In a principal ideal domain, "prime" and "irreducible" are the same

pid_implies_prime_equals_irreducible.json

https://arbital.com/p/pid_implies_prime_equals_irreducible

by Patrick Stevens Jul 28 2016

Principal ideal domains have a very useful property that we don't need to distinguish between the informal notion of "prime" (i.e. "irreducible") and the formal notion.

[summary: While there are two distinct but closely related notions of "prime" and "irreducible", the two ideas are actually the same in a certain class of ring. This result is basically what enables the Fundamental Theorem of Arithmetic.]

Let $~$R$~$ be a ring which is a PID, and let $~$r \not = 0$~$ be an element of $~$R$~$. Then $~$r$~$ is irreducible if and only if $~$r$~$ is prime.

In fact, it is easier to prove a stronger statement: that the following are equivalent. %%note:Every proof known to the author is of this shape either implicitly or explicitly, but when it's explicit, it should be clearer what is going on.%%

  1. $~$r$~$ is irreducible.
  2. $~$r$~$ is prime.
  3. The generated [ideal_ring_theory ideal] $~$\langle r \rangle$~$ is [maximal_ideal maximal] in $~$R$~$.

[todo: motivate the third bullet point]

Proof

$~$2 \Rightarrow 1$~$

A proof that "prime implies irreducible" appears on the page for irreducibility.

$~$3 \Rightarrow 2$~$

We wish to show that if $~$\langle r \rangle$~$ is maximal, then it is prime. (Indeed, $~$r$~$ is prime if and only if its generated ideal is [prime_ideal prime].)

An ideal $~$I$~$ is maximal if and only if the [quotient_ring quotient] $~$R/I$~$ is a field. ([ideal_maximal_iff_quotient_is_field Proof.])

An ideal $~$I$~$ is [prime_ideal prime] if and only if the quotient $~$R/I$~$ is an Integral domain. ([ideal_prime_iff_quotient_is_integral_domain Proof.])

All fields are integral domains. (A proof of this appears on the page on integral domains.)

Hence maximal ideals are prime.

$~$1 \Rightarrow 3$~$

Let $~$r$~$ be irreducible; then in particular it is not invertible, so $~$\langle r \rangle$~$ isn't simply the whole ring.

To show that $~$\langle r \rangle$~$ is maximal, we need to show that if it is contained in any larger ideal then that ideal is the whole ring.

Suppose $~$\langle r \rangle$~$ is contained in the larger ideal $~$J$~$, then. Because we are in a principal ideal domain, $~$J = \langle a \rangle$~$, say, for some $~$a$~$, and so $~$r = a c$~$ for some $~$c$~$. It will be enough to show that $~$a$~$ is invertible, because then $~$\langle a \rangle$~$ would be the entire ring.

But $~$r$~$ is irreducible, so one of $~$a$~$ and $~$c$~$ is invertible; if $~$a$~$ is invertible then we are done, so suppose $~$c$~$ is invertible.

Then $~$a = r c^{-1}$~$. We have supposed that $~$J$~$ is indeed larger than $~$\langle r \rangle$~$: that there is $~$j \in J$~$ which is not in $~$\langle r \rangle$~$. Since $~$j \in J = \langle a \rangle$~$, we can find $~$d$~$ (say) such that $~$j = a d$~$; so $~$j = r c^{-1} d$~$ and hence $~$j \in \langle r \rangle$~$, which is a contradiction.