Section 4.3 Ring Homomorphisms
A homomorphism between rings is a function which preserves the operations of addition and multiplication in the ring. More specifically, if \(R\) and \(S\) are rings, then a ring homomorphism is a map \(\phi : R \rightarrow S\) satisfying
for all \(a, b \in R\text{.}\) If \(\phi \colon R \rightarrow S\) is a one-to-one and onto homomorphism, then \(\phi\) is called an isomorphism of rings and \(R\) and \(S\) are called isomorphic.
Example 4.19.
For any integer \(n\) we can define a ring homomorphism \(\phi : {\mathbb Z} \rightarrow {\mathbb Z}_n\) by \(a \mapsto a \pmod{n}\text{.}\) This is indeed a ring homomorphism, since
and
The kernel of the homomorphism \(\phi\) is \(n {\mathbb Z}\text{.}\)
Example 4.20.
Let \(C[a, b]\) be the ring of continuous real-valued functions on an interval \([a,b]\) as in Example 4.5. For a fixed \(\alpha \in [a, b]\text{,}\) we can define a ring homomorphism \(\phi_{\alpha} : C[a, b] \rightarrow {\mathbb R}\) by \(\phi_{\alpha} (f ) = f( \alpha)\text{.}\) This is a ring homomorphism since
Ring homomorphisms of the type \(\phi_{\alpha}\) are called evaluation homomorphisms.
In the next proposition we will examine some fundamental properties of ring homomorphisms.
Proposition 4.21.
Let \(\phi : R \rightarrow S\) be a ring homomorphism.
\(\phi(R)\) is a subring of \(S\text{.}\)
If \(R\) is a commutative ring, then \(\phi(R)\) is a commutative ring.
\(\phi( 0 ) = 0\text{.}\)
\(\phi(-a) = -\phi(a)\) for all \(a \in R\text{.}\)
If \(R\) is a field, then \(\phi(R)\) is a field.
Proof.
We prove parts (3) and (4) and leave the other parts as an exercise.
For (3):
and so by adding \(-\phi(0)\) to both sides, we have that \(\phi(0) = 0\text{.}\)
To prove (4),
which shows that \(\phi(-a)\) is the additive inverse of \(\phi(a)\text{.}\)
We give an example of an isomorphism between two rings.
Example 4.22.
Let \(R\) be the subset of \(\mathbb M_2(\mathbb R)\) consisting of matrices
One can check that \(R\) is a subring of \(\mathbb M_2(\mathbb R)\text{.}\)
We claim that the ring of complex numbers \(\mathbb C\) is isomorphic to \(R\text{.}\) We define the function \(\phi\) to be:
To check that \(\phi\) is an isomorphism, we need to show that it preserves addition:
For multiplication, we multiply before the function \(\phi\text{:}\)
and after:
which both give the same matrix. Therefore, \(\phi\) is an isomorphism and \(R\) and \(\mathbb C\) are isomorphic.