Ring Homomorphisms

In the previous paragraph we studied the notion of isomorphism between rings. We turn our attention now to ring homomorphisms. A ring homomorphism is a mapping from one ring to another that preserves both ring operations but which is not bijective.

4.2.0 Definition. Give two rings (R, +, ·) and (R', +, ·) we define homomorphism between R and R' a mapping between φ between R and R' such that

Examples of rings homorphisms

1. The mapping φ: ℤ₆ ⟶ ℤ₆ defined by

   φ([a]) = [3][a]

   It follows from

   φ([a] + [b]) = [3][a + [a]]
   =[3][a] + [3][b]
   =φ([a]) + φ(b])

   that φ preserves addition. For multiplication, we have

   φ([a][b]) = φ([ab]) = [3][ab] = [3ab]

   and

   φ([a]) φ[b]) = [3][a] [3][b] = [9][ab] = [9ab] = [3ab]

   since [9] = [3] in ℤ₆. Thus φ is a homomorphism. It can be verified that the image/range is φ(ℤ₆) = {[0],[3]}. So we see that φ is neither onto nor one-to-one.is thus a ring homorphism.

2. φ: ℤ ⟶ ℤ such that φ(z) = 2z, is not an homomorphism between rings because even if addition is preserved, multiplication is not.

3. φ: ℤ ⟶ ℤ such that φ(z) = |z|, is not an homomorphism between rings because even if multiplication is preserved, addition is not.

When an homomorphism is injective is called monomorphism, when it is onto is called epimorphism.

4.2.1 Proposition. If φ is a homomorphism from the ring R to the ring R', then the image of the zero of R is the zero of R'.

Proof. Indeed φ(0_R) = φ(0_R + 0_R) = φ(0_R) + φ(0_R) and it follows φ(0_R) = 0_R'.□

4.2.2 Corollary. φ(−a) = −φ(a), ∀ a ∈ R.

If two rings R and R' have a unit, in general the φ(1_R) ≠ 1_R'. For example, let

$R=\{\left(\begin{array}{cc}r& 0\\ 0& 0\end{array}\right)|r\in <!--\; \in \; -->\mathbb{R}\}$

and R' is the ring of 2 x 2 matrix with elements in ℝ. Both rings have a unit, nonetheless considering the application

$\mathrm{\varphi <!--\; \varphi \; -->}:\left(\begin{array}{cc}r& 0\\ 0& 0\end{array}\right)⟼<!--\; ⟼\; -->\left(\begin{array}{cc}r& 0\\ 0& 0\end{array}\right)$

an homorphism between R and R' which does not map the unit of R to the unit of R' which are respectively

$$\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)and\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$

The only elements of R' that are hit by φ are of the form [r 0; 0 0], hence the map is not onto. Nevertheless the following is true.

4.2.3 Proposition. Let φ a non-null homomorphism between two rings R and R' with unity. If R' is an integral domain or φ is an epimorphism, then the image of the unity of R is the unit of R'.

Proof. If φ is surjective every r' ∈ R' is equal to φ(r) for some r ∈ R, thus

r' = φ(r) = φ(r ⋅ 1_R) = φ(r) φ(1_R) = r' φ(1_R)

hence φ(1_R) represents the right unity for R'. Analogously we can notice it is also a left unity. If R' is an integral domain φ(r) = φ(r) φ(1_R) implies that (with φ(r) ≠ 0) by the cancellation property φ(1_R) = 1_R'.□

4.2.4 Definition. If φ is a homomorphism from the ring R to the ring R', the kernel of φ is the subset of the elements of R which are mapped onto the zero element of R'

Ker φ := {r ∈ R | φ(r) = 0_R'} □

Alternatively, the kernel of φ is

φ⁻¹(0) = {r ∈ R: φ(r) = 0};

Lemma 4.2.5. Let φ a non-null homomorphism between two rings R and R'. Then φ is injective iff Ker φ = {0}.

Proof. Suppose φ is injective, and let a ∈ Ker φ. Then φ(a) = 0_R' = φ(0_R) (by Proposition 4.2.1), hence a = 0_R by injectivity. Conversely, if Ker φ = {0_R} and a, b ∈ R with φ(a) = φ(b), then 0_R' = φ(a) − φ(b) = φ(a − b), which means a − b ∈ Ker φ. It follows that a − b = 0, and thus a = b. □

As you can notice Ker φ is a subring; By multiplying an element of Ker φ by an element of R from right or left the result is again an element of Ker φ. Indeed let k ∈ Ker φ and r ∈ R, we have

φ(k⋅r) = φ(k) ⋅ φ(r) = 0 ⋅ φ(r) = 0

from which follows k⋅r ∈ Ker φ. Left multiplication leads to r⋅k ∈ Ker φ.

Example 4.2.6. The kernel of φ: ℤ → ℤ/nℤ sending a to [a] is {kn| k ∈ ℤ}, in other words, the set of all multiples of n. This is because [kn] = [0].   ■

4.2.7 Proposition. The homomorphism φ is a monomorphism (injective) if and only if Ker φ = {0_R}

Proof. Let Ker φ = {0_R} and a,b ∈ R, φ(a) = φ(b). Now φ(a − b) = φ(a) − φ(b) = 0. So a − b ∈ Ker φ = {0_R}, i.e. a − b = 0 or a = b. Hence φ is injective. So φ is a monomorphism, from R, onto φ(R).   □

The property of the kernel of a ring homomorphism −that is closed under multiplication by arbitrary elements of the ring−is abstracted in the concept of an ideal.

4.2.6 Definition. An ideal of an arbitrary ring (R, + , ⋅ ) is a subset I of R that is iteself a group under addition and is such that if r ∈ R and a ∈ I, then ra ∈ I.
We call left ideal, I, of R is an additive subgroup such that ∀a ∈ I and r ∈ R, ar ∈ I.
Similarly a right ideal is an additive subgroup I of R such that ∀a ∈ I and r ∈ R, ra ∈ I.
If I is both a left and right ideal of R is is called a two-sided ideal, or simply an ideal of R and it is indicated as

I ⊲̲ R □

Examples of ideals

• Every ring R has two obviuos ideals, namely, {0}, and R itself.

• The even integers form an ideal in the ring ℤ of all integers; it is usually denoted by 2ℤ. This is because the sum of any even integers is even, and the product of any integer with an even integer is also even. Similarly, the set of all integers divisible by a fixed integer n is an ideal denoted nℤ.

• The polynomials 𝕂[x] with zero constant term.

4.2.7 Lemma. If φ: R ⟶ R' is a homomorphim, then Ker φ is an ideal of R.

A left or right ideal is a subring of R but the viceversa is not true consider for example S_a = {x ∈ R | ax = xa} is a subring but not an ideal.

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