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Proof

(related to Proposition: Cancellation of Congruences With Factor Co-Prime To Module, Field $\mathbb Z_p$)

• Let $m > 1$ be a positive integer and $a,b,c$ be integers.
• By hypothesis, we have $m\mid (ac-bc)=c(a-b).$
• Also by hypothesis, $c$ and $m$ are co-prime, therefore from divisors of a product of two factors, co-prime to one factor, it follows that $m\mid (a-b).$
• Therefore $a(m)=b(m).$
• It remains to be shown that if $m=p$ is a prime number, then $\mathbb Z_p$ is a finite field.
  • We have seen that $Z_p$ is a commutative ring.
  • If for two elements $a(p),b(p)$ we have $a(p)\cdot b(p)\equiv 0(p),$ then we have $p\mid ab.$
  • From the Euclidean lemma, it follows that $p\mid a$ or $p\mid b.$
  • Therefore $a(p)=0(p)$ or $b(p)=0(p).$
  • Thus, $\mathbb Z_p$ is a commutative ring with $0(p)$ as its only zero divisor, in other words it is a integral domain.
  • Since finite integral domains are fields, $\mathbb Z_p$ is finite field.
• In particular, for every nonzeror $a(p)\in\mathbb Z_p,$ there is an multiplicative inverse element $a^{-1}(p)\in\mathbb Z_p$ and the congruence $(ax)(p)\equiv b(p)$ has the unique solution $x(p)\equiv (b\cdot a^{-1})(p).$
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References

Bibliography

1. Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927
2. Kraetzel, E.: "Studienbücherei Zahlentheorie", VEB Deutscher Verlag der Wissenschaften, 1981