MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $S$ be a non-empty closed subscheme of $P^1_K$, where $K$ is a number field. Assume that the cardinality of $S$ is finite.

Is $S$ closed under the action of the absolute Galois group of the field of rational numbers ?

I would like to know this because a certain theorem I want to apply requires this condition.

share|cite|improve this question

If you mean to use the action of the absolute Galois group of $K$, then this is true for closed subschemes of $P^n_K$, pretty much by definition. A closed subscheme of $P^n_K={\rm Proj}K[x_0,...,x_n]$ is described by a homogeneous ideal $I$, and the closed subscheme is equal to $W={\rm Proj}K[x_0,...,x_n]/I$. (More precisely, $W$ embeds naturally into $P_K^n$.) Then ${\rm Gal}(\bar K/K)$ acts on $W$, since the ideal $I$ is Galois invariant. So for example, ${\rm Gal}(\bar K/K)$ clearly acts on the geometric points $W(\bar K)$ of $W$. (But maybe I'm missing some subtlety here.)

But if you really mean to look at the action of ${\rm Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$, then it's certainly false. For example, take $K=\mathbb{Q}(i)$ and $S$ to be the subscheme consisting of the single point $[i,1]\in P^1_K$. (Formally, $S$ is associated to the ideal generated by the polynomial $x-iy$, which defines a closed subscheme of $P_K^1$.) The ${\rm Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ orbit of $S$ consists of two points.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.