Rings and domains · Definition

Unit of a ring

modern algebrarings

Units are the elements by which you can «divide» inside a ring, and they measure how far the ring is from being a field.

Definition

A unit of a ring \(R\) with identity \(1\) is an element \(u\) that has a multiplicative inverse \(v\) within \(R\).

\[uv=vu=1\]

If \(u\) is a unit, its inverse \(v\) is unique and is written \(u^{-1}\). The elements \(1\) and \(-1\) are always units, and zero never is except in the trivial ring. The product of two units is a unit, since \(uv\) has inverse \(v^{-1}u^{-1}\), and the inverse of a unit is also a unit; this is why the set \(R^\times\) of units is a group under multiplication. The sum of units, however, need not be a unit.

Guiding examples: in \(\mathbb{Z}\) the units are \(\pm1\); in \(\mathbb{Z}_n\), the \(a\) with \(\gcd(a,n)=1\), and their number is \(\varphi(n)\) (for instance, \(\mathbb{Z}_{12}\) has four); in \(\mathbb{Z}[i]\) they are \(\pm1,\pm i\); and in \(\mathbb{Z}[\sqrt2]\) there are infinitely many, such as the powers of \(1+\sqrt2\). A field is exactly a commutative ring in which every nonzero element is a unit.

A typical mistake is to confuse «unit» with the identity element \(1\). The identity is one particular unit, but a ring can have many more. Another mistake is to think that an element with no inverse is always a zero divisor; in \(\mathbb{Z}\), \(2\) is not a unit and it is not a zero divisor either. That difference is studied in the next concept, the zero divisor.

Units underlie later notions: two elements are associates when they differ by a unit, an irreducible element cannot be factored except with units, and unique factorization is stated «up to units». So it is worth knowing how to compute them in each concrete ring.

Think about it

If \(u\) is a unit and \(ua=ub\), what follows?

That \(a=b\): multiply on the left by \(u^{-1}\) to get \(a=b\). This is the cancellation law for units, which does not hold for arbitrary elements of the ring.

What is the group of units of a field \(F\)?

All of \(F\setminus\{0\}\), under multiplication; for example, \(\mathbb{Q}^\times=\mathbb{Q}\setminus\{0\}\). That is precisely the definition of a field.

Exercises 10

What comes next

References

  • David S. Dummit, Richard M. Foote, Abstract Algebra, 3rd edition, Wiley, 2004. §7.1
  • Joseph A. Gallian, Contemporary Abstract Algebra, 9th edition, Cengage, 2017. cap. 12
  • I. N. Herstein, Abstract Algebra, 3rd edition, Wiley, 1996. §3.1
  • John B. Fraleigh, A First Course in Abstract Algebra, 7th edition, Pearson, 2002. §18