Unformatted text preview: 1.9. COUNTING 61 Lemma 1.9.8. Let p be a prime number.
ÂÃ
p
(a) If 0 < k < p , then
is divisible by p .
k
(b) For all integers a and b , .a C b/p Á ap C b p .mod p/. Proof. For part (a), we have pŠ D ÂÃ
p
kŠ .p
k k /Š. Now p divides pŠ,
ÂÃ
p
but p does not divide kŠ or .n k /Š. Consequently, p divides
.
ÂÃ k
Pp
p knk
The binomial theorem gives .a C b/p D
ab
. By
k D0 k
part (a), all the terms for 0 < k < p are divisible by p , so .a C b/p is
congruent modulo p to the sum of the terms for k D 0 and k D p , namely
to ap C b p .
I
I hope you already discovered the ﬁrst part of the next lemma while
doing the exercises for Section 1.7.
Proposition 1.9.9.
(a) Let n
2 be a natural number. An element Œa 2 Zn has a
multiplicative inverse if, and only if, a is relatively prime to n.
(b) If p is a prime, then every nonzero element of Zp is invertible. Proof. If a is relatively prime to n, there exist integers s; t such that as C
nt D 1. But then as Á 1 .mod n/, or ŒaŒs D Œ1 in Zn . On the other
hand, if a is not relatively prime to n, then a and n have a common divisor
k > 1. Say kt D a and ks D n. Then as D k t s D nt . Reducing
modulo n gives ŒaŒs D Œ0, so Œa is a zero divisor in Zn , and therefore
not invertible.
If p is a prime, and Œa ¤ 0 in Zp , then a is relatively prime to p , so
Œa is invertible in Zp by part (a).
I Proposition 1.9.10. (Fermat’s little theorem). Let p be a prime number.
(a) For all integers a, we have ap Á a .mod p/.
(b) If a is not divisible by p , then ap 1 Á 1 .mod p/. ...
View
Full Document
 Fall '08
 EVERAGE
 Algebra, Integers, Counting, Prime number, zp

Click to edit the document details