2Rings

IB Groups, Rings and Modules



2.2 Homomorphisms, ideals, quotients and isomorphisms
Just like groups, we will come up with analogues of homomorphisms, normal
subgroups (which are now known as ideals), and quotients.
Definition
(Homomorphism of rings)
.
Let
R, S
be rings. A function
φ
:
R S
is a ring homomorphism if it preserves everything we can think of, i.e.
(i) φ(r
1
+ r
2
) = φ(r
1
) + φ(r
2
),
(ii) φ(0
R
) = 0
S
,
(iii) φ(r
1
· r
2
) = φ(r
1
) · φ(r
2
),
(iv) φ(1
R
) = 1
S
.
Definition
(Isomorphism of rings)
.
If a homomorphism
φ
:
R S
is a bijection,
we call it an isomorphism.
Definition (Kernel). The kernel of a homomorphism φ : R S is
ker(φ) = {r R : φ(r) = 0
S
}.
Definition (Image). The image of φ : R S is
im(φ) = {s S : s = φ(r) for some r R}.
Lemma. A homomorphism φ : R S is injective if and only if ker φ = {0
R
}.
Proof.
A ring homomorphism is in particular a group homomorphism
φ
:
(
R,
+
,
0
R
)
(
S,
+
,
0
S
) of abelian groups. So this follows from the case of
groups.
In the group scenario, we had groups, subgroups and normal subgroups,
which are special subgroups. Here, we have a special kind of subsets of a ring
that act like normal subgroups, known as ideals.
Definition (Ideal). A subset I R is an ideal, written I C R, if
(i)
It is an additive subgroup of (
R,
+
,
0
R
), i.e. it is closed under addition and
additive inverses. (additive closure)
(ii) If a I and b R, then a · b I. (strong closure)
We say I is a proper ideal if I 6= R.
Note that the multiplicative closure is stronger than what we require for
subrings for subrings, it has to be closed under multiplication by its own
elements; for ideals, it has to be closed under multiplication by everything in
the world. This is similar to how normal subgroups not only have to be closed
under internal multiplication, but also conjugation by external elements.
Lemma. If φ : R S is a homomorphism, then ker(φ) C R.
Proof.
Since
φ
: (
R,
+
,
0
R
)
(
S,
+
,
0
R
) is a group homomorphism, the kernel is
a subgroup of (R, +, 0
R
).
For the second part, let
a ker
(
φ
),
b R
. We need to show that their
product is in the kernel. We have
φ(a · b) = φ(a) · φ(b) = 0 · φ(b) = 0.
So a · b ker(φ).
Example.
Suppose
I C R
is an ideal, and 1
R
I
. Then for any
r R
, the
axioms entail 1
R
· r I. But 1
R
· r = r. So if 1
R
I, then I = R.
In other words, every proper ideal does not contain 1. In particular, every
proper ideal is not a subring, since a subring must contain 1.
We are starting to diverge from groups. In groups, a normal subgroup is a
subgroup, but here an ideal is not a subring.
Example.
We can generalize the above a bit. Suppose
I C R
and
u I
is
a unit, i.e. there is some
v R
such that
uv
= 1
R
. Then by strong closure,
1
R
= u · v I. So I = R.
Hence proper ideals are not allowed to contain any unit at all, not just 1
R
.
Example.
Consider the ring
Z
of integers. Then every ideal of
Z
is of the form
nZ = ·· , 2n, n, 0, n, 2n, ···} Z.
It is easy to see this is indeed an ideal.
To show these are all the ideals, let
I C Z
. If
I
=
{
0
}
, then
I
= 0
Z
. Otherwise,
let
n N
be the smallest positive element of
T
. We want to show in fact
I
=
nZ
.
Certainly nZ I by strong closure.
Now let m I. By the Euclidean algorithm, we can write
m = q · n + r
with 0
r < n
. Now
n, m I
. So by strong closure,
m, qn I
. So
r
=
m q · n I
. As
n
is the smallest positive element of
I
, and
r < n
, we must
have r = 0. So m = q · n nZ. So I nZ. So I = nZ.
The key to proving this was that we can perform the Euclidean algorithm on
Z
. Thus, for any ring
R
in which we can “do Euclidean algorithm”, every ideal
is of the form
aR
=
{a · r
:
r R}
for some
a R
. We will make this notion
precise in later.
Definition (Generator of ideal). For an element a R, we write
(a) = aR = {a · r : r R} C R.
This is the ideal generated by a.
In general, let a
1
, a
2
, ··· , a
k
R, we write
(a
1
, a
2
, ··· , a
k
) = {a
1
r
1
+ ··· + a
k
r
k
: r
1
, ··· , r
k
R}.
This is the ideal generated by a
1
, ··· , a
k
.
We can also have ideals generated by infinitely many objects, but we have to
be careful, since we cannot have infinite sums.
Definition
(Generator of ideal)
.
For
A R
a subset, the ideal generated by
A
is
(A) =
(
X
aA
r
a
· a : r
a
R, only finitely-many non-zero
)
.
These ideals are rather nice ideals, since they are easy to describe, and often
have some nice properties.
Definition
(Principal ideal)
.
An ideal
I
is a principal ideal if
I
= (
a
) for some
a R.
So what we have just shown for
Z
is that all ideals are principal. Not all
rings are like this. These are special types of rings, which we will study more in
depth later.
Example. Consider the following subset:
{f R[X] : the constant coefficient of f is 0}.
This is an ideal, as we can check manually (alternatively, it is the kernel of the
“evaluate at 0” homomorphism). It turns out this is a principal ideal. In fact, it
is (X).
We have said ideals are like normal subgroups. The key idea is that we can
divide by ideals.
Definition
(Quotient ring)
.
Let
I C R
. The quotient ring
R/I
consists of the
(additive) cosets
r
+
I
with the zero and one as 0
R
+
I
and 1
R
+
I
, and operations
(r
1
+ I) + (r
2
+ I) = (r
1
+ r
2
) + I
(r
1
+ I) · (r
2
+ I) = r
1
r
2
+ I.
Proposition. The quotient ring is a ring, and the function
R R/I
r 7→ r + I
is a ring homomorphism.
This is true, because we defined ideals to be those things that can be
quotiented by. So we just have to check we made the right definition.
Just as we could have come up with the definition of a normal subgroup
by requiring operations on the cosets to be well-defined, we could have come
up with the definition of an ideal by requiring the multiplication of cosets is
well-defined, and we will end up with the strong closure property.
Proof.
We know the group (
R/I,
+
,
0
R/I
) is well-defined, since
I
is a (normal)
subgroup of R. So we only have to check multiplication is well-defined.
Suppose
r
1
+
I
=
r
0
1
+
I
and
r
2
+
I
=
R
0
2
+
I
. Then
r
0
1
r
1
=
a
1
I
and
r
0
2
r
2
= a
2
I. So
r
0
1
r
0
2
= (r
1
+ a
1
)(r
2
+ a
2
) = r
1
r
2
+ r
1
a
2
+ r
2
a
1
+ a
1
a
1
.
By the strong closure property, the last three objects are in
I
. So
r
0
1
r
0
2
+
I
=
r
1
r
2
+ I.
It is easy to check that 0
R
+
I
and 1
R
+
I
are indeed the zero and one, and
the function given is clearly a homomorphism.
Example.
We have the ideals
nZ C Z
. So we have the quotient ring
Z/nZ
. The
elements are of the form m + nZ, so are just
0 + nZ, 1 + nZ, 2 + nZ, ··· , (n 1) + nZ.
Addition and multiplication is just what we are used to it is addition and
multiplication modulo n.
Note that it is easier to come up with ideals than normal subgroups we
can just pick up random elements, and then take the ideal generated by them.
Example.
Consider (
X
)
C C
[
X
]. What is
C
[
X
]
/
(
X
)? Elements are represented
by
a
0
+ a
1
X + a
2
X
2
+ ··· + a
n
X
n
+ (X).
But everything but the first term is in (
X
). So every such thing is equivalent to
a
0
+ (
X
). It is not hard to convince yourself that this representation is unique.
So in fact C[X]/(X)
=
C, with the bijection a
0
+ (X) a
0
.
If we want to prove things like this, we have to convince ourselves this
representation is unique. We can do that by hand here, but in general, we want
to be able to do this properly.
Proposition
(Euclidean algorithm for polynomials)
.
Let
F
be a field and
f, g F[X]. Then there is some r, q F[X] such that
f = gq + r,
with deg r < deg g.
This is like the usual Euclidean algorithm, except that instead of the absolute
value, we use the degree to measure how “big” the polynomial is.
Proof. Let deg(f) = n. So
f =
n
X
i=0
a
i
X
i
,
and a
n
6= 0. Similarly, if deg g = m, then
g =
m
X
i=0
b
i
X
i
,
with b
m
6= 0. If n < m, we let q = 0 and r = f, and done.
Otherwise, suppose n m, and proceed by induction on n.
We let
f
1
= f a
n
b
1
m
X
nm
g.
This is possible since
b
m
6
= 0, and
F
is a field. Then by construction, the
coefficients of X
n
cancel out. So deg(f
1
) < n.
If n = m, then deg(f
1
) < n = m. So we can write
f = (a
n
b
1
m
X
nm
)g + f
1
,
and
deg
(
f
1
)
< deg
(
f
). So done. Otherwise, if
n > m
, then as
deg
(
f
1
)
< n
, by
induction, we can find r
1
, q
1
such that
f
1
= gq
1
+ r
1
,
and deg(r
1
) < deg g = m. Then
f = a
n
b
1
m
X
nm
g + q
1
g + r
1
= (a
n
b
1
m
X
nm
+ q
1
)g + r
1
.
So done.
Now that we have a Euclidean algorithm for polynomials. So we should be
able to show that every ideal of
F
[
X
] is generated by one polynomial. We will
not prove it specifically here, but later show that in general, in every ring where
the Euclidean algorithm is possible, all ideals are principal.
We now look at some applications of the Euclidean algorithm.
Example.
Consider
R
[
X
], and consider the principal ideal (
X
2
+ 1)
C R
[
X
].
We let R = R[X]/(X
2
+ 1).
Elements of R are polynomials
a
0
+ a
1
X + a
2
X
2
+ ··· + a
n
X
n
| {z }
f
+(X
2
+ 1).
By the Euclidean algorithm, we have
f = q(X
2
+ 1) + r,
with
deg
(
r
)
<
2, i.e
r
=
b
0
+
b
1
X
. Thus
f
+ (
X
2
+ 1) =
r
+ (
X
2
+ 1). So every
element of R[X]/(X
2
+ 1) is representable as a + bX for some a, b R.
Is this representation unique? If
a
+
bX
+ (
X
2
+ 1) =
a
0
+
b
0
X
+ (
X
2
+ 1),
then the difference (
a a
0
) + (
b b
0
)
X
(
X
2
+ 1). So it is (
X
2
+ 1)
q
for some
q
.
This is possible only if
q
= 0, since for non-zero
q
, we know (
X
2
+ 1)
q
has degree
at least 2. So we must have (
a a
0
) + (
b b
0
)
X
= 0. So
a
+
bX
=
a
0
+
b
0
X
. So
the representation is unique.
What we’ve got is that every element in
R
is of the form
a
+
bX
, and
X
2
+ 1 = 0, i.e.
X
2
=
1. This sounds like the complex numbers, just that we
are calling it X instead of i.
To show this formally, we define the function
φ : R[x]/(X
2
+ 1) C
a + bX + (X
2
+ 1) 7→ a + bi.
This is well-defined and a bijection. It is also clearly additive. So to prove this
is an isomorphism, we have to show it is multiplicative. We check this manually.
We have
φ((a + bX + (X
2
+ 1))(c + dX + (X
2
+ 1)))
= φ(ac + (ad + bc)X + bdX
2
+ (X
2
+ 1))
= φ((ac bd) + (ad + bc)X + (X
2
+ 1))
= (ac bd) + (ad + bc)i
= (a + bi)(c + di)
= φ(a + bX + (X
2
+ 1))φ(c + dX + (X
2
+ 1)).
So this is indeed an isomorphism.
This is pretty tedious. Fortunately, we have some helpful results we can use,
namely the isomorphism theorems. These are exactly analogous to those for
groups.
Theorem
(First isomorphism theorem)
.
Let
φ
:
R S
be a ring homomorphism.
Then ker(φ) C R, and
R
ker(φ)
=
im(φ) S.
Proof. We have already seen ker(φ) C R. Now define
Φ : R/ ker(φ) im(φ)
r + ker(φ) 7→ φ(r).
This well-defined, since if
r
+
ker
(
φ
) =
r
0
+
ker
(
φ
), then
r r
0
ker
(
φ
). So
φ(r r
0
) = 0. So φ(r) = φ(r
0
).
We don’t have to check this is bijective and additive, since that comes for
free from the (proof of the) isomorphism theorem of groups. So we just have to
check it is multiplicative. To show Φ is multiplicative, we have
Φ((r + ker(φ))(t + ker(φ))) = Φ(rt + ker(φ))
= φ(rt)
= φ(r)φ(t)
= Φ(r + ker(φ))Φ(t + ker(φ)).
This is more-or-less the same proof as the one for groups, just that we had a
few more things to check.
Since there is the first isomorphism theorem, we, obviously, have more
coming.
Theorem
(Second isomorphism theorem)
.
Let
R S
and
J CS
. Then
J RCR
,
and
R + J
J
= {r + J : r R}
S
J
is a subring, and
R
R J
=
R + J
J
.
Proof. Define the function
φ : R S/J
r 7→ r + J.
Since this is the quotient map, it is a ring homomorphism. The kernel is
ker(φ) = {r R : r + J = 0, i.e. r J} = R J.
Then the image is
im(φ) = {r + J : r R} =
R + J
J
.
Then by the first isomorphism theorem, we know
R J C R
, and
R+J
J
S
, and
R
R J
=
R + J
J
.
Before we get to the third isomorphism theorem, recall we had the subgroup
correspondence for groups. Analogously, for I C R,
{subrings of R/I} {subrings of R which contain I}
L
R
I
{x R : x + I L}
S
I
R
I
I C S R.
This is exactly the same formula as for groups.
For groups, we had a correspondence for normal subgroups. Here, we have a
correspondence between ideals
{ideals of R/I} {ideals of R which contain I}
It is important to note here quotienting in groups and rings have different
purposes. In groups, we take quotients so that we have a simpler group to work
with. In rings, we often take quotients to get more interesting rings. For example,
R
[
X
] is quite boring, but
R
[
X
]
/
(
X
2
+ 1)
=
C
is more interesting. Thus this ideal
correspondence allows us to occasionally get interesting ideals from boring ones.
Theorem
(Third isomorphism theorem)
.
Let
I C R
and
J C R
, and
I J
.
Then J/I C R/I and
R
I
J
I
=
R
J
.
Proof. We define the map
φ : R/I R/J
r + I 7→ r + J.
This is well-defined and surjective by the groups case. Also it is a ring homo-
morphism since multiplication in
R/I
and
R/J
are “the same”. The kernel
is
ker(φ) = {r + I : r + J = 0, i.e. r J} =
J
I
.
So the result follows from the first isomorphism theorem.
Note that for any ring
R
, there is a unique ring homomorphism
Z R
, given
by
ι : Z R
n 0 7→ 1
R
+ 1
R
+ ··· + 1
R
| {z }
n times
n 0 7→ (1
R
+ 1
R
+ ··· + 1
R
| {z }
n times
)
Any homomorphism
Z R
must be given by this formula, since it must send the
unit to the unit, and we can show this is indeed a homomorphism by distributivity.
So the ring homomorphism is unique. In fancy language, we say
Z
is the initial
object in (the category of) rings.
We then know ker(ι) C Z. Thus ker(ι) = nZ for some n.
Definition
(Characteristic of ring)
.
Let
R
be a ring, and
ι
:
Z R
be the
unique such map. The characteristic of
R
is the unique non-negative
n
such
that ker(ι) = nZ.
Example.
The rings
Z, Q, R, C
all have characteristic 0. The ring
Z/nZ
has
characteristic n. In particular, all natural numbers can be characteristics.
The notion of the characteristic will not be too useful in this course. How-
ever, fields of non-zero characteristic often provide interesting examples and
counterexamples to some later theory.