Z-p-is-a-field-p-is-prime-

Question Number 175049 by mnjuly1970 last updated on 17/Aug/22

$$ \\ $$$$\:\:\:\:\:\:\mathbb{Z}_{\:{p}} \:,\:{is}\:\:{a}\:\:{field}\:…\left(\:\:{p}\:{is}\:{prime}\:\right) \\ $$$$\:\:\:\:\:\: \\ $$

Commented by kaivan.ahmadi last updated on 17/Aug/22

we show that every element of Z_p has inverse. let [a]∈Z_p ,0<a<p since (a,p)=1⇒∃r,s∈Z s.t ar+ps=1⇒ [a].[r]=[ar]+[0]=[ar]+[ps]= [ar+ps]=[1]⇒[a]^(−1) =[r].■

$${we}\:{show}\:{that}\:{every}\:{element} \\ $$$${of}\:\mathbb{Z}_{{p}} \:{has}\:{inverse}. \\ $$$${let}\:\left[{a}\right]\in\mathbb{Z}_{{p}} \:,\mathrm{0}<{a}<{p} \\ $$$${since}\:\left({a},{p}\right)=\mathrm{1}\Rightarrow\exists{r},{s}\in\mathbb{Z}\:{s}.{t} \\ $$$${ar}+{ps}=\mathrm{1}\Rightarrow \\ $$$$\left[{a}\right].\left[{r}\right]=\left[{ar}\right]+\left[\mathrm{0}\right]=\left[{ar}\right]+\left[{ps}\right]= \\ $$$$\left[{ar}+{ps}\right]=\left[\mathrm{1}\right]\Rightarrow\left[{a}\right]^{−\mathrm{1}} =\left[{r}\right].\blacksquare \\ $$

Commented by mnjuly1970 last updated on 17/Aug/22

$${thanks}\:{alot} \\ $$

Answered by mnjuly1970 last updated on 17/Aug/22

Z_p is an integer domain because. a^− , b^− ∈ Z_p , a^− b^− =0 ⇒ a^ b ≡^p 0 p∣ ab ⇒^(p is prime) p∣a or p∣ b a≡^p 0 or b≡^p 0 each finite integer domain is a field ...

$$\:\:\mathbb{Z}_{{p}} \:{is}\:{an}\:{integer}\:{domain}\: \\ $$$$\:\:\:{because}. \\ $$$$\:\:\:\overset{−} {{a}}\:,\:\overset{−} {{b}}\:\in\:\mathbb{Z}_{{p}} \:,\:\:\overset{−} {{a}}\overset{−} {{b}}\:=\mathrm{0}\:\Rightarrow\:\overset{\:} {{a}b}\:\overset{{p}} {\equiv}\mathrm{0} \\ $$$$\:\:\:\:\:{p}\mid\:{ab}\:\overset{{p}\:{is}\:{prime}} {\Rightarrow}{p}\mid{a}\:\:{or}\:\:{p}\mid\:{b}\:\: \\ $$$$\:\:\:\:\:\:{a}\overset{{p}} {\equiv}\mathrm{0}\:{or}\:\:{b}\overset{{p}} {\equiv}\:\mathrm{0}\:\: \\ $$$$\:\:\:\:\:\:{each}\:\:{finite}\:{integer}\:{domain} \\ $$$$\:\:\:\:\:\:{is}\:\:{a}\:{field}\:… \\ $$

Commented by kaivan.ahmadi last updated on 19/Aug/22

ok its true. but do you know why each finite integer domain is a field? because every element in it has an inverse. let R={0,1,a_1 ,...,a_n } so a_i R=R⇒∃a_j ≠a_i s.t a_i a_j =1⇒a_i ^(−1) =a_j (or a_i =1).

$${ok}\:{its}\:{true}.\:{but}\:{do}\:{you}\:{know} \\ $$$${why}\:{each}\:{finite}\:{integer}\:{domain} \\ $$$${is}\:{a}\:{field}?\:{because}\:{every}\:{element} \\ $$$${in}\:{it}\:{has}\:{an}\:{inverse}. \\ $$$${let}\:{R}=\left\{\mathrm{0},\mathrm{1},{a}_{\mathrm{1}} ,…,{a}_{{n}} \right\} \\ $$$${so}\:{a}_{{i}} {R}={R}\Rightarrow\exists{a}_{{j}} \neq{a}_{{i}} \:{s}.{t} \\ $$$${a}_{{i}} {a}_{{j}} =\mathrm{1}\Rightarrow{a}_{{i}} ^{−\mathrm{1}} ={a}_{{j}} \:\left({or}\:{a}_{{i}} =\mathrm{1}\right). \\ $$

Commented by mnjuly1970 last updated on 17/Aug/22

$$\:\:\:\:{thx}\:{sir} \\ $$

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