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“mcs” — 2017/3/3 — 11:21 — page 340 — #348<br />

340<br />

Chapter 9<br />

Number Theory<br />

A sum or product of numbers in ZΠp 5 is in ZΠp 5, and since ZΠp 5 is a<br />

subset of the complex numbers, all the usual rules <strong>for</strong> addition and multiplication<br />

are true <strong>for</strong> it. But some weird things do happen. For example, the prime 29 has<br />

factors:<br />

(a) Find x; y 2 ZŒ p 5 such that xy D 29 and x ¤ ˙1 ¤ y.<br />

On the other hand, the number 3 is still a “prime” even in ZŒ p 5. More precisely,<br />

a number p 2 ZΠp 5 is called irreducible over ZΠp 5 iff when xy D p<br />

<strong>for</strong> some x; y 2 ZŒ p 5, either x D ˙1 or y D ˙1.<br />

Claim. The numbers 3; 2 C p 5, and 2<br />

p<br />

5 are irreducible over ZŒ<br />

p<br />

5.<br />

In particular, this Claim implies that the number 9 factors into irreducibles over<br />

ZΠp 5 in two different ways:<br />

3 3 D 9 D .2 C p 5/.2<br />

p<br />

5/:<br />

So ZΠp 5 is an example of what is called a non-unique factorization domain.<br />

To verify the Claim, we’ll appeal (without proof) to a familiar technical property<br />

of complex numbers given in the following Lemma.<br />

Definition. For a complex number c D r C si where r; s 2 R and i is p 1, the<br />

norm jcj of c is p r 2 C s 2 .<br />

Lemma. For c; d 2 C,<br />

jcdj D jcj jdj :<br />

(b) Prove that jxj 2 ¤ 3 <strong>for</strong> all x 2 ZŒ p 5.<br />

(c) Prove that if x 2 ZŒ p 5 and jxj D 1, then x D ˙1.<br />

(d) Prove that if jxyj D 3 <strong>for</strong> some x; y 2 ZŒ p 5, then x D ˙1 or y D ˙1.<br />

Hint: jzj 2 2 N <strong>for</strong> z 2 ZΠp 5.<br />

(e) Complete the proof of the Claim.<br />

Problems <strong>for</strong> Section 9.6<br />

Practice Problems<br />

Problem 9.26.<br />

Prove that if a b .mod 14/ and a b .mod 5/, then a b .mod 70/.

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