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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
\begin{align*} \phi( a + b ) & = \phi( a ) + \phi(b)\\ \phi( a b ) & = \phi( a) \phi(b)\\ \phi(1_R) & = 1_S \end{align*}
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
\begin{align*} \phi( a + b ) & = (a + b) \pmod{n}\\ & = a \pmod{n} + b \pmod{n}\\ & = \phi( a ) + \phi(b) \end{align*}
and
\begin{align*} \phi( a b ) & = ab \pmod{n}\\ & = a \pmod{n}\cdot b \pmod{n}\\ & = \phi( a ) \phi(b). \end{align*}
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
\begin{gather*} \phi_{\alpha}( f + g ) = (f + g)( \alpha) = f(\alpha) + g(\alpha) = \phi_{\alpha}( f ) + \phi_{\alpha}(g )\\ \phi_{\alpha}( f g ) = (f g)( \alpha) = f(\alpha) g(\alpha) = \phi_{\alpha}( f ) \phi_{\alpha}(g ). \end{gather*}
Ring homomorphisms of the type \(\phi_{\alpha}\) are called evaluation homomorphisms.
In the next proposition we will examine some fundamental properties of ring homomorphisms.

Proof.
We prove parts (3) and (4) and leave the other parts as an exercise.
For (3):
\begin{equation*} \phi(0) = \phi(0+0) = \phi(0) + \phi(0)\text{,} \end{equation*}
and so by adding \(-\phi(0)\) to both sides, we have that \(\phi(0) = 0\text{.}\)
To prove (4),
\begin{equation*} \phi(-a) + \phi(a) = \phi(-a + a) = \phi(0) = 0 \end{equation*}
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
\begin{equation*} R = \left\{ \begin{pmatrix} a & b \\ -b & a \end{pmatrix} : a, b \in \mathbb R \right\}\text{.} \end{equation*}
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:
\begin{equation*} \phi(a+bi) = \begin{pmatrix} a & b \\ -b & a \end{pmatrix}\text{.} \end{equation*}
To check that \(\phi\) is an isomorphism, we need to show that it preserves addition:
\begin{align*} \phi\big((a+bi) + (a'+b'i)\big) &= \phi\big((a+a') + (b+b')i\big)\\ &= \begin{pmatrix}a + a' & b + b' \\ -b - b' & a + a'\end{pmatrix}\\ &= \begin{pmatrix}a & b \\ -b & a \end{pmatrix} + \begin{pmatrix}a' & b' \\ -b' & a' \end{pmatrix}\\ &= \phi(a + bi) + \phi(a' + b'i). \end{align*}
For multiplication, we multiply before the function \(\phi\text{:}\)
\begin{align*} \phi\big((a+bi)(a' + b'i)\big) &= \phi(aa' + ab'i + ba'i - bb')\\ &= \phi((aa' -bb') + (ab' + ba')i)\\ &= \begin{pmatrix}aa' - bb' & a b' + ba' \\ -ab' -b a' & aa' + bb'\end{pmatrix} \end{align*}
and after:
\begin{align*} \phi(a+bi)\phi(a+b'i) &= \begin{pmatrix}a & b \\ -b & a \end{pmatrix} \begin{pmatrix}a' & b' \\ -b' & a' \end{pmatrix}\\ &= \begin{pmatrix}aa' - bb' & a b' + ba' \\ -ba' -a b' & aa' - bb'\end{pmatrix}\text{,} \end{align*}
which both give the same matrix. Therefore, \(\phi\) is an isomorphism and \(R\) and \(\mathbb C\) are isomorphic.
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