How to do \widthof with a symbol

Question

I have the following code:

\begin{align*}
c_s &\equiv b_s^2\\
&\equiv (c_{s-1}^{(2^{k_{s-1}-k_s-1})})^2 \\
&\equiv c_{s-1}^{(2^{k_{s-1}-k_s})} \\
&\equiv c_{s-2}^{(2^{k_{s-2}-k_{s-1}} \cdot 2^{k_{s-1}-k_s})}\\
&\mathrel{\makebox[\widthof{=}]{\vdots}}\\
&\equiv c_0^{(2^{k_{0}-k_{1}} \cdot \cdots \cdot 2^{k_{s-1}-k_s})} \pmod p
\end{align*}

On the third last line, I want to change the argument to \widthof to be \equiv, but doing this gives me compile error. What do I need to do to achieve the desired result?

Answer 1

You could use

\widthof{$\equiv$}

but there's a slicker way with mathtools:

\documentclass{article}
\usepackage{amsmath,mathtools}

\begin{document}

\begin{align*}
c_s &\equiv b_s^2\\
&\equiv (c_{s-1}^{(2^{k_{s-1}-k_s-1})})^2 \\
&\equiv c_{s-1}^{(2^{k_{s-1}-k_s})} \\
&\equiv c_{s-2}^{(2^{k_{s-2}-k_{s-1}} \cdot 2^{k_{s-1}-k_s})}\\
&\vdotswithin{\equiv}\\
&\equiv c_0^{(2^{k_{0}-k_{1}} \cdot \cdots \cdot 2^{k_{s-1}-k_s})} \pmod{p}  
\end{align*}

\end{document}

The space might seem too big, so \mathtools also provides \shortvdotswithin:

\documentclass{article}
\usepackage{amsmath,mathtools}
\mathtoolsset{shortvdotsadjustabove=3pt} % I don't like the default

\begin{document}

\begin{align*}
c_s &\equiv b_s^2\\
&\equiv (c_{s-1}^{(2^{k_{s-1}-k_s-1})})^2 \\
&\equiv c_{s-1}^{(2^{k_{s-1}-k_s})} \\
&\equiv c_{s-2}^{(2^{k_{s-2}-k_{s-1}} \cdot 2^{k_{s-1}-k_s})}\\
&\shortvdotswithin{\equiv}
&\equiv c_0^{(2^{k_{0}-k_{1}} \cdot \cdots \cdot 2^{k_{s-1}-k_s})} \pmod{p}
\end{align*}

\end{document}

Note that there is no \\ after \shortvdotswithin{\equiv}.

Be careful that it's \pmod{p} and not \pmod p. The latter seems to work, but try \pmod 11 and you'll see.

Answer 2

The argument of \widthof (package calc) is set in text mode. Mathematical symbols need math mode:

\widthof{$\equiv$}

Full example:

\documentclass{article}
\usepackage{calc}
\usepackage{amsmath}
\begin{document}
\begin{align*}
c_s &\equiv b_s^2\\
&\equiv (c_{s-1}^{(2^{k_{s-1}-k_s-1})})^2 \\
&\equiv c_{s-1}^{(2^{k_{s-1}-k_s})} \\
&\equiv c_{s-2}^{(2^{k_{s-2}-k_{s-1}} \cdot 2^{k_{s-1}-k_s})}\\
&\mathrel{\makebox[\widthof{$\equiv$}]{\vdots}}\\
&\equiv c_0^{(2^{k_{0}-k_{1}} \cdot \cdots \cdot 2^{k_{s-1}-k_s})} \pmod p
\end{align*}
\end{document}