Solved: False - Product of Two Nonunits in Z_n Cannot be Unit

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Homework Help Overview

The discussion revolves around a ring theory problem concerning the properties of units and nonunits in the ring Z_n. The original poster poses a true or false question regarding whether the product of two nonunits can be a unit.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the definitions of units and nonunits in Z_n, questioning the conditions under which their product might yield a unit. Some participants consider the relationship between the product and its relative primality to n.

Discussion Status

There is an ongoing exploration of the implications of the definitions involved. Some participants suggest that if the product of two nonunits were a unit, it would lead to contradictions regarding the nature of the elements involved. The discussion reflects a mix of interpretations and reasoning without a clear consensus.

Contextual Notes

Participants are considering the specific properties of Z_n and the definitions of units and nonunits within this context. There is an acknowledgment of the structure of multiplication in commutative rings and its relevance to the problem.

ehrenfest
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[SOLVED] ring theory problem

Homework Statement


True or false. The product of two nonunits in Z_n may be a unit.


Homework Equations





The Attempt at a Solution


If a and b are two non units in Z_n, and ab <= n, then the result is clear, since ab would not be relatively prime to n. But what about if ab > n ?
Obviously the units form a group under multiplication, but I don't see how that helps.
 
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i don't see how it matters we are in Z_n,

if a and b are not units in a commutative ring R, then ab is not a unit, if it was then ..some stuff..
 
Last edited:
If ab were a unit, then we would have abc=1, for some c. This implies that a(bc)=1 which implies that a is a unit.

Is that it?
 
of course yea! (just also notice by commutativity (bc)a = 1 too of course, but that's obvious)
 

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