## Return to Question

4 Galois action must be nice to define a k-variety

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 $X(\bar{k})$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)$. If L is a transcendental extension of k (or even a k-algebra which is not a field), is there a direct way of writing down the L-points of X which does not require going through the coordinate ring (or essentially equivalently, going through defining equations for X)? 3 Corrected error: irreducible -> reduced Let k be a perfect field. By a k-variety, I shall mean a geometrically irreducible reduced separated scheme of finite type over k. I think that is enough conditions that the following data determine an affine k-variety: 1. A subset$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})$When I say that these data determine an affine k-variety, I mean that there is a unique affine k-variety X whose$\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)?

2 added tag (arithmetic geometry)
1