Morphisms, ideals and quotient rings

MM329 Topics in Abstract Algebra 

Morphisms, Ideals and Quotient Rings 

Definitions 

Let R and S be rings. A morphism (or homomorphism) from R to S is a mapping 

θ : R → S such that, for all x, y ∈ R, 

A bijective morphism is an isomorphism. 

The kernel of the morphism θ : R → S is 

θ(x + y) = θ(x) + θ(y) and θ(xy) = θ(x)=θ(y). 

{ θ } 

ker θ = x∈ R: ( x) 

= 0 . 

Proposition 6 (Kernels and images) 

Let θ : R → S be a morphism of rings. Then 

(i) ker θ = { ∈ : θ( ) = } 

x R x 0 is a subring of R. 

(ii) im θ = { y ∈ S: y = θ( x) 

x ∈R} 

for some is a subring of S. 

Proof 

Exercise: 

use the subring test (Proposition 2) and compare with the group theory case (see 

MM225). 

Note 

If R and S are both rings with 1, denoted 1 R and 1 S respectively, and 

θ : R → S is a morphism, then it is not necessarily the case that θ(1 R ) = 1 S . 

Proposition 7 (Preservation of identities) 

Let R and S be rings with 1, denoted 1 R and 1 S respectively, and let θ : R → S be a 

surjective morphism. Then θ(1 R ) = 1 S . 

Proof: 

Let y ∈ S. Then, since θ is surjective, y = θ(x) for some x ∈ R. Now 

θ(1 R )y = θ(1 R )=θ(x) = θ(1 R x) = θ(x) = y. 

Hence θ(1 R ) is the identity element of S and, since identity elements are unique, it follows 

that θ(1 R ) = 1 S . 

Proposition 8 (Properties of kernels) 

Let θ : R → S be a morphism of rings and let K = ker θ . Then 

(i) (K, +) is a subgroup of (R, +) 

(ii) for all x ∈ K, r ∈ R, both xr and rx belong to K. 

MM329 Morphisms, ideals and quotient rings 

© John Taylor

Proof 

(i) Group theory. 

(ii) Let x ∈ K, r ∈ R. Then 

θ(xr) = θ(x)=θ(r) (since θ is a morphism) 

= 0=θ(r) (since x ∈ K so θ(x) = 0) 

= 0. 

Therefore xr ∈ K. 

The proof that θ(rx) = 0 (and hence rx ∈ K) is similar. 

Definition 

Let R be a ring. A subset I of R is an ideal of R, written I k R , if 

(I1) (I, +) is a subgroup of (R, +) 

(I2) x ∈ I, r ∈ R ⇒ xr ∈ I and rx ∈ I. 

A proper ideal is an ideal I not equal to {0} or R; this is denoted I 

i R . 

Notes 

1. Every ideal I of R is a subring of R. 

2. The kernel of a morphism θ : R → S is an ideal of R (proposition 8). 

3. (I, +) is a normal subgroup of (R, +) since (R, +) is Abelian. 

By note 3, if I is an ideal of R, then the set of (right) cosets forms a group under addition; 

RI, + . We shall write the cosets of this group additively; that is, 

this is the quotient group ( ) 

the coset containing r ∈ R will be denoted I + r or r + I: 

I + r = {x + r : x ∈ I} and RI = {I + r : r ∈ R}. 

Proposition 9 (Quotient rings) 

Let I be an ideal of a ring R. Then 

(i) RI is a ring 

(ii) the mapping π : R → RI, r I + r is a surjective morphism. 

Proof. 

Below. 

Example 12 = {0, 1, 2, …, 11}. 

Let I = {0, 3, 6, 9}. Then I is an ideal of 12 . 

Proof The set I contains all multiples of 3 in 12 . 

(I1) Clearly I ≠ ∅. The difference between any two multiples of 3 is also a 

multiple of 3; more precisely: x, y ∈ I ⇒ x – y ∈ I. 

MM329 Morphisms, ideals and quotient rings Page 2

It follows from the second subgroup test (MM225) that (I, +) is a subgroup 

of ( 12 , +). 

(I2) Let x be a multiple of 3; then any multiple of x is also a multiple of 3. 

More precisely, x ∈ I, r ∈ 12 ⇒ rx = xr ∈I. 

Therefore I is an ideal of 12 . 

The cosets are: 

Therefore: 

I = {0, 3, 6, 9} = I + 0 = I + 3 = I + 6 = I + 9 

I + 1 = {1, 4, 7, 10} = I + 4 = I + 7 = I + 10 

I + 2 = {2, 5, 8, 11} = I + 5 = I + 8 = I + 11. 

RI = {I, I + 1, I + 2}. 

The Cayley tables for RI under addition and multiplication are given below. The 

operations on cosets are: 

(I + r) + (I + s) = I + (r + s) 

(I + r)(I + s) = I + (rs) 

+ I I + 1 I + 2 

I 

I + 1 

I + 2 

× I I + 1 I + 2 

I 

I + 1 

I + 2 

The mapping π : R → RI defined in proposition 9 is: 

0 I, 1 I + 1, 2 I + 2, 

3 I, 4 I + 1, 5 I + 2, 

6 I, 7 I + 1, 8 I + 2, 

9 I, 10 I + 1, 11 I + 2, 


Proof of Proposition 9 

Suppose I is an ideal of R. 

(i) We know from group theory that ( RI, +) is an Abelian group. 

First we must show that multiplication on cosets is a well-defined operation. In other 

words, multiplication of cosets does not depend on the choice of labels for the cosets: 

if I + r 1 = I + r 2 and I + s 1 = I + s 2 then I + r 1 s 1 = I + r 2 s 2 . 

Suppose I + r 1 = I + r 2 and I + s 1 = I + s 2 . 

(ii) 

Then r 1 = r 2 + x for some x ∈ I and s 1 = s 2 + y for some y ∈ I. 

Now rs 11= ( r2 + x)( s2+ 

y) 

= rs 22+ ry 2 + xs2+ 

xy 

= rs + z 

22 

where z = r2y+ xs2 + xy. 

Now r 2 y ∈ I, xs 2 ∈ I and xy ∈ I (by property I2 of ideals). 

Therefore z ∈ I (by property I1 of ideals). It follows that I + r 1 s 1 = I + r 2 s 2 . Therefore 

multiplication of cosets is well-defined. 

Now the verification of axioms (R2) and (R3) is straightforward. Each follows from 

the corresponding axiom in R. For example, below is a verification of one of the 

distributive laws. 

Let I + r, I + s, I +t ∈ RI. Then 

( ) ( ) 

( I + r) ( I + s) + ( I + t) = ( I + r) I + ( s+ 

t) 

= I + r( s+ 

t) 

= I + ( rs+ 

rt) 

= ( I + rs) + ( I + rt) 

= ( I + r)( I + s) + ( I + r)( I + t). 

The other verifications are left as an exercise. 

Therefore RI is a ring. 

It is now easy to check that π is a morphism: 

π(r + s) = I + (r + s) = (I + r) + (I + s) = π(r) + π(s) 

and π(rs) = I + (rs) = (I + r)(I + s) = π(r)π(s). 

Since every element of RI is of the form I + r = π(r) for some r ∈ R, it follows 

that π is surjective. 

Proposition 10 

Let R be a ring with 1 and let I be an ideal of R. Then the quotient ring RI has identity 

I + 1. 

Proof 

By proposition 9 (ii), π is a surjective morphism. The result now follows from proposition 7. 



Let R be a ring with 1 and let I be an ideal of R. If 1 ∈ I then I = R. 

Proof 

Suppose I is an ideal of R and 1 ∈ I. Then, for all r ∈ R, r = 1.r ∈ I (by property I2 of 

ideals). Therefore I = R. 


A field has no proper ideals. 

Proof 

Let F be a field and let I ≠ {0} be an ideal of F. Then there exists x ∈ I, x ≠ 0. 

Since F is a field, x has an inverse x –1 ∈ F. 

Now x ∈ I and x –1 ∈ F so xx –1 ∈ I (by property I2 of ideals). Hence 1 ∈ I, so I = F, 

by proposition 11. 


Let θ : F → R be a ring morphism where F is a field. 

Them θ is either injective or trivial (maps everything to 0). 

Proof 

Since K = kerθ is an ideal of F, it follows from proposition 12 that K = {0} or K = F. 

Therefore θ is injective or trivial (respectively). 

Proposition 14 (First isomorphism theorem for rings) 

(i) Let θ : R → S be a surjective morphism of rings and let K = kerθ . Then 

RK 

≅ S. 

(ii) 

Let θ : R → S be a morphism of rings and let K = kerθ . Then 

RK≅ imθ . 

Proof Part (ii) follows from part (i), so we prove part (i) only. 

Suppose θ : R → S is a surjective morphism of rings and let K = kerθ . 

Define φ : RK→ S by φ : K+ 

r θ( r) 

. 

The mapping φ is illustrated in the following diagrams. 

R 

θ 

S 

r 

θ 

θ(r) 

π φ =====π φ 

RK I + r 


From the first isomorphism theorem for groups, φ is an isomorphism of the additive groups: 

( RK, + ) ≅ ( S, + ). It only remains to show that φ is a morphism of rings; i.e. that φ 

preserves multiplication. 

φ( ( I + r)( I + s) ) = φ( I + rs) 

= θ( rs) ( definition of φ) 

= θ()() s θ s ( θ is a morphism) 

= φ( I + r) φ( I + s). 

Therefore φ is an isomorphism of rings, so RK 

≅ S. 


A field has no ‘proper’ quotient rings. More precisely, if F is a field and FI is a quotient 

FI≅ 

F or FI≅ 

0 . 

ring, then { } 

Proof 

This follows immediately from proposition 12.

Morphisms, ideals and quotient rings

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