Prove that the set$ (Z_p,+_p,•_p)$ is a field

Question

For a prime p, prove that the set$ (Z_p,+_p,•_p)$ is a field. (Show only that every nonzero element has a multiplicative inverse)

How can I prove it, I know that field if ring commutative and has inverse but I do not know how to start, any hint or note help me . Thank you in advance

Answer 1

HINT: Apply the Extended Euclidean Algorithm to prove that for given $a \not = 0$ the equation $ax \equiv 1 \pmod p$ has a solution.

Answer 2

Let us denote by $\bar{n}$ the elements of $\mathbb{Z}_{p}$ (I think that such notation is the most common one).

You have to check that if $\bar{n}\in \mathbb{Z}_{p}$ is non-zero, then there exists another element of $\mathbb{Z}_{p}$, $\bar{m}$, such that $\bar{n}\bar{m}=\bar{1}$ in $\mathbb{Z}_{p}$.

Hence, let $\bar{n}\in \mathbb{Z}_{p}$ be non-zero. By definition, that means that $n$ is not divisible by $p$ in $\mathbb{Z}$. Then, $\gcd(n,p)=1$, because $p$ is a prime number and $p$ does not divide $n$.

By Bezout, there exist $a,b\in \mathbb{Z}$ such that $1=\gcd(n,p)=an+bp$ in $\mathbb{Z}$.

Now, by doing that equality mod $p$, $\bar{1}=\overline{an+bp}=\overline{an}+\overline{bp}=\bar{a}\bar{n}+\bar{b}\bar{p}$. Now, observe that $\bar{p}=\bar{0}$ in $\mathbb{Z}_{p}$, so $$\bar{1}=\bar{a}\bar{n},$$ so $\bar{a}$ is the inverse element of $\bar{n}$ in $\mathbb{Z}_{p}$.

Answer 3

Let's show that it's an integral domain, i.e there are no non-zero zero-divisors. This is easy, since if $ab \equiv 0 \pmod p$, then by Euclid's lemma $p|a$ or $p|b$.

Then let's show that a commutative finite integral domain is in fact a field. This follows from the fact that the sequence $x^k, k=1,2,3\dots$ must have a repeated element because of finiteness. So $x^n=x^m$ for some $n>m$. Now we can cancel (in integral domain) so for every non-zero $x$ we get

$$x*x^{n-m-1} = x^{n-m} = 1$$ So $x$ has the inverse $x^{n-m-1}$.