Rings and domains · Definition

Characteristic of a ring

modern algebrarings

Before you start you need to know

The characteristic sums up in one number how repeated addition of the identity behaves, and with it the whole ring.

Definition

The characteristic of a ring \(R\) with identity \(1\) is the least positive integer \(n\) such that adding the identity to itself \(n\) times gives zero; if none exists, it is 0.

\[\operatorname{char}(R)=\min\{\,n>0 : n\cdot1=0\,\}\]

The characteristic is the additive order of 1 when that order is finite, and 0 when 1 has infinite order. So \(\mathbb{Z}_n\) has characteristic \(n\), while \(\mathbb{Z}\), \(\mathbb{Q}\), \(\mathbb{R}\) and \(\mathbb{Z}[x]\) have characteristic 0. In a product \(R\times S\) the characteristic is the least common multiple of those of \(R\) and \(S\), not their product.

If the characteristic is \(n>0\), then \(n\cdot a=(n\cdot1)a=0\) for all \(a\in R\): the number \(n\) annihilates every element, so the additive order of any \(a\) divides \(n\). Thus the characteristic is read off from 1 but governs the whole ring, and in characteristic \(n\) the elements \(k\cdot1\) depend only on \(k\) modulo \(n\).

If a ring without zero divisors had composite characteristic \(n=ab\) with \(1<a,b<n\), then \((a\cdot1)(b\cdot1)=0\) with both factors nonzero, which is impossible. Hence the characteristic of an integral domain is 0 or a prime \(p\); in the latter case the domain contains a copy of \(\mathbb{Z}_p\). Finite fields always have prime characteristic.

A frequent mistake is to confuse the characteristic with the number of elements: the field \(\mathbb{F}_4\) has 4 elements and characteristic 2. This notion will be key for classifying fields: the characteristic is 0 (and they contain \(\mathbb{Q}\)) or a prime \(p\) (and they contain \(\mathbb{Z}_p\)).

Think about it

Why can the characteristic of a finite field not be 0?

With characteristic 0 the multiples \(n\cdot1\) would all be distinct, giving infinitely many elements.

If \(R\) has characteristic 2, what is \(a+a\) for any \(a\)?

Zero: \(a+a=2a=(2\cdot1)a=0\), so every element is its own negative.

Exercises 10

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. 13
  • I. N. Herstein, Abstract Algebra, 3rd edition, Wiley, 1996. §3.2
  • John B. Fraleigh, A First Course in Abstract Algebra, 7th edition, Pearson, 2002. §19