# Spacing in proof environment with \item \documentclass[letterpaper,12pt]{article}
\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.

• Don't use \text{dim}, but \dim. The main problem with your text is that proof is not supposed to be used inside a list. Sep 27 '20 at 9:07

Some observations:

• There's a lone \item directive in the first proof environment that doesn't do much. This \item directive happens not to generate an error or warning message simply because the proof environment itself is set up as a trivlist environment. I think you should remove that \item directive and replace it with a simple paragraph break.

• The \newline directive in the third proof environment creates an odd-looking paragraph. I would display the subsequent equation; your readers may appreciate the gesture.

• I think it looks odd to embed the answer to the second question -- which, after all, starts with "Give an example of ..." -- in a proof environment. I'd omit the proof wrapper.

• Since you're loading the enumitem package, you could assign the option nosep to the level-2 enumerate environment. \documentclass[letterpaper,12pt]{article}
\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}


• Thanks for your answer, this is quite helpful. I liked the spacing between the statement and proof in each of a), b), c), is there a way to have the same spacing between the statement and proof for single-part questions without using [nosep]? Using nosep changes the spacing for the first part, but not using nosep leaves the space too big for single-part proofs. Is there reason the spacing is not the same for sub-proofs within a certain question compared to single-answer proofs?
– A.B.
Sep 26 '20 at 3:16
• @A.B. - I must admit that I don't fully understand your terminology; e.g., what's a sub-proof? The spacing between items in level-1 and level-2 enumerate environments is different by design. This spacing doesn't depend on whether or not any of the items contain proof environments. Are you asking to to make the inter-item distance the same for level-1 and level-2 enumerate environments?
– Mico
Sep 26 '20 at 4:54
• I am trying to make the vertical distance between the word "Proof" and the statement of the proposition above it, the same regardless of whether there are multiple proofs in the question (level-2) or just one proof (level-1). Assuming you use the proof environment in question 2, I just want to make the space between the word "Proof" and the proposition to be the same as for each proof in question 1. So I think the answer to your last question is yes. Sorry for the confusion.
– A.B.
Sep 26 '20 at 16:53
• @A.B. - Still not sure I understand your objective. Please try the following: Affix [nosep,parsep=0.333\baselineskip] to each \begin{enerumate} statement. Does this meet your objective?
– Mico
Sep 26 '20 at 21:06
• That doesn't work unfortunately. I added a picture to my question. I'm trying to make the vertical white space in both highlighted spaces the same. In particular, I think there is too much space in the second one, so I want the vertical space in Q2 to be what the current vertical space is in Q1 (both highlighted).
– A.B.
Sep 26 '20 at 23:25

From your MWE understood that you need to add more vertical space at the end of the proof envirionment, if I'm correct, please try with the modified MWE:

\documentclass[letterpaper,12pt]{article}
\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}


Please correct me, if my understanding was wrong....

• Sorry, I was a bit unclear about what I was was asking. I'm happy with all the spaces between the questions in question 1. What I would like to do is make the vertical space between the statement of question 1 and part a), the same as the vertical space between the statement of question 2 and the beginning of the proof. Personally I think it would look better with that size consistency, but perhaps there is a reason it is the way it is. Does the output I gave you look "standard?"
– A.B.
Sep 25 '20 at 5:19