# Why is this cubic polynomial generic for cyclic field extensions?

[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?

• 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...
– Eoin
May 26 '17 at 17:35
• ... because any curve of genus 0 has a point of degree at most two, see for example exercise 2.A.
– Eoin
May 26 '17 at 17:37
• I am hoping for a complete answer, I got stuck in the very same part. Sep 30 '20 at 16:01

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.

• This answer 100% deserves the bounty, let me just read into it making sure it answers all my questions. Thank you very much! @Ben Oct 2 '20 at 16:35
• Yes, thanks Ben for the detailed answer, and thanks also @AlonYariv for reinvigorating interest in this question. Oct 5 '20 at 17:52