Question-algebra-Count-the-number-of-zero-divisor-of-ring-Z-45-m-n-

Question Number 175132 by mnjuly1970 last updated on 20/Aug/22

≺ Q u e s t i o n - a l g e b r a ≻

C o u n t t h e n u m b e r o f^{″} z e r o d i v i s o r^{″}

o f r i n g, (Z_{45},, \oplus) m . n

- - - - - - - -

Commented by kaivan.ahmadi last updated on 20/Aug/22

w e m u s t f i n d t h e n u m b e r

o f d i g i t s k s u c h t h a t

(k, 45) \neq 1 .

\emptyset (45) = ϕ (3^{2} \times 5) = 45 (1 - \frac{1}{3}) (1 - \frac{1}{5})

= 45 (\frac{2}{3}) (\frac{4}{5}) = 24

\Rightarrow n u m b e r o f z e r o d i v i s o r =

44 - 24 = 20

o r K = {3, 5, 6, 9, 10, 12, 15, 18, 20, 21

, 24, 25, 27, 30, 33, 35, 36, 39, 40, 42}

f o r e a c h k \in K; (45, k) \neq 1 a n d

∣ K ∣= 20 .

Commented by mnjuly1970 last updated on 21/Aug/22

t h a n k y o u s o m u c h . .

i n f a c t

{[n - (φ (n) + 1)] = e a s y}

= 45 - 25 = 20

Commented by kaivan.ahmadi last updated on 20/Aug/22

y e s . a n d s o i f Z_{n} i s a f i e l d

i t h a s n o z e r o d i v i s o r .

s i n c e n = p = p r i m e a n d

p - 1 - ϕ (p) =

p - 1 - p (1 - \frac{1}{p}) = p - 1 - p + 1 = 0

Commented by mnjuly1970 last updated on 21/Aug/22

z e n d e h b a s h i d \dots