# Is the free abelian group on an affine scheme represented by an ind-scheme?

Let $X/\mathbb C$ be an affine scheme of finite type, and let $\mathbb Z[X(\mathbb C)]$ be the free abelian group generated by the $X(\mathbb C)$.

Can the elements of $\mathbb Z[X(\mathbb C)]$ be identified with the $\mathbb C$-points of an ind-scheme $\mathcal X$ over $\mathbb C$ so that the natural map $(X^n\times X^m)(\mathbb C)\to\mathbb Z[X(\mathbb C)]$ given by $(x_1,\ldots,x_n,y_1,\ldots,y_m)\mapsto\sum x_i-\sum y_i$ lifts to a map $X^n\times X^m\to\mathcal X$?

There is an obvious strategy to construct such an ind-scheme, but I don't know whether it works. For ease of notation, let us observe that there is an exact sequence: $$0\to\mathbb Z[X(\mathbb C)]_0\to\mathbb Z[X(\mathbb C)]\xrightarrow\epsilon\mathbb Z\to 0$$ where $\epsilon$ is simply the sum of all the coefficients. Thus it suffices to describe $\mathbb Z[X(\mathbb C)]_0$ as the $\mathbb C$-points of an ind-scheme over $\mathbb C$. Now we look at the map $(X^n\times X^n)(\mathbb C)\to\mathbb Z[X(\mathbb C)]_0$. If we are lucky (this is probably where $X$ being affine is helpful), then perhaps there is a scheme $Y_n$ and a surjection $X^n\times X^n\to Y_n$ so that the map above factors as: $$(X^n\times X^n)(\mathbb C)\to Y_n(\mathbb C)\to\mathbb Z[X(\mathbb C)]_0$$ where the second map is injective. Here the coordinate ring of $Y_n$ should be a subring of the coordinate ring of $X^n\times X^n$, and it is not immediately clear that the natural choice would be of finite type. Hopefully there are then natural inclusions $Y_1\hookrightarrow Y_2\hookrightarrow\cdots$ giving the desired ind-scheme.

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