Algebra

Rings and fields

Algebraic structures with two operations, addition and multiplication, generalizing the integers and rationals.

IntuitionWhen you can add, subtract, and multiply — and when you can also divide

In the integers $\mathbb{Z}$ , you can add, subtract, and multiply without ever leaving $\mathbb{Z}$ , and multiplication distributes over addition via $a(b + c) = ab + ac$ — that two-operation package is what algebraists call a ring. Division, however, fails in $\mathbb{Z}$ : $1/2$ is not an integer. Moving to the rationals $\mathbb{Q}$ (or to clock arithmetic $\mathbb{Z}/p\mathbb{Z}$ modulo a prime $p$ ) restores division by every nonzero element via $a \cdot a^{-1} = 1$ , turning the ring into a field.

In the ring

\mathbb{Z}/9\mathbb{Z}

(

m=9

), multiplying by

a=3

sends

3

and

6

0

(zero divisors!), whereas when

m

is prime every nonzero

a

is invertible and

\mathbb{Z}/m\mathbb{Z}

is a field.

SchoolAxioms of rings, integral domains, and fields

Definition: Ring, integral domain, and field

A ring $(R,+ ,\cdot)$ is a set where $(R,+)$ is an abelian group (with zero element $0$ and negatives $-a$ ), multiplication $\cdot$ is associative with identity $1$ , and multiplication distributes over addition: $a(b + c) = ab + ac$ and $(a+b)c = ac + bc$ . A commutative ring ( $ab=ba$ ) with $1 \neq 0$ is an integral domain if it has no zero divisors ( $ab=0 \implies a=0$ or $b=0$ ), and is a field if every $a \neq 0$ has a multiplicative inverse $a^{-1}$ satisfying $a \cdot a^{-1} = 1$ .

a(b + c) = ab + ac, \qquad (a + b)c = ac + bc

Inside any commutative ring $R$ , an ideal $I \subseteq R$ is a subgroup under addition that absorbs multiplication by arbitrary ring elements ( $r \in R, x \in I \implies rx \in I$ ). Quotienting by the principal ideal $n\mathbb{Z}$ inside $\mathbb{Z}$ produces the residue ring $\mathbb{Z}/n\mathbb{Z}$ , whose arithmetic is ordinary addition and multiplication modulo $n$ .

\mathbb{Z}/n\mathbb{Z} = \{\bar{0}, \bar{1}, \dots, \overline{n-1}\}, \qquad \bar{a}\,\bar{b} = \overline{ab \bmod n}

Comparison of standard rings and fields
Structure	Key multiplicative property
Integers $\mathbb{Z}$	Integral domain, not a field: only $\pm 1$ have multiplicative inverses
Rationals $\mathbb{Q}$ , reals $\mathbb{R}$ , complexes $\mathbb{C}$	Infinite fields: every nonzero element has an inverse
$\mathbb{Z}/n\mathbb{Z}$ with $n$ composite (e.g. $n=6$ )	Commutative ring with zero divisors ( $\bar{2}\cdot\bar{3}=\bar{0}$ in $\mathbb{Z}/6\mathbb{Z}$ ); not a field
$\mathbb{Z}/p\mathbb{Z}$ with $p$ prime	Finite field of $p$ elements: every $\bar{a} \neq \bar{0}$ is invertible

UndergraduateWhen a ring becomes a field

Theorem: Every finite integral domain is a field

If $D$ is a finite integral domain, then every nonzero element of $D$ has a multiplicative inverse, so $D$ is a field.

Why is it true?

On a finite set, a map that never sends two different inputs to the same output is forced by the pigeonhole principle to hit every possible output — including $1$ .

Proof

Let $a \in D$ be any nonzero element, and consider the left-multiplication map $L_a : D \to D$ defined by $L_a(x) = ax$ .

We first show $L_a$ is injective. Suppose $L_a(x) = L_a(y)$ , so $ax = ay$ , i.e. $a(x - y) = 0$ . Since $D$ is an integral domain and $a \neq 0$ , there are no zero divisors, which forces $x - y = 0$ and hence $x = y$ .

Because $D$ is a finite set, any injective map $D \to D$ is automatically surjective (by the pigeonhole principle: $|L_a(D)| = |D|$ ). In particular, the multiplicative identity $1 \in D$ must lie in the image of $L_a$ , so there exists $b \in D$ with $L_a(b) = ab = 1$ . Thus $b = a^{-1}$ , and since $a \neq 0$ was arbitrary, $D$ is a field.

Theorem: Criterion for

\mathbb{Z}/p\mathbb{Z}

to be a field

For an integer $p \ge 2$ , the residue ring $\mathbb{Z}/p\mathbb{Z}$ is a field if and only if $p$ is a prime number.

Why is it true?

When $p$ is prime, no nonzero number smaller than $p$ shares a factor with $p$ , so Bézout's identity always manufactures an inverse modulo $p$ ; when $p$ is composite, its factors multiply to $p \equiv 0$ and destroy invertibility.

Proof

( $\Leftarrow$ ) Suppose $p$ is prime, and let $\bar{a} \in \mathbb{Z}/p\mathbb{Z}$ be any nonzero residue class, represented by an integer $a$ with $1 \le a \le p - 1$ . Since $p$ is prime and $p \nmid a$ , the greatest common divisor of $a$ and $p$ is $\gcd(a,p) = 1$ .

By Bézout's identity, there exist integers $u, v \in \mathbb{Z}$ such that $au + pv = 1$ . Reducing both sides modulo $p$ kills the multiple $pv$ , leaving $\bar{a}\,\bar{u} = \bar{1}$ in $\mathbb{Z}/p\mathbb{Z}$ . Thus $\bar{u} = \bar{a}^{-1}$ is the multiplicative inverse of $\bar{a}$ , and since every nonzero element is invertible, $\mathbb{Z}/p\mathbb{Z}$ is a field.

( $\Rightarrow$ ) Conversely, suppose $p$ is composite, so $p = bc$ for integers $1 < b, c < p$ . Because $b$ and $c$ are strictly between $0$ and $p$ , their residue classes $\bar{b}$ and $\bar{c}$ are both nonzero in $\mathbb{Z}/p\mathbb{Z}$ , yet their product is $\bar{b}\,\bar{c} = \overline{bc} = \bar{p} = \bar{0}$ .

Thus $\bar{b}$ is a zero divisor, and no zero divisor can have a multiplicative inverse: if $\bar{b}$ had an inverse $\bar{b}^{-1}$ , multiplying $\bar{b}\,\bar{c} = \bar{0}$ on the left by $\bar{b}^{-1}$ would force $\bar{c} = \bar{0}$ , a contradiction. Hence $\mathbb{Z}/p\mathbb{Z}$ is not a field when $p$ is composite.

UndergraduateReal-World Applications and Worked Examples

Finite fields and residue rings are the working engine of modern public-key cryptography, digital signatures, and the Reed-Solomon error-correcting codes inside QR codes and solid-state drives.

Example: Computing a modular inverse via Bézout's identity

In the prime field $\mathbb{Z}/17\mathbb{Z}$ , find the multiplicative inverse of $\bar{5}$ , and use it to solve the linear congruence $5x \equiv 3 \pmod{17}$ .

Solution

Apply the Euclidean algorithm to $17$ and $5$ : $17 = 3 \cdot 5 + 2$ , then $5 = 2 \cdot 2 + 1$ . Back-substituting $2 = 17 - 3 \cdot 5$ gives $1 = 5 - 2(17 - 3 \cdot 5) = 7 \cdot 5 - 2 \cdot 17$ .

Reducing this Bézout identity modulo $17$ yields $\bar{7} \cdot \bar{5} = \bar{1}$ in $\mathbb{Z}/17\mathbb{Z}$ , so $\bar{5}^{-1} = \bar{7}$ (indeed $5 \cdot 7 = 35 = 2 \cdot 17 + 1 \equiv 1 \pmod{17}$ ).

Multiplying $5x \equiv 3 \pmod{17}$ on both sides by $7$ immediately isolates $x$ : $x \equiv 7 \cdot 3 = 21 \equiv 4 \pmod{17}$ . Checking: $5 \cdot 4 = 20 \equiv 3 \pmod{17}$ .

Example: Erasure recovery over a prime field (Reed-Solomon idea)

Two data symbols $m_0 = 2, m_1 = 5 \in \mathbb{Z}/7\mathbb{Z}$ are encoded as the line $P(t) = m_0 + m_1 t \pmod{7}$ and sent as the four evaluations $(P(0), P(1), P(2), P(3)) = (2, 0, 5, 3)$ . During transmission the first two values are lost, leaving only $P(2)=5$ and $P(3)=3$ . Recover $m_0$ and $m_1$ .

Solution

In the field $\mathbb{Z}/7\mathbb{Z}$ we have the two linear equations $m_0 + 2m_1 \equiv 5 \pmod{7}$ and $m_0 + 3m_1 \equiv 3 \pmod{7}$ .

Subtracting the first equation from the second eliminates $m_0$ directly: $(3 - 2)m_1 \equiv 3 - 5 = -2 \equiv 5 \pmod{7}$ , so $m_1 = 5$ . (Even if the step difference had been $k \not\equiv 0$ rather than $1$ , we could always divide by $k$ because $\mathbb{Z}/7\mathbb{Z}$ is a field.)

Substituting $m_1 = 5$ back into $m_0 + 2m_1 \equiv 5 \pmod{7}$ gives $m_0 \equiv 5 - 10 = -5 \equiv 2 \pmod{7}$ , recovering $(m_0, m_1) = (2, 5)$ exactly from any two surviving packets — the core principle of Reed-Solomon erasure coding.

What is the multiplicative inverse of $\bar{4}$ in the prime field $\mathbb{Z}/11\mathbb{Z}$ ?

Why is the ring $\mathbb{Z}/9\mathbb{Z}$ not a field, even though $9 = 3^2$ is a prime power?

Which number-theoretic tool directly produces the multiplicative inverse of $\bar{a} \neq \bar{0}$ in $\mathbb{Z}/p\mathbb{Z}$ when $p$ is prime?

Why is the ring of integers $\mathbb{Z}$ an integral domain even though it is not a field?

References

David S. Dummit, Richard M. Foote (2004). Abstract Algebra (3rd ed.)
Michael F. Atiyah, Ian G. Macdonald (1969). Introduction to Commutative Algebra