The Characteristic of an integral domain
We analyzed different properties of rings: the commutativity of multiplication, the existence of 1, the existence of zero divisors, being principal ring. If a ring differs from another for just one of these properties they cannot be isomorphic. We now study another properties which allows us to further characterize rings.
Remember, that in a ring R, the writing
n ⋅ a, n ∈ ℕ, a ∈ R
means the sum a, n times, i.e.
n ⋅ a := a + a + ... + a
n times
We studied rings in which if we add an element to itself a certain number of times, we obtain the zero of the ring. The characteristic of ℤn is n; For example in ℤ5, if we add five times an element to itself we obtain the zero, whilst in ℤ there is no such n > 0 ∈ ℕ, such that n ⋅ z = 0, ∀z ∈ ℤ. We give the following definition
4.14.1 Definition. Let R a ring. If there is no positive integer n such that n ⋅ a = 0, for every a ∈ R, we say that the characteristic of R is 0. If instead there exists such a n, then let m the least positive integer such that m ⋅ a = 0, ∀a ∈ R, we say the characteristic of R is m. □
4.14.2 Examples.
The ring ℤ, has characteristic zero since nx = 0 for all x ∈ ℤ requires n = 0. For the same reason, the fields ℝ, ℂ and ℚ have all characteristic zero. ■
Consider the ring ℤ6. For the various elements of ℤ6, we have
1[0] = [0] 6[1] = [0] 3[2] = [0]
2[3] = [0] 3[4] = [0] 6[5] = [0].Although smaller positive integers work for some individual elements of ℤ6, the smallest positive integer m such that m[a] = [0] for all [a] ∈ ℤ6 is m = 6. Thus ℤ6 has characteristic 6. This example generalizes readily, and we see that ℤn has characteristic n. ■
ℤn[x] of rational functions over ℤn is infinite and has characteristic p. ■
𝕂[x] and Gauss integers ℤ[i] have characteristic zero. ■
4.14.3 Proposition. The characteristic of an Integral domain, R, is either 0 or a prime number.
Proof. Suppose that the characteristic, m of R, is neither zero nor a prime number, so suppose m = rs. Then, for every a ∈ R, 0 = m ⋅ a = (rs)a, from which (rs)a ⋅ a = 0. Since (rs)ab = (ra)(sb), we have 0 = (r ⋅ a)(s ⋅ a), which is the product of two elements of R, equal to zero. Since R is an integral domain, one of the two factors is non-zero. But this contradicts that m is minimal. Thus m must be prime. □
In a ring with identity the proof would be even simpler. Let m a non-trivial factorization m = rs. Then
(r ⋅ 1)(s ⋅1) = (1 + ··· + 1)(1 + ··· + 1) = (1 + 1 + ··· + 1) = (rs)1 = 0
(where we have added 1 to itself r times, s times, and then rs times, in the previous computation). But because a field has no zero divisors, this means that either r ⋅ 1 or s ⋅ 1 is zero, and this contradicts that m is minimal, because we would have s ⋅ a = s ⋅ 1 ⋅ a = 0 ⋅ a = 0 for all a ∈ R. So in a field R with identity, the characteristic is the least m such that m · 1 = 0.
The rings, ℤ and ℤ5[X] are both infinite integral domains which differ by the characteristic, then they cannot be isomorphic. ■
From now on we indicate with p = char F, the positive characteristic of an integral domain. The above definitions hold in particular in the case of the integral domain being a field.
Remark 4.14.4. A finite field, F, has a prime number characteristic, as 1 ∈ F, the elements 1F, 1F + 1F = 2 ⋅ 1F, 1F + 1F + 1F = 3 ⋅ 1F, and in general n ⋅ 1F, n ∈ ℕ, all belong to F. There a two cases to consider:
The elements n ⋅ 1F, n ∈ ℕ, are all distinct. In this case, the field F is infinite;
The elements n ⋅ 1F, n ∈ ℕ, are not all distinct. In this case, there exists r,s ∈ ℕ with r > s such that r ⋅ 1F = s ⋅ 1F, and therefore, (r − s) ⋅ 1F = 0, where r − s is a positive integer. Hence there exist a least positive integer p such that p ⋅ 1F = 0. In this case, the field F is finite.
Nota bene. A finite field, has finite (positive) characteristic, but the converse is not true, if the characteristic of a field is finite the field could not be finite. For example ℤp which is a infinite field has characteristic p. Thus the fields of real, complex and rational numbers, as well as the ring of the integers have characteristic 0. ■
4.14.5 Definition. We call prime subfield or fundamental subfield, P of a field F, the intersection of all subfields of F.
Since F is a field, there exists at least a subfield of F, and the intersection of an arbitrary number of subfields is never empty since every subfield contains 0 and 1.
4.14.6 Theorem. A field F with characteristic zero has prime subfield isomorphic to ℚ. If F has characteristic p, the primary subfield is isomorphic to ℤp.
Proof. Consider first the set D ⊂ F, of multiples of 1F, that is
D := {m · 1F | m ∈ ℤ}
It is easy to see that D is a ring, contained in the prime subfield P of F. This is called the fundamental subring. We define the following map
φ: ℤ ⟶ F
z ⟼ z · 1F
φ is a ring homomorphism, for we have
φ(z + z') = (z + z') · 1F = z · 1F + z'· 1F = φ(z) + φ(z')
φ(zz') = zz' · 1F = z · 1F · z' · 1F = φ(z) · φ(z')
By definition φ(ℤ) = D; Moreover Kerφ = {z ∈ ℤ | z · 1F = 0F}. Thus,
If char F = 0, then Kerφ = {0}, that is φ is injective and D ≃ ℤ. The fundamental subfield is the smallest subfield containing ℤ, i.e. ℚ.
If char F = p, then Kerφ = pℤ, because p represents the least positive integer in Kerφ.
By the fundamental homomorphism theorem for rings, D ≃ ℤ/pℤ, which is a field (= fundamental subfield). □
Thus every field with characteristic 0 includes the rational field, that is it can be thought as extension of ℚ, whilst every field with characteristic p includes the field ℤp, i.e. it is an extension of ℤp.
In ℤp, p · 1 = 0, and so ℤp, does not contain a copy of ℤ.
Examples 4.14.7. Calculate the characteristic of 5ℤ, ℤ x 3ℤ and ℤ12 x ℤ6. More generally, what is the characteristic of ℤm x ℤn.
Char(5ℤ) = 0; Char(ℤ x 3ℤ) = 0; Char(ℤ12 x ℤ6) = 3. We use the operations defined on the direct sum. The ring has unity (1,1): the characteristic is the smallest positive integer k such that k(1,1) = (k[1]m, k [1]n) = 0. We want the integer k such that [1]m summed k times gives as result [0]m, and such that [1]n summed k times gives [0]n. We know that if we sum up [1]m, m times and [1]n, n times we get [0]m and [0]n respectively. So we have to find the least common multiple k of m and n, thus lcm(m,n) is the characteristic of ℤm x ℤn.