Let k be a perfect field. By a k-variety, I shall mean a geometrically reduced separated scheme of finite type over k. I think that is enough conditions that the following data determine an affine k-variety:
- A subset $X(\bar{k})$ of $\bar{k}^n$ which is defined by polynomials
- A continuous action of $\mathop{\mathrm{Gal}}(\bar{k}/k)$ on $X(\bar{k})$, such that each $\sigma \in \mathop{\mathrm{Gal}}(\bar{k}/k)$ acts as $\sigma \circ f$ where f is a $\bar{k}$-regular map
When I say that these data determine an affine k-variety, I mean that there is a unique affine k-variety X whose $\bar{k}$-points are $X(\bar{k})$ with the correct Galois action.
Given these data, I want to work out the functor of points of X (which I consider to have domain the category of k-algebras). You can do that by following through the proof that these data determine a k-variety: first construct the coordinate ring A of X, as the Galois-fixed points of the ring of regular functions $X(\bar{k}) \to \bar{k}$; then $X(R) = \mathop{\mathrm{Hom}}(A, R)$ for any k-algebra R.
But if L is an algebraic extension of k, then there is a much simpler way of working out the L-points of X: just take the subset of $X(\bar{k})$ fixed by $\mathop{\mathrm{Gal}}(\bar{k}/L)$.
If L is a transcendental extension of k (or even a k-algebra which is not a field), is there a direct way of writing down the L-points of X which does not require going through the coordinate ring (or essentially equivalently, going through defining equations for X)?
Best Answer
Despite what you learn in logic classes, there's truth to the conventional wisdom that you can't prove a negative. That is, it's a lot harder to explain why something necessarily can't work than why it can. (One almost invariably settles for: "it cannot work under the following explicit conditions, plus possibly others that I have left implicit.")
With that proviso, my preliminary answer is no. The data of the coordinate ring is of course equivalent to that of the set of polynomials $\{P_i\}$ in (1). Thus your question sounds to me like asking: is there some way to dispense with condition (1)? Of course not: just because the set is Galois invariant doesn't mean it has any kind of algebraic structure (e.g. take $k = \overline{k}$ and we are merely saying that not just any old subset of affine space defines an affine variety).
Moreover, I don't see any shortcut around actually using the data of (1) and (2) to compute the coordinate ring. This is a very basic Galois descent argument involving Hilbert 90 applied to the ideal of $\overline{k}[x_1,\ldots,x_n]$ defined via (1).