2Rings

IB Groups, Rings and Modules



2.3
Integral domains, field of factions, maximal and prime
ideals
Many rings can be completely nothing like
Z
. For example, in
Z
, we know that if
a, b 6
= 0, then
ab 6
= 0. However, in, say,
Z/
6
Z
, we get 2
,
3
6
= 0, but 2
·
3 = 0. Also,
Z
has some nice properties such as every ideal is principal, and every integer
has an (essentially) unique factorization. We will now classify rings according to
which properties they have.
We start with the most fundamental property that the product of two non-
zero elements are non-zero. We will almost exclusively work with rings that
satisfy this property.
Definition (Integral domain). A non-zero ring
R
is an integral domain if for all
a, b R, if a · b = 0
R
, then a = 0
R
or b = 0
R
.
An element that violates this property is known as a zero divisor.
Definition (Zero divisor). An element
x R
is a zero divisor if
x 6
= 0 and
there is a y 6= 0 such that x · y = 0 R.
In other words, a ring is an integral domain if it has no zero divisors.
Example. All fields are integral domains, since if
a · b
= 0, and
b 6
= 0, then
a = a ·(b · b
1
) = 0. Similarly, if a 6= 0, then b = 0.
Example. A subring of an integral domain is an integral domain, since a zero
divisor in the small ring would also be a zero divisor in the big ring.
Example. Immediately, we know
Z, Q, R, C
are integral domains, since
C
is a
field, and the others are subrings of it. Also,
Z
[
i
]
C
is also an integral domain.
These are the nice rings we like in number theory, since there we can sensibly
talk about things like factorization.
It turns out there are no interesting finite integral domains.
Lemma. Let
R
be a finite ring which is an integral domain. Then
R
is a field.
Proof. Let a R be non-zero, and consider the ring homomorphism
a · : R R
b 7→ a · b
We want to show this is injective. For this, it suffices to show the kernel is trivial.
If
r ker
(
a ·
), then
a · r
= 0. So
r
= 0 since
R
is an integral domain. So the
kernel is trivial.
Since
R
is finite,
a ·
must also be surjective. In particular, there is an
element
b R
such that
a · b
= 1
R
. So
a
has an inverse. Since
a
was arbitrary,
R is a field.
So far, we know fields are integral domains, and subrings of integral domains
are integral domains. We have another good source of integral domain as follows:
Lemma. Let R be an integral domain. Then R[X] is also an integral domain.
Proof.
We need to show that the product of two non-zero elements is non-zero.
Let f, g R[X] be non-zero, say
f = a
0
+ a
1
X + ··· + a
n
X
n
R[X]
g = b
0
+ b
1
X + ··· + b
m
X
m
R[X],
with
a
n
, b
m
6
= 0. Then the coefficient of
X
n+m
in
fg
is
a
n
b
m
. This is non-
zero since
R
is an integral domain. So
fg
is non-zero. So
R
[
X
] is an integral
domain.
So, for instance, Z[X] is an integral domain.
We can also iterate this.
Notation. Write
R
[
X, Y
] for (
R
[
X
])[
Y
], the polynomial ring of
R
in two vari-
ables. In general, write R[X
1
, ··· , X
n
] = (···((R[X
1
])[X
2
]) ···)[X
n
].
Then if R is an integral domain, so is R[X
1
, ··· , X
n
].
We now mimic the familiar construction of
Q
from
Z
. For any integral
domain
R
, we want to construct a field
F
that consists of “fractions” of elements
in
R
. Recall that a subring of any field is an integral domain. This says the
converse every integral domain is the subring of some field.
Definition (Field of fractions). Let
R
be an integral domain. A field of fractions
F of R is a field with the following properties
(i) R F
(ii)
Every element of
F
may be written as
a ·b
1
for
a, b R
, where
b
1
means
the multiplicative inverse to b 6= 0 in F .
For example, Q is the field of fractions of Z.
Theorem. Every integral domain has a field of fractions.
Proof.
The construction is exactly how we construct the rationals from the
integers as equivalence classes of pairs of integers. We let
S = {(a, b) R × R : b 6= 0}.
We think of (a, b) S as
a
b
. We define the equivalence relation on S by
(a, b) (c, d) ad = bc.
We need to show this is indeed a equivalence relation. Symmetry and reflexivity
are obvious. To show transitivity, suppose
(a, b) (c, d), (c, d) (e, f),
i.e.
ad = bc, cf = de.
We multiply the first equation by f and the second by b, to obtain
adf = bcf, bcf = bed.
Rearranging, we get
d(af be) = 0.
Since
d
is in the denominator,
d 6
= 0. Since
R
is an integral domain, we must
have
af be
= 0, i.e.
af
=
be
. So (
a, b
)
(
e, f
). This is where being an integral
domain is important.
Now let
F = S/
be the set of equivalence classes. We now want to check this is indeed the field
of fractions. We first want to show it is a field. We write
a
b
= [(
a, b
)]
F
, and
define the operations by
a
b
+
c
d
=
ad + bc
bd
a
b
·
c
d
=
ac
bd
.
These are well-defined, and make (
F,
+
, ·,
0
1
,
1
1
) into a ring. There are many
things to check, but those are straightforward, and we will not waste time doing
that here.
Finally, we need to show every non-zero element has an inverse. Let
a
b
6
= 0
F
,
i.e.
a
b
6=
0
1
, or a ·1 6= b · 0 R, i.e. a 6= 0. Then
b
a
F is defined, and
b
a
·
a
b
=
ba
ba
= 1
F
.
So
a
b
has a multiplicative inverse. So F is a field.
We now need to construct a subring of
F
that is isomorphic to
R
. To do so,
we need to define an injective isomorphism φ : R F . This is given by
φ : R F
r 7→
r
1
.
This is a ring homomorphism, as one can check easily. The kernel is the set of
all
r R
such that
r
1
= 0, i.e.
r
= 0. So the kernel is trivial, and
φ
is injective.
Then by the first isomorphism theorem, R
=
im(φ) F .
Finally, we need to show everything is a quotient of two things in
R
. We
have
a
b
=
a
1
·
1
b
=
a
1
·
b
1
1
,
as required.
This gives us a very useful tool. Since this gives us a field from an integral
domain, this allows us to use field techniques to study integral domains. Moreover,
we can use this to construct new interesting fields from integral domains.
Example. Consider the integral domain
C
[
X
]. Its field of fractions is the field
of all rational functions
p(X)
q(X)
, where p, q C[X].
To some people, it is a shame to think of rings as having elements. Instead,
we should think of a ring as a god-like object, and the only things we should
ever mention are its ideals. We should also not think of the ideals as containing
elements, but just some abstract objects, and all we know is how ideals relate to
one another, e.g. if one contains the other.
Under this philosophy, we can think of a field as follows:
Lemma. A (non-zero) ring
R
is a field if and only if its only ideals are
{
0
}
and
R.
Note that we don’t need elements to define the ideals
{
0
}
and
R
.
{
0
}
can be
defined as the ideal that all other ideals contain, and
R
is the ideal that contains
all other ideals. Alternatively, we can reword this as R is a field if and only if
it has only two ideals” to avoid mentioning explicit ideals.
Proof.
(
) Let
I C R
and
R
be a field. Suppose
x 6
= 0
I
. Then as
x
is a unit,
I = R.
(
) Suppose
x 6
= 0
R
. Then (
x
) is an ideal of
R
. It is not
{
0
}
since it
contains
x
. So (
x
) =
R
. In other words 1
R
(
x
). But (
x
) is defined to be
{x · y
:
y R}
. So there is some
u R
such that
x · u
= 1
R
. So
x
is a unit.
Since x was arbitrary, R is a field.
This is another reason why fields are special. They have the simplest possible
ideal structure.
This motivates the following definition:
Definition (Maximal ideal). An ideal
I
of a ring
R
is maximal if
I 6
=
R
and
for any ideal J with I J R, either J = I or J = R.
The relation with what we’ve done above is quite simple. There is an easy
way to recognize if an ideal is maximal.
Lemma. An ideal I C R is maximal if and only if R/I is a field.
Proof. R/I
is a field if and only if
{
0
}
and
R/I
are the only ideals of
R/I
. By
the ideal correspondence, this is equivalent to saying
I
and
R
are the only ideals
of R which contains I, i.e. I is maximal. So done.
This is a nice result. This makes a correspondence between properties of
ideals I and properties of the quotient R/I. Here is another one:
Definition (Prime ideal). An ideal
I
of a ring
R
is prime if
I 6
=
R
and whenever
a, b R are such that a · b I, then a I or b I.
This is like the opposite of the property of being an ideal being an ideal
means if we have something in the ideal and something outside, the product is
always in the ideal. This does the backwards. If the product of two random
things is in the ideal, then one of them must be from the ideal.
Example. A non-zero ideal nZ C Z is prime if and only if n is a prime.
To show this, first suppose
n
=
p
is a prime, and
a · b pZ
. So
p | a · b
. So
p | a or p | b, i.e. a pZ or b pZ.
For the other direction, suppose
n
=
pq
is a composite number (
p, q 6
= 1).
Then n nZ but p 6∈ nZ and q 6∈ nZ, since 0 < p, q < n.
So instead of talking about prime numbers, we can talk about prime ideals
instead, because ideals are better than elements.
We prove a result similar to the above:
Lemma. An ideal I C R is prime if and only if R/I is an integral domain.
Proof.
Let
I
be prime. Let
a
+
I, b
+
I R/I
, and suppose (
a
+
I
)(
b
+
I
) = 0
R/I
.
By definition, (
a
+
I
)(
b
+
I
) =
ab
+
I
. So we must have
ab I
. As
I
is prime,
either
a I
or
b I
. So
a
+
I
= 0
R/I
or
b
+
I
= 0
R/I
. So
R/I
is an integral
domain.
Conversely, suppose
R/I
is an integral domain. Let
a, b R
be such that
ab I
. Then (
a
+
I
)(
b
+
I
) =
ab
+
I
= 0
R/I
R/I
. Since
R/I
is an integral
domain, either
a
+
I
= 0
R/I
or
b
+
I
= 0
R/i
, i.e.
a I
or
b I
. So
I
is a prime
ideal.
Prime ideals and maximal ideals are the main types of ideals we care about.
Note that every field is an integral domain. So we immediately have the following
result:
Proposition. Every maximal ideal is a prime ideal.
Proof. I C R
is maximal implies
R/I
is a field implies
R/I
is an integral domain
implies I is prime.
The converse is not true. For example,
{
0
} Z
is prime but not maximal.
Less stupidly, (
X
)
Z
[
X, Y
] is prime but not maximal (since
Z
[
X, Y
]
/
(
X
)
=
Z
[
Y
]). We can provide a more explicit proof of this, which is essentially the
same.
Alternative proof.
Let
I
be a maximal ideal, and suppose
a, b 6∈ I
but
ab I
.
Then by maximality,
I
+ (
a
) =
I
+ (
b
) =
R
= (1). So we can find some
p, q R
and n, m I such that n + ap = m + bq = 1. Then
1 = (n + ap)(m + bq) = nm + apm + bqn + abpq I,
since n, m, ab I. This is a contradiction.
Lemma. Let
R
be an integral domain. Then its characteristic is either 0 or a
prime number.
Proof.
Consider the unique map
φ
:
Z R
, and
ker
(
φ
) =
nZ
. Then
n
is the
characteristic of R by definition.
By the first isomorphism theorem,
Z/nZ
=
im
(
φ
)
R
. So
Z/nZ
is an
integral domain. So nZ C Z is a prime. So n = 0 or a prime number.