验证环境中的间距

Question 1

问题在于牢不可破的最终展示。

事实证明，如果你

删除\smallskip定义和定理之间的
\qedhere在最终显示的最后一行中使用（可能你还有其他步骤需要分析，在这种情况下删除\qedhere），
在最终显示的第三行中使用\biggland代替and ，\biggr\left\right

那么文本可以容纳两页。但这是特别指定. 在其他情况下，您可能需要明智地添加\displaybreak，以允许较长的多行显示跨页显示。

请注意，\smallskip在和在类似定理的环境之前。

您的代码中还存在其他不一致之处：定义

\DeclareMathOperator{\s}{\oplus}
\DeclareMathOperator{\summand}{\subseteq^{\oplus}}

是错误的，因为符号不是运算符。定义

\newcommand{\s}{\oplus} % binary operation
\newcommand{\summand}{\subseteq^{\oplus}} % binary relation

之后\,Im颇为可疑。

我展示了一个编辑版本。它仍然有一个需要解决的溢出框。\textit最好使用而不是\emph。此外，您可以使用enumitem设施来避免手动编号列表。

\documentclass[12pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[
  left=1in, 
  right=1in, 
  top=1.5in, 
  bottom=1in,
  headsep=20pt,
  heightrounded, % <-- because of flushbottom
]{geometry}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amsthm}
\usepackage{enumitem}
\usepackage{mathrsfs}

\newtheorem{theorem}{Theorem}
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\DeclareMathOperator{\im}{Im} % No \,
\DeclareMathOperator{\rad}{rad}
%\DeclareMathOperator{\s}{\oplus} %????
%\DeclareMathOperator{\summand}{\subseteq^{\oplus}} %???
\newcommand{\s}{\oplus}
\newcommand{\summand}{\subseteq^{\oplus}}

\begin{document}
\flushbottom

\begin{definition}
Let $M$ be a right $R$-module, a submodule $A\subset M$ is called \emph{small} 
(written $A \ll M$) if there is no proper submodule $B\subset M$ with $A+B=M$. 
Equivalently, whenever $A+B=M$ then $B=M$. The radical of $M$, denoted $\rad(M)$, is 
defined to be the sum of small submodules of $M$. 
\end{definition}

\begin{theorem}[\textsc{Properties Of Small Submodules And The Radical}]
Let $M$ and $N$ be any right modules, and $f:M\to N$ any homomorphism. 
\begin{enumerate}[nolistsep,label=\upshape(\arabic*)]
\item If $A\ll M$, then $f(A) \ll f(N)$.
\item If $A \ll X$ and $X \leq M$, then $A \ll M$.
\item $f\bigl( \rad(M)\bigr) \subseteq \rad(N)$.
\item $\rad(X)\subseteq \rad(M)$ for every submodule $X \leq M$. 
\item If $M= \bigoplus_{i\in I}M_i$, then $\rad(M)=\bigoplus_{i\in I}\rad(M_i)$. 
\item If $M$ is semisimple, then $\rad(M)=0$. In particular, if $R$ is 
      semisimple then $\rad(M_R)=0$. 
\item $\rad(M)=\bigcap \lbrace E : E < M \text{ maximal} \rbrace$. 
\item If $A \summand M$, then $\rad(A)=A\cap \rad(M)$. 
\item $\rad\bigl(M/\rad(M) \bigr)=0$.
\item $\rad(M)=\bigcap \lbrace B \subseteq M : (M/B)_R \text{ semisimple} \rbrace$. 
\item $J(R)= \bigcap \bigl\lbrace \mathcal{R}(S_R) : S_R \text{ simple} 
      \bigr\rbrace$, where $\mathcal{R}(S_R)$ stands for the right annihilator of $S_R$.
       In particular, $J(R)$ is a two-sided ideal of $R$.
\item $M J(R) \subseteq \rad(M)$. The equality holds if $M$ is projective. 
\item A finite sum of small submodules of $M$ is always small in $M$.
\item If $M$ is finitely generated, then $\rad(M) \ll M$. In particular, $J(R) \ll R$.
\item If $M\neq 0$ is finitely generated, then $\rad(M)\neq M$. 
\item If $C\leq M$, then $(C+\rad(M))/C \subseteq \rad(M/C)$.
\end{enumerate}
\end{theorem}

\begin{proof}\leavevmode
\begin{enumerate}[nolistsep,label=\upshape(\arabic*)]
\item Suppose that $f(A)+B=f(M)$ for some submodule $B\subseteq f(M)$. Set 
$X=f^{-1}(B)$. We claim that $A+X=M$. If $m\in M$, then $f(m)\in f(M)=f(A)+B$ and 
hence $f(m)=f(a)+b$ for some $a\in A; b\in B$, implying $f(m-a)=b\Longleftrightarrow 
m-a\in f^{-1}(b)\subseteq f^{-1}(B)=X \Longrightarrow m-a=x;\, x\in X \Longrightarrow 
m=a+x$. Therefore, $M=A+X$, as claimed. Since $A \ll M$, $X=M$. Now, 
$f(M)=f(X)=f\bigl(f^{-1}(B) \bigr)\subseteq B$, proving $B=f(M)$. Consequently, $f(A) 
\ll f(M)$.

\item Let $M=A+A'$ for some submodule $A'\subseteq M$. Then $X=A+(A'\cap X)$. 
Since $A \ll X$, $A'\cap X = X$ which gives $X\subseteq A'$. So, $A\subseteq X 
\subseteq A'$ and thus $A'=A+A'=M$, as needed. Accordingly, $A \ll M$.

\item By (1) and (2), $f\bigl( \rad(M) \bigr)=f\left(\sum_{A\ll M}A 
\right)=\sum_{A \ll M}f(A) \subseteq \sum_{B\ll f(M)}B\subseteq \sum_{B\ll 
N}B=\rad(N)$.

\item Let $X$ be any submodule of $M$. Consider the injection $\iota:X 
\hookrightarrow M$. By (3), $\rad(X)=\iota\bigl(\rad(X) \bigr)\subseteq \rad(M)$.

\item Consider the canonical projection $\pi_i:M\to M_i$ for each $i\in I$. By 
(3), $\pi_i\bigl(\rad(M)\bigr)\subseteq \rad(M_i)$. Consequently, $\rad(M) \subseteq 
\sum_{i\in I} \pi_i\bigl(\rad(M) \bigr) \subseteq \sum_{i\in I}\rad(M_i)$. Indeed, by 
(4), $\sum_{i\in I}\rad(M_i) \subseteq \rad(M)$. Hence, $\rad(M)=\bigoplus_{i\in 
I}\rad(M_i)$. 

\item Notice first that $\rad(S)=0$ for any simple right $R$-module $S$. If $M$ 
is semisimple, then $S=\bigoplus_{\lambda\in \Lambda}S_{\lambda}$ with each 
$S_{\lambda}$ simple. By (5), $\rad(M)=\bigoplus_{\lambda\in 
\Lambda}\rad(S_{\lambda})=0$. If $R$ is a semisimple ring, then any right $R$-module 
is semisimple and hence its radical is zero.

\item First, we show that $\rad(M)\subseteq \bigcap \bigl\lbrace E: E< M \text{ 
maximal} \bigr\rbrace$. Let $E$ be any essential submodule of $M$ and consider the 
canonical projection $\pi:M\to M/L$. By (3), $\rad(M)+L=\pi\bigl(\rad(M) \bigr) 
\subseteq \rad(M/L)=L$. Hence, $\rad(M)\subseteq L$, proving $\rad(M)\subseteq 
\bigcap \bigl\lbrace E: E< M \text{ maximal} \bigr\rbrace$. Second, we show that 
$\bigcap \bigl\lbrace E: E < M \text{ maximal} \bigr\rbrace \subseteq \rad(M)$. Let 
$x\in \bigcap \bigl\lbrace E: E< M \text{ maximal} \bigr\rbrace$ be arbitrary. We 
prove that $\rad(xR)=xR$. If not, then there exists a maximal submodule $L$ of $xR$ 
containing $\rad(xR)$. By the hypothesis, $x\in L$. But then $L=xR$ which 
contradicts the maximality of $L$. Thus, we must have $\rad(xR)=xR$. So, $x\in 
\rad(xR)\subseteq \rad(M)$. 

\item If $A\summand M$, then $M=A\s B$ for some $B\leq M$. By (5), 
  $\rad(M)=\rad(A) \s \rad(B)$. If $x\in \rad(M)\cap A$, then $x=y+z$ with $y\in 
  \rad(A)$ and $z\in \rad(B)$. Equivalently, $z=x-y\in A\cap \rad(B)=0$. As a 
  consequence, $x=y\in \rad(A)$. This proves that $\rad(A)=A\cap \rad(M)$. 

\item For clarity, we split the proof of this item into the following steps:
  \begin{enumerate}[nolistsep,label=(\alph*)]
  \item Given a right module $A$ and a submodule $B\subseteq A$, we show that 
  the collection $\mathscr{L}$ of all maximal submodules of $A/B$ is 
  $\mathcal{M}:=\lbrace C/B : B \subset C \,\, \& \,\, C \subset A \text{ maximal} 
  \rbrace$. Let $C/B$ be any maximal submodule of $A/B$ and put $C\subseteq L 
  \subseteq A$. Therefore, $C/B \subseteq L/B \subseteq A/B$ and hence, by the 
  maximality of $C/B$, either $C/B=L/B$ or $L/B=A/B$. Consequently, either $C=L$ or 
  $L=A$, proving the maximality of $C$ in $A$. Hence, $\mathscr{L}\subseteq 
  \mathcal{M}$. Conversely, let $C$ be any maximal submodule of $A$ containing $B$. 
  We show that $C/B$ is maximal in $A/B$. If $C/B \subseteq D/B \subseteq A/B$, then 
  $C\subseteq D \subseteq A$ and therefore, by the maximality of $C\subset A$, either 
  $C=D$ or $D=A$. That is, either $C/B=D/B$ or $D/B=A/B$, proving the maximality of 
  $C/B\subset A/B$. This proves that $\mathcal{M} \subseteq \mathscr{L}$. 

  \item Given a right module $A$, a submodule $B\subseteq A$, and a collection 
  $\lbrace A_i \rbrace_{i\in I}$ of submodules of $A$ all of which containing $B$, we 
  show that $\bigcap_{i\in I}A_i/B = \left(\bigcap_{i\in I}A_i \right)/B$. Indeed, 
  $\left(\bigcap_{i\in I}A_i \right)/B \subseteq \bigcap_{i\in I}A_i/B$. Conversely, 
  let $x\in \bigcap_{i\in I}A_i/B$ be arbitrary. Let $i_0\in I$ be fixed. Since $x\in 
  A_{i_0}/B$, $x=a+B$ for some $a\in A_{i_0}$. Given $i\in I$ (not necessarily 
  $i_0$), $x=a_i+B$ for some $a_i\in A$. Hence, $a+B=a_i+B \Longleftrightarrow a- 
  a_i=b\in B \Longleftrightarrow a=a_i+b \Longleftrightarrow a\in A_i$. That is, 
  $a\in \bigcap_{i\in I}A_i$ and so $x\in \bigcap_{i\in I}A_i/B$, proving 
  $\left(\bigcap_{i\in I}A_i\right)/B \subseteq \bigcap_{i\in I}A_i/B$. 

  \item We are now in position to prove our statement. By (a) and (b), 
  \begin{align*}
  \rad(M/\rad(M))
  &= \bigcap_{\substack{E\subset M/\rad(M) \\ \text{maximal} }} E \\
  &= \bigcap_{\substack{A\subset M \\ \text{maximal} }} A/\rad(M) \\
  &= \biggl(\, \bigcap_{\substack{A\subset M \\ \text{maximal} }} A  \biggr)\bigg/\rad(M) \\
  &= \rad(M) / \rad(M)  \\
  &= 0.\qedhere
  \end{align*} 
  \end{enumerate}
  \end{enumerate}
  \end{proof}

  \end{document}

检查差异。

Answer

问题在于牢不可破的最终展示。

事实证明，如果你

删除\smallskip定义和定理之间的
\qedhere在最终显示的最后一行中使用（可能你还有其他步骤需要分析，在这种情况下删除\qedhere），
在最终显示的第三行中使用\biggland代替and ，\biggr\left\right