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Thread: urgent help about field

  1. #1
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    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.
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  2. #2
    Lord of certain Rings
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    If n is composite then there exists a,b both less than n such that n = ab. Thus $\displaystyle a,b \in Z_n, ab = 0\text{ mod n} $

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

    To prove $\displaystyle \mathbb{Z}_n$ 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 $\displaystyle a > 0 \in \mathbb{Z}_n, S=\{a,2a,3a...,(n-1)a\}$ must be a permutation of $\displaystyle \{1,2,3...,(n-1)\}$.

    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 $\displaystyle S \subset \mathbb{Z}_{n-1}$ and S has n distinct elements. Thus there must exist a "ja" in S such that $\displaystyle ja = 1$. That $\displaystyle j = a^{-1}$
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  3. #3
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    Here is another way to prove that $\displaystyle \mathbb{Z}_p$ is a field.

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

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

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

    In order to prove your result you just need to show that $\displaystyle \mathbb{Z}_p$ is an integral domain.
    This is similar to what Isomorphism did, but this is more abstract.
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  4. #4
    Super Member Gamma's Avatar
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    Alternative way of finding the inverse

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

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