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.**

The proof reads as follows:

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

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

Then $\displaystyle 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

Re: Prime elements in integral domains

$\displaystyle p = ab = prb \implies rb=1$

You use the fact that an integral domain has a cancellation law.

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

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 $\displaystyle \mathbb{Z}$. If $\displaystyle 2a=2b$, then obviously $\displaystyle a=b$. But $\displaystyle 2$ does not have an inverse in $\displaystyle \mathbb{Z}$.