Ring Homomorphism: Show Phi(1) Is Unity for R

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Homework Help Overview

The problem involves a ring homomorphism phi between two rings R and R', focusing on the conditions under which phi(1) serves as a unity in R'. The context includes discussions about the properties of rings, subrings, and the implications of having no zero divisors in R'.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of phi(1) being a unity for the subring phi[R] and question whether phi(1) must also act as a unity for all elements in R'. There are inquiries about the nature of identities in subrings and how the lack of zero divisors influences the situation.

Discussion Status

Participants are actively engaging with the problem, raising questions about the definitions and properties of rings and subrings. Some have suggested exploring the properties of the unit in a subring, while others have provided examples and attempted to clarify the implications of the conditions given in the problem.

Contextual Notes

There is a mention of a variant definition of a ring that does not require the presence of a unity, which may affect the interpretation of the problem. Additionally, the discussion includes references to specific examples, such as the ring of 2x2 matrices, to illustrate points about zero divisors and identities.

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Homework Statement


Let R and R' are rings and phi: R to R' is a ring homomorphism such that phi[R] is not identically 0'. Show that if R has unity 1 and R' has no 0 divisors, then phi(1) is a unity for R'.

Homework Equations


The Attempt at a Solution



Its relatively obviously why phi(1) has to be unity for the subring phi[R]. I don't see why phi(1)r' has to be r' for every r' in R'.
 
Last edited:
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CAN subring have an identity other than the ring identity?
 
I don't know.

If S is a subring of R and 1 is a unity of S, then is it true that 1r=r for an r in R? I don't see why?

How do I use the fact that R' has no 0 divisors.
 
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Recall that ehrenfest is not using the usual notion of a 'ring'; he uses a variant that omits '1' from the definition. So if the ring and subring have '1', his definition of subring doesn't require them to be the same.

There is a simple example of a subring with a different unit: consider the ring of 2x2 matrices, and the subring of those whose upper-left entry is the only nonzero.
 
And in that example, there are zero divisors. Use the fact that lack of zero divisors admits a cancellation rule, i.e., r's' = r't' => s' = t'.

At least, I think that's right.
 
A ring <R,+,*> is a set R together with two binary operations + and * such that the following axioms are satisfied:
1) <R,+> is an abelian group
2) Multiplication is associative
3) For all a,b,c in R, a*(b+c)=(b+c)*a=a*b+a*c

A subring is a subset of R that is also a ring under + and *.

Now, how to solve the problem...

EDIT: sorry I didn't read the post above this when I wrote that
 
If nothing springs to mind, then just try exploring. e.g. what properties does the unit of a subring have?
 
Here is the proof that if R is a ring with no 0 divisors and S is a nonzero subring of R with unity 1, then 1 MUST be a unity of R:

11=1 implies that r11=r1 implies that r1=r by cancellation

11=1 implies that 11r=1r implies that 1r=r by cancellation

Is that right?
 

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