nLab unit

For other kinds of units see also unit of an adjunction and unit of a monad. Different (but related) is physical unit.

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Idea

Considering a ring $R$, then by the unit element or 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 nonunital ring.

But more generally 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.

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 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 ${\cdot}$.

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 left unit if and only if $u$ has a left inverse, and $u$ is a 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

$$x = (x u^{-1}) u$$

(on the left) or

$$x = u (u^{-1} x)$$

(on the right). Conversely, a unit must have an inverse, since there must a solution to

$$1 = a u$$

(on the left) or

$$1 = u a$$

(on the right).

The collection of all units in a unital ring form a group, the group of units.

In a commutative ring (or rig), a unit is an element of $R$ that has an inverse, period. Of course, a commutative ring $R$ is a field just when every non-zero element is a unit.

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.

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 left unit if, for every $x$, there is an $a$ such that

$$x = a u ;$$

and $u$ is a right unit if, for every $x$, there is an $a$ such that

$$x = u a .$$

Units in nonassocative rings or magmas

In a 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.

Units in modules

If $R$ is a ring (or rig) and $M$ an $R$-module, then a 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 $R$ qua ring is the same as a unit in $R$ qua (left or right) $R$-module.

Units of measurement

In physics, the quantities of a given dimension generally form an $\mathbb{R}$-line, a $1$-dimensional real vector space. Since $\mathbb{R}$ is a field, any non-zero element is a unit, called in this context a 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 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.

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.

Response: I believe it also works in the non-commutative case, but with a more complicated proof.
Suppose $R$ is a not-necessarily-commutative rng with a unit $u$;
first, observe that $u$ is neither a left nor a right zero divisor, as the equations $0=x\cdot u$ and $0=u\cdot x$ both have $0$ as a solution, and that must be unique.
Now, for any $a,b\in R$ denote by ${}_a u^{-1}$ the unique element s.t. $a= {}_a u^{-1} \cdot u$, and denote by $u^{-1}_b$ the unique element s.t. $b= u\cdot u^{-1}_b$;
we want to show that $u^{-1}_u= {}_u u^{-1}$ and that is the identity of $R$.
First, notice that

$$a\cdot u^{-1}_u= {}_a u^{-1}\cdot u \cdot u^{-1}_u= {}_a u^{-1} \cdot u = a$$

and

$${}_u u^{-1} \cdot b= {}_u u^{-1} \cdot u\cdot u^{-1}_b=u\cdot u^{-1}_b=b,$$

for all $a,b\in R$;
therefore, we just need to show that $u^{-1}_u= {}_u u^{-1}$.
To accomplish this, first notice that, on the one hand, one has $u = {}_u u^{-1}\cdot u$;
on the other hand, one has $u\cdot u^{-1}_u \cdot u=u\cdot u$, hence $u\cdot (u^{-1}_u\cdot u-u)=0$, which must imply $u= u^{-1}_u\cdot u$ since $u$ is not a zero divisor.
By uniqueness of the solution $x$ to $u=x\cdot u$, we deduce that $u^{-1}_u= {}_u u^{-1}$.

Addendum: Having uniqueness for the solutions is essential in order for the converse to hold.
This is because, if a rng has an identity, the units are not zero divisors, therefore the equations of the form $a=x\cdot u$ and $a=u\cdot x$ all have a unique solution;
therefore, by contrapositive, if some of these equations have multiple solutions, then the rng has no identity.

It is this meaning of ‘unit’ which gives rise to the unit of an adjunction.

Related concepts

• group of units

• unit object

• unit of an adjunction

• unit of a monad

• monad terminology

• exponential ring

References

See also

• Wikipedia, Unit (Ring theory)