S. D. V. Stephens

notes / Integrated Mathematics / chapter 18: Groups, Rings and Fields

18.1 Integral Domains: Definition and Elementary Properties

In progress: these notes stop partway.

This page states the postulates for an integral domain and derives the first rules of algebra from them.

The postulates

Definition 18.1.1. Let be a set of elements for which the sum and the product of any two elements and (distinct or not) of are defined. Then is called an integral domain if the following postulates (i)–(ix) hold:

  1. (i) Closure. If and are in , then and are in ;
  2. (ii) Uniqueness. If and in , then and ;
  3. (iii) Commutativity. For all and in , and ;
  4. (iv) Associativity. For all , and in , and ;
  5. (v) Distributivity. For all , and in , ;
  6. (vi) Zero / identity. contains an element such that for all in ;
  7. (vii) Unity. contains an element such that for all in ;
  8. (viii) Additive invertibility. For each in , the equation has a solution in ;
  9. (ix) Cancellation law. If and , then .

Elementary properties

Note.

  • Reflexive law: .
  • Symmetric law: if , then .
  • Transitive law: if and , then .

Proposition 18.1.2 (Rule 1). for all in .

Proof. For all in :

Statement Reason
1. commutativity, mult.
2. distributivity
3. 1, 2, transitivity
4. , commutativity, mult.
5. 4, uniqueness (of addition)
6. 3, 5, transitivity

∎

Proposition 18.1.3 (Rule 2). For all in , and .

Proof. For all in :

Statement Reason
1. commutativity, addn.
2. zero / identity
3. 1, 2, transitivity

and

Statement Reason
1. commutativity, mult.
2. unity
3. 1, 2, transitivity

∎

Proposition 18.1.4 (Rule 3 (Uniqueness of zero / identity)). If , then .

Proof.

Statement Reason
1. hypothesis
2. zero
3. symmetric law (with 1, 2)
4. cancellation for addition (Rule 4)

∎

Proposition 18.1.5 (Rule 4). For all in , implies .

Proof. Let be a solution of (additive invertibility).

Statement Reason
1. comm. addn., trans. law, addn. invert.
2. hypothesis, uniqueness
3. assoc. addn.
4. addn. inv.
5. zero (Rule 2)

∎

Proposition 18.1.6 (Rule 5). For each , contains one and only one solution of the equation .

Proof. Assume, for purposes of contradiction, that there is a second, unique solution to , so that . Then since and , we have , and by cancellation (Rule 4), implies . We see that and are the same, and thus is not unique. ∎

Proposition 18.1.7 (Rule 6). For each and in , there exists one and only one in such that .

Proof. If , then . Since , we have . We see that is unique by application of Rules 4 and 5, which is trivial. ∎

Proposition 18.1.8 (Rule 7). For all in , .

We continue our rules and proofs.

Exercises

Exercise 18.1.9. Prove that the following hold in any integral domain:

(a) ;

(b) .

Show solution

(a) so .

(b) By associativity, at every step:

Exercise 18.1.10. If in has the property that for all in , then .

Show solution

Assume, for contradiction, that . Then gives , and by the cancellation law, implies . But we assumed , so we have a contradiction.

Remark (Correction). The cancellation law (ix) needs a nonzero factor, and the argument leaves unspecified. Take , which is nonzero by (vii): the hypothesis gives , and cancelling the factor gives .

Exercise 18.1.11. .

Show solution

Let and . If , then and . We know that the inverse of is and that , so gives , and gives . From here notice our result is , and by uniqueness .

Remark (Correction). An integral domain need not contain multiplicative inverses, so the steps and are unavailable. The result follows directly from postulate (vii): holds for every , in particular for , which gives .

Exercise 18.1.12. For all in , .

Show solution

Attempt, writing as and setting :

Remark (Correction). The chain starts from , which is the statement to be proved, ends with the identity , and uses , which an integral domain need not have; nothing is established. A correct proof: by Rule 1 and Rule 7, so by Rule 5. Applying this twice gives , where by commutativity and the same identity, and by Rule 5 because .