将$g^3$表成$\theta$的分式（不含$t$）

hbghlyj

hbghlyj 发表于 2024-6-5 08:01
则 $g^3\in\mathbb C(\theta)$
猜測$\mathbb C(g^3)=\mathbb C(\theta)$

令$\omega=\frac{1}{2}(-1+i \sqrt{3})$
$$g=\frac{t+\omega^2}{t+\omega}$$
$$\theta=\frac{t^3-3t+1}{t(t-1)}$$

• 如何将$g^3$表成$\theta$的分式（不含$t$）
• 如何将$\theta$表成$g^3$的分式（不含$t$）

hbghlyj

计算得到非常简单的结果：
$$g^3=\frac{\theta+3\omega^2}{\theta+3\omega}$$

fullsimplify[(t + ω^2)^3/(t + ω)^3 - (θ+3ω^2)/(θ +3ω)] for θ=(t^3 - 3 t + 1)/(t (t - 1)),ω=(-1)^(2/3)

hbghlyj

如果$\tau,\theta\inC(t),\;\mathbb C(\tau)=\mathbb C(\theta)$，那么$\exists a,b,c,d\inC:\tau=\frac{a\theta+b}{c\theta+d}$对所有$t$恒成立

Czhang271828

hbghlyj 发表于 2024-6-6 02:09如果 $\tau,\theta\inC(t),\;\mathbb C(\tau)=\mathbb C(\theta)$, 那么 $\exists a,b,c,d\inC:\tau=\frac{a\theta+b}{c\theta+d}$ 对所有 $t$ 恒成立.

给定任意域 $k$, 那么
$$
\mathrm{PGL}_2(k)=\mathrm{Aut}(\mathbb A^1)\cong \mathrm{Gal}(k(x)/k)
$$
是正确的. 特别地, $\mathrm{PSL}_2(k)$ 对应保持上半平面的那些自同构, $k^\times$ 对应保持零点的那些自同构.

hbghlyj

Czhang271828 发表于 2024-6-6 03:37
给定任意域 $k$, 那么
$\mathrm{PGL}_2(k)=\mathrm{Aut}(\mathbb A^1)\cong \mathrm{Gal}(k(x)/k)$

找到了：math.stackexchange.com/questions/1842130

Page 647 of Dummit&Foote Abstract algebra
This also provides further examples of the Galois correspondence which can be written out completely for small values of $q$. For instance, if $q=|\mathbb{F}|=2$, $\mathrm{PGL}_2(\mathbb{F})$ is nonabelian of order $6$, hence is isomorphic to $S_3$, and has the following lattice of subgroups:
\[
\xymatrix{
    &1\ar@{-}[d] \ar@{-}[dr]\ar@{-}[drr] & \\
     &\left<\pmatrix{0&1\\1&0}\right>\ar@{-}[dd]&\left<\pmatrix{1&0\\1&1}\right>\ar@{-}[ddl] &\left<\pmatrix{1&1\\0&1}\right>\ar@{-}[ddll] \\
\left<\pmatrix{0&1\\1&1}\right>\ar@{-}[uur] \ar@{-}[dr] & \\
    & PGL(2,\mathbb{F}_2) &
}
\]
The field $\mathbb{F}(t)$ is of degree 6 over the fixed field $K$ of $\operatorname{Aut}(\mathbb{F}(t) / \mathbb{F})$ and the lattice of subfields $K \subseteq L \subseteq \mathbb{F}(t)$ is dual to the lattice of subgroups of $S_3$. The fixed field of a cyclic subgroup $\langle\sigma\rangle$ is easily found (via the preceding theorem) by finding a rational function $r$ in $t$ which is fixed by $\sigma$ such that $[\mathbb{F}(t): \mathbb{F}(r)]=|\sigma|$. For example, if $\sigma: t \mapsto 1 /(1+t)$, then $\sigma$ has order 3. The rational function
\[
r=t+\sigma(t)+\sigma^2(t)=\frac{t^3+t+1}{t(t+1)}
\]
is fixed by $\sigma$ and $[\mathbb{F}(t): \mathbb{F}(r)]=3$ (by part (2) of the theorem). Since $\mathbb{F}(r)$ is contained in the fixed field of $\langle\sigma\rangle$ and the degree of $\mathbb{F}(t)$ over the fixed field is $3$, $\mathbb{F}(r)$ is the fixed field of $\langle\sigma\rangle$. In this way one can explicitly describe the lattice of all subfields of $\mathbb{F}(t)$ containing $K$ shown in Figure 6.
\[
\xymatrix{
    &\mathbb F_2(t)\ar@{-}[d] \ar@{-}[dr]\ar@{-}[drr] & \\
     & \mathbb{F}_2\left(\frac{t^2+1}{t}\right)\ar@{-}[dd]& \mathbb{F}_2\left(t^2+t\right)\ar@{-}[ddl] & \mathbb{F}_2\left(\frac{t^2}{t+1}\right)\ar@{-}[ddll] \\
\mathbb{F}_2\left(\frac{t^3+t+1}{t^2+t}\right)\ar@{-}[uur] \ar@{-}[dr] & \\
    & \mathbb{F}_2\left(\frac{\left(t^3+t+1\right)\left(t^3+t^2+1\right)}{\left(t^2+t\right)^2}\right)&
}
\]

hbghlyj

Czhang271828 发表于 2024-6-6 11:07
P 和 A 的 function field 是相同的

P是什麼呢？

好像(1) Lüroth's theorem没有給出證明？


(2)的證明在Exercise 18 of Section 13.2.

(2)好像很簡單？deg(P(X)-tQ(x))是max(deg(P),deg(Q))很顯然吧

hbghlyj

相关：第585页题 29, 30

第653页题 9

subgroups of A₄ appears in Section 3.5).

青青子衿

\begin{align*}
g^3=1-\frac{27- 9\sqrt{3}\,i+6\sqrt{3}\,i\,\theta}{2 \left(9-3\theta+\theta^2\right)}
\end{align*}

青青子衿

hbghlyj 发表于 2024-6-6 01:49

令$\omega=\frac{1}{2}(-1+i \sqrt{3})$
$$g=\frac{t+\omega^2}{t+\omega}$$
$$\theta=\frac{t^3-3t+1}{t(t-1)}$$
如何将$g^3$表成$\theta$的分式（不含$t$）
如何将$\theta$表成$g^3$的分式（不含$t$）

\begin{align*}
g^3&=1-\frac{3 (2\omega +1)}{3 \omega+\theta}\\
&=1-\frac{3 (3-3\omega+(2\omega+1)\theta)}{9-3 \theta+\theta^2}\\
&=1-\frac{27- 9\sqrt{3}\,i+6\sqrt{3}\,i\,\theta}{2 \left(9-3\theta+\theta^2\right)}\\
\theta&=\frac{3(1 +\omega +\omega\,\!g^3 )}{1-g^3}
\end{align*}

2. GroebnerBasis[{g - (t + \[Omega]^2)/(t + \[Omega]), \[Theta] - (
3.    t^3 - 3 t + 1)/(t (t - 1)), 1 + \[Omega] + \[Omega]^2}, {t}]
4. g^3 - (1 - (3 (1 + 2 \[Omega]))/(3 \[Omega] + \[Theta])) /. {g -> (
5.      t + \[Omega]^2)/(t + \[Omega]), \[Theta] -> (t^3 - 3 t + 1)/(
6.      t (t - 1))} /. \[Omega] -> (-1 + I Sqrt[3])/2 // Factor
7. g^3 - (1 - (3 (3 - 3 \[Omega] + (2 \[Omega] + 1) \[Theta]))/(
8.       9 - 3 \[Theta] + \[Theta]^2)) /. {g -> (t + \[Omega]^2)/(
9.      t + \[Omega]), \[Theta] -> (t^3 - 3 t + 1)/(
10.      t (t - 1))} /. \[Omega] -> (-1 + I Sqrt[3])/2 // Factor
11. \[Theta] - (3 (1 + \[Omega] + \[Omega]*g^3))/(
12.     1 - g^3) /. {g -> (t + \[Omega]^2)/(t + \[Omega]), \[Theta] -> (
13.      t^3 - 3 t + 1)/(t (t - 1))} /. \[Omega] -> (-1 + I Sqrt[3])/
14.    2 // Factor

复制代码

hbghlyj

hbghlyj 发表于 2024-6-7 18:09
第653页题 9
$\ldots$Recall that $G \cong P G L_2\left(\mathbb{F}_3\right)$

怎麼看出$S_4 \cong P G L_2\left(\mathbb{F}_3\right)$呢？

Czhang271828

hbghlyj 发表于 2024-6-10 20:50
怎麼看出$S_4 \cong P G L_2\left(\mathbb{F}_3\right)$呢？

对有限域 $\mathbb F_q$, 鉴于 $\mathrm{PGL}$ 的定义是一维射影空间之间的置换, 从而有
$$
\varphi: \mathrm{PGL}_2(\mathbb F_q)\to S_{q+1}=\mathrm{Aut}(P\mathbb F_q^1).
$$
再证明 $\varphi$ 是单射. 若存在 $M\in \mathrm{GL}_2(\mathbb F_q)$ 使得 $[M]\in \ker (\varphi)$, 此时 $M\binom{1}{0}=\binom{x}{0}$, $M\binom{0}{1}=\binom{0}{y}$, 以及 $M\binom{1}{1}=\binom{z}{z}$. 这表明 $M$ 形如单位元的数乘, 亦即 $[M]=[I]$.

$\varphi$ 是 $(q+1)q(q-1)$-元群到 $(q+1)!$ 元群的单射. $q=3$ 时为同构, $q=4$ 时 $\mathrm{im}(\varphi)$ 视同正规子群 $A_5\lhd S_5$, 等等.

hbghlyj

Czhang271828 发表于 2024-7-5 06:14
对有限域 $\mathbb F_q$, 鉴于 $\mathrm{PGL}$ 的定义是一维射影空间之间的置换, 从而有
$$
\varphi: \mathrm{PGL}_2(\mathbb F_q)\to S_{q+1}=\mathrm{Aut}(P\mathbb F_q^1).
$$

例如：$\mathbb F_3$上的一维射影空间有$q+1=4$个：
$\{\binom{0}{0},\binom{0}{1},\binom{0}{2}\}$
$\{\binom{0}{0},\binom{1}{0},\binom{2}{0}\}$
$\{\binom{0}{0},\binom{1}{1},\binom{2}{2}\}$
$\{\binom{0}{0},\binom{1}{2},\binom{2}{1}\}$
它们的置换有$4!$个。
groupprops.subwiki.org/wiki/PGL(2,3)_is_isomorphic_to_S4
math.stackexchange.com/a/4874465/