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Intro to Abstract Algebra Fall 2025

Blake Farman

Chapter 14 Integral Domains

Section 14.1 Zero Divisors

Definition 14.1. Zero Divisor.

Assume \(R\) is a ring and \(a \in R \setminus \{0\}\text{.}\)
  1. We say that \(a\) is a left zero divisor if there exists \(b \in R \setminus \{0\}\) such that \(ab = 0\text{.}\)
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  2. We say \(a\) is a right zero divisor if there exists \(b \in R \setminus \{0\}\) such that \(ba = 0\text{.}\)
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  3. We say \(a\) is a zero divisor if it is both a left and right zero divisor.
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Remark 14.2.

If \(R\) is a commutative ring, \(a \in R\) is a left zero divisor if and only if it is also a right zero divisor. Hence the distinction between left and right zero divisor only matters for noncommutative rings.
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Example 14.3.

When \(n\) is a composite number, there exists a divisor \(d \in \Z\) of \(n\) such that \(1 \lt d \lt n\text{.}\) Let \(k = n/d\text{.}\) Then \(\overline{d} \neq 0\) and \(\overline{k} \neq 0\text{,}\) but
\begin{equation*} \overline{d}\,\overline{k} = \overline{dk} = \overline{n} = \overline{0}\text{.} \end{equation*}
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Proof.

We handle the left case and note the right case follows from a symmetric argument.
  1. Assume \(a\) is a left unit. By assumption, there exists \(b \in R\) such that \(ba = 1\text{.}\) By definition, \(a0 = 0\text{.}\) Conversely, suppose there exists \(c \in R\) such that \(ac = 0\text{.}\) Then
    \begin{equation*} c = 1c = (ba)c = b(ac) = b0 = 0 \end{equation*}
    Therefore if \(a\) is a left unit, then \(ax = 0\) if and only if \(x = 0\text{.}\)
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  2. Assume \(a\) is a left zero divisor. By assumption, \(a \neq 0\) and there exists \(b \in R \setminus \{0\}\) such that \(ab = 0\text{.}\) Suppose to the contrary that \(a\) is a left unit. By definition, there exists \(c \in R\) such that \(ca = 1\text{.}\) Hence
    \begin{equation*} 0 = c(0) = c(ab) = (ca)b = 1b = b \neq 0\text{,} \end{equation*}
    a contradiction. Therefore if \(a\) is a left zero divisor, then \(a\) is not a left unit.
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Proof.

We handle the left case and note the right case follows from a symmetric argument. Assume \(a \in R \setminus \{0\}\) is not a left zero divisor; that is, for all \(x \in R \setminus\{0\}\text{,}\) \(ax \neq 0\text{.}\) If \(ab = ac\text{,}\) then
\begin{equation*} 0 = ab - ac = a(b - c) \end{equation*}
implies \(b - c = 0\) . Therefore \(b = c\text{.}\)
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It is common to refer to these two implications as the cancellation laws
Left Cancellation
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If \(ab = ac\text{,}\) then \(b = c\text{.}\)
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Right Cancellation
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If \(ba = ca\text{,}\) then \(b = c\)
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Proof.

If \(R\) has no left (respectively right) zero divisors, then Proposition 14.5 implies left (respectively right) cancellation holds. Conversely, assume \(R\) has left (respectively right) cancellation. For all \(a,b \in R\text{,}\) if \(ab = 0\) (respectively\(ba = 0\)), then \(b = 0\) by left (respectively right) cancellation. Therefore left (respectively right) cancellation holds if and only if \(R\) has no left (respectively right) zero divisors.
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For the final claim, assume \(R\) has no left or right zero divisors. Assume there exist \(a,b \in R \setminus\{0\}\) such that \(ab = 0\text{.}\) By definition, \(a\) is a left zero divisor and \(b\) is a right zero divisor, contrary to the assumption that \(R\) has no left or right zero divisors. Hence for all \(a,b \in R\text{,}\) if \(ab = 0\text{,}\) then \(a = 0\) or \(b = 0\text{.}\)
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Conversely, assume \(R\) has the property that for all \(a,b \in R\text{,}\) if \(ab = 0\text{,}\) then \(a = 0\) or \(b = 0\text{.}\) Suppose for contradiction that \(a\) is a left (respectively right) zero divisor. By definition, \(a \neq 0\) and there exists \(b \in R \setminus \{0\}\) such that \(ab = 0\) (respectively \(ba = 0\)), contrary to the assumption that \(ab = 0\) implies \(a = 0\) or \(b = 0\text{.}\) Hence \(R\) has no left or right zero divisors. Therefore \(R\) has no left or right zero divisors if and only if for all \(a,b \in R\text{,}\) if \(ab = 0\text{,}\) then \(a = 0\) or \(b = 0\text{.}\)
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Section 14.2 Integral Domains

Definition 14.7.

Assume \(R\) is a commutative, unital ring and \(0 \neq 1\text{.}\) We say \(R\) is a integral domain if \(R\) has no zero divisors.
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Example 14.8.

In a field, \(F\text{,}\) every non-zero element is a unit. Hence every field is a domain by Proposition 14.4. In particular, \(\Q\text{,}\) \(\R\text{,}\) \(\C\) are all integral domains.
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Proof.

Let \(S\) be a subring of a domain \(R\text{.}\) Assume \(a,b \in S\) satisfy \(ab = 0\text{.}\) Since the product \(ab\) as elements of \(S\) coincides with the product \(ab\) as elements of \(R\text{,}\) either \(a = 0\) or \(b = 0\text{.}\) Therefore \(S\) is an integral domain.
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Example 14.10.

The ring \(\Z\) is an integral domain because it is a subring of the field \(\Q\text{.}\)
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Proof.

Assume \(F\) is a domain with finitely many elements. Let \(a \in F \setminus \{0\}\) be given. The function
\begin{align*} \phi \colon F \amp\to F\\ x \amp\mapsto ax \end{align*}
is a morphism of groups because for all \(x_1, x_2 \in F\)
\begin{equation*} \phi(x_1 + x_2) = a(x_1 + x_2) = ax_1 + ax_2 = \phi(x_1) + \phi(x_2)\text{.} \end{equation*}
Since \(F\) is assumed to be a domain,
\begin{equation*} 0 = f(x) = ax \iff x = 0 \end{equation*}
implies \(\phi\) is injective by Theorem 6.17. Hence \(\phi\) is surjective because \(F\) is finite. In particular, there exists \(b \in F\) such that \(ab = f(b) = 1\text{.}\) Therefore every non-zero element of \(F\) is a unit and thus \(F\) is a field.
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Proof.

By Theorem 14.11, \(\Z/n\Z\) is an integral domain if and only if it is a field. We have seen that if \(n\) is a prime number, then \(\Z/n\Z\) is a field Example 13.30. For the converse, we observe that Example 14.3 establishes the contrapositive: if \(n\) is composite, then \(\Z/n\Z\) is not a domain.
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Section 14.3 The Characteristic of a Ring

Definition 14.13. Characteristic.

Assume \(R\) is a ring. Recall from Remark 7.11 that for all \(r \in R\text{,}\) for all \(n \in \N\text{,}\)
\begin{equation*} nr = \sum_{i=1}^n r = \underbrace{r + r + \cdots + r}_n\text{.} \end{equation*}
Define the set
\begin{equation*} C = \left\{n \in \N \;\middle\vert\; \forall r \in R, nr = 0\right\}\text{.} \end{equation*}
  1. If \(C \neq \emptyset\text{,}\) then the characteristic of \(R\) is
    \begin{equation*} \operatorname{Char}(R) = \min C\text{.} \end{equation*}
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  2. If \(C = \emptyset\text{,}\) then \(\operatorname{Char}(R) = 0\text{.}\)
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Example 14.14.

The rings \(\Z\text{,}\) \(\Q\text{,}\) \(\R\text{,}\) and \(\C\) all have characteristic zero.
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Example 14.15.

For \(n \in \N\text{,}\) the ring \(\Z/n\Z\) has characteristic \(n\text{.}\)
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Proof.

First we prove there exists \(n \in \N\) such that for all \(r \in R\text{,}\) \(nr = 0\) if and only if there exists \(n \in \N\) such that \(n1_R = 0\text{.}\) If there exists \(n \in \N\) such that for all \(r \in R\text{,}\) \(nr = 0\text{,}\) then \(n1_R = 0\text{.}\)
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Conversely, assume there exists \(n \in \N\) such that \(n1_R = 0\text{.}\) For every \(r \in R\text{,}\)
\begin{equation*} nr = \sum_{i=1}^n r = \sum_{i=1}^n 1_R r = \left(\sum_{i=1}^n 1_R\right) r = \left(n1_R\right)r = 0r = 0\text{.} \end{equation*}
Hence there exists \(n \in \N\) such that for all \(r \in R\text{,}\) \(nr = 0\text{.}\)
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This establishes the equivalence
\begin{equation*} \left\{n \in \N \;\middle\vert\; \forall r \in R, nr = 0\right\} = \left\{n \in \N \;\middle\vert\; n1_R = 0\right\} \end{equation*}
and the result now follows from Definition 14.13
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Proof.

Let \(n \in \Z\) be given. Following the notation in Remark 7.11, define the function
\begin{align*} \phi \colon \Z \amp\to R\\ x \amp \mapsto x1_R \end{align*}
This is a morphism of rings because for all \(m,n \in \Z\text{,}\)
\begin{equation*} \phi(m + n) = (m+n)1_R = m1_R + n1_R = \phi(m) + \phi(n) \end{equation*}
and
\begin{align*} \phi(mn) \amp= (mn)1_R\\ \amp= m(n1_R)\\ \amp= m\phi(n)\\ \amp= \sum_{i=1}^m\phi(n)\\ \amp= \sum_{i=1}^m 1_R\phi(n)\\ \amp= \left(\sum_{i=1}^m 1_R\right) \phi(n)\\ \amp= \phi(m)\phi(n)\text{.} \end{align*}
Hence for every ring \(R\text{,}\) there exists a morphism \(\phi \colon \Z \to R\text{.}\)
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To see this morphism is unique, assume \(\psi \colon \Z \to R\) is a morphism of rings. By Definition 13.5, \(\psi(1) = 1_R\text{.}\) Hence for every \(n \in \N\text{,}\)
\begin{equation*} \psi(n) = \psi\left(\sum_{i=1}^n 1\right) = \sum_{i=1}^n \psi(1) = \sum_{i=1}^n 1_R = \phi(n)\text{.} \end{equation*}
Therefore \(\phi\) is the unique morphism of rings \(\Z \to R\text{.}\)
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Exercises 14.4 Exercises for Undergrads & Grads

Units and Zero Divisors.

For each of the following rings, classify each element as a unit, zero divisor, or neither.
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\(\Z/8\Z\)
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Solution.
  • The element \(\overline{0}\) is neither a zero divisor nor a unit.
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  • The units are \(\overline{1}\text{,}\) \(\overline{3}\text{,}\) \(\overline{5}\text{,}\) and \(\overline{7}\text{.}\)
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  • The zero divisors are \(\overline{2}\text{,}\) \(\overline{4}\text{,}\) and \(\overline{6}\)
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2.
\(\Z/3\Z \times \Z/3\Z\)
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Solution.
  • The element \((\overline{0}, \overline{0})\) is neither a unit nor a zero divisor.
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  • The elements \((\overline{1}, \overline{1})\text{,}\) \((\overline{1}, \overline{2})\text{,}\) \((\overline{2}, \overline{1})\text{,}\) and \((\overline{2}, \overline{2})\) are units.
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  • The elements \((\overline{0}, \overline{1})\text{,}\) \((\overline{0}, \overline{2})\text{,}\) \((\overline{1},\overline{0})\text{,}\) \((\overline{2},\overline{0})\) are zero divisors.
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3.
\(\Z/4\Z \times \Z/3\Z\)
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Solution.
  • The element \((\overline{0}, \overline{0})\) is neither.
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  • The elements \((\overline{1},\overline{1})\text{,}\) \((\overline{1},\overline{2})\text{,}\) \((\overline{3},\overline{1})\text{,}\) and \((\overline{3},\overline{2})\) are units.
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  • The elements \((\overline{0},\overline{1})\text{,}\) \((\overline{0},\overline{2})\) \((\overline{1},\overline{0})\text{,}\) \((\overline{2},\overline{0})\text{,}\) \((\overline{3},\overline{0})\text{,}\) \((\overline{2},\overline{1})\text{,}\) and \((\overline{2},\overline{2})\text{,}\) are zero divisors
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4.

Assume \(R\) is a commutative ring of characteristic 2. Prove that for all \(a,b \in R\text{,}\)
\begin{equation*} (a + b)^2 = a^2 + b^2\text{.} \end{equation*}
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