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[EDIT: There doesn't seem to be any interest in answering this question, so could anyone just provide me a reference for understanding (2), and if possible (1)? Hopefully that would be enough to help me move forward on this.]

On page 1 of Serre's Topics in Galois Theory, he shows that $\mathbb{Q}$ admits a generic $\mathbb{Z}/3\mathbb{Z}$ extension in the following way:

$F(x) = x^3 - Tx^2 + (T-3)x + 1$ generates a $G = \mathbb{Z}/3\mathbb{Z}$-extension $Y$ of $\mathbb{P}^1$, since the function $T = \frac{x^3-3x+1}{x^2-x} \in \mathbb{Q}[x]$ is invariant under the automorphism of $Y$ of order 3 given by $\sigma:= x \mapsto \frac{1}{1-x}$ generating $G$. We think of this cover as a quotient $Y = \mathbb{P}^1 \to \mathbb{P}^1/G$.

Then he states that any extension $L/K$ with group $G = \mathbb{Z}/3\mathbb{Z}$ induces a homomorphism $\phi:Gal(\bar{K}/K) \to G \to \mathrm{Aut}(Y)$, which can be viewed as a $1$-cycle of $Gal(\bar{K}/K$ with values in $\mathrm{Aut}(Y)$.

Now here is the part I don't understand. The extension $L/K$ is given by the pullback of a rational point on $\mathbb{P}^1/G$ if and only if the twist of $Y$ by this cocycle has a rational point not invariant by $\sigma$. This is a general property of Galois twists. But this twist has a rational point over a cubic extension of $K$, and every curve of genus $0$ which has a point over an odd degree extension is a projective line, and so there is at least one more point distinct from the ones fixed by $\sigma$.

1) How can I understand or describe more explicitly the twist of $Y$ given by the cocycle coming from $\phi$?

2) (Serre says this is a well-known fact of Galois twists) Why is the extension $L/K$ is given by the pullback of a rational point on $\mathbb{P}^1/G$ if and only if the twist of $Y$ by this cocycle has a rational point not invariant by $\sigma$? I understand the basic concept that a cover of a curve induces a cover over each of its points, and the fiber corresponds to a field extension only if it is connected.

3) Why does this twist have a rational point over a cubic extension of $K$?

4) Why is a curve of genus $0$ with a point over an odd degree extension a projective line?

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  • $\begingroup$ Caution: It's not $\Bbb{Q}[x]$. 1) You can try to write down cocycles using the cocycle notation but it will probably be hard. 2) I don't really want to think about this anymore. 3) I think you have to find one. 4) Any curve of genus 0 is either a twist of the projective line or is the projective line. In general these are varieties under the action of a projective general linear group. If there is a rational point, it is the projective line. More generally if there is a 0-cycle of degree 1, it has a rational point. In this case it's enough to check only odd degree extensions... $\endgroup$
    – Eoin
    May 26, 2017 at 17:35
  • $\begingroup$ ... because any curve of genus 0 has a point of degree at most two, see for example exercise 2.A. $\endgroup$
    – Eoin
    May 26, 2017 at 17:37
  • $\begingroup$ I am hoping for a complete answer, I got stuck in the very same part. $\endgroup$
    – Alon Yariv
    Sep 30, 2020 at 16:01

1 Answer 1

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First let's set some notation, $G_K = \mathrm{Gal}(\bar{K}/K)$ and $C = \mathbb{Z} / 3 \mathbb{Z}$ and $A = \mathrm{Aut}(Y)$.

Make sure to note that the varieties $Y = \mathbb{P}^1$ and $X = \mathbb{P}^1 / C$ are defined over $K$ but they have points with coefficients in $\bar{K}$. Notice also that $C$ acts on $Y$ via the automorphism $\sigma : Y \to Y$ where $\sigma : x \mapsto \frac{1}{1 - x}$ which is defined over $K$.

Now before we really get started, let's understand why $\varphi : G_K \to C \to A$ is a cocycle. $A$ is a $G_K$-group meaning a group with a $G_K$-action compatible with the group structure. Since $G_K$ acts on $Y$, we can describe the $G_K$-action on $A$ by conjugation $f \mapsto {}^\tau f = \tau \circ f \circ \tau^{-1}$ for $f \in A$ and $\tau \in G_K$. A $1$-cocycle (or crossed homomorphism) should be a map $\varphi : G \to A$ satisfying $\varphi(\tau_1 \tau_2) = \varphi(\tau_1) \: {}^{\tau_1} \varphi(\tau_2)$ but our map $\varphi : G_K \to A$ is actually a group homomorphism. How can this make sense? Well, because $\sigma$ is defined over $K$, we know that ${}^{\tau} \sigma = \sigma$ for all $\tau \in G_K$ and thus $\varphi$ is indeed a $1$-cocycle.

Serre's book on Galois Cohomology (section 5 on nonabelian cohomology) is a good reference for twisting.

(1) For a $1$-cocycle $\varphi : G_K \to A$ we can define a twist ${}_{\varphi} A$ as the group $A$ but with a twisted $G_K$-action defined by, $$ \tau \cdot f = \varphi(\tau) \circ {}^{\tau} f $$ (note this is no longer a $G_K$-group because the $G_K$-action is not compatible with the multiplcation structure but it is a an $A$-torsor or $A$ principal homogeneous space in Serre's notation). This a $G_K$-action because, $$ (\tau_1 \tau_2) \cdot f = \varphi(\tau_1 \tau_2) \circ {}^{\tau_1 \tau_2} f = \varphi(\tau_1) \circ {}^{\tau_1} \varphi(\tau_2) \circ {}^{\tau_1 \tau_2} f = \varphi(\tau_1) \circ {}^{\tau_1} (\varphi(\tau_2) \circ {}^{\tau_2} f) \\ = \tau_1 \cdot (\tau_2 \cdot f) $$ Also to be an action we require that $e \cdot f = f$ so we must have $\varphi(e) = \mathrm{id}$ which is called $\varphi$ being normalized (note that your cocycle is normalized).

Now it turns out that ${}_{\varphi} A$ will correspond to some variety ${}_{\varphi} Y$ defined over $K$ such that $\mathrm{Iso}({}_{\varphi} Y, Y) = {}_{\varphi} A$ with the conjugation $G_K$-action. Explicily, on $\bar{K}$-points ${}_{\varphi} Y = Y$ but where $G_K$ acts on $Y$ via automorphisms, $\rho(\tau) : x \mapsto \varphi(\tau)({}^{\tau} x)$. Let's check my claim. Given an isomorphsm $f : {}_{\varphi} Y \to Y$ let's consider $$ \tau \cdot f = \rho(\tau) \circ f \circ \tau^{-1} = \varphi(\tau) \circ \tau \circ f \circ \tau^{-1} = \varphi(\tau) \circ {}^{\tau} f$$ (a note on notation: $\tau$ as a function on ${}_{\varphi} Y = Y$ is the original Galois action on $\overline{K}$-points) and clearly $A$ acts on ${}_{\varphi} A$ simply-transitively on the right so $\mathrm{Iso}({}_{\varphi} Y, Y) = {}_{\varphi} A$ is an $A$-torsor. However, as constructed, ${}_{\varphi} Y$ is not defined over $K$. When $Y$ is quasi-projective the quotient $({}_{\varphi} Y) / G_K$ gives a $K$-variety which when base-changed to $\overline{K}$ agrees with ${}_{\varphi} Y$.

(2) The map $Y \to X = \mathbb{P}^1 / C$, over a $K$-point (i.e. rational point) $p$, has fiber $Y_p$ which is a rational $C$-torsor i.e. three $\overline{K}$-points related by the action of $C$ on $Y$ (at least when $T^2 - 3 T + 9 \neq 0$. To remove this case we need to stipulate that $\sigma$ cannot fix the fiber). These might be $K$-points or they might be defined over some cubic extension so let $L_p$ be the extension at $p \in (\mathbb{P}^1 / C)(K)$. From the Galois correspondence $L_p = L$ iff $G_K$ acts on the fiber transitively through the quotient $G_K \to C$ (in particular is fixed by $\mathrm{Gal}(L/K)$). In this case, we can choose the map $C \to A$ such that a fixed generator $\tau_0 \in \mathrm{Gal}(L/K)$ acts on the fiber via ${}^{\tau_0} x = \sigma^{-1}(x)$.

Using the (defined over $\bar{K}$) isomorphism ${}_{\varphi} Y \to Y$ we get a map $f : {}_{\varphi} Y \to X = \mathbb{P}^1 / C$. The fiber over $p$ is the same $C$-torsor but where $G_K$ acts via $x \mapsto \varphi(\tau)({}^{\tau} x)$. Therefore $L_p = L$ iff $f^{-1}(p)$ is fixed by $G_K$ because in this case for $x \in f^{-1}(p)$ we have ${}^{\tau} x = \varphi(\tau)^{-1}(x)$ since the action $x \mapsto {}^{\tau} x$ factors through $G_K \to C$ and agrees with $\varphi^{-1} : G_K \to C \to A$ on a generator of $C$. Since $K$-points of ${}_{\varphi} Y$ are the fixed points of the $G_K$-action, we have $p \in (\mathbb{P}/G)(K)$ with $L_p = L$ iff $f^{-1}(p)$ are $K$-points.

Finally, such a $p$ exists iff there is a $K$-point $q \in ({}_{\varphi} Y)(K)$ whose orbit under $C$ is nontrivial since then $f^{-1}(f(q))$ equals the $C$-orbit of $q$ which is comprised of $K$-points ($\sigma$ commutes with $\varphi(\tau)$ since $C$ is abelian so $\sigma : {}_{\varphi} Y \to {}_{\varphi} Y$ commutes with the $G_K$-action i.e. is defined over $K$ and thus preserves $K$-points of ${}_{\varphi} Y$) and thus $f(q) \in (\mathbb{P}^1/G)(K)$ is the desired point.

(3) If $K$ contains the $6^{\mathrm{th}}$-roots of unity then $x = \zeta_6$ is a $K$-point of $Y$ fixed by $\sigma$ and thus a $K$-point of ${}_{\varphi} Y$. Otherwise, consider the cubic extension $K' = K[\zeta_6]$. Since $\sigma$ fixes $x = \zeta_6$ we get a $K'$-point of ${}_{\varphi} Y$.

(4) Any smooth curve $X$ of genus zero (${}_{\varphi} Y$ is smooth because ${}_{\varphi} Y \cong \mathbb{P}^1$ over $\bar{K}$) is a conic in $\mathbb{P}^2_K$ (consider the anti-canonical embedding). For any finite extension $F / K$, an $F$-point of $X$ defines a projection map $X \to \mathbb{P}^1$ which is an isomorphism defined over $F$. In particular, if $F = K$ we find that $X$ is $K$-isomorphic to $\mathbb{P}^1$. Let $[F : K]$ be odd. You can show that an $F$ point satisfying a quadric equation is actually in $K$ by degree arguments.

Alternatively, an $F$-point $p \in Z$ defines a divisor $[p]$ of degree $n = [F : K] = 2 m + 1$. Let $K_X$ be the canonical divisor. Then $D = [p] + m K_X$ has degree $1$ since $\deg{K_X} = -2$. By Riemann-Roch, $$ \dim H^0(X, \mathcal{O}_{X}(D)) - \dim H^0(X, \mathcal{O}_{X}(K - D)) = \deg{D} + 1 - g = 2 $$ but $K - D$ has negative degree so $\mathcal{O}_{X}(K - D)$ has no sections and thus, $$ \dim H^0(X, \mathcal{O}_{X}(D)) = 2 $$ Therefore, $D$ defines a $K$-isomorphism $X \to \mathbb{P}^1$. This also shows that the linear system $|D|$ is positive dimensional so $D \sim [q]$ for some effective divisor of degree $1$ implying that $q \in X$ is a $K$-point.

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  • $\begingroup$ This answer 100% deserves the bounty, let me just read into it making sure it answers all my questions. Thank you very much! @Ben $\endgroup$
    – Alon Yariv
    Oct 2, 2020 at 16:35
  • $\begingroup$ Yes, thanks Ben for the detailed answer, and thanks also @AlonYariv for reinvigorating interest in this question. $\endgroup$ Oct 5, 2020 at 17:52

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