# arrow, diagrams.sty (Paul Taylor)

My code follows:

\documentclass[]{book}

\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}


I used the tag \rDashto two places, but it came with two dashes in one place and three dashes in another place, refer the marked sample for clear understanding. I need to fix three dashes in all places. How to fix it? Advise...

• Are you tied to diagrams.sty? As far as I can see, \rDashto uses \leaders, so it is quite normal that the arrow is printed differently according to its position on the line. – egreg Dec 28 '18 at 11:38
• @egreg Yes, I meant exactly what you mean. But I need to fix it with 3 dashes, is it possible? – MadyYuvi Dec 28 '18 at 11:43
• I don't think so. If you enclose the diagram in a box (with \mbox) you get two dashes. I abandoned diagrams.sty several years ago: it's poorly documented and difficult to customize, its code is obfuscated. – egreg Dec 28 '18 at 11:50
• By the way, \root\ell\of\gamma should be \sqrt[\ell]{\gamma} and ... should always be \dots. – egreg Dec 28 '18 at 11:52
• FWIW, here's a picture (click) of what I get with tikz-cd. – egreg Dec 28 '18 at 11:54

The diagrams.sty package is difficult to customize and its code is obfuscated, so it is really hard to change its output.

I can suggest you to use tikz-cd instead.

\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}


Some things to note:

1. I changed all ... to \dots;
2. \root\ell\of\gamma is foreign syntax, the proper LaTeX syntax is \sqrt[\ell]{\gamma};
3. it's better to use \colon for maps, instead of :.
• Many thanks, your hat is superb, Happy New Year... – MadyYuvi Dec 28 '18 at 13:44
• In addition to @egreg's excellent advice, I would also point out the Taylor's diagrams package contains a time bomb which always seems to kick in when it is least expected and wanted. Moreover, if for any reason you have to submit a TeX file that uses diagrams.sty to a publisher, it may pose problems on their end. Just a thought... – sgmoye Dec 28 '18 at 13:44
• @sgmoye Yes, I agreed your words, will take care...thanks for the notification...Happy New Year.... – MadyYuvi Dec 29 '18 at 4:48