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Let $X$ be a "nice" algebraic curve over some field $K$, say characteristic zero.

The twists of $X$, i.e., the curves $Y$ over $K$ such that $X_{\overline K} \cong Y_{\overline K}$, are in bijection with the continuous cohomology pointed set $$H^1(\mathrm{Gal}(\overline{K}/K),X).$$

I was just wondering about an analogous question for "sections" of curves.

Now, let $x\in X(K)$. Define a twist of $x$ to be a point $y\in X(K)$ such that $x_{\overline K}$ is "isomorphic" to $y_{\overline K}$. (I don't know if it makes sense to talk about "isomorphic" sections. But what I mean is that there is an automorphism $\sigma$ of $\overline K$ such that $x_{\overline K}\circ \sigma = y_{\overline K}$.

Q1. Are twists of $x$ in bijection with some "cohomology set"?

Q2. Do there even exist any non-trivial twists of $x$? I get a feeling that $x_{\overline K}\circ \sigma$ always descends to $x$ forcing $x=y$.

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up vote 3 down vote accepted

Galois acts trivially on $x$ if $x \in X(K)$. (Concretely, a quasiprojective curve is defined over $K$ if and only if you can find equations over $K$ for it, and the Galois action just acts coordinatewise on solutions to those equations, so that the action is trivial on $K$-points.) So the definition of twist already implies there aren't any.

You might wonder if two $K$-point in projective space (of any dimension) could become the same over a common field extension (the point being that their coordinates could differ by a scalar) it's an exercise in Chapter I or II of Silverman to show that this can't happen, and one does use a bijection with a cohomology group - one which is known by Hilbert 90 to be zero - to show it.

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For the record - I love Silverman's book - which I think you are reading based on the question - and I wouldn't know anything about arithmetic today without it. But one downside of learning algebraic geometry via his approach to elliptic curves is that the notion of "defined over" in Silverman, while logically equivalent to the scheme-theoretic definition, is linguistically quite different. I got confused many times over the years because of this and will definitely be confused about it again before I die. – user29743 Apr 26 '12 at 16:38
Thanks for this great answer! I actually haven't ever read Silverman ever myself...I'm more of a Liu guy. Anyway, it's definitely the notion of "defined over" which confuses me often. In your last paragraph you mention that distinct points of $\mathbf{P}^1(K)$ remain distinct if we consider them as points in $\mathbf{P}^1(L)$. What if we replace $\mathbf{P}^1$ by another curve? – Harry Apr 26 '12 at 17:03
The point is that it's true in any projective space, so it's true for any quasi-projective variety - actually any locally quasi-projective variety, i.e. any variety at all. – user29743 Apr 26 '12 at 17:48

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