Primes and Associates in Rings

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SUMMARY

The discussion centers on the property of prime elements in commutative rings, specifically whether if \( a \) is a prime element and \( a \) and \( b \) are associates, then \( b \) is also prime. The conclusion reached is that this statement is true, as demonstrated through the proof that if \( b = au \) for some unit \( u \), then \( a \) divides \( xy \) implies \( a \) divides either \( x \) or \( y \), leveraging the definition of prime elements. The discussion also references the irreducibility of elements in rings such as \( \mathbb{Z}[5i] \).

PREREQUISITES
  • Understanding of commutative rings with identity
  • Knowledge of prime and irreducible elements in ring theory
  • Familiarity with units in rings
  • Basic proficiency in algebraic manipulation and proof techniques
NEXT STEPS
  • Study the properties of prime and irreducible elements in commutative rings
  • Explore the structure of the ring \( \mathbb{Z}[5i] \) and its units
  • Learn about the concept of associates in ring theory
  • Investigate counterexamples in ring theory to understand the limitations of prime elements
USEFUL FOR

Mathematics students, particularly those studying abstract algebra, and educators looking to deepen their understanding of prime elements and their properties in commutative rings.

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Homework Statement


Let a, b be members of a commutative ring with identity R. If a is a a prime and a, b are associates then b is also prime. True/False


Homework Equations


Definitions: a is prime if a|xy implies a|x or a|y
a and b are associates if there exists a unit u s.t a=bu


The Attempt at a Solution


First off I'm not sure whether its even true or not; so I've tried to both prove it and find a counter example:
Attempt at a proof:
let b|xy.
Then if b = au for some unit u, au|xy, that is for some k in R auk = xy.
so ak divides xy(u^(-1))
but k does not necessarily have an inverse so this doesn't really get me anywhere.

I've shown in an earlier question that if a is irreducible and a, b are associates then b is irreducible. Prime implies irreducible but not vice versa so if there is a counter example it will depend on an element of R that is irreducible but not prime. I know examples of these can be found in rings such as Z(5i) (the set of numbers a+5bi for a, b in Z). But I don't know how to find units of this ring.
Consider (a+5bi)(c+5di) = 1.
This gives the simultaneous equations 1+25bd-ac = 0 = ad+bc
But so far I have not managed to find any integer solutions (and to be honest I don't know any way of doing this other than trial and error).

Thanks for any help!
 
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Zoe-b said:
First off I'm not sure whether its even true or not; so I've tried to both prove it and find a counter example:
Attempt at a proof:
let b|xy.
Then if b = au for some unit u, au|xy, that is for some k in R auk = xy.

OK, so stop here. We know that auk=xy. So this implies that [itex]ak=u^{-1}xy[/itex]. Thus a divides [itex]u^{-1}xy[/itex]. Now use that a is prime.
 
Thank you! Done it now :)
 

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