1. ## urgent help about field

Prove that Z_
n has zero divisors if n is composite, and that Z_n is a field

if
n is prime.

2. If n is composite then there exists a,b both less than n such that n = ab. Thus $a,b \in Z_n, ab = 0\text{ mod n}$

To give a concrete example, $3,2 \in \mathbb{Z}_6, 3.2 = 0$

To prove $\mathbb{Z}_n$ is a field whenever n is prime, one really needs to worry only about checking whether every non zero element has a multiplicative inverse. The rest of the axioms are easy to verify.

We claim that if $a > 0 \in \mathbb{Z}_n, S=\{a,2a,3a...,(n-1)a\}$ must be a permutation of $\{1,2,3...,(n-1)\}$.

Observe that if two elements, say ia and ja, in S are equal then n|(i-j)a. But n is prime, which means n|a or n|(i-j). if i is not equal to j, both are positive and less than n. Thus both the conditions are impossible. Hence i=j and this establishes all the elements of S are unique.

Now observe that $S \subset \mathbb{Z}_{n-1}$ and S has n distinct elements. Thus there must exist a "ja" in S such that $ja = 1$. That $j = a^{-1}$

3. Here is another way to prove that $\mathbb{Z}_p$ is a field.

Definition: A commutative unitary ring $R$ with $1\not = 0$ (i.e. it is not trivial) is called an integral domain iff $ab = 0 \implies a=0 \text{ or }b=0$ (this is referring to as having "no zero divisors").

Theorem: If $F$ is a finite integral domain then $F$ is a field.

Proof: Let $F = \{ a_1,...,a_n\}$. And let $a\in F^{\times}$ i.e. a non-zero element. Consider, $aa_1,aa_2,...,aa_n$, we argue that all these elements are distinct. Say $aa_i = aa_j \implies a(a_i - a_j) = 0$ and so $a=0$ or $a_i - a_j=0$. It cannot be the first case, $a=0$, therefore $a_i = a_j$. We have shown that $aa_1,aa_2,...,aa_n$ are distinct. It follows by the pigeonhole principle that this list is a permutation of the $\{ a_1,...,a_n\}$ and therefore $aa_i = 1$ for some $i$. We have shown that each non-zero element has a multiplicative inverse. Thus, $F$ is a field.

In order to prove your result you just need to show that $\mathbb{Z}_p$ is an integral domain.
This is similar to what Isomorphism did, but this is more abstract.

4. ## Alternative way of finding the inverse

If $a \in \mathbb{Z}_p$ $(a \not = 0)$ where $p$ is prime we know $gcd(a,p)=1$ thus, by the Euclidean Algorithm we are guaranteed the existence of $x,y \in \mathbb{Z}$ such that $ax+py=1$ then since $p|py$ we see that $ax \cong 1 (mod\ p)$ and thus $x$ is the inverse of $a$ in $\mathbb{Z}_p$

Just thought a more constructive way of actually finding the inverse might be of interest.