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\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*{unit} \begin{quote}% For other kinds of units see also \emph{[[unit of an adjunction]]} and \emph{[[unit of a monad]]}. Different (but related) is \emph{[[physical unit]]}. \end{quote} \vspace{.5em} \hrule \vspace{.5em} \hypertarget{units}{}\section*{{Units}}\label{units} \noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak \noindent\hyperlink{definitions}{Definitions}\dotfill \pageref*{definitions} \linebreak \noindent\hyperlink{units_in_rings}{Units in rings}\dotfill \pageref*{units_in_rings} \linebreak \noindent\hyperlink{units_in_monoids}{Units in monoids}\dotfill \pageref*{units_in_monoids} \linebreak \noindent\hyperlink{units_in_rngs_or_semigroups}{Units in rngs or semigroups}\dotfill \pageref*{units_in_rngs_or_semigroups} \linebreak \noindent\hyperlink{units_in_nonassocative_rings_or_magmas}{Units in nonassocative rings or magmas}\dotfill \pageref*{units_in_nonassocative_rings_or_magmas} \linebreak \noindent\hyperlink{units_in_modules}{Units in modules}\dotfill \pageref*{units_in_modules} \linebreak \noindent\hyperlink{units_of_measurement}{Units of measurement}\dotfill \pageref*{units_of_measurement} \linebreak \noindent\hyperlink{identities_as_units}{Identities as units}\dotfill \pageref*{identities_as_units} \linebreak \noindent\hyperlink{related_concepts}{Related concepts}\dotfill \pageref*{related_concepts} \linebreak \noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak \hypertarget{idea}{}\subsection*{{Idea}}\label{idea} Considering a [[ring]] $R$, then by \emph{the unit element} or \emph{the multiplicative unit} one usually means the [[neutral element]] $1 \in R$ with respect to [[multiplication]]. This is the sense of ``unit'' in terms such as \emph{[[nonunital ring]]}. But more generally \emph{a unit element} in a unital (!) ring is any element that has an [[inverse element]] under [[multiplication]]. This concept generalizes beyond [[rings]], and this is what is discussed in the following. \hypertarget{definitions}{}\subsection*{{Definitions}}\label{definitions} Exactly what this means depends on context. A very general definition is this: Given [[sets]] $R$ and $M$, and a [[function]] ${\cdot}\colon R \times M \to M$, an [[element]] $u$ of $M$ is a \textbf{unit} (relative to the operation ${\cdot}$) if, given any element $x$ of $M$, there exists a unique element $a$ of $R$ such that $x = a \cdot u$. That is, every element of $M$ is a multiple (in a unique way) of $u$, where `multiple' is defined in terms of the operation ${\cdot}$. \hypertarget{units_in_rings}{}\subsubsection*{{Units in rings}}\label{units_in_rings} If $R$ is a [[ring]] (or [[rig]]), then $R$ comes equipped with a multiplication map ${\cdot}\colon R \times R \to R$. So $R$ can play the role of both $R$ and $M$ above, although there are two ways to do this: on the left and on the right. We find that $u$ is a \textbf{left unit} if and only if $u$ has a [[left inverse]], and $u$ is a \textbf{right unit} if and only if $u$ has a [[right inverse]]. First, an element $u$ with an inverse is a unit because, given any element $x$, we have \begin{displaymath} x = (x u^{-1}) u \end{displaymath} (on the left) or \begin{displaymath} x = u (u^{-1} x) \end{displaymath} (on the right). Conversely, a unit must have an inverse, since there must a solution to \begin{displaymath} 1 = a u \end{displaymath} (on the left) or \begin{displaymath} 1 = u a \end{displaymath} (on the right). In a [[commutative ring]] (or rig), a \textbf{unit} is an element of $R$ that has an [[inverse element|inverse]], period. Of course, a commutative ring $R$ is a [[field]] just when every non-[[zero]] element is a unit. \hypertarget{units_in_monoids}{}\subsubsection*{{Units in monoids}}\label{units_in_monoids} Notice that addition plays no role in the characterisation above of a unit in a ring. Accordingly, a unit in a [[monoid]] may be defined in precisely the same way. A [[group]] is precisely a monoid in which every element is a unit. \hypertarget{units_in_rngs_or_semigroups}{}\subsubsection*{{Units in rngs or semigroups}}\label{units_in_rngs_or_semigroups} In a [[rng]] (or, ignoring addition, in a [[semigroup]]), we cannot speak of inverses of elements. However, we can still talk about units; $u$ is a \textbf{left unit} if, for every $x$, there is an $a$ such that \begin{displaymath} x = a u ; \end{displaymath} and $u$ is a \textbf{right unit} if, for every $x$, there is an $a$ such that \begin{displaymath} x = u a . \end{displaymath} \hypertarget{units_in_nonassocative_rings_or_magmas}{}\subsubsection*{{Units in nonassocative rings or magmas}}\label{units_in_nonassocative_rings_or_magmas} In a [[nonassociative algebra|nonassociative ring]] (or, ignoring addition, in a [[magma]]), even if we have an identity element, an invertible element might not be a unit. So we must use the same explicit definition as in a rng (or semigroup) above. A [[quasigroup]] is precisely a magma in which every element is a two-sided unit. \hypertarget{units_in_modules}{}\subsubsection*{{Units in modules}}\label{units_in_modules} If $R$ is a [[ring]] (or [[rig]]) and $M$ an $R$-[[module]], then a \textbf{unit} in $M$ is an element $u \in M$ such that every other $x \in M$ can be written as $x = a u$ (or $x = u a$ for a right module) for some $a \in R$. This is the same as a [[generator]] of $M$ as an $R$-module. There is no need to distinguish left and right units unless $M$ is a [[bimodule]]. Note that a (left or right) unit in $R$ \emph{qua} ring is the same as a unit in $R$ \emph{qua} (left or right) $R$-module. \hypertarget{units_of_measurement}{}\subsubsection*{{Units of measurement}}\label{units_of_measurement} In [[physics]], the quantities of a given dimension generally form an $\mathbb{R}$-[[line]], a $1$-dimensional [[vector space]] over the [[real numbers]]. Since $\mathbb{R}$ is a field, any non-[[zero]] quantity is a unit, called in this context a \textbf{unit of measurement}. This is actually a special case of a unit in a module, where $R \coloneqq \mathbb{R}$ and $M$ is the line in question. Often (but not always) these quantities form an [[orientation|oriented]] line, so that nonzero quantities are either positive or negative. Then we usually also require a unit of measurement to be positive. In fact, for some dimensions, there is no physical meaning to a negative quantity, in which case the quantities actually form a module over the rig $\mathbb{R}_{\ge 0}$ and every nonzero element is ``positive.'' For example, the [[kilogram]] is a unit of mass, because any mass may be expressed as a real multiple of the kilogram. Further, it is a positive unit; the mass of any physical object is a nonnegative quantity (so that mass quantities actually form an $\mathbb{R}_{\ge 0}$-module) and may be expressed as a nonnegative real multiple of the kilogram. \hypertarget{identities_as_units}{}\subsection*{{Identities as units}}\label{identities_as_units} Often the term `unit' (or `unity') is used as a synonym for `[[identity element]]', especially when this identity element is denoted $1$. For example, a `ring with unit' (or `ring with unity') is a ring with an identity (used by authors who say `ring' for a rng). Of course, a rng with identity has a unit, since $1$ itself is a unit; conversely, a commutative rng with a unit must have an identity. I haven't managed to find either a proof or a counterexample to the converse (in the noncommutative case): that a rng with a unit must have an identity. Response: If $R$ is a rng with a unit $u$, then every element uniquely factors through $u$. In particular, $u$ itself does. $u = a u$, with $a$ unique. So $a$ is an identity. Reply: Why is $a$ an identity then? This works if the rng is commutative: given any $v$, write $v$ as $b u$, and then $a v = a (b u) = b (a u) = b u = v$. But without commutativity (and associativity), this doesn't work. It is this meaning of `unit' which gives rise to the [[unit of an adjunction]]. \hypertarget{related_concepts}{}\subsection*{{Related concepts}}\label{related_concepts} \begin{itemize}% \item [[unit object]] \item [[unit of an adjunction]] \item [[unit of a monad]] \end{itemize} \hypertarget{references}{}\subsection*{{References}}\label{references} See also \begin{itemize}% \item Wilkipedia, \emph{} \end{itemize} [[!redirects unit]] [[!redirects units]] [[!redirects unit of measurement]] [[!redirects unit of measurements]] [[!redirects units of measurement]] [[!redirects units of measurements]] \end{document}