m \times \mathbb G_m$.

3 added three lines to my "edit"