Section5.3Ring 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 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 \}\text{.} \end{equation*}
Example5.20.

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)\text{.} \end{align*}

The kernel of the homomorphism $\phi$ is $n {\mathbb Z}\text{.}$

Example5.21.

Let $C[a, b]$ be the ring of continuous real-valued functions on an interval $[a,b]$ as in Example 5.5. 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 )\text{.} \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 $r \in R\text{.}$

Example5.23.

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, $I = R\text{.}$

Example5.24.

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.

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

Example5.26.

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 5.25, these are the only ideals of ${\mathbb Z}\text{.}$

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\text{.} \end{equation*}
Remark5.28.

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.

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 5.29 is called the factor or quotient ring. Just as with group homomorphisms and normal subgroups, there is a relationship between ring homomorphisms and ideals.

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)\text{,} \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 4. 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.

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 )\text{.} \end{align*}

1.

Which of the following is true of the kernel of a ring homomorphism? Briefly explain.

1. The kernel is the set of all things sent to 0.

2. The kernel is the set of all things sent to 1.

3. The kernel is the set of all things sent to either 0 or 1.

4. The kernel is the set of all things sent to both 0 and 1.

2.

What is an ideal of a ring? How do ideals relate to subrings?

3.

If $R$ is a ring and $I$ is an ideal, what sort of thing is $R/I\text{?}$ What do its elements look like and what operation(s) exist for it?

4.

After reading the section, what questions do you still have? Write at least one well formulated question (even if you think you understand everything).

ExercisesPractice Problems

1.

Let $\phi:R \to S$ be a surjective (onto) ring homomorphism. Prove that if $R$ and $S$ both have unities $1_R$ and $1_S$ then $\phi(1_R) = 1_S\text{.}$

2.

Let $R$ be a ring with unity and $\phi:R \to S$ a (not necessarily surjective) homomorphism. Could $1$ be in the kernel? Show that if it is, that $\phi$ is the trivial homomorphism $\phi(x) = 0$ for all $x \in R\text{.}$

3.

Consider the ring $R = \Z_3 \times \Z_3\text{.}$ This contains 9 elements, each of the form $(a,b)$ where $a, b \in \Z_3\text{.}$

1. Find the two non-trivial ideas of $R$ and the subring of $R$ that is not an ideal. Call this non-ideal subring $B\text{.}$

1. Give an example of what goes wrong if you try to form the quotient ring $R/B\text{.}$

4.

Give an example of rings $R$ and $S$ and a function $\phi:R \to S$ which is a group homomorphism but not a ring homomorphism.

Hint

You know that if the kernel of $\phi$ happens to be an ideal, then $\phi$ will automatically be a ring homomorphism, by the Fundamental Homomorphism Theorem. So we must find a ring $R$ that has a subgroup (which would automatically be a normal subgroup) which is not an ideal.

5.

Here is a nice application of homomorphisms. For each $n\text{,}$ we consider the canonical homomorphism from $\Z$ to $\Z/n\Z\text{.}$ It is easier to think of this as the homomorphism that sends $\Z$ to $\Z_n\text{,}$ which maps $a \in \Z$ to $a \pmod{n}\text{.}$

1. Consider the equation $11x^2 - 8y^2 + 109 = 0\text{.}$ Note that $(7,9)$ is a solution to this equation. Explain how we can use this fact and a homomorphism to be sure that there is a solution in $\Z_5$ to the equation $x^2 + 2y^2 + 4 = 0\text{.}$

2. Prove that the equation $x^2 - 7y^2 - 24 = 0$ has no integer solutions. You might want to use a homomorphism from $\Z$ to $\Z_7\text{.}$

3. Prove that $x^2 + (x+1)^2 + (x+2)^2 = y^2$ has no integer solutions. Hint: $\Z_3\text{.}$

Hint

The point of all this is that if there is no solution to the corresponding equation in $\Z_n\text{,}$ then there is no solution in $\Z\text{.}$ When you apply the homomorphism to both sides of the equation, you use the homomorphism property to reduce every coefficient mod $n\text{.}$ And then you only have $n$ possible values for each variable. Better yet, if you pick $n$ that makes one of the coefficients 0, you only have to check the other one.

ExercisesCollected Homework

C1.

Let $F$ be a field and $R$ any ring containing more than one element.

1. Prove that $F$ has no nontrivial ideals.

2. Prove that if there is a surjective homomorphism $\phi:F \to R\text{,}$ then $F \cong R\text{.}$

Hint
1. The trivial ideals are $\{0\}$ and $F\text{.}$ What if an ideal $I$ contained some $a \ne 0\text{?}$

2. On a previous homework you proved that if the kernel of a homomorphism is the identity, then the homomorphism is injective. This holds for ring homomorphisms as well and you may use that fact here.

Bonus.

Make up a problem similar to 5(b) or 5(c) from the practice problem above and show how to solve it.

C2.

Consider the ring $\R[x]$ of polynomials in the variable $x$ with real coefficients. We can divide polynomials using long division, which results in some remainder.

1. What is the remainder when you divide $3x^4 - 7x^3 + x^2 - 3x + 7$ by $x^2 + 1\text{?}$ (Perform the long division, but you only need to submit the results: quotient and remainder.)

2. What do you get when you substitute $x^2 = -1$ into $3x^4 - 7x^3 + x^2 - 3x + 7\text{?}$

3. Coincidence? What sort of polynomial will you always get as a remainder when dividing a polynomial by$x^2 + 1\text{?}$ What sort of polynomial will you always get when substituting $x^2 = -1$ into a polynomial?

4. Consider the quotient ring $\R[x]/\langle x^2 + 1\rangle\text{.}$ Use an evaluation homomorphism (with $a = \sqrt{-1}$) to prove that this quotient ring is isomorphic to $\bC$ (the complex numbers, $\bC = \{a+bi \st a,b \in \R; i^2 = -1\}$).