我的代码如下：
\documentclass[]{book}
\usepackage[silent,nohug,heads=LaTeX,midshaft]{diagrams}
\newarrow{Dashto}{}{dash}{}{dash}>
\newcommand{\undera}{\underline{a}}
\newcommand{\Spec}{\mathop{\mathrm{Spec}}}
\begin{document}
By the Chain Lemma (2),
the function field of $S(\root{\ell}\of \gamma)$ splits $\undera$.
If $X$ is a norm variety for $\undera$,
there is a finite field extension
$F'$ of $k(S(\root{\ell}\of \gamma))$ of degree prime to $\ell$
and an $F'$-point $\Spec(F')\to X$. Forming $\tilde{S}$ as in
Lemma~3, this $F'$-point extends to a rational map
$\phi:\tilde{S}(\root{\ell}\of \gamma)\rDashto^{}X$.
Recall that the cyclic group $C_\ell=\langle\sigma\rangle$ acts on
$X^\ell$ by $\sigma(x_1,...,x_\ell)=(x_2,...,x_\ell,\break x_1)$, and that
$C^\ell X$ denotes the geometric quotient variety $X^\ell/C_\ell$.
Let $\sigma$ be a generator of $C_\ell$,
and let $\phi:\tilde{S}(\root{\ell}\of \gamma)\rDashto^{}X$
be the rational map mentioned above.
Choosing an isomorphism $C_\ell\cong\mu_\ell$, the rational maps
$\phi\sigma^i$ assemble to form a $C_\ell$-equivariant rational map
$g=(\phi,\phi\sigma,...,\phi\sigma^{\ell-1})$
from $\tilde{S}(\root{\ell}\of \gamma)$ to $X^\ell$.
\end{document}
我在两个地方使用了标签\rDashto，但其中一个地方带有两个破折号，另一个地方带有三个破折号，请参考标记的示例以更清楚地理解。我需要修复所有地方的三个破折号。如何修复？建议...
答案1
该diagrams.sty包难以定制，并且其代码被混淆，因此很难改变其输出。
我可以建议你改用tikz-cd。
\documentclass[]{book}
\usepackage{amsmath}
\usepackage{tikz-cd}
\newcommand{\undera}{\underline{a}}
\DeclareMathOperator{\Spec}{Spec}
\newcommand{\ratmap}[3]{%
\begin{tikzcd}[
arrow style=tikz,
>=LaTeX,
ampersand replacement=\&,
sep=scriptsize,
cramped,
]
#1\colon #2 \arrow[r,dashed] \& #3
\end{tikzcd}%
}
\begin{document}
By the Chain Lemma (2),
the function field of $S(\sqrt[\ell]{\gamma})$ splits $\undera$.
If $X$ is a norm variety for $\undera$,
there is a finite field extension
$F'$ of $k(S(\sqrt[\ell]{\gamma}))$ of degree prime to $\ell$
and an $F'$-point $\Spec(F')\to X$. Forming $\tilde{S}$ as in
Lemma~3, this $F'$-point extends to a rational map
\ratmap{\phi}{\tilde{S}}{X}.
Recall that the cyclic group $C_\ell=\langle\sigma\rangle$ acts on
$X^\ell$ by $\sigma(x_1,\dots,x_\ell)=(x_2,\dots,\allowbreak x_\ell, x_1)$,
and that $C^\ell X$ denotes the geometric quotient variety $X^\ell/C_\ell$.
Let $\sigma$ be a generator of $C_\ell$,
and let \ratmap{\phi}{\tilde{S}}{X}
be the rational map mentioned above.
Choosing an isomorphism $C_\ell\cong\mu_\ell$, the rational maps
$\phi\sigma^i$ assemble to form a $C_\ell$-equivariant rational map
$g=(\phi,\phi\sigma,\dots,\phi\sigma^{\ell-1})$
from $\tilde{S}(\sqrt[\ell]{\gamma})$ to $X^\ell$.
\end{document}
需要注意的事项：
- 我全部
...改为\dots； \root\ell\of\gamma是外来语法，正确的 LaTeX 语法是\sqrt[\ell]{\gamma}；- 最好用于
\colon地图，而不是:。