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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 $X\times A^1$.

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I believe the following is an elementary example: Let $X$ be the complement of a point in a compact an affine smooth automorphism-free curve of geometric genus at least twoone. 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 $$p\circ f: L_y\rightarrow X\times A^1 \rightarrow stackrel{p}{\rightarrow} X$$ 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 $$p\circ \phi=p\circ f\circ s: X\rightarrow X$$ is an injective map, and hence, must be the identityan automorphism. Thus, $f$ $(\phi^{-1}\times 1)\circ f:L\rightarrow X\times A^1 \stackrel{\phi^{-1}\times 1}{\rightarrow} X\times A^1$$is a fiber-preserving map . preserving the fibers of the projections to X. Now let$$q:X\times A^1 \rightarrow A^1$$be the other projection, and h=q\circ h=q\circ (\phi^{-1}\times 1)\circ f\circ s. s.$$ So$h(y)$is the image in$A^1$under$(\phi^{-1}\times 1)\circ f$of the origin of$L_y$. Then we We use this function to get a fiber preserving 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 origin origins of each fiber. It must then be fiber-wise an isomorphism of vector spaces. Thus, it is an isomorphism of line bundles. 2 edited body I believe the following is an elementary example: Let$X$be the complement of a point in a compact smooth automorphism-free curve of genus at least onetwo. Let$L$be a non-trivial algebraic line bundle on$X$(easy to produce such things). Then$L$is analytically trivial because$X$is Stein. 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 line bundles. Proof: For any fiber$L_y$of$L$, if we consider the composite $$p\circ f: L_y\rightarrow X\times A^1 \rightarrow X$$ 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 $$p\circ f\circ s: X\rightarrow X$$ is an injective map, and hence, must be the identity. Thus,$f$is a fiber-preserving map. Now let $$q:X\times A^1 \rightarrow A^1$$ be the other projection, and$h=q\circ f\circ s$. So$h(y)$is the image in$A^1$of the origin of$L_y$. Then we 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, $$g\circ f: L\simeq X\times A^1$$ is a fiber-preserving isomorphism of varieties that furthermore preserves the origin of each fiber. It must then be fiber-wise an isomorphism of vector spaces. Thus, it is an isomorphism of line bundles.

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