[![Vertical space in Q2 should be same as in Q1](https://i.stack.imgur.com/y9Dlk.png)](https://i.stack.imgur.com/y9Dlk.png)

```
    \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}

```

I am trying to write up some proofs, and I noticed that there is more vertical space between the end of question 1 and the beginning of question 2 than there is between the different parts of question 1. Since they are all using the same proof environment, I was wondering why this is the case. It's possible that this is conventional but I don't think I've seen this kind of spacing before in other papers, so I was wondering whether to try to adjust it or leave it as is.