We know that if is a field and if
is a variable over
then
is a PID and a non-zero ideal
of
is maximal if and only if
is prime if and only if
is generated by an irreducible element of
If
is a PID which is not a field, then
could have prime ideals which are not maximal. For example, in
the ideal
is prime but not maximal. In this two-part post, we will find prime and maximal ideals of
when
is a PID.
Notation. Throughout this post, is a PID and
is the polynomial ring in the variable
over
Given a prime element
we will denote by
the natural ring homomorphism
Definition Let be a prime element of
An element
is called irreducible modulo
if
is irreducible in
Let
be the natural ring homomorphism. Then, since
an element
is irreducible modulo
if and only if
is irreducible in
Note that
is a field because
is a PID.
Problem 1. Prove that if is prime and if
is irreducible modulo
then
is a maximal ideal of
If
then
is a prime but not a maximal ideal of
Solution. Clearly So
is a maximal ideal of
because
is irreducible in
and
is a field. So
is a maximal ideal of
If
then
and so
is a domain which implies that
is prime. Finally,
is not maximal because, for example,
Problem 2. Prove that a non-zero ideal of
is prime if and only if either
for some irreducible element
or
for some prime
and some
which is either zero or irreducible modulo
Solution. If is irreducible, then
is a prime ideal of
because
is a UFD. If
or
is irreducible modulo a prime
then
is a prime ideal of
by Problem 1.
Conversely, suppose that is a non-zero prime ideal of
We consider two cases.
Case 1. : Let
Then
is clearly not a unit because then
wouldn’t be a proper ideal of
So, since
and
is a prime ideal of
there exists a prime divisor
of
such that
So
and hence
is a prime ideal of
Thus we have two possibilities. The first possibility is that
which gives us
and therefore
The second possibility is that
for some irreducible element
which gives us
because
Case 2. : Let
be the field of fractions of
and put
Then
is a non-zero prime ideal of
because
is a prime ideal of
Note that
So, since
is a PID,
for some irreducible element
Obviously, we can write
where
and
is irreducible and the gcd of the coefficients of
is one. Thus
and, since
we have
for some
and
But then
and so
because
is prime and
Hence
We will be done if we prove that
To prove this, let
So
for some
Therefore, since the gcd of the coefficients of
is one, we must have
by Gauss’s lemma. Hence
and the solution is complete.
See the next part here!