# Prime elements in integral domains

• March 13th 2013, 10:57 PM
Bernhard
Prime elements in integral domains
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.

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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.

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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
• March 13th 2013, 11:19 PM
Gusbob
Re: Prime elements in integral domains
$p = ab = prb \implies rb=1$
You use the fact that an integral domain has a cancellation law.
• March 14th 2013, 01:03 AM
Bernhard
Re: Prime elements in integral domains
Thanks

I was just imagining that they were pre-multiplying both sides by the inverse of p, but of course that assumes p has an inverse, and we do not necessarily have a field - only an integral domain.

Presumably the cancellation law can operate without p having an inverse???

Peter
• March 14th 2013, 03:06 AM
Gusbob
Re: Prime elements in integral domains
Quote:

Originally Posted by Bernhard
Presumably the cancellation law can operate without p having an inverse???

Yes. Consider the integral domain $\mathbb{Z}$. If $2a=2b$, then obviously $a=b$. But $2$ does not have an inverse in $\mathbb{Z}$.