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Section16.3Ring 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 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 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 of a ring homomorphism to be the set

\begin{equation*} \ker \phi = \{ r \in R : \phi( r ) = 0 \}. \end{equation*}
Example16.20

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 $n {\mathbb Z}\text{.}$

Example16.21

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 .

In the next proposition we will examine some fundamental properties of ring homomorphisms. The proof of the proposition is left as an exercise.

Proposition16.22

Let $\phi : R \rightarrow S$ be a ring homomorphism.

  1. If $R$ is a commutative ring, then $\phi(R)$ is a commutative ring.

  2. $\phi( 0 ) = 0\text{.}$

  3. Let $1_R$ and $1_S$ be the identities for $R$ and $S\text{,}$ respectively. If $\phi$ is onto, then $\phi(1_R) = 1_S\text{.}$

  4. If $R$ is a field and $\phi(R) \neq \{ 0 \}\text{,}$ then $\phi(R)$ is a field.

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 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 $r \in R\text{.}$

Example16.23

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

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, $I = R\text{.}$

Example16.24

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 .

Theorem16.25

Every ideal in the ring of integers ${\mathbb Z}$ is a principal ideal.

Proof

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 $I = \langle n \rangle\text{.}$

Example16.26

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 ${\mathbb Z}\text{.}$

Proposition16.27

The kernel of any ring homomorphism $\phi : R \rightarrow S$ is an ideal in $R\text{.}$

Proof

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*}
Remark16.28

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 . We can also consider ; 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 and , respectively. Of course, in a commutative ring any ideal must be two-sided. In this text we will concentrate on two-sided ideals.

Theorem16.29

Let $I$ be an ideal of $R\text{.}$ The factor group $R/I$ is a ring with multiplication defined by

\begin{equation*} (r + I)(s + I) = rs + I. \end{equation*}
Proof

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 or . Just as with group homomorphisms and normal subgroups, there is a relationship between ring homomorphisms and ideals.

Theorem16.30

Let $I$ be an ideal of $R\text{.}$ The map $\phi : R \rightarrow R/I$ defined by $\phi( r ) = r + I$ is a ring homomorphism of $R$ onto $R/I$ with kernel $I\text{.}$

Proof

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 or . 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.

Theorem16.31First Isomorphism Theorem

Let $\psi : R \rightarrow S$ be a ring homomorphism. Then $\ker \psi$ is an ideal of $R\text{.}$ If $\phi : R \rightarrow R/\ker \psi$ is the canonical homomorphism, then there exists a unique isomorphism $\eta: R/\ker \psi \rightarrow \psi(R)$ such that $\psi = \eta \phi\text{.}$

Proof

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*}
Theorem16.32Second Isomorphism Theorem

Let $I$ be a subring of a ring $R$ and $J$ an ideal of $R\text{.}$ Then $I \cap J$ is an ideal of $I$ and

\begin{equation*} I / I \cap J \cong (I+ J) /J. \end{equation*}
Theorem16.33Third Isomorphism Theorem

Let $R$ be a ring and $I$ and $J$ be ideals of $R$ where $J \subset I\text{.}$ Then

\begin{equation*} R/I \cong \frac{R/J}{I/J}. \end{equation*}
Theorem16.34Correspondence Theorem

Let $I$ be an ideal of a ring $R\text{.}$ Then $S \mapsto S/I$ is a one-to-one correspondence between the set of subrings $S$ containing $I$ and the set of subrings of $R/I\text{.}$ Furthermore, the ideals of $R$ containing $I$ correspond to ideals of $R/I\text{.}$