Prime elements in integral domains

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SUMMARY

The discussion centers on Proposition 10 from Dummit and Foote's "Abstract Algebra," which states that in an integral domain, a prime element is always irreducible. The proof demonstrates that if a prime ideal (p) can be expressed as a product ab, then one of the factors must belong to the ideal, leading to the conclusion that p is irreducible. A key aspect highlighted is the reliance on the integral domain's property of having no zero divisors, which allows for the application of the cancellation law during the proof.

PREREQUISITES
  • Understanding of integral domains in abstract algebra
  • Familiarity with prime ideals and their properties
  • Knowledge of the cancellation law in integral domains
  • Basic concepts of unique factorization domains
NEXT STEPS
  • Study the properties of unique factorization domains in detail
  • Explore the implications of the cancellation law in integral domains
  • Review examples of prime elements in various integral domains
  • Investigate the differences between prime and irreducible elements in rings
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This discussion is beneficial for students and researchers in abstract algebra, particularly those focusing on integral domains, prime ideals, and unique factorization properties.

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In Dummit and Foote, Section 8.3 on Unique Factorization Domains, Proposition 10 reads as follows:

Proposition 10: In an integral domain a prime element is always irreducible.

The proof reads as follows:

===========================================================

Suppose (p) is a non-zero prime ideal and p = ab.

Then ab = p \in (p), so by definition of prime ideal, one of a or b, say a, is in (p).

Thus a = pr for some r.

This implies p = ab = prb and so rb = 1 and b is a unit.

This shows that p is irreducible.

==============================================================

My question is as follows: Where in this proof do D&F use the fact that p is in an integral domain? (It almost reads as if this applies for any ring)

Peter
 
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In this step:
Peter said:
This implies p = ab = prb and so rb = 1 and b is a unit.

Since an integral domain has no zero divisors by definition there's a cancelation law which says:
Let R be an integral domain and a,b,c \in R. If a \neq 0 and ab=ac then b=c.
 

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