答案1
\documentclass{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{tikz-cd}
\begin{document}
\[
\begin{tikzcd}[column sep=0em]
& \text{Intermediate fields}\\[-4ex]
& \mathbb{Q}(\sqrt{3},\sqrt{5})\ar[d, dash]\ar[dr, dash]\ar[dl, dash]\\
\mathbb{Q}(\sqrt{3}) & \mathbb{Q}(\sqrt{5}) & \mathbb{Q}(\sqrt{15})\\
& \mathbb{Q}\ar[u, dash]\ar[ur, dash]\ar[ul, dash]\\
\end{tikzcd}\quad
\begin{tikzcd}[column sep=1.5em]
& \text{Subgroups}\\[-4ex]
& \{\iota\}\ar[d, dash]\ar[dr, dash]\ar[dl, dash]\\
\{\iota,\alpha\} & \{\iota,\tau\} & \{\iota,\beta\}\\
& \{\iota,\alpha,\tau,\beta\}\ar[u, dash]\ar[ur, dash]\ar[ul, dash]\\
\end{tikzcd}
\]
\end{document}
答案2
根据评论中非常好的用户给出的建议,这是我的简短建议。
\documentclass[a4paper,12pt]{article}
\usepackage{tikz-cd}
\usepackage[bb = boondox]{mathalfa}
\usepackage[margin=1cm]{geometry}
\usepackage{mathtools}
\begin{document}
\begin{tikzcd}
& \overset{\text{Intermediate fields}}{\mathbb{Q}(\sqrt 3, \sqrt 5)} \arrow[d, no head] \arrow[ld, no head] \arrow[rd, no head] & & & & \overset{\text{Subgroups}}{\{\iota\}} \arrow[ld, no head] \arrow[d, no head] \arrow[rd, no head] & \\
\mathbb{Q}(\sqrt 3)\arrow[rd, no head] & \mathbb{Q}(\sqrt{5}) \arrow[d, no head] & \mathbb{Q}(\sqrt{15}) \arrow[ld, no head] & & \{\iota,\alpha\} \arrow[rd, no head] & \{\iota,\tau\} \arrow[d, no head] & \{\iota,\beta\} \arrow[ld, no head] \\
& \mathbb{Q}& & & & \{\iota,\tau, \alpha, \beta\} &
\end{tikzcd}
\end{document}