三次代数整数与行列式

青青子衿

\begin{align*}
N\left(u+v\sqrt[3]{t}+w\sqrt[3]{t^2}\,\right)&=u^3+tv^3+t^2w^3-3tuvw\\
N\left(U+V\sqrt[3]{t}+W\sqrt[3]{t^2}\,\right)&=U^3+tV^3+t^2W^3-3tUVW\\
N(\alpha)N(\beta)&=N(\alpha\beta)
\end{align*}

The Euclidean Condition in Pure
Cubic and Complex Quartic Fields
\begin{gather*}
\begin{vmatrix}
u & v & w \\
tw & u & v \\
tv & tw & u \\
\end{vmatrix}\begin{vmatrix}
U & V & W \\
tW & U & V \\
tV & tW & U \\
\end{vmatrix}=\begin{vmatrix}
x & y & z \\
tz & x & y \\
ty & tz & x \\
\end{vmatrix}\\
\\
\qquad\left\{
\begin{split}
x&=uU+t(wV+vW)\\
y&=vU+uV+twW\\
z&=wU+vV+uW\\
\end{split}
\right.
\end{gather*}

Czhang271828

这是 $\mathbb Q(\sqrt[3]{t})$ 上代数有理数的行列式, 当然是个乘法同态.

hbghlyj

Czhang271828 发表于 2023-1-19 12:03
这是 $\mathbb Q(\sqrt[3]{t})$ 上代数有理数的行列式, 当然是个乘法同态.

Field norm

青青子衿

Czhang271828 发表于 2023-1-19 19:03
这是 $\mathbb Q(\sqrt[3]{t})$ 上代数有理数的行列式, 当然是个乘法同态.

\begin{align*}
\det(\boldsymbol{N}_1)&=\begin{vmatrix}
a^2 u & a^2 v & a^2 w \\
-a d w & a (a u-c w) & a (a v-b w) \\
-d (a v-b w) & -c (a v-b w)-a d w & a (a u-c w)-b (a v-b w) \\
\end{vmatrix}\\
\\
\det(\boldsymbol{N}_2)&=\begin{vmatrix}
a^2 U & a^2 V & a^2 W \\
-a d W & a (a U-c W) & a (a V-b W) \\
-d (a V-b W) & -c (a V-b W)-a d W & a (a U-c W)-b (a V-b W) \\
\end{vmatrix}\\
\\
\det(\boldsymbol{N}_1\boldsymbol{N}_2)&=\begin{vmatrix}
a^2x & a^2 y & a^2z \\
-a d z & a (a x-c z) & a (a y-b z) \\
-d (a y-b z) & -c (a y-b z)-a d z & a (a x-c z)-b (a y-b z) \\
\end{vmatrix}
\end{align*}

\begin{align*}
\left\{
\begin{split}
x&=a^2 u U-a d w V - d (a v-b w) W\\
y&=a^2 v U+a (a u-c w) V - (c (a v-b w)+a d w) W\\
z&=a^2 w U+a (a v-b w) V+ (a (a u-c w)-b (a v-b w)) W
\end{split}
\right.
\end{align*}

青青子衿

青青子衿 发表于 2023-1-20 17:46
\begin{align*}
\left\{
\begin{split}
x&=a^2 u U-a d w V - d (a v-b w) W\\
y&=a^2 v U+a (a u-c w) V - (c (a v-b w)+a d w) W\\
z&=a^2 w U+a (a v-b w) V+ (a (a u-c w)-b (a v-b w)) W
\end{split}
\right.
\end{align*}

\begin{align*}
D_1&=\begin{vmatrix}
\left(b^2-a c\right) u+a c v & -a c (u-v) \\
a c (u-v) & a c u+\left(b^2-a c\right)v \\
\end{vmatrix}\\
\\
D_2&=\begin{vmatrix}
\left(b^2-a c\right) U+a c V & -a c (U-V) \\
a c (U-V) & a c U+\left(b^2-a c\right)V \\
\end{vmatrix}\\
\\
D_1D_2&=\begin{vmatrix}
\left(b^2-a c\right) x+a c y & -a c (x-y) \\
a c (x-y) & a c x+\left(b^2-a c\right)y \\
\end{vmatrix}\\
\\
&\quad\left\{\begin{split}
x&=\left(b^2-a c\right)u U+a c (u V+v U-v V)\\
y&=\left(b^2-a c\right)v V+a c (u V+v U-u U)
\end{split}\right.
\end{align*}


\begin{align*}
\color{black}{
\begin{align*}
\color{black}{\displaystyle{\Large\int}_{0}^{1}\frac{\mathrm{d}x}{\displaystyle\int_{0}^{1}\frac{\sqrt{x\left(1-x\right)}\left(1+x\right)\,}{\sqrt{\left(1-t^{2}\right)\left(1-t^{2}x^{2}\right)}}\mathrm{d}t}=\sqrt{2}\ln\left(1+\sqrt{2}\right)}
\end{align*}
}
\end{align*}