Unital rings, homomorphisms, etc

In summary, In problem 1, one constructs an equivalence relation ~ on a multiplicative submonoid R x S by (a,s)~(b,t) if there is a r in S with rat = rbs. This allows one to make RxS/~ into a commutative, unital ring. In problem 2, one defines a homomorphism psi: RxS/~ --> T of unital rings, and shows that there is only one such homomorphism.
  • #1
calvino
108
0
I've already completed 1), but it's necessary for one to know it for question 2). I'm pretty sure that I've found my homomorphism in 2, but I don't know whether or not is unique. How do I show a homomorphism is unique in this case?

Problem 1: Let R be a commutative unital ring, and let S be a multiplicative submonoid of R. Define an equivalence relation ~ on R x S by (a,s)~(b,t) if there is r in S with rat = rbs. Let a/s denote the ~-equivalence class of (a,s). Show that with

a/s + b/t = (at+bs)/st and (a/s)(b/t) = ab/st

one can make RxS/~ into a commutative, unital ring, and that j(a) = a/1 defines a homomorphism j of unital rings from R into RxS/~ that maps S into the group of invertible elements of RxS/~

Problem 2 (=continuation of Problem 1): Let R and S be as above, and let phi: R --> T be a homomorphism that maps S into the group of invertible elements of the commutative unital ring T. Show that there is unique homomorphism psi: RxS/~ ---> T of unital rings with psi.j = phi.
 
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  • #2
The condition psi(j(r))=phi(r) tells you that psi(r/1)=phi(r). What must psi(1/s) be?
 
  • #3
i understand how to define the function (i think). Am i suppose to see it's uniqueness, naturally?
 
  • #4
How did you find the homomorphism? What steps did you take? At each step, can you argue that there is no other choice you could have made that would still leave you with a function satisfying the necessary conditions? If so, you have uniqueness.
 
  • #5
can we use the fundamental theorm of homomorphisms?

i didn't. i simple constructed a diagram of the homomorphisms involved and showed that since psi.j is in a sense doing the same things as phi is, then they method is unique.
 
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  • #6
I have no idea what "since psi.j is in a sense doing the same things as phi is, then they method is unique" means. Like I was saying before, psi(r/1)=phi(r). Can you show you only have one choice for psi(1/s)? Then psi(r/s)=psi(r/1)psi(1/s) is unique.
 

What is a unital ring?

A unital ring is a mathematical structure consisting of a set, an addition operation, and a multiplication operation, which satisfy certain properties. The set is closed under both operations and the multiplication operation is associative and distributive over addition. Additionally, there exists an identity element for the multiplication operation, known as the unity or identity element.

What is a homomorphism?

A homomorphism is a function between two algebraic structures that preserves the operations of the structures. In the context of unital rings, a homomorphism is a function that maps elements from one ring to another in a way that preserves addition and multiplication.

What is the difference between a ring homomorphism and a ring isomorphism?

A ring homomorphism is a function that preserves the operations of addition and multiplication between two rings, while a ring isomorphism is a bijective homomorphism. This means that a ring isomorphism is not only a homomorphism, but also has an inverse function that is also a homomorphism.

How can unital rings be used in real-world applications?

Unital rings have many applications in mathematics and other fields, such as computer science and physics. In computer science, unital rings are used in coding theory and cryptography. In physics, they are used in the study of symmetry and group theory.

What is the significance of the unity element in a unital ring?

The unity element, also known as the identity element, is important in unital rings because it is the element that maintains the structure of the ring. It acts as the neutral element for the multiplication operation, and without it, the ring would not be considered a unital ring.

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