Espaçamento no ambiente de prova com \ item

Question

    \documentclass[letterpaper,12pt]{article}
    \usepackage{empheq}  % loads »mathtools«, which in turn loads »amsmath«
    \usepackage{enumitem}
    \usepackage{graphicx}
    \usepackage{fancyvrb}
    \usepackage{amsfonts}
    \usepackage{epsfig}
    \usepackage{amssymb}
    \usepackage{amsmath}
    \usepackage{amsthm}
    \usepackage{verbatim}
    \usepackage[paper=letterpaper,left=0.5in,right=0.75in,top=0.75in,bottom=0.75in]{geometry}
    \newcommand{\Z}{\mathbb{Z}}
    \newcommand{\Q}{\mathbb{Q}}
    \newcommand{\R}{\mathbb{R}}
    \begin{document}
    
    \begin{enumerate}
        \item %Question 1
        \begin{enumerate}
            \item $[F:K] = 1$ if and only if $F=K$.
            \begin{proof}
            Suppose that $[F:K] = 1$. Then $\text{dim}_KF=1$, so there exists a basis $\{u\}$ of $F$ over $K$ consisting of a single element $u \in F$. Now, let $x \in F$. Then $x = au$ for some $a \in K$. In particular, we can write the multiplicative identity $1$ of both $F$ and $K$ as $1=bu$ for some $b \in K$. So $u = b^{-1} \in K$ since $K$ is a field. Hence $x=ab^{-1} \in K$ since $a,b \in K$ and $K$ is a field and so closed under multiplication. Therefore, $F \subseteq K$, and since we know that $K \subseteq F$, we have $F=K$ as desired.
            \item Conversely, suppose that $F=K$. We claim that \{1\} is a basis for $F$ as an $F$-vector space. We see that the set \{1\} is linearly independent since if $a\cdot1=0$, we must have $a=0$. We also see that \{1\} spans $F$ since each $f \in F$ can be written as $f=f\cdot1$. Hence \{1\} is a basis for $F$ as an $F$-vector space, and so $[F:K] = [F:F] = \text{dim}_FF=1$.
            \end{proof}
    
            \item If $[F:K]$ is prime, then there are no intermediate fields between $F$ and $K$.
            \begin{proof}
            Suppose, towards a contradiction, that $[F:K]$ is prime and that there exists some intermediate field $E$ between $F$ and $K$. Then we have $[F:K] = [F:E][E:K]=p$ for some prime $p$. Since $[F:K]$ is prime, then we must have one of $[F:E]$ or $[E:K]$ equal to 1 (and the other equal to $p$). If either $[F:E]$ or $[E:K]$ is equal to 1, then part a) implies that $E=F$ or $E=K$. But in either case, this contradicts the fact that $E$ is between $F$ and $K$. Therefore, there must be no intermediate fields between $F$ and $K$.
            \end{proof}
    
            \item If $u \in F$ has degree $n$ over $K$, then $n$ divides $[F:K]$.
            \begin{proof}
            Suppose that $u \in F$ has degree $n$ over $K$. Then $[K(u):K] = n$. Since $\{u\} \subseteq F$ and $F$ is a field extension of $K$, then $K(u)$ is a subfield of $F$. Now, \newline $[F:K]=[F:K(u)][K(u):K]=[F:K(u)]\cdot n$. Hence $n$ divides $[F:K]$ as desired.
            \end{proof}
        \end{enumerate}
            \item Give an example of a finitely generated field extension, which is not finite dimensional.
            \begin{proof}
            Let $K = \Q$, $E=\Q(\pi)$, and $F=\R$. Then $F$ is a field extension of $E$ and $E$ is a field extension of $K$. In particular, $E$ is a finitely generated transcendental field extension of $K$, since $\pi \in E$ is transcendental over $K$. We know that if $A$ is a finite dimensional field extension of $B$, then $A$ is a finitely generated algebraic field extension of $B$. Since $E$ is a transcendental field extension of $K$, the contrapositive of the previous statement implies that $E$ is not finite dimensional. Hence $\Q(\pi)$ is a finitely generated field extension of $\Q$ which is not finite dimensional, and we have found our example.
            \end{proof}
        \end{enumerate}
    \end{document}
   

Estou tentando escrever algumas provas e percebi que há mais espaço vertical entre o final da pergunta 1 e o início da pergunta 2 do que entre as diferentes partes da pergunta 1. Uma vez que todas estão usando o mesmo ambiente de prova , Eu queria saber por que esse é o caso. É possível que seja convencional, mas não acho que já tenha visto esse tipo de espaçamento antes em outros artigos, então estava pensando se devo tentar ajustá-lo ou deixá-lo como está.

Answer 1

Algumas observações:

• Há uma \itemdiretiva solitária no primeiro proofambiente que não faz muito. Essa \itemdiretiva não gera uma mensagem de erro ou aviso simplesmente porque o proofpróprio ambiente está configurado como um trivlistambiente. Acho que você deveria remover essa \itemdiretiva e substituí-la por uma simples quebra de parágrafo.

• A \newlinediretiva no terceiro proofambiente cria um parágrafo de aparência estranha. Gostaria de exibir a equação subsequente; seus leitores podem apreciar o gesto.

• Acho estranho incorporar a resposta à segunda pergunta - que, afinal, começa com "Dê um exemplo de ..." - em um proofambiente. Eu omitiria o proofinvólucro.

• Já que você está carregando o enumitempacote, pode atribuir a opção nosepao enumerateambiente de nível 2 .

    \documentclass[letterpaper,12pt]{article}
    \usepackage{empheq}  % loads »mathtools«, which in turn loads »amsmath«
    \usepackage{enumitem}
    \usepackage{graphicx}
    \usepackage{fancyvrb}
%%%%    \usepackage{amsfonts} % is loaded by 'amssymb'
%%%%    \usepackage{epsfig} % is superseded by 'graphicx'
    \usepackage{amssymb,amsmath,amsthm}
    \usepackage{verbatim}
    \usepackage[paper=letterpaper,
      left=0.5in,right=0.75in,vmargin=0.75in]{geometry}
    \newcommand{\Z}{\mathbb{Z}}
    \newcommand{\Q}{\mathbb{Q}}
    \newcommand{\R}{\mathbb{R}}
    \begin{document}
    
    \begin{enumerate}
        \item %Question 1
        \begin{enumerate}[nosep]
            \item $[F:K] = 1$ if and only if $F=K$.
            \begin{proof}
            Suppose that $[F:K] = 1$. Then $\dim_K F=1$, so there exists a basis $\{u\}$ of $F$ over~$K$ consisting of a single element $u \in F$. Now, let $x \in F$. Then $x = au$ for some $a \in K$. In particular, we can write the multiplicative identity $1$ of both $F$ and $K$ as $1=bu$ for some $b \in K$. So $u = b^{-1} \in K$ since $K$ is a field. Hence $x=ab^{-1} \in K$ since $a,b \in K$ and $K$ is a field and so closed under multiplication. Therefore, $F \subseteq K$, and since we know that $K \subseteq F$, we have $F=K$ as desired.
            %%\item % <-- why?
 
            Conversely, suppose that $F=K$. We claim that $\{1\}$ is a basis for $F$ as an $F$-vector space. We see that the set $\{1\}$ is linearly independent since if $a\cdot1=0$, we must have $a=0$. We also see that $\{1\}$ spans $F$ since each $f \in F$ can be written as $f=f\cdot1$. Hence $\{1\}$ is a basis for $F$ as an $F$-vector space, and so $[F:K] = [F:F] = \dim_FF=1$.
            \end{proof}
    
            \item If $[F:K]$ is prime, then there are no intermediate fields between $F$ and $K$.
            \begin{proof}
            Suppose, towards a contradiction, that $[F:K]$ is prime and that there exists some intermediate field $E$ between $F$ and $K$. Then we have $[F:K] = [F:E][E:K]=p$ for some prime $p$. Since $[F:K]$ is prime, then we must have one of $[F:E]$ or $[E:K]$ equal to~$1$ (and the other equal to $p$). If either $[F:E]$ or $[E:K]$ is equal to~$1$, then part a) implies that $E=F$ or $E=K$. But in either case, this contradicts the fact that $E$ is between $F$ and $K$. Therefore, there must be no intermediate fields between $F$ and $K$.
            \end{proof}
    
            \item If $u \in F$ has degree $n$ over $K$, then $n$ divides $[F:K]$.
            \begin{proof}
            Suppose that $u \in F$ has degree $n$ over~$K$. Then $[K(u):K] = n$. Since $\{u\} \subseteq F$ and~$F$ is a field extension of $K$, then $K(u)$ is a subfield of $F$. Now, %\newline
            \[
            [F:K]=[F:K(u)][K(u):K]=[F:K(u)]\cdot n\,.
            \] 
            Hence $n$ divides $[F:K]$ as desired.
            \end{proof}
        \end{enumerate}
        
        \item %Question 2
        Give an example of a finitely generated field extension which is not finite dimensional.
        
        %\begin{proof}
        Let $K = \Q$, $E=\Q(\pi)$, and $F=\R$. Then $F$ is a field extension of $E$ and $E$ is a field extension of $K$. In particular, $E$ is a finitely generated transcendental field extension of $K$, since $\pi \in E$ is transcendental over $K$. We know that if $A$ is a finite dimensional field extension of~$B$, then~$A$ is a finitely generated algebraic field extension of $B$. Since $E$ is a transcendental field extension of $K$, the contrapositive of the previous statement implies that $E$ is not finite dimensional. Hence $\Q(\pi)$ is a finitely generated field extension of $\Q$ which is not finite dimensional, and we have found our example.
        %\end{proof}
    \end{enumerate}
    \end{document}
   

Answer 2

Pelo MWEque você entendeu, é necessário adicionar mais espaço vertical ao final do ambiente de prova, se estou correto, tente com o modificado MWE:

\documentclass[letterpaper,12pt]{article}
\usepackage{empheq}  % loads »mathtools«, which in turn loads »amsmath«
\usepackage{enumitem}
\usepackage{graphicx}
\usepackage{fancyvrb}
\usepackage{amsfonts}
\usepackage{epsfig}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{verbatim}
\usepackage[paper=letterpaper,left=0.5in,right=0.75in,top=0.75in,bottom=0.75in]{geometry}
\usepackage{etoolbox}%%added

\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}

\AtEndEnvironment{proof}{\bigskip}%%added

\begin{document}

\begin{enumerate}
    \item %Question 1
    \begin{enumerate}
        \item $[F:K] = 1$ if and only if $F=K$.
        \begin{proof}
        Suppose that $[F:K] = 1$. Then $\text{dim}_KF=1$, so there exists a basis $\{u\}$ of $F$ over $K$ consisting of a single element $u \in F$. Now, let $x \in F$. Then $x = au$ for some $a \in K$. In particular, we can write the multiplicative identity $1$ of both $F$ and $K$ as $1=bu$ for some $b \in K$. So $u = b^{-1} \in K$ since $K$ is a field. Hence $x=ab^{-1} \in K$ since $a,b \in K$ and $K$ is a field and so closed under multiplication. Therefore, $F \subseteq K$, and since we know that $K \subseteq F$, we have $F=K$ as desired.
        \item Conversely, suppose that $F=K$. We claim that \{1\} is a basis for $F$ as an $F$-vector space. We see that the set \{1\} is linearly independent since if $a\cdot1=0$, we must have $a=0$. We also see that \{1\} spans $F$ since each $f \in F$ can be written as $f=f\cdot1$. Hence \{1\} is a basis for $F$ as an $F$-vector space, and so $[F:K] = [F:F] = \text{dim}_FF=1$.
        \end{proof}

        \item If $[F:K]$ is prime, then there are no intermediate fields between $F$ and $K$.
        \begin{proof}
        Suppose, towards a contradiction, that $[F:K]$ is prime and that there exists some intermediate field $E$ between $F$ and $K$. Then we have $[F:K] = [F:E][E:K]=p$ for some prime $p$. Since $[F:K]$ is prime, then we must have one of $[F:E]$ or $[E:K]$ equal to 1 (and the other equal to $p$). If either $[F:E]$ or $[E:K]$ is equal to 1, then part a) implies that $E=F$ or $E=K$. But in either case, this contradicts the fact that $E$ is between $F$ and $K$. Therefore, there must be no intermediate fields between $F$ and $K$.
        \end{proof}

        \item If $u \in F$ has degree $n$ over $K$, then $n$ divides $[F:K]$.
        \begin{proof}
        Suppose that $u \in F$ has degree $n$ over $K$. Then $[K(u):K] = n$. Since $\{u\} \subseteq F$ and $F$ is a field extension of $K$, then $K(u)$ is a subfield of $F$. Now, \newline $[F:K]=[F:K(u)][K(u):K]=[F:K(u)]\cdot n$. Hence $n$ divides $[F:K]$ as desired.
        \end{proof}
    \end{enumerate}
        \item Give an example of a finitely generated field extension, which is not finite dimensional.
        \begin{proof}
        Let $K = \Q$, $E=\Q(\pi)$, and $F=\R$. Then $F$ is a field extension of $E$ and $E$ is a field extension of $K$. In particular, $E$ is a finitely generated transcendental field extension of $K$, since $\pi \in E$ is transcendental over $K$. We know that if $A$ is a finite dimensional field extension of $B$, then $A$ is a finitely generated algebraic field extension of $B$. Since $E$ is a transcendental field extension of $K$, the contrapositive of the previous statement implies that $E$ is not finite dimensional. Hence $\Q(\pi)$ is a finitely generated field extension of $\Q$ which is not finite dimensional, and we have found our example.
        \end{proof}
    \end{enumerate}
\end{document}

Por favor, me corrija, se meu entendimento estava errado ....