Ring Homomorphism

• Def:
Consider two rings R and R0 . A mapping f : R → R0 is called a homomorphism if

1. f (a + b) = f (a) + f (b) and

2. f(ab) = f(a)f(b) ∀a, b ∈ R.

• Lemma 11:
If f : R → R0 is a homomorphism, then

1. f (0) = 0 and
2. f (−a) = −f (a) ∀a ∈ R.

• Note:
In a ring homomorphism f (1) = 10 may not hold.

• Def Kernel:
If f : R → R0 is a homomorphism then the kernel I(f ) is defined as the subset of R

I(f ) = {a ∈ R|f (a) = 00 ∈ R0 }

• Lemma 12:
If f : R → R0 is a homomorphism with kernel I(f ), then

1. I(f ) is a subgroup of hR, +i and

2. If a ∈ I(f ) then ∀ r ∈ R, ra and ar ∈ I(f ).

• The equivalent of Normal subgroup here are called an Ideal.

• Every normal subgroup N of a group G gives us a homomorphism from G onto G/N with
kernel N .

• So we define an ideal here using the property of the kernel of a homomorphism.

• Def Ideal:
A non-empty subset U of R is said to be an ideal of R if

1. U is a subgroup of hR, +i and

2. ∀ u ∈ U and r ∈ R, ur and ru ∈ U .

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 • Since U is a normal subgroup of hR, +i,
we can construct a quotient group R/U .
Consider the cosets a + U and b + U . We define the multiplication of these cosets as

(a + U )(b + U ) = (ab) + U

• With the multiplication defined in the additive quotient group R/U , we can verify that R/U
is a ring.

• R is homomorphic to R/U .
The homomorphic mapping φ : R → R/U is given by φ(a) = a + U

• Proposition 15:
Let R and R0 be rings and f : R → R0 be a homomorphism with kernel U . Then R0 is
isomorphic to R/U .

Proposition 15 can be generalized in the following way:

Let R and R0 be rings with a homomorphic mapping φ : R → R0 . Let W 0 be an ideal of R0 .
Consider a subset W of R defined as

W = {a ∈ R|φ(a) ∈ W 0 }

Then W is an ideal of R. This can be proved as follows:

Let a, b ∈ W . Then φ(a), φ(b) ∈ W 0 .
∴ φ(a + b) = φ(a) + φ(b) ∈ W 0 .
=⇒ a + b ∈ W .
∴ W is closed under + in R.
Also φ(−a) = −φ(a) ∈ W 0 since W 0 is a subgroup of hR0 , +i.
∴ −a ∈ W .
∴ W is a subgroup of hR, +i.
Also it is a normal subgroup of hR, +i.
Now let x ∈ R.
Then φ(xa) = φ(x)φ(a) ∈ W 0 since W 0 is an ideal in R0 .
∴ xa ∈ W .
∴ W is an ideal of R.

If U is the kernel of φ : R → R0 then U ⊆ W .

With W and W 0 so defined the homomorphism theorem on rings can be generalized as follows:

Theorem: With φ : R → R0 and W and W 0 defined as above R/W is isomorphic to R0 /W 0 .

Proof:
Toprove this we construct an isomorphism between R/W and R0 /W 0 .
Define a mapping ψ : R/W → R0 /W 0 as follows:

ψ(a + W ) = φ(a) + W 0

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Then

ψ((a + W ) + (b + W )) = ψ(a + b + W )
= φ(a + b) + W 0
= φ(a) + φ(b) + w0
= φ(a) + W 0 + φ(b) + W 0
= ψ(a + W ) + ψ(b + W )

Similarly ψ((a + W ) · (b + W )) = ψ(a + W ) · ψ(b + W ).

So ψ : R/W → R0 /W 0 is a homomorphism.
To show that this is an isomorphism we prove that the kernel of ψ is W .
The additive identity in R0 /W 0 is W 0 which is the zero element in R0 /W 0 .
Let ψ(a + W ) = W 0 .
Then φ(a) + W 0 = W 0 .
∴ φ(a) ∈ W 0 =⇒ a ∈ W .
∴ a + W = W.
So the only coset in G/W mapped to W 0 is W which is the 0 element in G/W .
∴ ker ψ = {W }.
∴ ψ : R/W → R0 /W 0 is an isomorphism.

This theorem establishes a one-one correspondence between the ideals of R0 and those ideals of R
that contain the kernel of the homomorphic mapping φ : R → R0 .
This theorem helps in proving many other interesting results on ideals. We look at some of them.

The only ideals of a field F are {0} and F .

It is easy to prove this. Consider a 6= 0 in an ideal I of F . Then a−1 a = e ∈ I. Now
xe = x ∈ I, ∀ x ∈ F . This implies I = F . This proves the statement.

Lemma: If R is a commutative ring with unity whose only ideals are {0} and R then R is a field.
The Lemma is the converse of the above statement.
Proof:
Consider a 6= 0 in R. Consider (a) = {ra|r ∈ R}.
It can be proved that (a) is an ideal of R. It is called the ideal generated by a.
Let e be the unity in R. Then a = ea ∈ (a).
Since (a) contains a non-zero element a, it is not the ideal {0}. So (a) = R. So there is an element
x ∈ R such that xa = e, the unity in R. So R contains the inverse of a. So R is a field.

Def: Maximal Ideal

M 6= R is a maximal ideal if for an ideal U ; M ⊂ U ⊂ R, either U = M or U = R.

Eg.: In the ring of integers the ideal (p)) generated by a prime p is maximal. The ideal generated
by a composite number say (12) is contained in the ideals (3). It is also contained in the ideals (4),
(2) and (6).

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Theorem: If R is a commutative ring with unit element and M is an ideal of R, then M is maximal
iff R/M is a field.
Proof:
only if part
M is a maximal ideal. There is a natural homomorphism φ : R → R/M . The kernel of this
homomorphism is M .
There is a one-one correspondence between the ideals W of R that contain M and the ideals W 0
of R/M .
M ⊂ W ⊂ R has only two possibilities: W = M or W = R.
W = M =⇒ W 0 = {00 } = {M } ∈ R/M .
W = R =⇒ W 0 = R/M .
So R/M has only two ideals {00 } and R/M . Hence by the previous lemma R/M is a field.
if part
Here we assume that R/M is a field.
Then the only ideals of R/M are {00 = {M }} and R/M .
There is a one-one correspondence between the ideals of R/M and those in R containing M . So
we have only two such ideals W in R, M ⊂ W ⊂ R.
∴ W = M or W = R, =⇒ M is a maximal.

Eg: Zp ' Z/hpi.

1 The Field of quotients of an integral domain

Z is an integral domain with the usual addition and multiplication. Using the integers we can con-
struct the rational numbers Q that form a field under the addition and multiplication defined in the
rational numbers. Z can then shown to be isomorphic to a subset of Q.
Since Z is the representative example of an integral domain, we naturally explore whether this
construction can be generalized to all integral domains. We show how to do this in this section.
We are accustomed to view a rational number pq as a fraction, understood as q|p. In order to gener-
alize the construct of rational numbers it will be beneficial to view the number pq as just an ordered
pair (p, q) where p is an integer and q is a non zero integer.

Def:
A ring R can be imbedded in a ring R0 if there is an isomorphism of R into R0 .

Theorem: Every Integral domain can be imbedded in a field.

Proof:
Let D be an Integral domain. We consider the set D × D which consists of ordered pairs (a, b)
where a, b ∈ D.
Just like the rational number 29 is the same as 18
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we may have the pair (a, b) to be equivalent to
(c, d). We would like to put all such members of D × D under the same category. This is best
attained by establishing an equivalence relation on D × D. We establish a relation as follows:
(a, b) ∼ (c, d) iff ad = bc.
This is an equivalence relation as can be easily verified.

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We consider the equivalence classes of D×D under this relation. If (a, b) belongs to an equivalence
class we can represent this class by the symbol [a, b].
Our candidate for the required field is this collection of equivalence classes on D × D.
The addition and multiplication of two elements [a, b] and [c, d] is defined as

[a, b] + [c, d] = [ad + bc, bd] and [a, b][c, d] = [ac, bd]

These operations make sense as b 6= 0, d 6= 0 =⇒ bd 6= 0. Since the elements are equivalence

classes which can be represented with any elements in them, if we have [a, b] = [a0 , b0 ] and [c, d] =
[c0 , d0 ] then it is easy to verify that [a, b] + [c, d] = [a0 , b0 ] + [c0 , d0 ] and [a, b][c, d] = [a0 , b0 ][c0 , d0 ].
The addition and multiplications so defined are commutative, which follows from the commutativ-
ity of D under these two operations. They are associative too.
[0, a], a 6= 0 is the additive identity and [−a, b] is the additive inverse of [a, b]. So D × D is an
abelian group under addition.
For non zero a [a, a] is the multiplicative identity, which can be written as [1, 1]. For a 6= 0, the
multiplicative inverse of [a, b] is [b, a] which belongs to the ring.
So D × D without the additive identity forms a group under multiplication. Hence it is a field.
This is called the field of quotients of an Integral domain.
The integral domain D has its isomorphic image inside its field of quotients. The mapping is given
as f (a) = [a, 1]. Verify that this is an isomorphism.