MHB Understanding Irreducible Elements in Integral Domains - Peter's Questions

[Math Amateur]

Joseph A. Gallian, in his book, "Contemporary Abstract Algebra" (Fifth Edition) defines an irreducible element in a domain as follows ... (he also defines associates and primes but I'm focused on irreducibles) ...
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I am trying to get a good sense of this definition ...

My questions are as follows:

(1) Why are we dealing with a definition restricted to an integral domain ... why can't we deal with a general ring ... presumably we don't want zero divisors ... but why ...

(2) What is the logic or rationale for excluding a unit ...that is why is a unit not allowed to be an irreducible element ..

(3) We read that for an irreducible element $$a$$, if $$a = bc$$ then $$b$$ or $$c$$ is a unit ... ... why is this ... ... ? ... ... presumably for an irreducible we want to avoid a situation where $$a$$ has a "genuine" factorisation ... ... but how does $$b$$ or $$c$$ being a unit achieve this ...Hope someone can help ...

Peter

 

[caffeinemachine]

(1) Why are we dealing with a definition restricted to an integral domain ... why can't we deal with a general ring ... presumably we don't want zero divisors ... but why ...

I do not see any problem with defining the notion of an irreducible in an arbitrary ring. Though we may not get any nice theorems concerning irreducibles in an arbitrary ring.

(2) What is the logic or rationale for excluding a unit ...that is why is a unit not allowed to be an irreducible element ..

It seems like a matter of convention. I would suggest not sweating it and moving on.

(3) We read that for an irreducible element $$a$$, if $$a = bc$$ then $$b$$ or $$c$$ is a unit ... ... why is this ... ... ? ... ... presumably for an irreducible we want to avoid a situation where $$a$$ has a "genuine" factorisation ... ... but how does $$b$$ or $$c$$ being a unit achieve this ...

I think you have answered your own question. If $a=ub$ for some unit $u$, then this does not look like a "genuine factorization" of $a$. For one can always write $a=u(a/u)$ for a unit $u$. So if we allow units to appear in genuine factorizations then nothing would be irreducible.