Commutative Ring Definition: What is a Commutative Ring?

  • Thread starter Thread starter Miike012
  • Start date Start date
  • Tags Tags
    Definition
AI Thread Summary
A commutative ring is defined as a set where the sum and product of any two elements are well-defined. The discussion highlights confusion regarding the set A, defined as non-zero integers, which fails to meet the criteria for a commutative ring due to the absence of an additive identity. Specifically, since 0 is excluded from set A, it cannot satisfy the necessary axioms of a ring. The importance of the definition lies in understanding these foundational properties, which are crucial for mathematical structures. Overall, the set A does not qualify as a commutative ring due to its failure to include essential elements and operations.
Miike012
Messages
1,009
Reaction score
0
Commutative Ring: Let R be a set of elements a,b,c ... for which the sum (a + b) and the product (ab) of any two elements a and b of R are defined is called a commutative ring

This is my understanding, tell me if I am wrong...

f(x) = 1/x
Domain of f: { x | x =/ 0 } = A

Thus any two integers (a + b) = 0 and (ab) = 0 are not in the commutative ring... commutative ring being A
Is this correct?
 
Last edited:
Physics news on Phys.org
Why are you saying that for every two integers a,b in A a+b = 0 holds?

This set is not even a ring (let alone a commutative ring) because it fails to hold a certain axiom. Any guesses which one? Can you prove it?
 
I just opened the book.. Never took a class on this... so I am confused ab out why this definition is important.
And I was hoping for an example for the definition...

It also goes on to say that it must follow 8 rules
 
http://en.wikipedia.org/wiki/Ring_(mathematics )
 
Last edited by a moderator:
Miike012 said:
Commutative Ring: Let R be a set of elements a,b,c ... for which the sum (a + b) and the product (ab) of any two elements a and b of R are defined is called a commutative ring

This is my understanding, tell me if I am wrong...

f(x) = 1/x
Domain of f: { x | x =/ 0 } = A

Thus any two integers (a + b) = 0 and (ab) = 0 are not in the commutative ring... commutative ring being A
Is this correct?
A "ring" consists of a set of objects together with two operations, * and +, satisfying a lot of requirements- mostly those for the integers with regular addition and subtraction. You have defined your set, A, the set of all nonnegative numbers, but have not defined addition or subtraction. If you intended "regular" addition of numbers, this is not a ring because it does not have an "additive identity"- you have specifically excluded 0 which is the identity for "regular" addition.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top