Rings and fields – Binary Operations (Senior 4 Mathematics) | Ganzaa

Binary Operations

Lessons

Senior 4 · Mathematics

Binary Operations

4 lessons

Lesson 4 of 4

Rings and fields

25 min

Two operations at once

Look at the barcode area of an older book and you may find a 10-digit ISBN such as 0-306-40615-2. The last digit is not part of the book's number: it is a check digit, chosen so that a weighted sum of all ten digits is a multiple of 11. If a shop assistant mistypes any single digit, the sum stops being a multiple of 11 and the computer refuses the number. Why does this always work, and why 11 and not 10? The answer needs a set with two operations, addition and multiplication, that behave as well as they do for real numbers. Such a set is a field, and arithmetic modulo 11 is one.

Definition

Ring: a set $R$ with two binary operations, written $+$ and $\times$ , is a ring $(R, +, \times)$ if

$(R, +)$ is an abelian group (its identity is written 0);
$\times$ is associative;
$\times$ is distributive over $+$ on both sides: $a (b + c) = a b + a c$ and $(b + c) a = b a + c a$ .

The ring is commutative if $a b = b a$ for all $a, b$ , and is a ring with unity if $\times$ has an identity element, written 1.

Examples: $(\mathbb{Z}, +, \times)$ , $(\mathbb{Q}, +, \times)$ , $(\mathbb{R}, +, \times)$ and $(\mathbb{Z}_n, +_n, \times_n)$ are commutative rings with unity. The even integers $2\mathbb{Z}$ form a commutative ring without unity, since no even number $u$ satisfies $2 u = 2$ . $(\mathbb{N}, +, \times)$ is not a ring: $(\mathbb{N}, +)$ has no inverses.

Definition

Zero divisor: in a commutative ring, a non-zero element $a$ is a zero divisor if there is a non-zero $b$ with $a b = 0$ .

Integral domain: a commutative ring with unity $1 \ne 0$ and no zero divisors.

Example: in $\mathbb{Z}_6$ , $2 \times_6 3 = 0$ although neither factor is 0, so 2 and 3 are zero divisors and $\mathbb{Z}_6$ is not an integral domain. $\mathbb{Z}$ is an integral domain: a product of non-zero integers is never 0.

Worked example 1: solving equations in $\mathbb{Z}_6$

Solve in $\mathbb{Z}_6$ : (a) $2 x = 4$ (b) $2 x = 3$ . Compare with real numbers.

Solution The row of 2 in the $\times_6$ table is $2 \times 0, 2 \times 1, \ldots, 2 \times 5 = 0, 2, 4, 0, 2, 4$ .

4 appears in the columns of 2 and 5: $x = 2$ or $x = 5$ . (Check: $2 \times 5 = 10 = 6 + 4$ .)
3 never appears in that row: no solution.

Over $\mathbb{R}$ , $2 x = b$ always has exactly one solution, $x = \frac{b}{2}$ . In $\mathbb{Z}_6$ it can have two solutions or none, because 2 has no multiplicative inverse there. That is what a field forbids.

Definition

Field: a commutative ring with unity $1 \ne 0$ in which every non-zero element $a$ has a multiplicative inverse $a^{- 1}$ , with $a \times a^{-1} = 1$ . Equivalently, $(K, +)$ and $(K \setminus \{0\}, \times)$ are both abelian groups, and $\times$ distributes over $+$ .

Examples: $\mathbb{Q}$ and $\mathbb{R}$ are fields. $\mathbb{Z}$ is not, since 2 has no inverse in $\mathbb{Z}$ . In a field, $a x = b$ with $a \ne 0$ always has exactly one solution, $x = a^{-1}b$ .

Important

Theorem. $(\mathbb{Z}_n, +_n, \times_n)$ is a field if and only if $n$ is prime.

Why. (1) A field has no zero divisors: if $a b = 0$ and $a \ne 0$ , then $b = a^{-1}(ab) = a^{-1} \times 0 = 0$ . If $n = p q$ with $1 \lt p, q \lt n$ , then $p \times_n q = 0$ , so $\mathbb{Z}_n$ is not a field. (2) If $n$ is prime and $a \ne 0$ , the products $a \times_n 1, a \times_n 2, \ldots, a \times_n (n - 1)$ are all different (if two were equal, $n$ would divide $a (b - c)$ , which is impossible for a prime $n$ that divides neither $a$ nor $b - c$ ). So they are the $n - 1$ non-zero elements in some order, and one of them is 1: $a$ has an inverse. $\blacksquare$

The field $\mathbb{Z}_5$

$\times_5$	1	2	3	4
1	1	2	3	4
2	2	4	1	3
3	3	1	4	2
4	4	3	2	1

Every row contains 1 exactly once, so every non-zero element has an inverse: $1^{-1} = 1$ , $2^{-1} = 3$ , $3^{-1} = 2$ , $4^{-1} = 4$ . The structures of this unit fit inside each other:

Worked example 2: equations in the field $\mathbb{Z}_5$

Solve in $\mathbb{Z}_5$ : (a) $3 x + 4 = 1$ (b) $x^{2} = 4$ .

Solution

Add the additive inverse of 4, which is 1: $3 x = 1 + 1 = 2$ . Multiply by $3^{-1} = 2$ : $x = 2 \times 2 = 4$ . Check: $3 \times 4 + 4 = 16 = 15 + 1$ , so $1$ . ✓
Square each element: $1^{2} = 1$ , $2^{2} = 4$ , $3^{2} = 9 = 4$ , $4^{2} = 16 = 1$ . So $x = 2$ or $x = 3$ . (Note $3 = - 2$ in $\mathbb{Z}_5$ : the two roots are $\pm 2$ , just as over $\mathbb{R}$ .)

Worked example 3: the ISBN-10 check digit

For an ISBN-10 $d_1 d_2 \ldots d_{10}$ , the rule is that $10d_1 + 9d_2 + 8d_3 + \cdots + 2d_9 + 1d_{10}$ is a multiple of 11; the symbol X stands for a check digit of 10. (a) Find the check digit of 0-306-40615-?. (b) Show that typing the fourth digit, 6, as 8 would be detected.

Solution

$0 (10) + 3 (9) + 0 (8) + 6 (7) + 4 (6) + 0 (5) + 6 (4) + 1 (3) + 5 (2) = 130$ . In $\mathbb{Z}_{11}$ , $130 = 11 \times 11 + 9$ is 9, so we need $9 + d_{10} = 0$ , giving $d_{10} = 2$ . The ISBN is 0-306-40615-2, and the full sum is $132 = 12 \times 11$ . ✓
The fourth digit has weight 7, so the sum changes by $7 \times (8 - 6) = 14$ to 146, and $146 = 13 \times 11 + 3$ is not a multiple of 11: detected. In general a single error changes the sum by $w \times d$ with $1 \le w \le 10$ and $d$ a non-zero difference of digits. Since $\mathbb{Z}_{11}$ is a field it has no zero divisors, so $w \times d \ne 0$ in $\mathbb{Z}_{11}$ : every single error is caught. With 10 instead of 11 this fails ( $2 \times 5 = 10 \equiv 0$ ).

Tip

To decide whether $\mathbb{Z}_n$ is a field, just ask whether $n$ is prime. To find an inverse in $\mathbb{Z}_p$ , look for 1 in the element's row of the multiplication table, or try multiples: in $\mathbb{Z}_7$ , $3 \times 5 = 15 = 14 + 1$ , so $3^{-1} = 5$ .

Socratica: Ring Definition (expanded). Watch for the ring axioms, then for the examples that are rings with and without a unity.

✦

Practice

Show all working in modular arithmetic.

Show that $(2\mathbb{Z}, +, \times)$ is a commutative ring without unity.
Is $(\mathbb{N}, +, \times)$ a ring? Explain.
Write the rows of 2, 3 and 4 in the $\times_6$ table and list the zero divisors of $\mathbb{Z}_6$ .
In $\mathbb{Z}_7$ , find the inverses of 3 and of 4.
Solve $4 x + 3 = 2$ in $\mathbb{Z}_7$ .
Explain why $\mathbb{Z}_8$ is not a field.
Find the check digits of the ISBN-10s that begin (a) 0-571-08989 (b) 0-486-47416.
Prove that in a field, $a b = 0$ implies $a = 0$ or $b = 0$ .

Show answersHide answers

Sums, differences and products of even integers are even; $(2\mathbb{Z}, +)$ is an abelian group (identity 0, inverse $- a$ ); $\times$ is associative, commutative and distributive, inherited from $\mathbb{Z}$ . No unity: $2 u = 2$ needs $u = 1$ , which is odd.
No: 1 has no additive inverse in $\mathbb{N}$ , so $(\mathbb{N}, +)$ is not a group.
2: 0, 2, 4, 0, 2, 4; 3: 0, 3, 0, 3, 0, 3; 4: 0, 4, 2, 0, 4, 2. Zero divisors: 2, 3 and 4 ( $2 \times 3 = 0$ , $4 \times 3 = 0$ ).
$3 \times 5 = 15 = 1$ , so $3^{-1} = 5$ ; $4 \times 2 = 8 = 1$ , so $4^{-1} = 2$ .
$4 x = 2 - 3 = - 1 = 6$ , so $x = 6 \times 4^{-1} = 6 \times 2 = 12 = 5$ . Check: $4 \times 5 + 3 = 23 = 21 + 2$ . ✓
$2 \times_8 4 = 0$ : zero divisors, so 2 has no inverse (8 is not prime).
(a) Weighted sum 226 $= 20 \times 11 + 6$ ; $6 + d = 11$ gives check digit 5. (b) Sum 232 $= 21 \times 11 + 1$ ; $1 + d = 11$ gives $d = 10$ , written X.
If $a \ne 0$ , multiply $a b = 0$ by $a^{- 1}$ : $b = a^{-1} \times 0 = 0$ .

Quick check

Which of these is a field?

Key points

A ring: $(R, +)$ an abelian group, $\times$ associative and distributive over $+$ .
A field: a commutative ring with unity in which every non-zero element has a multiplicative inverse; a field has no zero divisors.
$\mathbb{Z}_n$ is a field exactly when $n$ is prime.
ISBN-10 check digits work modulo 11 because $\mathbb{Z}_{11}$ is a field.

1Two operations at once

2The field Z5 \mathbb{Z}_5 Z5​

Two operations at once

The field $\mathbb{Z}_5$