Understanding Irreducible Elements in Integral Domains - Peter's Questions

In summary, Gallian defines an irreducible element in a domain as one that cannot be factored into non-units. This definition is specific to integral domains and not general rings. The rationale for excluding units as irreducibles is a matter of convention and helps avoid situations where a does not have a genuine factorization.
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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 \(\displaystyle a\), if \(\displaystyle a = bc\) then \(\displaystyle b\) or \(\displaystyle c\) is a unit ... ... why is this ... ... ? ... ... presumably for an irreducible we want to avoid a situation where \(\displaystyle a\) has a "genuine" factorisation ... ... but how does \(\displaystyle b\) or \(\displaystyle c\) being a unit achieve this ...Hope someone can help ...

Peter
 
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Peter said:
(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.

Peter said:
(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.

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

What are irreducible elements in integral domains?

Irreducible elements in integral domains are elements that cannot be broken down any further into products of two non-units. They are considered the building blocks of the domain and play a crucial role in understanding the structure of the domain.

How are irreducible elements different from prime elements?

Irreducible elements and prime elements are often confused, but they have a subtle difference. While irreducible elements cannot be factored further, prime elements are those that cannot be divided by any other non-unit element without leaving a remainder.

Why is it important to study irreducible elements in integral domains?

Understanding irreducible elements in integral domains is important because they allow us to factor elements and find their prime factorizations. This is crucial in many areas of mathematics, including number theory and abstract algebra, and has practical applications in cryptography and coding theory.

Can there be more than one irreducible element in an integral domain?

Yes, an integral domain can have multiple irreducible elements. In fact, every non-unit element in an integral domain can be written as a product of irreducible elements, and this factorization is unique up to order and units.

How do irreducible elements relate to the concept of divisibility?

If an element in an integral domain is divisible by an irreducible element, then it must be a multiple of that irreducible element. This is because an irreducible element cannot be broken down any further, so it must be a factor of the element. However, the converse is not always true as an element can be divisible by a non-irreducible element without being a multiple of it.

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