\documentclass[12pt,titlepage]{article}

\usepackage{amsmath}
\usepackage{mathrsfs}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{mathtools}
\usepackage{graphicx}
\usepackage{color}
\usepackage{ucs}
\usepackage[utf8x]{inputenc}
\usepackage{xparse}
\usepackage{hyperref}

%----Macros----------
%
% Unresolved issues:
%
%  \righttoleftarrow
%  \lefttorightarrow
%
%  \color{} with HTML colorspec
%  \bgcolor
%  \array with options (without options, it's equivalent to the matrix environment)

% Of the standard HTML named colors, white, black, red, green, blue and yellow
% are predefined in the color package. Here are the rest.
\definecolor{aqua}{rgb}{0, 1.0, 1.0}
\definecolor{fuschia}{rgb}{1.0, 0, 1.0}
\definecolor{gray}{rgb}{0.502, 0.502, 0.502}
\definecolor{lime}{rgb}{0, 1.0, 0}
\definecolor{maroon}{rgb}{0.502, 0, 0}
\definecolor{navy}{rgb}{0, 0, 0.502}
\definecolor{olive}{rgb}{0.502, 0.502, 0}
\definecolor{purple}{rgb}{0.502, 0, 0.502}
\definecolor{silver}{rgb}{0.753, 0.753, 0.753}
\definecolor{teal}{rgb}{0, 0.502, 0.502}

% Because of conflicts, \space and \mathop are converted to
% \itexspace and \operatorname during preprocessing.

% itex: \space{ht}{dp}{wd}
%
% Height and baseline depth measurements are in units of tenths of an ex while
% the width is measured in tenths of an em.
\makeatletter
\newdimen\itex@wd%
\newdimen\itex@dp%
\newdimen\itex@thd%
\def\itexspace#1#2#3{\itex@wd=#3em%
\itex@wd=0.1\itex@wd%
\itex@dp=#2ex%
\itex@dp=0.1\itex@dp%
\itex@thd=#1ex%
\itex@thd=0.1\itex@thd%
\advance\itex@thd\the\itex@dp%
\makebox[\the\itex@wd]{\rule[-\the\itex@dp]{0cm}{\the\itex@thd}}}
\makeatother

% \tensor and \multiscript
\makeatletter
\newif\if@sup
\newtoks\@sups
\def\append@sup#1{\edef\act{\noexpand\@sups={\the\@sups #1}}\act}%
\def\reset@sup{\@supfalse\@sups={}}%
\def\mk@scripts#1#2{\if #2/ \if@sup ^{\the\@sups}\fi \else%
  \ifx #1_ \if@sup ^{\the\@sups}\reset@sup \fi {}_{#2}%
  \else \append@sup#2 \@suptrue \fi%
  \expandafter\mk@scripts\fi}
\def\tensor#1#2{\reset@sup#1\mk@scripts#2_/}
\def\multiscripts#1#2#3{\reset@sup{}\mk@scripts#1_/#2%
  \reset@sup\mk@scripts#3_/}
\makeatother

% \slash
\makeatletter
\newbox\slashbox \setbox\slashbox=\hbox{$/$}
\def\itex@pslash#1{\setbox\@tempboxa=\hbox{$#1$}
  \@tempdima=0.5\wd\slashbox \advance\@tempdima 0.5\wd\@tempboxa
  \copy\slashbox \kern-\@tempdima \box\@tempboxa}
\def\slash{\protect\itex@pslash}
\makeatother

% math-mode versions of \rlap, etc
% from Alexander Perlis, "A complement to \smash, \llap, and lap"
%   http://math.arizona.edu/~aprl/publications/mathclap/
\def\clap#1{\hbox to 0pt{\hss#1\hss}}
\def\mathllap{\mathpalette\mathllapinternal}
\def\mathrlap{\mathpalette\mathrlapinternal}
\def\mathclap{\mathpalette\mathclapinternal}
\def\mathllapinternal#1#2{\llap{$\mathsurround=0pt#1{#2}$}}
\def\mathrlapinternal#1#2{\rlap{$\mathsurround=0pt#1{#2}$}}
\def\mathclapinternal#1#2{\clap{$\mathsurround=0pt#1{#2}$}}

% Renames \sqrt as \oldsqrt and redefine root to result in \sqrt[#1]{#2}
\let\oldroot\root
\def\root#1#2{\oldroot #1 \of{#2}}
\renewcommand{\sqrt}[2][]{\oldroot #1 \of{#2}}

% Manually declare the txfonts symbolsC font
\DeclareSymbolFont{symbolsC}{U}{txsyc}{m}{n}
\SetSymbolFont{symbolsC}{bold}{U}{txsyc}{bx}{n}
\DeclareFontSubstitution{U}{txsyc}{m}{n}

% Manually declare the stmaryrd font
\DeclareSymbolFont{stmry}{U}{stmry}{m}{n}
\SetSymbolFont{stmry}{bold}{U}{stmry}{b}{n}

% Manually declare the MnSymbolE font
\DeclareFontFamily{OMX}{MnSymbolE}{}
\DeclareSymbolFont{mnomx}{OMX}{MnSymbolE}{m}{n}
\SetSymbolFont{mnomx}{bold}{OMX}{MnSymbolE}{b}{n}
\DeclareFontShape{OMX}{MnSymbolE}{m}{n}{
    <-6>  MnSymbolE5
   <6-7>  MnSymbolE6
   <7-8>  MnSymbolE7
   <8-9>  MnSymbolE8
   <9-10> MnSymbolE9
  <10-12> MnSymbolE10
  <12->   MnSymbolE12}{}

% Declare specific arrows from txfonts without loading the full package
\makeatletter
\def\re@DeclareMathSymbol#1#2#3#4{%
    \let#1=\undefined
    \DeclareMathSymbol{#1}{#2}{#3}{#4}}
\re@DeclareMathSymbol{\neArrow}{\mathrel}{symbolsC}{116}
\re@DeclareMathSymbol{\neArr}{\mathrel}{symbolsC}{116}
\re@DeclareMathSymbol{\seArrow}{\mathrel}{symbolsC}{117}
\re@DeclareMathSymbol{\seArr}{\mathrel}{symbolsC}{117}
\re@DeclareMathSymbol{\nwArrow}{\mathrel}{symbolsC}{118}
\re@DeclareMathSymbol{\nwArr}{\mathrel}{symbolsC}{118}
\re@DeclareMathSymbol{\swArrow}{\mathrel}{symbolsC}{119}
\re@DeclareMathSymbol{\swArr}{\mathrel}{symbolsC}{119}
\re@DeclareMathSymbol{\nequiv}{\mathrel}{symbolsC}{46}
\re@DeclareMathSymbol{\Perp}{\mathrel}{symbolsC}{121}
\re@DeclareMathSymbol{\Vbar}{\mathrel}{symbolsC}{121}
\re@DeclareMathSymbol{\sslash}{\mathrel}{stmry}{12}
\re@DeclareMathSymbol{\bigsqcap}{\mathop}{stmry}{"64}
\re@DeclareMathSymbol{\biginterleave}{\mathop}{stmry}{"6}
\re@DeclareMathSymbol{\invamp}{\mathrel}{symbolsC}{77}
\re@DeclareMathSymbol{\parr}{\mathrel}{symbolsC}{77}
\makeatother

% \llangle, \rrangle, \lmoustache and \rmoustache from MnSymbolE
\makeatletter
\def\Decl@Mn@Delim#1#2#3#4{%
  \if\relax\noexpand#1%
    \let#1\undefined
  \fi
  \DeclareMathDelimiter{#1}{#2}{#3}{#4}{#3}{#4}}
\def\Decl@Mn@Open#1#2#3{\Decl@Mn@Delim{#1}{\mathopen}{#2}{#3}}
\def\Decl@Mn@Close#1#2#3{\Decl@Mn@Delim{#1}{\mathclose}{#2}{#3}}
\Decl@Mn@Open{\llangle}{mnomx}{'164}
\Decl@Mn@Close{\rrangle}{mnomx}{'171}
\Decl@Mn@Open{\lmoustache}{mnomx}{'245}
\Decl@Mn@Close{\rmoustache}{mnomx}{'244}
\makeatother

% Widecheck
\makeatletter
\DeclareRobustCommand\widecheck[1]{{\mathpalette\@widecheck{#1}}}
\def\@widecheck#1#2{%
    \setbox\z@\hbox{\m@th$#1#2$}%
    \setbox\tw@\hbox{\m@th$#1%
       \widehat{%
          \vrule\@width\z@\@height\ht\z@
          \vrule\@height\z@\@width\wd\z@}$}%
    \dp\tw@-\ht\z@
    \@tempdima\ht\z@ \advance\@tempdima2\ht\tw@ \divide\@tempdima\thr@@
    \setbox\tw@\hbox{%
       \raise\@tempdima\hbox{\scalebox{1}[-1]{\lower\@tempdima\box
\tw@}}}%
    {\ooalign{\box\tw@ \cr \box\z@}}}
\makeatother

% \mathraisebox{voffset}[height][depth]{something}
\makeatletter
\NewDocumentCommand\mathraisebox{moom}{%
\IfNoValueTF{#2}{\def\@temp##1##2{\raisebox{#1}{$\m@th##1##2$}}}{%
\IfNoValueTF{#3}{\def\@temp##1##2{\raisebox{#1}[#2]{$\m@th##1##2$}}%
}{\def\@temp##1##2{\raisebox{#1}[#2][#3]{$\m@th##1##2$}}}}%
\mathpalette\@temp{#4}}
\makeatletter

% udots (taken from yhmath)
\makeatletter
\def\udots{\mathinner{\mkern2mu\raise\p@\hbox{.}
\mkern2mu\raise4\p@\hbox{.}\mkern1mu
\raise7\p@\vbox{\kern7\p@\hbox{.}}\mkern1mu}}
\makeatother

%% Fix array
\newcommand{\itexarray}[1]{\begin{matrix}#1\end{matrix}}
%% \itexnum is a noop
\newcommand{\itexnum}[1]{#1}

%% Renaming existing commands
\newcommand{\underoverset}[3]{\underset{#1}{\overset{#2}{#3}}}
\newcommand{\widevec}{\overrightarrow}
\newcommand{\darr}{\downarrow}
\newcommand{\nearr}{\nearrow}
\newcommand{\nwarr}{\nwarrow}
\newcommand{\searr}{\searrow}
\newcommand{\swarr}{\swarrow}
\newcommand{\curvearrowbotright}{\curvearrowright}
\newcommand{\uparr}{\uparrow}
\newcommand{\downuparrow}{\updownarrow}
\newcommand{\duparr}{\updownarrow}
\newcommand{\updarr}{\updownarrow}
\newcommand{\gt}{>}
\newcommand{\lt}{<}
\newcommand{\map}{\mapsto}
\newcommand{\embedsin}{\hookrightarrow}
\newcommand{\Alpha}{A}
\newcommand{\Beta}{B}
\newcommand{\Zeta}{Z}
\newcommand{\Eta}{H}
\newcommand{\Iota}{I}
\newcommand{\Kappa}{K}
\newcommand{\Mu}{M}
\newcommand{\Nu}{N}
\newcommand{\Rho}{P}
\newcommand{\Tau}{T}
\newcommand{\Upsi}{\Upsilon}
\newcommand{\omicron}{o}
\newcommand{\lang}{\langle}
\newcommand{\rang}{\rangle}
\newcommand{\Union}{\bigcup}
\newcommand{\Intersection}{\bigcap}
\newcommand{\Oplus}{\bigoplus}
\newcommand{\Otimes}{\bigotimes}
\newcommand{\Wedge}{\bigwedge}
\newcommand{\Vee}{\bigvee}
\newcommand{\coproduct}{\coprod}
\newcommand{\product}{\prod}
\newcommand{\closure}{\overline}
\newcommand{\integral}{\int}
\newcommand{\doubleintegral}{\iint}
\newcommand{\tripleintegral}{\iiint}
\newcommand{\quadrupleintegral}{\iiiint}
\newcommand{\conint}{\oint}
\newcommand{\contourintegral}{\oint}
\newcommand{\infinity}{\infty}
\newcommand{\bottom}{\bot}
\newcommand{\minusb}{\boxminus}
\newcommand{\plusb}{\boxplus}
\newcommand{\timesb}{\boxtimes}
\newcommand{\intersection}{\cap}
\newcommand{\union}{\cup}
\newcommand{\Del}{\nabla}
\newcommand{\odash}{\circleddash}
\newcommand{\negspace}{\!}
\newcommand{\widebar}{\overline}
\newcommand{\textsize}{\normalsize}
\renewcommand{\scriptsize}{\scriptstyle}
\newcommand{\scriptscriptsize}{\scriptscriptstyle}
\newcommand{\mathfr}{\mathfrak}
\newcommand{\statusline}[2]{#2}
\newcommand{\tooltip}[2]{#2}
\newcommand{\toggle}[2]{#2}

% Theorem Environments
\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{prop}{Proposition}
\newtheorem{cor}{Corollary}
\newtheorem*{utheorem}{Theorem}
\newtheorem*{ulemma}{Lemma}
\newtheorem*{uprop}{Proposition}
\newtheorem*{ucor}{Corollary}
\theoremstyle{definition}
\newtheorem{defn}{Definition}
\newtheorem{example}{Example}
\newtheorem*{udefn}{Definition}
\newtheorem*{uexample}{Example}
\theoremstyle{remark}
\newtheorem{remark}{Remark}
\newtheorem{note}{Note}
\newtheorem*{uremark}{Remark}
\newtheorem*{unote}{Note}

%-------------------------------------------------------------------

\begin{document}

%-------------------------------------------------------------------


\section*{splitting field}

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{definition}{Definition}\dotfill \pageref*{definition} \linebreak
\noindent\hyperlink{construction}{Construction}\dotfill \pageref*{construction} \linebreak
\noindent\hyperlink{splitting_fields_of_sets_of_polynomials}{Splitting fields of sets of polynomials}\dotfill \pageref*{splitting_fields_of_sets_of_polynomials} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{definition}{}\subsection*{{Definition}}\label{definition}

Let $k$ be a [[field]], and $p \in k[x]$ a [[monic polynomial]] of degree $n$. A \emph{splitting field} for $p$ is a [[field extension]] $i: k \hookrightarrow E$ such that, regarding $p$ as a [[polynomial]] in $E[x]$ by applying the map

\begin{displaymath}
k[x] \cong k \otimes_k k[x] \stackrel{i \otimes 1}{\to} E \otimes_k k[x] \cong E[x],
\end{displaymath}
the polynomial factors or ``splits'' as a product of linear factors (possibly repeated):

\begin{displaymath}
p(x) = (x - r_1)(x - r_2)\ldots (x - r_n)
\end{displaymath}
with all the [[roots]] $r_i$ lying in $E$, and the smallest subfield of $E$ containing $k$ and these roots is $E$ itself (so that the roots generate $E$: $E = k(r_1, \ldots, r_n)$.

More generally, given a set $S \subseteq k[x]$ of monic polynomials, an extension field $E/k$ for which each $p \in S$ splits over $E$, and which as a field is generated over $k$ by the set of accompanying roots, is called a splitting field for $S$.

\hypertarget{construction}{}\subsection*{{Construction}}\label{construction}

The existence of a splitting field for a single polynomial $p \in k[x]$ can be proven by [[induction]] on $n = \deg(p)$. As $k[x]$ is a [[unique factorization domain]], $p$ is a product of irreducible polynomials $p_1 p_2 \ldots p_k$. The [[ideal]] $(p_1)$ is [[maximal ideal|maximal]] and so $E = k[x]/(p_1)$ is an [[finite-dimensional vector space|finite]] [[field extension]] of $k$ where the [[coset]] $r_1 \coloneqq x + (p_1) \in E$ is a [[root]] of $p_1(x) \in E[x]$, and $E = k(r_1)$. Then $q(x) = p(x)/(x - r_1) \in E[x]$ has degree $n-1$, and thus has a splitting field $F$ by induction, whence $q(x)$ splits as $(x - r_2)\ldots (x-r_n)$ in $F[x]$, and then $p(x) = (x - r_1)(x - r_2) \ldots (x - r_n)$ in $F[x]$. Finally we have the generation condition: $F = E(r_2, \ldots, r_n) = k(r_1)(r_2, \ldots, r_n) = k(r_1, r_2, \ldots, r_n).$

What is not clear from this construction is that splitting fields are unique (up to [[isomorphism]]). This is the case however:

\begin{theorem}
\label{unique}\hypertarget{unique}{}
Splitting fields of a polynomial are unique up to (non-unique) isomorphism.

\end{theorem}
\begin{proof}
(After \hyperlink{Con}{Conrad}.) Suppose $E = k(r_1, \ldots, r_n)$ and $E' = k(r_1', \ldots, r_n')$ are two splitting fields of a polynomial $p \in k[x]$. The [[tensor product]] $E \otimes_k E'$ is a [[coproduct]] of $E$ and $E'$ in the category of [[commutative algebras]] over $k$; let $i: E \to E \otimes_k E'$ and $i': E' \to E \otimes_k E'$ be the coproduct [[coprojection|injections]]. Observe that as a $k$-algebra, $E \otimes_k E'$ is generated by the $i(r_j)$ together with the $i'(r_j')$.

Now let $m$ be any maximal ideal of $E \otimes_k E'$ (note that $E \otimes_k E'$ is finite-dimensional over $k$, so the existence of $m$ does not involve any choice principles), and consider the composite

\begin{displaymath}
\phi = (E \stackrel{i}{\to} E \otimes_k E' \stackrel{quot}{\to} E \otimes_k E'/m).
\end{displaymath}
As is the case for any homomorphism between fields, this composite $\phi$ is an [[injective map]]. Moreover, letting $\rho_j$ denote the coset $i(r_j) + m$ and $\rho_j'$ the coset $i'(r_j') + m$, the polynomial $p(x)$ splits over the field $E \otimes_k E'/m$ as

\begin{displaymath}
\prod_j (x - \rho_j) = \prod_k (x - \rho_k')
\end{displaymath}
and so by [[unique factorization domain|unique factorization]] of polynomials, for each $\rho_k'$ there is a $\rho_j = \phi(r_j)$ with $\rho_k' = \rho_j$. This means the $\phi(r_j)$ exhaust the generators $\rho_j, \rho_k'$ of $E \otimes_k E'/m$. In other words, $\phi$ is also surjective and provides an isomorphism $E \cong E \otimes_k E'/m$. By symmetry we also have an isomorphism $E' \cong E \otimes_k E'/m$, and thus $E \cong E'$.

\end{proof}
\hypertarget{splitting_fields_of_sets_of_polynomials}{}\subsubsection*{{Splitting fields of sets of polynomials}}\label{splitting_fields_of_sets_of_polynomials}

A splitting field for a finite set of polynomials $\{p_1, \ldots, p_k\}$ is just a splitting field of the product $p_1 p_2 \ldots p_k$.

Now let $S \subseteq k[x]$ be an arbitrary set of monic polynomials. For each $p \in S$, let $n_p$ denote the degree of $p$, and let us introduce a set of indeterminates $X_p = \{x_{p, 1}, \ldots, x_{p, n_p}\}$. Let $X$ be the coproduct $\sum_{p \in S} X_p$, and form the [[polynomial|polynomial algebra]] $R = k[X]$. For each $p \in S$ let us write

\begin{displaymath}
p(t) - \prod_{i = 1}^{n_p} (t - x_{p, i}) = \sum_{j = 0}^{n_p - 1} s_{p, j} t^j
\end{displaymath}
for some elements $s_{p, j} \in R$. We claim that the set $\Sigma$ of all $s_{p, j}$ with $p$ ranging over $S$ generates a proper ideal in $R$. For given an $R$-linear combination $g = g_1 s_1 + \ldots + g_k s_k$ with $s_i \in \Sigma$, where $s_i$ is associated with a polynomial $p_i$, let $E$ be the splitting field of $p = p_1 \ldots p_k$, and define a homomorphism $k[X] \to E$ that sends distinct elements of the form $x = x_{p_i, j}$ to distinct roots $r_j$ of $p_i$, and otherwise sends $x = x_{f, j}$ with $f \notin \{p_1, \ldots, p_k\}$ to $0$. Then $g$ is sent to $0$ in $E$, hence $g$ cannot be $1$.

Let $\mathfrak{p}$ be a [[prime ideal]] containing the ideal generated by $\Sigma$; the existence of such $\mathfrak{p}$ is guaranteed under the [[ultrafilter principle]]. The ring $R/\mathfrak{p}$ is an [[integral domain]]; by construction every $p \in S$ splits over $R/\mathfrak{p}$ and the roots of such $p$ generate $R/\mathfrak{p}$ as a $k$-algebra. These roots thus generate its [[field of fractions]] $F$ as a field and $F$ is therefore a splitting field for $S$.

In particular, taking $S$ to be the set of all monic polynomials in $k[x]$, this gives an efficient construction of the [[algebraic closure]] of $k$ that invokes not the full strength of the [[axiom of choice]] (in the form of [[Zorn's lemma]]), but only of the weaker [[ultrafilter principle]].

\begin{theorem}
\label{}\hypertarget{}{}
Any two splitting fields of a given set $S$ of polynomials are isomorphic.

\end{theorem}
Again we prove this only under the assumption of the [[ultrafilter principle]], and not under the assumption of full [[axiom of choice]].

\begin{proof}
(After \hyperlink{AC}{Caicedo}.) Let $E, F$ be splitting fields for $S$; without loss of generality we may suppose $S$ is closed under multiplication of polynomials. For $p \in S$ let $E_p, F_p$ denote the corresponding subfields obtained by adjoining all roots of $p$ that occur in $E, F$ to the ground field $k$. Any isomorphism $E \to F$ restricts to an isomorphism $E_p \to F_p$. Notice that the $E_p$ form a directed system of fields, with an inclusion morphism $E_p \to E_q$ if $p$ divides $q$, and $E$ is the [[colimit]] of the $E_p$ over this directed system. We are required to show that the set of isomorphisms $Iso(E, F)$ is nonempty; we have restriction maps $Iso(E, F) \to Iso(E_p, F_p)$ which exhibit $Iso(E, F)$ as the [[inverse limit]] of $Iso(E_p, F_p)$ over the corresponding system, with transition maps $Res_{p q}: Iso(E_q, F_q) \to Iso(E_p, F_p)$ for $p|q$ also given by restriction. So we are required to show that this inverse limit is [[inhabited set|inhabited]].

By Theorem \ref{unique}, the set of isomorphisms $Iso(E_p, F_p)$ is inhabited, and it is also [[finite set|finite]] (being a [[torsor]] of the group $Aut(E_p)$ which has cardinality at most $n!$ where $n$ is the number of roots). It can then be proven from the [[ultrafilter theorem]] that $\Phi = \prod_{p \in S} Iso(E_p, F_p)$ is nonempty; see \href{/nlab/show/compactness+theorem#surj}{here} for a proof. Similarly, the subset

\begin{displaymath}
\Phi_{p q} \coloneqq \{\phi = (\phi_p)_{p \in S} \in \Phi: Res_{p q}(\phi_q) = \phi_p\}
\end{displaymath}
is also nonempty, as is any finite intersection of such subsets $\Phi_{p q}$ (which actually is another $\Phi_{p' q'}$, by considering the product of $p$`s and $q$'s).

Now $\Phi$ topologized as a product of finite discrete spaces is a [[compact Hausdorff space]]; this version of the [[Tychonoff theorem]] also follows from the ultrafilter theorem. We conclude from compactness that the full intersection $\bigcap_{p|q} \Phi_{p q}$ is inhabited. But this intersection is just the inverse limit which is $Iso(E, F)$, which concludes the proof.

\end{proof}
An alternative proof based on a similar pattern of reasoning, but more explicitly in the language of [[model theory]] and the [[compactness theorem]], was given by \hyperlink{JDH}{Joel David Hamkins}, in an answer following Caicedo's at MathOverflow.

\hypertarget{references}{}\subsection*{{References}}\label{references}

\begin{itemize}%
\item Bernhard Banaschewski, \emph{Algebraic closure without choice}, Mathematical Logic Quarterly, Vol. 38 Issue 1 (1992), 383--385. (doi: 10.1002/malq.19920380136, \href{http://onlinelibrary.wiley.com/doi/10.1002/malq.19920380136/pdf}{paywall})

\end{itemize}
\begin{itemize}%
\item Keith Conrad, \emph{Isomorphisms of Splitting Fields}, Galois theory notes (\href{http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/splittingisom.pdf}{pdf}).

\end{itemize}
\begin{itemize}%
\item Andres Caicedo (http://mathoverflow.net/users/6085/andres-caicedo), \emph{Is the statement that every field has an algebraic closure known to be equivalent to the ultrafilter lemma?}, URL (version: 2010-11-19): \href{http://mathoverflow.net/q/46568}{http://mathoverflow.net/q/46568}

\end{itemize}
\begin{itemize}%
\item Joel David Hamkins (http://mathoverflow.net/users/1946/joel-david-hamkins), \emph{Is the statement that every field has an algebraic closure known to be equivalent to the ultrafilter lemma?}, URL (version: 2010-11-20): \href{http://mathoverflow.net/q/46729}{http://mathoverflow.net/q/46729}

\end{itemize}


\end{document}