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:


1. A subset $X(\bar{k})$ of $\bar{k}^n$ which is defined by polynomials
2. 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)?



Points of a variety defined by Galois descent
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). 


