Irreducibles and Primes in Integral Domains ....

In summary: I think fields are already an example. Also matrix rings might do.But let's see how Euclidean rings work. Since Euclidean rings are principal ideal domains, which are unique factorization domains, and we are only allowed a finite number of primes, all elements have to look like ##\varepsilon \cdot p^{m_1}_1 \ldots p^{m_n}_n## with a unit ##\varepsilon## and ##(m_1 ,\ldots , m_n) \in \mathbb{N}_0^n##. I guess ##n=1## is a reasonable assumption here to start a search for an example. So all
  • #1
Math Amateur
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I am reading "Introductory Algebraic Number Theory"by Saban Alaca and Kenneth S. Williams ... and am currently focused on Chapter 1: Integral Domains ...

I need some help with understanding Example 1.4.1 ...

Example 1.4.1 reads as follows:
?temp_hash=d57e0fe7fa044114b40957202f2e6a5a.png


In the above text by Alaca and Williams we read the following:

"... ... From the first of these, as ##2## is irreducible in ##\mathbb{Z} + \mathbb{Z} \sqrt{ -5 }##, it must be the case that ##\alpha \sim 1## or ##\alpha \sim 2##. ... ...
My question is as follows ... how does ##2## being irreducible imply that ##\alpha \sim 1## or ##\alpha \sim 2##. ... ...?
Hope someone can help ...

Peter================================================================================NOTEThe notation ##\alpha \sim 1## is Alaca and Williams notation for ##\alpha## and ##1## being associates ...

Alaca's and Williams' definition of and properties of associates in an integral domain are as follows:
?temp_hash=d57e0fe7fa044114b40957202f2e6a5a.png
 

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  • #2
Math Amateur said:
My question is as follows ... how does ##2## being irreducible imply that##\alpha \sim 1## or ##\alpha \sim 2##. ... ...?
##\alpha \mid 2## means ## \alpha \cdot q=2## for some ##q##. Now irreducibility of ##2## implies either ##\alpha ## or ##q## is a unit. If ##\alpha ## is a unit, then ##\langle 2, 1+\sqrt{-5} \rangle = \langle \alpha \rangle = \mathbb{Z}+ \mathbb{Z}\sqrt{-5}## which is wrong since it doesn't contain, e.g. ##3##. Now if ##q## is a unit, then ##\alpha = 2 \cdot q^{-1} \mid 1+\sqrt{-5}## which means ##2 \mid 1+\sqrt{-5}## which is also wrong.

It is simply the definitions that are used. You should try to find those things by yourself. It's always a matter of practice. If you will have done it several times, you will start to "see" things more quickly. As a general advice: always try to use what's given, here that ##\alpha \mid 2## and ##2## cannot properly be written as ##2=a\cdot b##. Part of the fun reading those books is to read them as a book of riddles where you're supposed to solve them. A bit like solving crosswords if you like.
 
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  • #3
Does every infinite ring (meaning infinite as a set) have infinitely-many primes?
 
  • #4
WWGD said:
Does every infinite ring (meaning infinite as a set) have infinitely-many primes?
I guess you're not satisfied by ##R := (\mathbb{Z}; x\cdot y = 0 \;\forall \; x,y \in \mathbb{Z})## or ##R := \mathbb{R}##.
 
  • #5
fresh_42 said:
I guess you're not satisfied by ##R := (\mathbb{Z}; x\cdot y = 0 \;\forall \; x,y \in \mathbb{Z})## or ##R := \mathbb{R}##.
Yes, I would prefer a non-trivial multiplication. Ah, I see, good point with the Reals, I guess we need to EDIT exclude certain types of multiplication actions ( non-transitive?).EDIT 2: Maybe it is more meaningful to work with Euclidean rings here?
 
  • #6
WWGD said:
Yes, I would prefer a non-trivial multiplication. Ah, I see, good point with the Reals, I guess we need to EDIT exclude certain types of multiplication actions ( non-transitive?).EDIT 2: Maybe it is more meaningful to work with Euclidean rings here?
I think fields are already an example. Also matrix rings might do.

But let's see how Euclidean rings work. Since Euclidean rings are principal ideal domains, which are unique factorization domains, and we are only allowed a finite number of primes, all elements have to look like ##\varepsilon \cdot p^{m_1}_1 \ldots p^{m_n}_n## with a unit ##\varepsilon## and ##(m_1 ,\ldots , m_n) \in \mathbb{N}_0^n##. I guess ##n=1## is a reasonable assumption here to start a search for an example. So all elements are of the form ##\varepsilon \cdot p^m## which I think is impossible in an integral domain with unit. (Just scratched a few lines with ##1+1=p^k\, , \,1+1+1=p^l## etc.). Maybe there's an induction argument to rule out the cases ##n>1##, too. And this without using the full advantages of a Euclidean ring. So my guess is, it's impossible, but I don't have a slim argument at hand.
 
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  • #7
How about the usual proof that there are inifnitely many primes: Multiply all (if finite), add ##1## and all primes are units. This leaves us with rings without any primes, e.g. fields.
 

What is the difference between irreducibles and primes in integral domains?

In an integral domain, an element is considered irreducible if it cannot be factored into two non-unit elements. On the other hand, an element is considered prime if it divides any product of elements, it must divide at least one of the factors. In other words, primes are a stronger version of irreducibles.

How can we determine if an element in an integral domain is irreducible?

To determine if an element is irreducible, we can use the following criteria:

  • If the element is a unit, it is not irreducible.
  • If the element is a product of two non-unit elements, it is not irreducible.
  • If the element cannot be factored into two non-unit elements, it is irreducible.

What is the difference between irreducibles and primes in polynomial rings?

In polynomial rings, the definitions of irreducibles and primes are the same as in integral domains. However, in polynomial rings, the degree of the polynomial also plays a role. In general, an element in a polynomial ring is considered irreducible if it cannot be factored into two lower degree polynomials, and it is considered prime if it divides any product of polynomials, it must divide at least one of the factors.

How are irreducibles and primes important in number theory?

Irreducibles and primes play a crucial role in number theory, as they help us understand the structure of numbers and their properties. They are used in the fundamental theorem of arithmetic, which states that every positive integer can be uniquely expressed as a product of primes. This theorem allows us to study the divisibility and factorization properties of numbers.

Can an element be both irreducible and prime?

Yes, an element in an integral domain (or polynomial ring) can be both irreducible and prime. In fact, in a unique factorization domain, an element is prime if and only if it is irreducible. However, in general, an element can be prime without being irreducible, as in the case of polynomials in polynomial rings.

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