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Rings, Ideals & Fields

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Meaning & notation

Rings support addition and multiplication; fields also support division by nonzero elements. Ideals make quotient arithmetic consistent.

Ring convention: Here rings have an abelian additive group, associative multiplication, identity $\displaystyle 1 \ne 0$ , and both distributive laws. Multiplication need not commute; ring homomorphisms preserve 1.
Unit, integral domain, and field: A unit has a two-sided multiplicative inverse. An integral domain is a commutative ring with no product of two nonzero elements equal to zero. A field is a commutative ring whose every nonzero element is a unit.
Two-sided ideal: An ideal I is an additive subgroup such that ri and ir belong to I whenever i belongs to I and r belongs to the ring. These absorption conditions allow multiplication of residue classes.

Conditions before calculation

Under our convention $\displaystyle 1 \ne 0$ , use a proper ideal I for $\displaystyle \frac{R}{I}$ . Proper ideals contain neither 1 nor any other unit.
For integer $\displaystyle n \ge 2$ , the ring $\mathbb{Z}/n\mathbb{Z}$ is a field exactly when n is prime. A class [a] is a unit exactly when $\displaystyle \operatorname{gcd}\, \left(a, n\right) = 1$ .
Every finite integral domain is a field. Finiteness is essential: Z has no nonzero zero divisors but its element 2 has no integer multiplicative inverse.

Why the method works

1. Distinguish the two inverses

The additive inverse of a always exists and sums with a to zero. A multiplicative inverse must instead produce one; the additive group axioms do not supply it.

a+(-a)=0,\qquad aa^{-1}=a^{-1}a=1

2. Explain ideal absorption

If representatives change by ideal elements, the product changes by sums containing those ideal elements. Absorption keeps that difference in I, making the quotient product independent of representatives.

(a+i)(b+j)-ab=aj+ib+ij\in I

3. Explain the finite-domain theorem

For nonzero a in a domain, $\displaystyle ax = ay$ implies $\displaystyle x = y$ because $\displaystyle a\, \left(x - y\right) = 0$ . On a finite set an injective self-map is surjective, so multiplication by a reaches 1 and supplies an inverse.

Worked examples

EXAMPLE 1

Find the additive and multiplicative inverses of [7] in $\mathbb{Z}/11\mathbb{Z}$ .

The additive inverse must make a sum divisible by 11; $\displaystyle 7 + 4 = 11$ .
For multiplication, the integer identity $\displaystyle 8 \cdot 7 - 5 \cdot 11 = 1$ exhibits the required inverse modulo 11.
$[7]+[4]=[0],\qquad[7][8]=[1]$
Since 11 is prime, every nonzero class has a multiplicative inverse; here [4] and [8] play different roles.

Conclusion: The additive inverse is [4]; the multiplicative inverse is [8].

EXAMPLE 2

Explain why $\mathbb{Z}/10\mathbb{Z}$ is a quotient ring but not a field.

The multiples 10Z form an additive subgroup of Z. Multiplying a multiple of 10 by any integer gives another multiple, so it is an ideal.
It is proper because 1 is not a multiple of 10. Its ten cosets inherit well-defined addition and multiplication.
The nonzero classes [2] and [5] multiply to [0]. If [2] had an inverse, multiplying this equation by it would force $\displaystyle \left[5\right] = \left[0\right]$ , a contradiction.

Conclusion: The quotient is a commutative ring with zero divisors, so it is not a field.

Pitfalls & counterexamples

Tempting claim: Every nonzero ring element can be canceled from a product.

Correction: Cancellation needs an additional condition, such as being a unit or working in an integral domain.

Test the claim: Modulo 10, $\displaystyle \left[2\right]\, \left[1\right] = \left[2\right]\, \left[6\right]$ , but [1] and [6] are different classes.

Check your understanding

Can a proper ideal in a ring contain 1?

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Compare your reasoning

No.

If 1 belongs to I, then every $\displaystyle r = r \cdot 1$ belongs to I by absorption, so I equals the whole ring.

Connect this idea

Ideals enable quotient rings. Fields support linear algebra; zero divisors explain failed cancellation.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.