1
Integral Domains
As we saw in the example of the ring ℤ6 , in some rings it’s possible to have two
non-zero elements 𝑎, 𝑏 such that 𝑎𝑏 = 0. In ℤ6 , (2)(3) = 0. In elementary
algebra you learn to solve quadratic equations by factoring.
Ex. Solve 𝑥 2 − 3𝑥 + 2 = 0 where 𝑥 ∈ ℝ.
𝑥 2 − 3𝑥 + 2 = (𝑥 − 2)(𝑥 − 1) = 0
We conclude that either 𝑥 − 2 = 0 or 𝑥 − 1 = 0 and the solutions
are 𝑥 = 2, 1.
But this relies on the idea that if 𝑎𝑏 = 0 then 𝑎 = 0 or 𝑏 = 0 (or both). This is
true for a field like ℝ or ℂ, but for a general ring 𝑎𝑏 = 0 does not imply 𝑎 = 0
or 𝑏 = 0. Thus in a general ring we may get more that 2 solutions to a quadratic
equation.
Ex. Solve 𝑥 2 − 3𝑥 + 2 = 0 where 𝑥 ∈ ℤ6 .
We start by factoring:
𝑥 2 − 3𝑥 + 2 = (𝑥 − 2)(𝑥 − 1) = 0.
Notice that in ℤ6 there are several pairs of non-zero factors whose
product is 0.
(2)(3) = (3)(2) = (3)(4) = (4)(3) = 0.
, 2
So, for example if 𝑥 − 2 = 2 and 𝑥 − 1 = 3 (both must happen) has a
solution in ℤ6 (which it does, 𝑥 = 4) then 4 is a solution to the quadratic
equation in ℤ6 .
Notice 42 − 3(4) + 2 = 16 − 12 + 2 = 6 = 0 (𝑚𝑜𝑑 6).
So in this case, since the factors are 𝑥 − 2 and 𝑥 − 1, the pair of
factors whose product is 0 must differ by 1. The other pair that works in
ℤ6 is 𝑥 − 2 = 3 and 𝑥 − 1 = 4, i.e. 𝑥 = 5.
Since (𝑥 − 2)(𝑥 − 1) = 0 also has 𝑥 = 2, 𝑥 = 1 as solutions we have
the full set of solutions in ℤ6 is 𝑥 = 1, 2, 4, 5.
Def. If 𝑎 and 𝑏 are non-zero elements of a ring 𝑅 such that 𝑎𝑏 = 0, then 𝑎 and
𝑏 are called zero divisors.
Theorem: In the ring ℤ𝑛 , the zero divisors are those non-zero elements
that are not relatively prime to 𝑛.
Corollary: If 𝑝 is a prime number, then ℤ𝑝 has no zero divisors.
Theorem: The cancellation laws (e.g. if 𝑏𝑎 = 𝑐𝑎 then 𝑏 = 𝑐) hold in a
ring 𝑅 if, and only if, 𝑅 has no zero divisors.
Integral Domains
As we saw in the example of the ring ℤ6 , in some rings it’s possible to have two
non-zero elements 𝑎, 𝑏 such that 𝑎𝑏 = 0. In ℤ6 , (2)(3) = 0. In elementary
algebra you learn to solve quadratic equations by factoring.
Ex. Solve 𝑥 2 − 3𝑥 + 2 = 0 where 𝑥 ∈ ℝ.
𝑥 2 − 3𝑥 + 2 = (𝑥 − 2)(𝑥 − 1) = 0
We conclude that either 𝑥 − 2 = 0 or 𝑥 − 1 = 0 and the solutions
are 𝑥 = 2, 1.
But this relies on the idea that if 𝑎𝑏 = 0 then 𝑎 = 0 or 𝑏 = 0 (or both). This is
true for a field like ℝ or ℂ, but for a general ring 𝑎𝑏 = 0 does not imply 𝑎 = 0
or 𝑏 = 0. Thus in a general ring we may get more that 2 solutions to a quadratic
equation.
Ex. Solve 𝑥 2 − 3𝑥 + 2 = 0 where 𝑥 ∈ ℤ6 .
We start by factoring:
𝑥 2 − 3𝑥 + 2 = (𝑥 − 2)(𝑥 − 1) = 0.
Notice that in ℤ6 there are several pairs of non-zero factors whose
product is 0.
(2)(3) = (3)(2) = (3)(4) = (4)(3) = 0.
, 2
So, for example if 𝑥 − 2 = 2 and 𝑥 − 1 = 3 (both must happen) has a
solution in ℤ6 (which it does, 𝑥 = 4) then 4 is a solution to the quadratic
equation in ℤ6 .
Notice 42 − 3(4) + 2 = 16 − 12 + 2 = 6 = 0 (𝑚𝑜𝑑 6).
So in this case, since the factors are 𝑥 − 2 and 𝑥 − 1, the pair of
factors whose product is 0 must differ by 1. The other pair that works in
ℤ6 is 𝑥 − 2 = 3 and 𝑥 − 1 = 4, i.e. 𝑥 = 5.
Since (𝑥 − 2)(𝑥 − 1) = 0 also has 𝑥 = 2, 𝑥 = 1 as solutions we have
the full set of solutions in ℤ6 is 𝑥 = 1, 2, 4, 5.
Def. If 𝑎 and 𝑏 are non-zero elements of a ring 𝑅 such that 𝑎𝑏 = 0, then 𝑎 and
𝑏 are called zero divisors.
Theorem: In the ring ℤ𝑛 , the zero divisors are those non-zero elements
that are not relatively prime to 𝑛.
Corollary: If 𝑝 is a prime number, then ℤ𝑝 has no zero divisors.
Theorem: The cancellation laws (e.g. if 𝑏𝑎 = 𝑐𝑎 then 𝑏 = 𝑐) hold in a
ring 𝑅 if, and only if, 𝑅 has no zero divisors.
