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Prime elements in integral domains

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
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:

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

Then [TEX] ab = p \in (p) [/TEX], 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.

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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
 

Siron

Active member
Jan 28, 2012
150
In this step:
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 [tex]R[/tex] be an integral domain and [tex]a,b,c \in R[/tex]. If [tex]a \neq 0[/tex] and [tex]ab=ac[/tex] then [tex]b=c[/tex].