Question

Prove that a non-zero finite integral domain is a field.​

Answer 1

Answer:

Every finite integral domain is a field. The only thing we need to show is that a typical element a ≠ 0 has a multiplicative inverse. ... Then 0 = am - an = am(1 - an-m). Since there are no zero-divisors we must have am ≠ 0 and hence 1 - an-m = 0 and so 1 = a(an-m-1) and we have found a multiplicative inverse for a.

Step-by-step explanation:

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Answer 2

Theorem: Every finite integral domain is a field.

Proof: Let,

S be a finite integral domain. Because

S is finite, we may list its elements. Let us say

S := \{0,1,a_1, a_2, \ldots, a_n\} . To show that

S is a field, all we need to do is demonstrate that every nonzero element of

S is a unit (has a multiplicative inverse). To see that this must be true, take a nonzero element

s \in S . Consider the set

S' = \{s \cdot 1, sa_1, sa_2, \ldots, sa_n\} . Clearly

S' \subseteq S because

S is closed under multiplication. However, because

S is an integral domain it contains no zero divisors. Therefore, all elements of

S' are nonzero. Moreover, all of these elements must be distinct, because if

sa_i = sa_j for some

i \neq j then the cancelation law which holds in an integral domain demands that

a_i = a_j . We are forced to conclude that the elements of

S' must be a rearrangement of the nonzero elements of

S . In particular

sa_i = 1 for some

i \in \{1, \ldots, n\} , so

s has an inverse. Therefore, every nonzero element of

S is a unit, so

S is a field.

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