Results 1 to 4 of 4

Math Help - urgent help about field

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    18

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    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}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Gamma's Avatar
    Joined
    Dec 2008
    From
    Iowa City, IA
    Posts
    517

    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Ring, field, Galois-Field, Vector Space
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: November 15th 2012, 04:25 PM
  2. Splitting Field of a Polynomial over a Finite Field
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 1st 2011, 04:45 PM
  3. Field of char p>0 & splitting field
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 22nd 2009, 01:20 AM
  4. urgent about field
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 19th 2008, 03:38 PM
  5. URGENT! Find splitting field and roots
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 17th 2008, 05:25 PM

Search Tags


/mathhelpforum @mathhelpforum