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In the ring of integers of an algebraic number field, we say that an algebraic integer $p$ is prime if, whenever $p$ divides product of two algebraic integers, $p$ divides one of them. An algebraic integer is said to be irreducible if it can not be a product of two algebraic integers, which are not units.

I came across the following theorem: (1) The ring of integers of any algebraic number field contains infinitely many irreducible elements. (2) In the ring of algebraic integers of an algebraic number field, every prime element is irreducible (but not converse)".

To prove (1), I tried to use (2), and arrived at the following question.

Does the ring of integers of an algebraic number field contains infinitely many primes? If yes, does it follows from the argument of Euclid on the infiniteness of primes in $\mathbb{Z}$?

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  • $\begingroup$ See also math.stackexchange.com/questions/245487/…. $\endgroup$
    – Dietrich Burde
    Jul 15, 2014 at 10:03
  • $\begingroup$ Are you familiar with the ideal class group? My first instinct is that you would need to know some of its basic facts to leverage the properties of prime ideals that @Dietrich has suggested looking at. $\endgroup$
    – user14972
    Jul 15, 2014 at 10:04
  • $\begingroup$ I heard about that. It is a group defined by some equivalence relation on the set of ideals in ring of integers of a number field. Is it so? $\endgroup$
    – Groups
    Jul 15, 2014 at 10:06
  • $\begingroup$ @Groups: Right. The simple argument I can think of to show infinitely many prime elements would be to show there are infinitely many principal prime ideals, which are precisely those prime ideals that vanish in the class group. $\endgroup$
    – user14972
    Jul 15, 2014 at 10:13
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    $\begingroup$ If the ring is a UFD, then it does have infinitely many primes. There are infinitely many primes $p$ in $\mathbb{Z}$. Some of them are also prime in $\mathbb{Z}[x]$. And there are numbers with nonzero "extended" part that have a norm that is a prime $p$, then those numbers are prime but $p$ is composite. I don't have the theorem about those primes with nonzero "extended" part, so I'm not posting this as an answer. $\endgroup$
    – user153918
    Jul 16, 2014 at 16:50

2 Answers 2

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The answer is yes. One somewhat high-powered way to see this is as follows:

Any class in the ideal class group contains infinitely many prime ideals. (This is part of the generalization of Dirichlet's theorem on primes in arith. progressions to general number fields.)

In particular, the trivial class does. That is, there are infinitely many principal prime ideals, and the generators of these ideals are then the desired infinitely many prime elements.

I'm not that there is a simpler approach then this one, since the question really is very analogous to Dirichlet's theorem, which requires $L$-function machinery to prove. (And that is what is used to prove the result I am quoting too.)

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user160609 answers your first question, in what I think is the best possible way.

As for your other question: you cannot necessarily do a Euclid-style argument in a number ring, even to prove that there are infinitely many irreducible elements, because you cannot exclude the possibility that the product $p_1\dots p_n+1$ is a unit.

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