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Intuitively, if a complex affine variety $X$ is given by polynomials $$f_i(z_1,\ldots,z_n)=\sum_{j=0}^na_{ij}z_i^j,\quad i=1,\ldots,k,$$ for some $a_{ij}\in\mathbb{C}$ then we can perturbe the coefficients $a_{ij}$ very slightly so that they lie in $\mathbb{Q}(i)$ and the new variety is still isomorphic to $X$. Thus, we would conclude that the set of affine varieties, up to isomorphisms, is at most countable. Is that intuition correct? And can we make that argument rigorous?

Edit. We might encounter problems with singularities, but can we use that argument to at least show that there are countably many smooth affine varieties?

Background material: A complex affine variety is a subset $X$ of $\mathbb{C}^n$ (for some $n\geq1$) given by the zero set of a collection of finitely many polynomials $f_1,\ldots,f_k\in\mathbb{C}[z_1,\ldots,z_n]$. Two affine varieties are isomorphic if their coordinate ring $A(X):=\mathbb{C}[z_1,\ldots,z_n]/\langle f_1,\ldots,f_k\rangle$ are isomorphic.

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    @DaveL.Renfro If the intuition in the first sentence is true, then yes we can conclude countability (rigorously). But the question is precisely if that first sentence is true. – 2017-01-18
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    If you restrict to cubic curves in $\mathbb{C}^2$, all elliptic curves occur here and they are parametrized by their $j$-function. So, there are already uncountably many non-isomorphic cubic curves. – 2017-01-18
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    @Mohan Thanks. Are there uncountably many smooth cubic curves? – 2017-01-18
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    @SHP yes, each elliptic curve that Mohan discusses is smooth. So there are uncountably many non-isomorphic smooth cubic curves. – 2017-01-18

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