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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 $X(\bar{k})$

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)?

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)?

Let k be a perfect field. By a k-variety, I shall mean a geometrically irreducible 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)?

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 $X(\bar{k})$

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)?