Abstract Algebra: Ring Proof (Multiplicative Inverse)

Specifically, you should say "for all non-zero r in R, r is either a unit or a zero divisor." But that is not strictly necessary, because 0 is defined to be neither a unit nor a zero divisor.
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RJLiberator
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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?
 
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  • #2
RJLiberator said:

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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1. What is Abstract Algebra?

Abstract Algebra is a branch of mathematics that deals with the study of algebraic structures, such as groups, rings, and fields. It focuses on the properties and operations of these structures, rather than specific numbers or objects.

2. What is a Ring?

A Ring is an algebraic structure that consists of a set of elements and two binary operations, addition and multiplication. These operations must satisfy certain properties, such as closure, associativity, and distributivity. Rings can be either commutative or non-commutative.

3. What is a Multiplicative Inverse?

A Multiplicative Inverse, also known as a reciprocal, is a number that, when multiplied by another number, gives a product of 1. In a ring, every element (except 0) has a multiplicative inverse. This means that for any element a in the ring, there exists an element b such that a*b = b*a = 1.

4. How do you prove the existence of a Multiplicative Inverse in a Ring?

To prove the existence of a Multiplicative Inverse in a Ring, we must show that for every element a in the ring, there exists an element b such that a*b = b*a = 1. This can be done by using the properties of rings, such as closure and associativity, along with the definition of a Multiplicative Inverse.

5. What is the significance of the Multiplicative Inverse in Abstract Algebra?

The Multiplicative Inverse plays a crucial role in Abstract Algebra, as it allows us to solve equations and perform operations that would otherwise be impossible. It also helps us understand the structure and properties of rings, as well as their relationship to other algebraic structures.

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