Z_( p) , is a field ...( p is prime )


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

$Z_{p}, i s a f i e l d . . . (p i s p r i m e)$
Commented by kaivan.ahmadi last updated on 17/Aug/22

$w e s h o w t h a t e v e r y e l e m e n t$ $o f Z_{p} h a s i n v e r s e .$ $l e t [a] \in Z_{p}, 0 < a < p$ $s i n c e (a, p) = 1 \Rightarrow \exists r, s \in Z s . t$ $a r + p s = 1 \Rightarrow$ $[a] . [r] = [a r] + [0] = [a r] + [p s] =$ $[a r + p s] = [1] \Rightarrow {[a]}^{- 1} = [r] . ◼$
Commented by mnjuly1970 last updated on 17/Aug/22

$t h a n k s a l o t$
	Answered by mnjuly1970 last updated on 17/Aug/22

	$Z_{p} i s a n i n t e g e r d o m a i n$ $b e c a u s e .$ $\bar{a}, \bar{b} \in Z_{p}, \bar{a} \bar{b} = 0 \Rightarrow \overset{}{a b} \overset{p}{\equiv} 0$ $p ∣ a b \overset{p i s p r i m e}{\Rightarrow} p ∣ a o r p ∣ b$ $a \overset{p}{\equiv} 0 o r b \overset{p}{\equiv} 0$ $e a c h f i n i t e i n t e g e r d o m a i n$ $i s a f i e l d . . .$
	Commented by kaivan.ahmadi last updated on 19/Aug/22
	$ok its true. but do you know why each finite integer domain is a field? because every element in it has an inverse. let R={0,1,a_1 ,...,a_n } so a_i R=R⇒∃a_j ≠a_i s.t a_i a_j =1⇒a_i ^(−1) =a_j (or a_i =1).$
	$o k i t s t r u e . b u t d o y o u k n o w$ $w h y e a c h f i n i t e i n t e g e r d o m a i n$ $i s a f i e l d ? b e c a u s e e v e r y e l e m e n t$ $i n i t h a s a n i n v e r s e .$ $l e t R = {0, 1, a_{1}, . . ., a_{n}}$ $s o a_{i} R = R \Rightarrow \exists a_{j} \neq a_{i} s . t$ $a_{i} a_{j} = 1 \Rightarrow a_{i}^{- 1} = a_{j} (o r a_{i} = 1) .$
	Commented by mnjuly1970 last updated on 17/Aug/22

	$t h x s i r$
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