Rings and Homomorphism example

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Homework Help Overview

The discussion revolves around finding an example of a ring R and a function f: R → R that satisfies the condition f(a+b) = f(a)f(b) for all a, b in R, while ensuring that f(a) is not the zero element for any a in R. Participants are exploring the properties of rings and homomorphisms in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the possibility of using Zn as the ring and question what specific function could satisfy the given condition. There is also a suggestion to consider whether a division ring could be used. Some express uncertainty about how to approach the problem and whether to plug in values from the ring to identify patterns.

Discussion Status

Several participants are actively engaging with the problem, sharing their thoughts and uncertainties. Some have suggested using specific values or types of rings, while others are questioning the assumptions and definitions involved. There is no explicit consensus, but the discussion is generating ideas and potential directions for exploration.

Contextual Notes

Participants note that they have not yet covered division rings in their studies, which may limit their exploration. Additionally, there is a mention of the requirement that f(a) cannot be zero, which is a critical constraint in the discussion.

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Homework Statement


Give an example of a ring R and a function f: R---->R such that f(a+b)=f(a)f(b) for all a,b in R. and f(a) is not the zero element for all a in R. Is your function a homomorphism?


Homework Equations


Let R and S be rings. A function f:R----->S is said to be a homomorphism if
f(a+b)=f(a) + f(b) and f(ab)=f(a)f(b) for all a,b in R


The Attempt at a Solution



Not really sure where to start here,
I was thinking about using Zn as my ring, perhaps with n as a prime number, so that way f(a) wouldn't be zero for any a. but i don't know what my function would be to satisfy that. Any help would be greatly appreciated =)
 
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can the ring be a division ring?
 
I don't think so, because I don't think we've learned about division rings yet lol
 
dancergirlie said:
but i don't know what my function would be to satisfy that.
By "that", do you mean f(a+b)=f(a)f(b)?

That's an easy problem -- if you want f to satisfy that, and if you don't know a particular value of f, then the equation tells you how to compute it!
 
I don't think that is what I need to do. It asks to give a specific ring and function that fits those specifications. Like for ring Z,
f(x)=x^2, but that doesn't work.
 
If you're going to look for a ring and a function for which f(a+b)=f(a)f(b) is satisfied, you might as well use that equation to help you figure out what f should be. *shrug*
 
I'm not sure if I follow you here, should I just plug in values from the ring and see if I can notice a pattern? I am just completely lost with this problem, I don't even know which ring I should use.
 
Even if you don't notice a pattern, it will help you get started narrowing down a few specific values of f.

What ring to use? You had a few ideas you wanted to try, right? Do those! Or... you could start with values that every ring has. (e.g. 0, 1, 2...)
 
I seriously have been looking at this for at least an hour now and have made no progress... I have tried the ring of even numbers, just Z, the ring Z5. I just don't know what to do, I'm not exactly an expert at the whole rings thing yet.
 
  • #10
What kind of functions do you know that turn addition into multiplication (or vice versa)? That's what the question is asking you.
 
  • #11
Thank you so much! I didn't even think of the exponential functions. I got that my function was obviously not homomorphic since f(a+b) did not equal f(a)+f(b) and f(ab) was not equal to f(a)f(b). I think I was just too tired at the moment to think of that function. Thanks for the help everyone!
 
  • #12
That isn't quite the right reasoning for why such a map cannot be an homomorphism. Why can't f(a)f(b) equal f(a)+f(b)? Just because the expressions look different doesn't mean that they are, really. Just take the possibly illegal case of the ring with one element :0.

But the question is trivial since it provides a reason why f can't be a homomorphism in its own statement: you are told that f(a) is never 0, and homomorphisms send 0 to 0. Note that the question precludes the example of the ring with one element from being considered.
 

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