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Intro to Abstract Algebra Fall 2025

Blake Farman

Chapter 13 Rings and Fields

Section 13.1 Rings

Definition 13.1. Ring.

A ring is a triple \((R,+,\times)\) consisting
  • a set \(R\text{,}\)
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  • a binary operation \(+ \colon R \times R \to R\)—called addition—under which \(R\) is an abelian group, and
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  • a binary operation \(\times \colon R \times R \to R\)—called multiplication—under which \((R,\times)\) is a Semigroup
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such that the group and semigroup structures are compatible in the following sense.
Left Distibution
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For all \(a,b,c \in R\text{,}\) \(a \times (b + c) = a \times b + a \times c\text{,}\)
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Right Distribution
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For all \(a,b,c \in R\text{,}\) \((a + b) \times c = a \times c + b \times c\)
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Remark 13.2.

  1. Since \((R,+)\) is an additive group, we denote the identity by \(0\) and refer to this element as the additive identity
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  2. When the multiplication, \(\times\text{,}\) is clear from context, we write \(ab\) instead of \(a \times b\text{.}\)
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Definition 13.3. Commutative Ring.

We say that \(R\) is a commutative ring if multiplication of elements from \(R\) is a commutative binary operation.
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Definition 13.4. Unital Ring.

We say that a ring \((R,+,\times)\) is a unital ring if \((R,\times)\) is a Monoid.
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That is to say, there exists an element \(1 \in R\) that is the identity element for the binary operation \(\times \colon R \times R \to R\text{.}\) We refer to \(1\) as the multiplicative identity. It is also common to see \(1\) referred to by the name unity.
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Definition 13.5. Morphism of Rings.

Assume \(R\) and \(S\) are rings. A morphism of rings is a morphism of groups \(\phi \colon R \to S\) that also preserves the multiplication. That is to say, for all \(a,b \in R\text{,}\)
  • \(\phi(a+b) = \phi(a) + \phi(b)\text{,}\) and
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  • \(\phi(ab) = \phi(a)\phi(b)\text{.}\)
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If \(R\) and \(S\) are both unital rings, then we require that \(\phi(1_R) = 1_S\text{.}\)
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Section 13.2 Basic Examples

Example 13.6.

The zero ring, \(0\text{,}\) is the set \(\{0\}\text{.}\) It is trivially an abelian group under the addition \(0 + 0 = 0\) and trivially a ring under the multiplication \(0 \times 0 = 0\text{.}\)
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Example 13.7.

Each of \(\Z\text{,}\) \(\Q\text{,}\) \(\R\text{,}\) and \(\C\) have the structure of a commutative, unital ring given by the usual addition and multiplication operations. For each ring, the additive identity is \(0\) and the multiplicative identity is \(1\text{.}\)
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Example 13.8. Ring-Valued Functions.

Assume \(R\) is a ring. For any set \(X\text{,}\) the set of set functions
\begin{equation*} R^X = \left\{f \colon X \to R\right\} \end{equation*}
forms an abelian group under pointwise addition (see Example 2.28). We define a ring structure by defining the product of \(f, g \colon X \to R\) to be the set function
\begin{equation*} (fg)(x) = f(x)g(x)\text{.} \end{equation*}
This multiplication inherits associativity, left distribution, and right distribution from \(R\text{.}\)
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Note that if \(R\) is unital, then so is \(R^X\text{.}\) The multiplicative identity is the function
\begin{align*} \mathbb{1} \colon X \amp\to R\\ x \amp\mapsto 1 \end{align*}
which satisfies the property for all \(f \colon X \to R\text{,}\) \(f\mathbb{1} = f = \mathbb{1}f\) because for all \(x \in X\text{,}\)
\begin{align*} (f\mathbb{1})(x) \amp= f(x)\mathbb{1}(x) = f(x)1 = f(x),\, \text{and}\\ (\mathbb{1}f)(x) \amp= \mathbb{1}(x)f(x) = 1f(x) = f(x) \end{align*}
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Example 13.9. The Integers Modulo \(n\).

The additive abelian group \(\Z/n\Z\) is endowed with the structure of a ring via the multiplication
\begin{equation*} \overline{a}\overline{b} = \overline{ab}\text{.} \end{equation*}
First we prove this binary operation is well-defined. Let \(a,b,c,d \in \Z\) be given. Assume \(a \equiv c \pmod{n}\) and \(b \equiv d \pmod{n}\text{.}\) There exist \(k_1, k_2 \in \Z\) such that \(-a + c = k_1n\) and \(-b + d = k_2n\text{.}\) Hence
\begin{align*} -ab + cd \amp= -ab + bc - bc + cd\\ \amp= (-a + c)b + c(-b + d)\\ \amp= (bk_1)n + (ck_2)n\\ \amp= (bk_1 + ck_2)n \end{align*}
implies \(ab \equiv cd \pmod{n}\) and thus this binary operation is independent of the representatives chosen. Since \(\Z\) is a unital, commutative ring, this multiplication is associative, left distributive, right distributive, and has identity element \(\overline{1}\text{.}\) Indeed, for all \(\overline{a} \in \Z/n\Z\text{,}\)
\begin{equation*} \overline{1}\, \overline{a} = \overline{1a} = \overline{a} \end{equation*}
and for all \(\overline{a}, \overline{b}, \overline{c} \in \Z/n\Z\text{,}\)
\begin{align*} \overline{a}(\overline{b} + \overline{c}) \amp= \overline{a}\, \overline{b + c} = \overline{a(b+c)} = \overline{ab + ac} = \overline{ab} + \overline{ac} = \overline{a}\,\overline{b} + \overline{a}\,\overline{c}\\ (\overline{a} + \overline{b})\overline{c} \amp= \overline{a + b}\, \overline{c} = \overline{(a + b)c} = \overline{ac + bc} = \overline{ac} + \overline{bc} = \overline{a}\,\overline{c} + \overline{b}\,\overline{c} \end{align*}
Therefore \(\Z/n\Z\) forms a ring under addition and multiplication of representatives.
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Example 13.10. Products of Rings.

Assume \(I\) is a set and \(\{R_i\}_{i \in I}\) is a family of rings indexed by \(I\text{.}\) Just as for groups, we define the product of rings
\begin{equation*} \prod_{i \in I} R_i = \left\{(a_i)_{i \in I} \;\middle\vert\; a_i \in R_i\right\}\text{.} \end{equation*}
This is an abelian group under addition. Equipped with the component-wise product
\begin{equation*} (a_i)_{i \in I}(b_i)_{i \in I} = (a_ib_i)_{i \in I}\text{,} \end{equation*}
we obtain a ring structure on \(\prod_{i \in I}R_i\text{.}\)
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This ring satisfies the same universal mapping property as the product of sets or the product of groups. Simply replace morphism of groups with morphism of rings (see Definition 13.5) in Theorem 9.4.
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If for all \(i \in I\text{,}\) \(R_i\) is a unital ring, then the product is also unital with multiplicative identity \((1_{R_i})_{i \in I}\text{.}\)
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Example 13.11. The Ring of Square Matrices with Entries from a Ring.

Let \(R\) be a ring. Just as we view \(\R^n\) as the set of\(n\)-dimensional vectors with coordinates from \(\R\text{,}\) the ring
\begin{equation*} R^n = \prod_{i = 1}^n R = \underbrace{R \times R \times \cdots \times R}_n \end{equation*}
can be viewed as the set of \(n\)-dimensional vectors with coordinates from \(R\text{.}\)
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Doing the same with Definition 1.8, we can define \(M_n(R)\) to be all \(n \times n\) matrices with entries from \(R\text{.}\) We define addition and multiplication of matrices with entries from \(R\) exactly as we did in Example 1.9. The exact same proof from linear algebra, replacing \(\R\) by \(R\text{,}\) proves that \(M_n(R)\) is a ring.
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If \(R\) is unital, then so is \(M_n(R)\text{.}\) The multiplicative identity is the usual \(n \times n\) identity matrix,
\begin{equation*} \begin{pmatrix} 1 \amp 0 \amp \ldots \amp 0\\ 0 \amp 1 \amp \ldots \amp 0\\ \vdots \amp \vdots \amp \ddots \amp \vdots\\ 0 \amp 0 \amp \ldots \amp 1 \end{pmatrix} \end{equation*}
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Section 13.3 Basic Properties of Rings

Proof.

Let \(a \in R\) be given. Since multiplication is left distributive,
\begin{equation*} 0a + 0a = (0+0)a = 0a = 0 + 0a \end{equation*}
implies \(0a = 0\) by right cancellation. Similarly, right distrubtion allows us to write
\begin{equation*} a0 + a0 = a(0 + 0) = a0 = 0 + a0 \end{equation*}
and right cancellation implies \(a0 = 0\text{.}\)
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Proof.

Let \(a \in R\) be given. By right distribution and Proposition 13.12,
\begin{equation*} (-1)a + a = (-1)a + (1)a = (-1 + 1)a = 0a = 0\text{.} \end{equation*}
Similarly, by left distribution and Proposition 13.12,
\begin{equation*} a(-1) + a = a(-1) + a(1) = a(-1 + 1) = a0 = 0\text{.} \end{equation*}
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For the final claim, we observe that \(-(-a)\) is the additive inverse of \(-a\text{.}\) Since \(-a + a = 0\text{,}\) it follows from Corollary 2.32 that \(a = -(-a)\text{.}\)
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Proof.

This is an immediate consequence of associativity and Proposition 13.13:
\begin{equation*} -(ab) = (-1)(ab) = (-1a)b = (a(-1))b = a((-1)b) = a(-b)\text{.} \end{equation*}
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Proof.

This is immediate from Corollary 13.14 and \(-(-1) = 1\text{:}\)
\begin{align*} (-a)(-b) \amp= \left((-1)a\right)(-b)\\ \amp= \left(a(-1)\right)(-b)\\ \amp= a\left((-1)(-b)\right)\\ \amp= a\left[(-1)\left((-1)b\right)\right]\\ \amp= a\left[\left((-1)(-1)\right)b\right]\\ \amp= a(1b)\\ \amp= ab \end{align*}
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Section 13.4 Units

Definition 13.16. Units.

Assume \(R\) is a unital ring and \(a \in R \setminus \{0\}\text{.}\)
  1. We say \(a\) is a left unit if there exists \(b \in R\) such that \(ba = 1\)
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  2. We say \(a\) is a left unit if there exists \(b \in R\) such that \(ab = 1\text{.}\)
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  3. We say \(a\) is a unit if it is both a left unit and right unit.
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Remark 13.17.

If \(R\) is commutative, then the notion of left unit and right unit coincide because \(ab = ba\text{.}\)
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Proof.

Assume \(a\) is a left and right unit. By assumption, there exist elements \(b,c \in R\) such that \(ab = 1 = ca\text{.}\) Since multiplication is associative, we have
\begin{align*} b \amp= 1b\\ \amp= (ca)b\\ \amp= c(ab)\\ \amp= c(1)\\ \amp= c\text{.} \end{align*}
Since \(b\) and \(c\) were arbitrary, it follows that \(b = c\) is the unique element of \(R\) with the property that \(ab = ba = 1\text{.}\)
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Proof.

We must prove that \(R^\times\) is closed under the multiplication from \(R\) and the restriction of multiplication to \(R^\times\) is associative, has an identity element, and has inverses for every element.
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Closure: First, we observe that if \(a,b \in R\) are units, then their product is also a unit because
\begin{equation*} (ab)(b^{-1}a^{-1}) = a\left(bb^{-1}\right)a^{-1} = aa^{-1} = 1 \end{equation*}
and also
\begin{equation*} (b^{-1}a^{-1})(ab) = b^{-1}\left(a^{-1}a\right)b = b^{-1}b = 1\text{.} \end{equation*}
Hence \(R^\times \subseteq R\) is closed under multiplication.
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Associative: Let \(a,b,c \in \R^\times\) be given. Since \(a,b,c \in R\) and \(R\) is a monoid, \(a(bc) = (ab)c\text{.}\) Hence the restriction of multiplication to \(R^\times\) yields an associative operation.
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Identity: The multiplicative identity of \(R\) is a unit because \(1(1) = 1\text{.}\) Hence \(R^\times\) has a multiplicative identity, \(1 \in R^\times\text{.}\)
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Inverses: Let \(a \in R\) be given. Since \(a\) is a unit, there exists \(a^{-1} \in R\) by Proposition 13.18. We observe that \(a^{-1}\) is a unit because
\begin{equation*} a^{-1}a = 1 = aa^{-1}\text{.} \end{equation*}
Hence \(a^{-1} \in R^\times\text{.}\)
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Therefore if \((R,+,\times)\) is a ring, then \((R^\times, \times)\) is a group.
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Definition 13.20. Multiplicative Group of Units.

Assume \(R\) is a unital ring. The multiplicative group of units is the group \((R^\times,\times)\text{.}\)
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Remark 13.21.

When the ring structure on \(R\) is clear from context, we simply write \(R^\times\) for the multiplicative group of units.
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Example 13.22.

  • \(\displaystyle \Q^\times = \Q \setminus\{0\}\)
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  • \(\displaystyle \R^\times = \R \setminus\{0\}\)
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  • \(\displaystyle \C^\times = \C \setminus\{0\}\)
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Example 13.23.

The set \(\Z^\times\) consists of the elements \(n \in \Z\) such that there exists a solution in \(\Z\) to the equation \(nx = 1\text{.}\) If we pass to the rationals, then there exists a unique solution whenever \(n \neq 0\text{:}\) \(x = 1/n\text{.}\) However, \(1/n \in \Z\) if and only if \(n = 1\) or \(n = -1\text{.}\) Therefore \(\Z^\times = \{\pm 1\}\text{.}\)
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Since the group of units, \(\Z^\times\text{,}\) has order 2, it is cyclic by Corollary 10.20 and generated by \(-1\text{.}\) It is easy to see that
\begin{align*} \phi \colon \Z/2\Z \amp\to \Z^\times\\ 0 \amp\mapsto 1\\ 1 \amp\mapsto -1 \end{align*}
is an isomorphism between the additive group of integers modulo \(2\) and the multiplicative group of units in \(\Z\) because
\begin{align*} \phi(0 + 0) \amp= \phi(0) = 1 = 1(1) = \phi(0)\phi(0)\\ \phi(0 + 1) \amp= \phi(1) = -1 = 1(-1) = \phi(0)\phi(1)\\ \phi(1 + 1) \amp= \phi(0) = 1 = (-1)(-1) = \phi(1)\phi(1)\text{.} \end{align*}
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Example 13.24.

Fix an integer \(n\text{.}\) The group of units is slightly more nuanced for the integers modulo \(n\) than for any of \(\Z\text{,}\) \(\Q\text{,}\) \(\R\text{,}\) or \(\C\text{.}\) For all \(\overline{a} \in \Z/n\Z\text{,}\) \(\overline{a} \in \left(\Z/n\Z\right)^\times\) if and only if there exists \(\overline{b} \in \Z/n\Z\) such that \(\overline{ab} = 1\text{.}\) Chasing the definition of equivalence modulo \(n\text{,}\) there exists \(k \in \Z\) such that
\begin{equation*} ab = kn + 1 \iff ab + (-k)n = 1\text{.} \end{equation*}
By Proposition A.9, \(\overline{a} \in \left(\Z/n\Z\right)^\times\) if and only if \(\gcd(a,n) = 1\text{.}\)
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Remark 13.25.

Note that our characterization of units coincides with our characterization of generators for the additive cyclic group \(\Z/n\Z\text{.}\) That is, \(\overline{a} \in \left(\Z/n\Z\right)^\times\) if and only if \(\langle \overline{a} \rangle = \Z/n\Z\text{.}\)
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As a tie in to number theory, we note that the number of elements in \(\Z/n\Z\) is counted by Euler’s Totient Function
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, \(\phi\text{.}\) The value of \(\phi(n)\) is defined to be the number of positive integers up to \(n\) that are coprime to \(n\)—i.e. the greatest common divisor is 1.
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Section 13.5 Fields

Definition 13.26. Division Ring.

Assume \(R\) is a unital ring with \(0 \neq 1\text{.}\) We say \(R\) is a division ring if \(R^\times = R \setminus\{0\}\text{.}\)
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Definition 13.27. Field.

A commutative division ring, \(F\text{,}\) is called a field.
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Remark 13.28.

The stipulation \(0 \neq 1\) explicitly forbids the zero ring from being a division ring (hence also a field).
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Example 13.29.

The rings \(\Q\text{,}\) \(\R\text{,}\) and \(\C\) are all fields.
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Example 13.30. Finite Fields.

If \(p\) is a prime number, then every integer \(0 \lt a \lt p\) satisfies \(\operatorname{gcd}(a,p) = 1\text{.}\) Hence \(\left(\Z/p\Z\right)^\times = \Z/p\Z \setminus \{\overline{0}\}\) and \(\Z/p\Z\) is a field. We call this the finite field with \(p\) elements. It is common to write \(\F_p\) for \(\Z/p\Z\text{.}\)
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Section 13.6 Subrings and Subfields

Definition 13.31. Subring.

Assume \((R,+,\times)\) is a ring. We say \(S \subseteq R\) is a subring of \(R\) if \((S,+,\times)\) is a ring.
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Definition 13.32. Subfield.

Assume \(F\) is a field. We say \(E \subseteq F\) is a field if \(E\) is a subring of \(F\) and \(E\) is a field.
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Example 13.33.

  • \(\Z\) is a subring of \(\Q\text{.}\)
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  • \(\Q\) is a subfield of \(\R\)
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  • \(\R\) is a subfield of \(\C\)
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  • The zero ring is the only subring of \(\F_p\text{.}\)
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Exercises 13.7 Exercises for Undergrads & Grads

Identifying Rings.

Consider the following sets with associative operations. Determine whether this defines a ring structure. If so, indicate whether the ring is commutative or unital. If the ring is unital, is the ring a division ring or a field?
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The set \(n\Z\) with integer addition and multiplication.
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Solution.
We already know that \(n\Z \leq \Z\text{.}\) Observe \(n\Z\) is closed under integer multiplication because for all \(i,j \in \Z\text{,}\)
\begin{equation*} (ni)(nj) = n(ijn) \in n\Z\text{.} \end{equation*}
Since integer multiplication is commutative, associative, and distributive, it follows that \(n\Z\) is a commutative ring.
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This ring is unital if and only if there exists \(k \in \Z\) such that for all \(x \in n\Z\)
\begin{equation*} (nk)x = x \iff nk = 1 \iff n = k = \pm 1\text{.} \end{equation*}
This ring is never a field because, e.g., the element \(2n \in n\Z\) is invertible if and only if there exists \(x \in n\Z\) such that
\begin{equation*} (2n)x = 1 \iff \frac{1}{2} = nx \in n\Z\text{,} \end{equation*}
which is absurd.
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2.
The set \(\Z \times \Z\) with component-wise addition and component-wise multiplication.
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Solution.
We know this is a commutative, unital ring by Example 13.10. The multiplicative identity is \((1,1)\text{.}\) This is not a field because, for example, \((1,0) \in \Z \times \Z\) does not have a multiplicative inverse. We can prove this by contradiction: assume \((a,b) \in \Z \times \Z\) is the multiplicative inverse of \((1,0)\) and then observe that
\begin{equation*} (1,1) = (1,0)(a,b) = (a,0) \iff a = 1\ \text{and}\ 0 = 1 \end{equation*}
is absurd.
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3.
The set \(\Z[\sqrt{2}] = \left\{a + b\sqrt{2} \;\middle\vert\; a,b \in \Z\right\}\) with addition
\begin{equation*} a + b\sqrt{2} + c + d\sqrt{2} = (a + c) + (b + d)\sqrt{2} \end{equation*}
and multiplication
\begin{equation*} (a + b\sqrt{2})(c + d \sqrt{2}) = (ac + 2bd) + (ad + bc)\sqrt{2} \end{equation*}
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Solution.
Observe that \(0 = 0 + 0\sqrt{2} \in \Z[\sqrt{2}] \subseteq \R\) and for all \(a,b,c,d \in \Z\text{,}\)
\begin{equation*} (a + b\sqrt{2}) - (c + d\sqrt{2}) = (a - c) + (b - d)\sqrt{2} \in \Z[\sqrt{2}]\text{.} \end{equation*}
Hence \(\Z[\sqrt{2}]\) is a subgroup of the abelian group \(\R\) by the The Subgroup Criterion.
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Since \(\Z[\sqrt{2}]\) is, by definition, closed under the multiplication of real numbers, it follows that the multiplication is commutative, associative, and distributive. Moreover, \(1 = 1 + 0\sqrt{2} \in \Z[\sqrt{2}]\text{.}\) Hence \(\Z[\sqrt{2}]\) is a commutative, unital ring. This is not a field because the multiplicative inverse of \(2 = 2 + 0\sqrt{2}\) is \(1/2 = 1/2 + 0\sqrt{2} \not \in \Z[\sqrt{2}]\text{.}\)
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4.
The set \(\Q(\sqrt{2}) = \left\{a + b\sqrt{2} \;\middle\vert\; a,b \in \Q\right\}\) with the same addition and multiplication above.
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Solution.
By the exact same argument as above, \(\Q(\sqrt{2})\) is a commutative, unital ring. Moreover, \(\Q(\sqrt{2})\) a field because whenever \(a,b \in \Q\setminus\{0\}\text{,}\) \(a^2 - 2b^2 \neq 0\) and thus
\begin{align*} \frac{1}{a + b\sqrt{2}} \amp= \frac{1}{a + b\sqrt{2}}\left(\frac{a - b\sqrt{2}}{a - b\sqrt{2}}\right)\\ \amp= \frac{a - b\sqrt{2}}{a^2 - 2b^2}\\ \amp= \frac{a}{a^2 - 2b^2} + \frac{-b}{a^2 - 2b^2}\sqrt{2} \in \Q(\sqrt{2}) \end{align*}
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5. The Evaluation Morphism.

Let \(F = \left\{f \colon \R \to \R \;\middle\vert\; f\ \text{is a function}\right\}\text{.}\) This is a unital ring under pointwise addition and pointwise multiplication of functions
\begin{equation*} (f + g)(x) := f(x) + g(x) \quad\text{and}\quad (fg)(x) := f(x)g(x) \end{equation*}
with multiplicative identity the function \(f(x) = 1\text{.}\) Fix \(a \in \R\text{.}\) Prove the function
\begin{align*} \operatorname{Ev}_a \colon F \amp\to \R\\ f \amp\mapsto f(a) \end{align*}
is a morphism of rings. We call this the evaluation morphism.
Solution.
For all \(f,g \in F\text{,}\)
\begin{equation*} \operatorname{Ev}_a(f + g) = (f + g)(a) = f(a) + g(a) = \operatorname{Ev}_a(f) + \operatorname{Ev}_a(g) \end{equation*}
and
\begin{equation*} \operatorname{Ev}_a(fg) = (fg)(a) = f(a)g(a) = \operatorname{Ev}_a(f)\operatorname{Ev}_a(g)\text{.} \end{equation*}
Therefore \(\operatorname{Ev}_a\) is a morphism of rings.
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Since \(F\) and \(\R\) are both unital rings, it makes sense to ask whether this is a morphism of unital rings. Recall that the multiplicative identity, \(1_F : \R \to \R\text{,}\) is defined by the property for all \(x \in \R\text{,}\) \(1_F(x) = 1\text{.}\) The evaluation map therefore satisfies
\begin{equation*} \operatorname{Ev}_a(1_F) = 1_F(a) = 1\text{.} \end{equation*}
Therefore \(\operatorname{Ev}_a\) is a morphism of unital rings.
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