Abstract Algebra: Ring Proof (Multiplicative Inverse)

  • #1
RJLiberator
Gold Member
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Homework Statement


Suppose R is a commutative ring with only a finite number of elements and no zero divisors. Show that R is a field.

Homework Equations



Unit is an element in R which has a multiplicative inverse. If s∈R with r*s = 1.
A zero divisor is an element r∈R such that there exists s∈R and rs = 0 (or sr = 0).

The Attempt at a Solution



1. Since R is a commutative ring, we only need to prove one axiom, that is that it satisfies the multiplicative inverse for all values in R.
2. I have a theorem that we just went over that states:
Theorem: If R is a finite ring, then for all r ∈ R, r is either a unit or a zero divisor.
3. Since the question states that there is no zero divisors in this finite ring, we can use the theorem to state that they must all be units.
4. Noting that they are all units means they all have the stipulation that x*s = 1. Meaning they all have a multiplicative inverse.

Proof is done.

Is this a complete proof?
 

Answers and Replies

  • #2
Samy_A
Science Advisor
Homework Helper
1,242
510

Homework Statement


Suppose R is a commutative ring with only a finite number of elements and no zero divisors. Show that R is a field.

Homework Equations



Unit is an element in R which has a multiplicative inverse. If s∈R with r*s = 1.
A zero divisor is an element r∈R such that there exists s∈R and rs = 0 (or sr = 0).

The Attempt at a Solution



1. Since R is a commutative ring, we only need to prove one axiom, that is that it satisfies the multiplicative inverse for all values in R.
2. I have a theorem that we just went over that states:
Theorem: If R is a finite ring, then for all r ∈ R, r is either a unit or a zero divisor.
3. Since the question states that there is no zero divisors in this finite ring, we can use the theorem to state that they must all be units.
4. Noting that they are all units means they all have the stipulation that x*s = 1. Meaning they all have a multiplicative inverse.

Proof is done.

Is this a complete proof?
Yes. If I were in a pedantic mood, I could remark that you should exclude 0 in some statements.
 
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