Quick proof that every finite integral domain is a field.    I'm not sure if I'm missing anything, but here is my attempt.  Let $R$ be a finite integral domain. So it is unital and commutative. Since $R$ is finite, it is of the form $R= \{a_1,a_2,...,a_n\}.$ Let $0\neq a\in R.$ Then if $aa_i = aa_j,$ we must have $a_i=a_j$ (because $R$ has no zero divisors). Since $R$ is closed under multiplication, $R= \{aa_1,aa_2,...,aa_n\}.$ Since $R$ is unital, there exists $a_i\in R$ such that $aa_i=1.$ Thus $R$ is a field.
This is how I'm approaching it: let $R$ be a finite integral domain and I'm trying to show every element in $R$ has an inverse:   let $R-\{0\}=\{x_1,x_2,\ldots,x_k\}$,  then as $R$ is closed under multiplication $\prod_{n=1}^k\ x_i=x_j$,  therefore by canceling $x_j$ we get $x_1x_2\cdots x_{j-1}x_{j+1}\cdots x_k=1 $,  by commuting any of these elements to the front we find an inverse for first term, e.g. for $x_m$ we have $x_m(x_1\cdots x_{m-1}x_{m+1}\cdots x_{j-1}x_{j+1}\cdots x_k)=1$, where $(x_m)^{-1}=x_1\cdots x_{m-1}x_{m+1}\cdots x_{j-1}x_{j+1}\cdots x_k$.   As far as I can see this is correct, so we have found inverses for all $x_i\in R$ apart from $x_j$ if I am right so far. How would we find $(x_{j})^{-1}$?
Object referencing its own property on initialization Determining the default Method used in optimization and root-finding algorithms Is it possible to add new alphabets to unicode-math? In ordinary LaTeX, one can easily add new math alphabets by using \DeclareMathAlphabet. One is then able to use, say, cmsy, eus, and rsfs letters side-by-side. Unfortunately, there does not seem to be a similar mechanism in unicode-math, or at the very least \DeclareMathAlphabet doesn't do what it's supposed to do.  Question. How can I set up my own math alphabets in unicode-math?   Attempted workarounds  Since unicode-math has lots of math alphabets, one workaround might be to replace an unused alphabet with the desired one using \setmathfont[range=...-&gt;...]. This doesn't quite work right in my example:  \documentclass[12pt]{article} \usepackage{fontspec,xltxtra,xunicode} \defaultfontfeatures{Mapping=tex-text}  \usepackage{unicode-math}  \scrollmode  \setmathfont{xits-math.otf} \setmathfont[range={\mathscr,\mathbfscr},StylisticSet=0]{xits-math.otf} \setmathfont[range={\mathcal,\mathbfcal},StylisticSet=1]{xits-math.otf} \setmathfont[range={\mathtt-&gt;\mathcal}]{latinmodern-math.otf}  \begin{document} \[ \begin{array}{c} \mathcal{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \\ \mathscr{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \\ \mathtt{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \end{array} \]   I suspect the problem is that the codepoints of the calligraphic letters in Unicode are not contiguous.   Using \mathversion seems to work better:  \documentclass[12pt]{article} \usepackage{fontspec,xltxtra,xunicode} \defaultfontfeatures{Mapping=tex-text}  \usepackage{unicode-math}  \setmathfont{xits-math.otf}  \setmathfont[version=eucal]{latinmodern-math.otf} \newcommand{\eucal}[1]{\mathchoice %   {\mbox{\mathversion{eucal}$\displaystyle\mathscr{#1}$}} %   {\mbox{\mathversion{eucal}$\mathscr{#1}$}} %   {\mbox{\mathversion{eucal}$\scriptstyle\mathscr{#1}$}} %   {\mbox{\mathversion{eucal}$\scriptscriptstyle\mathscr{#1}$}}}  \setmathfont[range={\mathcal,\mathbfcal},StylisticSet=1]{xits-math.otf}  \begin{document} \[ \begin{array}{c} \mathcal{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \\ \mathscr{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \\ \eucal{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \end{array} \] \end{document}   But this feels very inefficient. 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