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Section 7.1 Ideals and Quotient Rings

An ideal in a ring \(R\) is a nonempty subset \(I\) of \(R\) such that if \(a\) and \(b\) are in \(I\text{,}\) then \(a+b\) is in \(I\text{,}\) and if \(a\) is in \(I\) and \(r\) is in \(R\text{,}\) then both \(ar\) and \(ra\) are in \(I\text{.}\) In other words, an ideal \(I\) is closed under addition and \(rI \subset I\) and \(Ir \subset I\) for all \(r \in R\text{.}\)

Example 7.1.

Every nonzero ring \(R\) has at least two ideals, \(\{ 0 \}\) and \(R\text{.}\) These ideals are called the trivial ideals.
Let \(R\) be a ring 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 7.2.

If \(a\) is any element in a commutative ring \(R\text{,}\) 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, 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 \(I = \langle n \rangle\text{.}\)

Example 7.4.

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 7.3, these are the only ideals of \({\mathbb Z}\text{.}\)
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*}

Proof.
Since \(\phi(0) = 0\text{,}\) we know that \(\ker\phi\) is nonempty. Second, if \(a, b\in \ker \phi\text{,}\) then
\begin{equation*} \phi(a+b) = \phi(a) + \phi(b) = 0 + 0 = 0, \end{equation*}
and so \(a+b \in \ker\phi\text{.}\) Now 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 7.6.

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.
Recall that we defined arithmetic modulo an integer \(n\) by declaring two integers to be equivalent if their difference was divisible by \(n\text{.}\) We now investigate a similar construction using ideals. Let \(I \subset R\) be an ideal, and we define elements \(r\) and \(s\) to be equivalent if \(r-s \in I\text{.}\) We leave it as an exercise to check that this defines an equivalence relation. Then, we let \(R/I\) denote the set of equivalence classes under this equivalence relation, and we denote the equivalence class containing \(r\) as \(r + I\text{.}\)

Proof.
We first show that the sum \((r+s)+I\) is independent of the choice of \(r\) and \(s\) from their equivalence classes. If \(r' \in r + I\) and \(s' \in s + I\text{,}\) then \(r' = r + a\) and \(s' = s + b\) for \(a,b \in I\text{.}\) Then,
\begin{equation*} r' + s' = r + a + s + b = (r + s) + (a + b), \end{equation*}
and \(a+b \in I\text{,}\) so \(r' + s' \in r + s + I\text{.}\)
Second, 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 properties of a ring.
The ring \(R/I\) in Theorem 7.7 is called the factor or quotient ring. There is a relationship between ring homomorphisms and ideals.

Proof.
Certainly \(\phi \colon R \rightarrow R/I\) is a surjective function. It remains to show that \(\phi\) works correctly under the ring operations. Let \(r\) and \(s\) be in \(R\text{.}\) Then
\begin{gather*} \phi(r) + \phi(s) = (r+I) + (s+I) = (r+s) + I = \phi(r+s),\\ \phi(r) \phi(s) = (r + I)(s+I) = rs + I = \phi(rs), \end{gather*}
which completes the proof of the theorem.
The map \(\phi \colon R \rightarrow R/I\) is often called the natural or canonical homomorphism. In ring theory we have isomorphism theorems relating ideals and ring homomorphisms. We will prove only the First Isomorphism Theorem for rings in this chapter and leave the proofs of the other two theorems as exercises.

Proof.
Let \(K = \ker \psi\text{.}\) We first define a function \(\eta \colon R/K \rightarrow \psi(R)\) by \(\eta(r + K) = \psi(r)\text{.}\) We must show that \(\eta\) is well-defined. If \(r' = r + k\) is a different element of the coset \(r+K\text{,}\) where \(k \in K\text{,}\) then
\begin{equation*} \psi(r') = \psi(r + k) = \psi(r) + \psi(k) = \psi(r) + 0 = \psi(r). \end{equation*}
Therefore, \(\eta(r)\) does not depend on the choice of representative from the coset \(r + K\text{.}\) Furthermore, \(\eta\) is uniquely defined since \(\psi = \eta\phi\text{.}\)
In order to show that \(\eta\) is a ring homomorphism, we need to show that it preserves addition and multiplication. Suppose that \(r\) and \(s\) are any two elements of \(R\text{.}\) Then,
\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*}
The calculation for multiplciation is similar:
\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*}
Therefore, \(\eta\) is a homomorphism.
It is clear that \(\eta\) is onto. To show that \(\eta\) is one-to-one, suppose that \(\eta(r + K) = \eta(r'+K)\text{.}\) Then \(\psi(r) = \psi(r')\) and so \(\psi(r-r') = \psi(r) - \psi(r') = 0\text{.}\) Therefore, \(r-r' \in K\text{,}\) so \(r\) and \(r'\) are in the same coset, which shows that \(\eta\) is one-to-one.
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