Let us recall that an element
is a zero divisor
if
and there exists
such that
and
.
Let us recall also that because the ring
has a unit
then there is a map

(1) 
which allows to talk of an integer as an element of
.
However a nonzero integer
may become
.
For instance every multiple of
becomes 0
in
.
Observe also the ring
may contain nonzero elements that are neither
zero divisor, nor units. For instance in the ring of square
matrices of order 2 with integer coefficients the matrix
is nonsingular but has no inverse (right or left).
But this is not a good example since we limit ourselves
to commutative rings. So consider instead the following example.
Example 1
Let
and let
be
the ring of univariate polynomials over
.
The ring
has units (for instance the constant polynomial
),
it has zero divisors (for instance the polynomial
since
) and also
elements like
which is not a unit nor
a zero divisor.
Why do we need to take such exotic examples?
Because among the rings you are familiar with, this is
the simplest example of ring having zero divisors, units
and nonzero elements that are not zero divisors or units.
Indeed our usual rings are
where
is a field and the residue class ring
(with
not necessarily prime).
The ring
has no zero divisors
whereas
has zero divisors iff
is not prime.
But for every non prime
the ring
consists only of zero, units and zero divisors.
This follows from the proposition below.
This is why we were led to the
example.
Proposition 1
Let
be an Euclidean domain and
an element of
with
and
.
Consider the residue class ring
where
denotes the ideal generated by
.
Then every element of
is either zero, a zero divisor or a unit.
Proof.
Studying the elements of
(for deciding if they
are null, zero divisors or units) can be done
by studying the
for every
.
So let
be in
.
If
is a multiple of
then
.
Assume from now on that
is not a a multiple of
.
Consider
the gcd of
and
together with the Bezout relation

(2) 
Observe that
cannot be zero
(otherwise
would be zero
and
would divide
and thus
).
If
then
and this shows
is a unit of
.
The same result holds if
is a unit.
Assume from now on that
is not a unit.
Since
divides
and
let
and
be such that

(3) 
Observe that
is not zero
(otheriwse
would write
leading to the contradiction
).
We have

(4) 
This shows that
is a zero divisor of
.
Yes, this was that simple!
Definition 1
Let
be a positive integer and
.

is a
th root of unity if
.

is a primitive
th root of unity if

,

is a unit in
,
 for every prime divisor
of
the element
is neither zero nor a zero divisor.
Remark 1
When
is an integral domain the last condition of Definition 1
becomes: for every prime divisor
of
the element
.
Observe also that a
th root of unity is necessarily a unit.
Example 2
Consider that
is the field
of complex numbers.
The number
is a
a primitive
th root of unity.
Example 3
In
we have
.
However
is not a primitive
th of unity, since
is not a unit in
.
Example 4
In
we have the following computation in AXIOM
(1) > R := PF(17)
(1) PrimeField 17
Type: Domain
(2) > w: R := 3
(2) 3
Type: PrimeField 17
(3) > [w^i for i in 0..16]
(3) [1,3,9,10,13,5,15,11,16,14,8,7,4,12,2,6,1]
Type: List PrimeField 17
(4) > u: R := 2
(4) 2
Type: PrimeField 17
(5) > [u^i for i in 0..16]
(5) [1,2,4,8,16,15,13,9,1,2,4,8,16,15,13,9,1]
Type: List PrimeField 17
The first list shows that
is a primitive
th root of unity.
However with
we have
since
.
Lemma 1
Let
be integers and let
be a primitive
th
root of unity. Then we have

is neither zero nor a zero divisor in
,

.
Proof.
It relies on the formula

(5) 
which holds for every
and every positive integer
.
Let us prove the first statement of the lemma.
Let
be the gcd of
and
.
Let
be such that

(6) 
Since
we have
.
Hence we can cancel a prime factor
of
such that

(7) 
Let

(8) 
in Relation (
5) leading to

(9) 
for some
.
Hence if
would be zero or a zero divisor
then so would be
which is false.
Now applying Relation (
5) with
and
implies that
divides
.
But with Relation (
6) we obtain

(10) 
Hence

(11) 
Therefore
cannot be zero or a zero divisor.
Now let us prove the second statement of the lemma.
By applying Relation (5)
with
and
we have

(12) 
Since
is neither zero nor a zero divisor we otain
the desired formula.
Marc Moreno Maza
20080107