lie-algebras-and-their-representations

diff --git a/sections/mathieu.tex b/sections/mathieu.tex
@@ -929,42 +929,43 @@ In a commutative domain, this can be achieved by tensoring our module by the
 field of fractions. However, \(\mathcal{U}(\mathfrak{g})\) is hardly ever
 commutative -- \(\mathcal{U}(\mathfrak{g})\) is commutative if, and only if
 \(\mathfrak{g}\) is Abelian -- and the situation is more delicate in the
-non-commutative case. For starters, a non-commutative ring \(R\) may not even
-have a ``field of fractions'' -- i.e. an over-ring where all elements of \(R\)
-have inverses. Nevertheless, it is possible to formally invert elements of
-certain subsets of \(R\) via a process known as \emph{localization}, which we
-now describe.
+non-commutative case. For starters, a non-commutative \(K\)-algebra \(A\) may
+not even have a ``field of fractions'' -- i.e. an over-ring where all elements
+of \(A\) have inverses. Nevertheless, it is possible to formally invert
+elements of certain subsets of \(A\) via a process known as
+\emph{localization}, which we now describe.
 
 \begin{definition}
-  Let \(R\) be a ring. A subset \(S \subset R\) is called \emph{multiplicative}
-  if \(s \cdot t \in S\) for all \(s, t \in S\) and \(0 \notin S\). A
-  multiplicative subset \(S\) is said to satisfy \emph{Ore's localization
-  condition} if for each \(r \in R\) and \(s \in S\) there exists \(u_1, u_2
-  \in R\) and \(t_1, t_2 \in S\) such that \(s r = u_1 t_1\) and \(r s = t_2
-  u_2\).
+  Let \(A\) be a \(K\)-algebra. A subset \(S \subset A\) is called
+  \emph{multiplicative} if \(s \cdot t \in S\) for all \(s, t \in S\) and \(0
+  \notin S\). A multiplicative subset \(S\) is said to satisfy \emph{Ore's
+  localization condition} if for each \(a \in A\) and \(s \in S\) there exists
+  \(b, c \in A\) and \(t, t' \in S\) such that \(s a = b t\) and \(a s = t'
+  c\).
 \end{definition}
 
 \begin{theorem}[Ore-Asano]
-  Let \(S \subset R\) be a multiplicative subset satisfying Ore's localization
-  condition. Then there exists a (unique) ring \(S^{-1} R\), with a canonical
-  ring homomorphism \(R \to S^{-1} R\), enjoying the universal property that
-  each ring homomorphism \(f : R \to T\) such that \(f(s)\) is invertible for
-  all \(s \in S\) can be uniquely extended to a ring homomorphism \(S^{-1} R
-  \to T\). \(S^{-1} R\) is called \emph{the localization of \(R\) by \(S\)},
-  and the map \(R \to S^{-1} R\) is called \emph{the localization map}.
+  Let \(S \subset A\) be a multiplicative subset satisfying Ore's localization
+  condition. Then there exists a (unique) \(K\)-algebra \(S^{-1} A\), with a
+  canonical algebra homomorphism \(A \to S^{-1} A\), enjoying the universal
+  property that each algebra homomorphism \(f : A \to B\) such that \(f(s)\) is
+  invertible for all \(s \in S\) can be uniquely extended to an algebra
+  homomorphism \(S^{-1} A \to B\). \(S^{-1} A\) is called \emph{the
+  localization of \(A\) by \(S\)}, and the map \(A \to S^{-1} A\) is called
+  \emph{the localization map}.
   \begin{center}
     \begin{tikzcd}
-      S^{-1} R \arrow[dotted]{rd} & \\
-      R \arrow{u} \arrow[swap]{r}{f} & T
+      S^{-1} A \arrow[dotted]{rd}    & \\
+      A \arrow{u} \arrow[swap]{r}{f} & B
     \end{tikzcd}
   \end{center}
 \end{theorem}
 
 If we identify an element with its image under the localization map, it follows
-directly from Ore's construction that every element of \(S^{-1} R\) has the
-form \(s^{-1} r_1\) for some \(s \in S\) and \(r_1 \in R\). Likewise, any
-element of \(S^{-1} R\) can also be written as \(r_2 t^{-1}\) for some \(t \in
-S\), \(r_2 \in R\).
+directly from Ore's construction that every element of \(S^{-1} A\) has the
+form \(s^{-1} a\) for some \(s \in S\) and \(a \in A\). Likewise, any element
+of \(S^{-1} A\) can also be written as \(b t^{-1}\) for some \(t \in S\), \(b
+\in A\).
 
 Ore's localization condition may seem a bit arbitrary at first, but a more
 thorough investigation reveals the intuition behind it. The issue in question
@@ -973,7 +974,7 @@ common denominators for granted. However, the existence of common denominators
 is fundamental to the proof of the fact the field of fractions is a ring -- it
 is used, for example, to define the sum of two elements in the field of
 fractions. We thus need to impose their existence for us to have any hope of
-defining consistent arithmetics in the localization of a ring, and Ore's
+defining consistent arithmetics in the localization of an algebra, and Ore's
 condition is actually equivalent to the existence of common denominators --
 see the discussion in the introduction of \cite[ch.~6]{goodearl-warfield} for
 further details.
@@ -983,10 +984,10 @@ be easier to check than Ore's -- known to imply Ore's condition. For
 instance\dots
 
 \begin{lemma}
-  Let \(S \subset R\) be a multiplicative subset generated by finitely many
+  Let \(S \subset A\) be a multiplicative subset generated by finitely many
   locally \(\operatorname{ad}\)-nilpotent elements -- i.e. elements \(s \in S\)
-  such that for each \(r \in R\) there exists \(n > 0\) such that
-  \(\operatorname{ad}(s)^n r = [s, [s, \cdots [s, r]]\cdots] = 0\). Then \(S\)
+  such that for each \(a \in A\) there exists \(n > 0\) such that
+  \(\operatorname{ad}(s)^n a = [s, [s, \cdots [s, a]]\cdots] = 0\). Then \(S\)
   satisfies Ore's localization condition.
 \end{lemma}
 
@@ -996,10 +997,10 @@ introduce one further construction, known as \emph{the localization of a
 module}.
 
 \begin{definition}
-  Let \(S \subset R\) be a multiplicative subset satisfying Ore's localization
-  condition and \(M\) be a \(R\)-module. The \(S^{-1} R\)-module \(S^{-1} M =
-  S^{-1} R \otimes_R M\) is called \emph{the localization of \(M\) by \(S\)},
-  and the homomorphism of \(R\)-modules
+  Let \(S \subset A\) be a multiplicative subset satisfying Ore's localization
+  condition and \(M\) be an \(A\)-module. The \(S^{-1} A\)-module \(S^{-1} M =
+  S^{-1} A \otimes_A M\) is called \emph{the localization of \(M\) by \(S\)},
+  and the homomorphism of \(A\)-modules
   \begin{align*}
     M & \to     S^{-1} M    \\
     m & \mapsto 1 \otimes m
@@ -1007,25 +1008,25 @@ module}.
   is called \emph{the localization map of \(M\)}.
 \end{definition}
 
-Notice that the \(S^{-1} R\)-module \(S^{-1} M\) has the natural structure of
-an \(R\)-module, where the action of \(R\) is given by the localization map \(R
-\to S^{-1} R\).
+Notice that the \(S^{-1} A\)-module \(S^{-1} M\) has the natural structure of
+an \(A\)-module, where the action of \(A\) is given by the localization map \(A
+\to S^{-1} A\).
 
 It is interesting to observe that, unlike in the case of the field of fractions
-of a commutative domain, in general the localization map \(R \to S^{-1} R\) --
-i.e. the map \(r \mapsto \frac{r}{1}\) -- may not be injective. For instance,
+of a commutative domain, in general the localization map \(A \to S^{-1} A\) --
+i.e. the map \(a \mapsto \frac{a}{1}\) -- may not be injective. For instance,
 if \(S\) contains a divisor of zero \(s\), its image under the localization map
-is invertible and therefore cannot be a divisor of zero in \(S^{-1} R\). In
-particular, if \(r \in R\) is nonzero and such that \(s r = 0\) or \(r s = 0\)
+is invertible and therefore cannot be a divisor of zero in \(S^{-1} A\). In
+particular, if \(a \in A\) is nonzero and such that \(s a = 0\) or \(a s = 0\)
 then its image under the localization map has to be \(0\). However, the
 existence of divisors of zero in \(S\) turns out to be the only obstruction to
 the injectivity of the localization map, as shown in\dots
 
 \begin{lemma}
-  Let \(S \subset R\) be a multiplicative subset satisfying Ore's localization
-  condition and \(M\) be a \(R\)-module. If \(S\) acts injectively in \(M\)
+  Let \(S \subset A\) be a multiplicative subset satisfying Ore's localization
+  condition and \(M\) be an \(A\)-module. If \(S\) acts injectively in \(M\)
   then the localization map \(M \to S^{-1} M\) is injective. In particular, if
-  \(S\) has no zero divisors then \(R\) is a subring of \(S^{-1} R\).
+  \(S\) has no zero divisors then \(A\) is a subalgebra of \(S^{-1} A\).
 \end{lemma}
 
 Again, in our case we are interested in inverting the actions of the