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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
        ≺  Question− algebra ≻                     Count the number of   ” zero divisor ”           of  ring ,   ( Z_( 45)  ,   , ⊕ )                ■  m.n                −−−−−−−−
$$ \\ $$$$\:\:\:\:\:\:\prec\:\:\boldsymbol{{Question}}−\:\boldsymbol{{algebra}}\:\succ \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{{Count}}\:\boldsymbol{{the}}\:\boldsymbol{{number}}\:\boldsymbol{{of}}\:\:\:''\:\boldsymbol{{zero}}\:\boldsymbol{{divisor}}\:'' \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{{of}}\:\:\boldsymbol{{ring}}\:,\:\:\:\left(\:\mathbb{Z}_{\:\mathrm{45}} \:,\: \:,\:\oplus\:\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare\:\:\boldsymbol{{m}}.\boldsymbol{{n}}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−−−−−−−−\:\:\:\: \\ $$
Commented by kaivan.ahmadi last updated on 20/Aug/22
we must find the number  of digits k such that  (k,45)≠1.  ∅(45)=φ(3^2 ×5)=45(1−(1/3))(1−(1/5))  =45((2/3))((4/5))=24  ⇒number of zero divisor=  44−24=20      or K={3,5,6,9,10,12,15,18,20,21  ,24,25,27,30,33,35,36,39,40,42}  for each k∈K ; (45,k)≠1 and  ∣K∣=20.
$${we}\:{must}\:{find}\:{the}\:{number} \\ $$$${of}\:{digits}\:{k}\:{such}\:{that} \\ $$$$\left({k},\mathrm{45}\right)\neq\mathrm{1}. \\ $$$$\emptyset\left(\mathrm{45}\right)=\phi\left(\mathrm{3}^{\mathrm{2}} ×\mathrm{5}\right)=\mathrm{45}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$$$=\mathrm{45}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\left(\frac{\mathrm{4}}{\mathrm{5}}\right)=\mathrm{24} \\ $$$$\Rightarrow{number}\:{of}\:{zero}\:{divisor}= \\ $$$$\mathrm{44}−\mathrm{24}=\mathrm{20} \\ $$$$ \\ $$$$ \\ $$$${or}\:{K}=\left\{\mathrm{3},\mathrm{5},\mathrm{6},\mathrm{9},\mathrm{10},\mathrm{12},\mathrm{15},\mathrm{18},\mathrm{20},\mathrm{21}\right. \\ $$$$\left.,\mathrm{24},\mathrm{25},\mathrm{27},\mathrm{30},\mathrm{33},\mathrm{35},\mathrm{36},\mathrm{39},\mathrm{40},\mathrm{42}\right\} \\ $$$${for}\:{each}\:{k}\in{K}\:;\:\left(\mathrm{45},{k}\right)\neq\mathrm{1}\:{and} \\ $$$$\mid{K}\mid=\mathrm{20}. \\ $$
Commented by mnjuly1970 last updated on 21/Aug/22
thank you so much..    in fact    {[n − (ϕ (n) +1 )]=easy}         =45−25=20
$${thank}\:{you}\:{so}\:{much}.. \\ $$$$\:\:{in}\:{fact} \\ $$$$\:\:\left\{\left[{n}\:−\:\left(\varphi\:\left({n}\right)\:+\mathrm{1}\:\right)\right]={easy}\right\} \\ $$$$\:\:\:\:\:\:\:=\mathrm{45}−\mathrm{25}=\mathrm{20} \\ $$$$\:\:\: \\ $$$$\: \\ $$
Commented by kaivan.ahmadi last updated on 20/Aug/22
yes. and so if Z_(n  ) is a field  it has no zero divisor.  since n=p=prime and  p−1−φ(p)=  p−1−p(1−(1/p))=p−1−p+1=0
$${yes}.\:{and}\:{so}\:{if}\:\mathbb{Z}_{{n}\:\:} {is}\:{a}\:{field} \\ $$$${it}\:{has}\:{no}\:{zero}\:{divisor}. \\ $$$${since}\:{n}={p}={prime}\:{and} \\ $$$${p}−\mathrm{1}−\phi\left({p}\right)= \\ $$$${p}−\mathrm{1}−{p}\left(\mathrm{1}−\frac{\mathrm{1}}{{p}}\right)={p}−\mathrm{1}−{p}+\mathrm{1}=\mathrm{0} \\ $$
Commented by mnjuly1970 last updated on 21/Aug/22
 zendeh bashid ...
$$\:{zendeh}\:{bashid}\:… \\ $$

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