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Section 16.3 Ring Homomorphisms and Ideals

In the study of groups, a homomorphism is a map that preserves the operation of the group. Similarly, a homomorphism between rings 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) \end{align*}

for all \(a, b \in R\text{.}\) If \(\phi : R \rightarrow S\) is a one-to-one and onto homomorphism, then \(\phi\) is called an isomorphism of rings.

The set of elements that a ring homomorphism maps to \(0\) plays a fundamental role in the theory of rings. For any ring homomorphism \(\phi : R \rightarrow S\text{,}\) we define the kernel of a ring homomorphism to be the set

\begin{equation*} \ker \phi = \{ r \in R : \phi( r ) = 0 \}. \end{equation*}
Example 16.20

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 16.21

Let \(C[a, b]\) be the ring of continuous real-valued functions on an interval \([a,b]\) as in Example 16.5. 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. The proof of the proposition is left as an exercise.

In group theory we found that normal subgroups play a special role. These subgroups have nice characteristics that make them more interesting to study than arbitrary subgroups. In ring theory the objects corresponding to normal subgroups are a special class of subrings called ideals. An ideal in a ring \(R\) is a subring \(I\) of \(R\) such that if \(a\) is in \(I\) and \(r\) is in \(R\text{,}\) then both \(ar\) and \(ra\) are in \(I\text{;}\) that is, \(rI \subset I\) and \(Ir \subset I\) for all \(r \in R\text{.}\)

Example 16.23

Every ring \(R\) has at least two ideals, \(\{ 0 \}\) and \(R\text{.}\) These ideals are called the trivial ideals.

Let \(R\) be a ring with identity and suppose that \(I\) is an ideal in \(R\) such that \(1\) is in \(I\text{.}\) Since for any \(r \in R\text{,}\) \(r1 = r \in I\) by the definition of an ideal, \(I = R\text{.}\)

Example 16.24

If \(a\) is any element in a commutative ring \(R\) with identity, then the set

\begin{equation*} \langle a \rangle = \{ ar : r \in R \} \end{equation*}

is an ideal in \(R\text{.}\) Certainly, \(\langle a \rangle\) is nonempty since both \(0 = a0\) and \(a = a1\) are in \(\langle a \rangle\text{.}\) The sum of two elements in \(\langle a \rangle\) is again in \(\langle a \rangle\) since \(ar + ar' = a(r + r')\text{.}\) The inverse of \(ar\) is \(-ar = a (-r) \in \langle a \rangle\text{.}\) Finally, if we multiply an element \(ar \in \langle a \rangle\) by an arbitrary element \(s \in R\text{,}\) we have \(s(ar) = a(sr)\text{.}\) Therefore, \(\langle a \rangle\) satisfies the definition of an ideal.

If \(R\) is a commutative ring with identity, then an ideal of the form \(\langle a \rangle = \{ ar : r \in R \}\) is called a principal ideal.

The zero ideal \(\{ 0 \}\) is a principal ideal since \(\langle 0 \rangle = \{ 0 \}\text{.}\) If \(I\) is any nonzero ideal in \({\mathbb Z}\text{,}\) then \(I\) must contain some positive integer \(m\text{.}\) There exists a least positive integer \(n\) in \(I\) by the Principle of Well-Ordering. Now let \(a\) be any element in \(I\text{.}\) Using the division algorithm, we know that there exist integers \(q\) and \(r\) such that

\begin{equation*} a = nq + r \end{equation*}

where \(0 \leq r \lt n\text{.}\) This equation tells us that \(r = a - nq \in I\text{,}\) but \(r\) must be \(0\) since \(n\) is the least positive element in \(I\text{.}\) Therefore, \(a = nq\) and \(I = \langle n \rangle\text{.}\)

Example 16.26

The set \(n {\mathbb Z}\) is ideal in the ring of integers. If \(na\) is in \(n{\mathbb Z}\) and \(b\) is in \({\mathbb Z}\text{,}\) then \(nab\) is in \(n {\mathbb Z}\) as required. In fact, by Theorem 16.25, these are the only ideals of \({\mathbb Z}\text{.}\)

We know from group theory that \(\ker \phi\) is an additive subgroup of \(R\text{.}\) Suppose that \(r \in R\) and \(a \in \ker \phi\text{.}\) Then we must show that \(ar\) and \(ra\) are in \(\ker \phi\text{.}\) However,

\begin{equation*} \phi(ar) = \phi(a) \phi(r) = 0 \phi(r) = 0 \end{equation*}

and

\begin{equation*} \phi(ra) = \phi(r) \phi(a) = \phi(r)0 = 0. \end{equation*}
Remark 16.28

In our definition of an ideal we have required that \(rI \subset I\) and \(Ir \subset I\) for all \(r \in R\text{.}\) Such ideals are sometimes referred to as two-sided ideals. We can also consider one-sided ideals; that is, we may require only that either \(rI \subset I\) or \(Ir \subset I\) for \(r \in R\) hold but not both. Such ideals are called left ideals and right ideals, respectively. Of course, in a commutative ring any ideal must be two-sided. In this text we will concentrate on two-sided ideals.

We already know that \(R/I\) is an abelian group under addition. Let \(r+I\) and \(s +I\) be in \(R/I\text{.}\) We must show that the product \((r + I)(s + I) = rs + I\) is independent of the choice of coset; that is, if \(r' \in r+I\) and \(s' \in s+I\text{,}\) then \(r's'\) must be in \(rs+I\text{.}\) Since \(r' \in r+I\text{,}\) there exists an element \(a\) in \(I\) such that \(r' = r + a\text{.}\) Similarly, there exists a \(b \in I\) such that \(s' = s + b\text{.}\) Notice that

\begin{equation*} r' s' = (r+a)(s+b) = rs + as + rb + ab \end{equation*}

and \(as + rb + ab \in I\) since \(I\) is an ideal; consequently, \(r' s' \in rs + I\text{.}\) We will leave as an exercise the verification of the associative law for multiplication and the distributive laws.

The ring \(R/I\) in Theorem 16.29 is called the factor or quotient ring. Just as with group homomorphisms and normal subgroups, there is a relationship between ring homomorphisms and ideals.

Certainly \(\phi : R \rightarrow R/I\) is a surjective abelian group homomorphism. It remains to show that \(\phi\) works correctly under ring multiplication. Let \(r\) and \(s\) be in \(R\text{.}\) Then

\begin{equation*} \phi(r) \phi(s) = (r + I)(s+I) = rs + I = \phi(rs), \end{equation*}

which completes the proof of the theorem.

The map \(\phi : R \rightarrow R/I\) is often called the natural or canonical homomorphism. In ring theory we have isomorphism theorems relating ideals and ring homomorphisms similar to the isomorphism theorems for groups that relate normal subgroups and homomorphisms in Chapter 11. We will prove only the First Isomorphism Theorem for rings in this chapter and leave the proofs of the other two theorems as exercises. All of the proofs are similar to the proofs of the isomorphism theorems for groups.

Let \(K = \ker \psi\text{.}\) By the First Isomorphism Theorem for groups, there exists a well-defined group homomorphism \(\eta: R/K \rightarrow \psi(R)\) defined by \(\eta(r + K) = \psi(r)\) for the additive abelian groups \(R\) and \(R/K\text{.}\) To show that this is a ring homomorphism, we need only show that \(\eta( (r + K)(s + K) ) = \eta(r + K) \eta( s + K)\text{;}\) but

\begin{align*} \eta( (r + K)( s +K )) & = \eta(r s +K )\\ & = \psi(r s)\\ & = \psi(r) \psi(s)\\ & = \eta( r + K ) \eta( s + K ). \end{align*}