Field Theory in Abstract Algebra 📂Abstract Algebra

Field Theory in Abstract Algebra

Definition ¹

If a ring $(R , + , \cdot)$ has an identity element $1 \in R$ for multiplication $\cdot$, $1$ is called a unity.
In a ring $R$ with a unity, the element $r \ne 0$ that has a multiplicative inverse is called a unit.
If every element other than $0$ is a unit in a ring $R$ with a unity, it is called a division ring.
A division ring $R$ that is commutative with respect to multiplication is called a field.

Explanation

In short, a field $(F , + , \cdot )$ is a commutative ring where every element except the additive identity $0 \in F$ has an inverse. Although it might seem complex when thought of in terms of abstract algebra, if you consider $\mathbb{R}$, which you learned in analysis, it can actually be seen as very fitting for ‘algebraic structures’.

Why are elements with inverses called units?

While the English term Unity for the identity element can be easily understood, many might find it hard to accept why elements with an inverse are called Units. The term unit is commonly translated as ‘unit’ and often used in the sense of ‘a standard measure for quantifying’. It seems unrelated to having an inverse, so why exactly define it as a ‘unit’? Here, I’d like to propose an interesting brain teaser.

In the early days of the development of algebra, research on integers was naturally quite active. The reason we refer to the set of integers as $\mathbb{Z}$ is because of the German Zahlring, with ‘Zahl-’ meaning ’number’, and ‘-ring’ translating to ring, as is well-known. Accepting that many concepts used in algebra originated from the field of number theory isn’t too difficult.

Now, consider the integer field $\mathbb{Z}$.

$\mathbb{Z}$ has infinitely many integers as elements. Among these, only $1$ serves as the multiplicative identity, and the elements that have an inverse are only $-1$ and $1$. For those familiar enough with mathematics to study abstract algebra, calling $-1$ and $1$ ‘units’ wouldn’t feel awkward. With this background, as one explores various algebraic structures beyond integers, it might have seemed appropriate to refer to these as units.

Coming up to $\mathbb{R}$, excluding $0$, every element $r \in \mathbb{R}$ has a multiplicative inverse $\displaystyle {{1} \over {r}} \in \mathbb{R}$, so every element except for $0$ is a unit. Thinking about it, you can multiply some number $a$ to $r$ to get the number you want $x$, so there’s no reason $r \ne 1$ can’t serve as a unit too. And this certain number $a$ is obviously $a = r^{-1}x$, as you couldn’t be certain without the existence of $r^{-1}$.