It is shown in Lang's Algebra (and many other books I assume) that:


if A if a principal entire ring, then A is a factorial ring.


The proof uses Zorn's Lemma. Is this theorem equivalent to the axiom of choice? 



Factorial Rings and The Axiom of Choice
Lang uses Zorn's lemma only in the step that nonzero nonunits in a PID admit irreducible factorizations (not the uniqueness of irreducible factorizations, once we know such factorizations exist). The way he uses Zorn's lemma, I think, is excessive. What follows is how I work out the existence of irreducible factorizations when I teach the abstract algebra class. 


Claim: In a PID which is not a field, any nonzero nonunit is a product of irreducibles.


We will use the following lemma (which is not Zorn's lemma).


Lemma: If $R$ is an integral domain and $a \in R$ is 
a nonzero nonunit which 
does not admit a factorization into irreducibles then 
there is a strict inclusion of principal ideals 
$(a) \subset (b)$ where $b$ is some other nonzero nonunit which 
does not admit a factorization into irreducibles.


Proof of lemma: By hypothesis $a$ is not irreducible, so (since it is 
neither 0 nor a unit either) there is some factorization
$a = bc$ where $b$ and $c$ are nonunits (and obviously are 
not 0 either). If both $b$ and $c$ admitted 
irreducible factorizations then so does $a$, so 
at least one of $b$ or $c$ has no irreducible factorization. 
Without loss of generality it is $b$ which has no 
irreducible factorization. Since $c$ is not a unit, the inclusion 
$(a) \subset (b)$ is strict. QED lemma.


Now we can prove the claim.


Proof of claim: Suppose there is an element $a$ in the PID which 
is not 0 or a unit and has no irreducible factorization. 
Then by the lemma there is a strict inclusion 
$$
(a) \subset (a\_1) 
$$
where $a\_1$ has no irreducible factorization.  

Then using $a\_1$ in the role of $a$ (and the lemma again) 
there is a strict inclusion 
$$
(a\_1) \subset (a\_2) 
$$
where $a\_2$ has no irreducible factorization. 
This argument (repeatedly applying the lemma to 
the generator of the next larger principal ideal) 
leads to an infinite increasing chain of principal ideals
$$
(a) \subset (a\_1) \subset (a\_2) \subset (a\_3) \subset \cdots 
$$
where all inclusions are strict. (At this step I suppose you may say we need the Axiom of Choice to get an infinite ascending chain, but it's only *countably* many choices, so really not the full thrust of Zorn's lemma and in any case it feels like a less pedantic use of Zorn's lemma than the way Lang does this.) Such a chain of ideals is impossible in a PID.


Indeed, suppose a PID contains an infinite strictly increasing chain of 
ideals:
$$
I\_0 \subset I\_1 \subset I\_2 \subset I\_3 \subset \cdots
$$
and set
$$
I = \bigcup\_{n \geq 0} I\_n.
$$
This union $I$ is an ideal. The reason is that 
the $I\_n$'s are strictly increasing, so any *finite* set 
of elements from $I$ lies in a common $I\_n$. 
Therefore $I$ is closed under addition and arbitrary multiplications from 
the ring since each $I\_n$ has these properties. Because we are in a PID, $I$ is principal: 
$I = (r)$ for some $r$ in the ring. But because 
$I$ is the union of the $I\_n$'s, $r$ is in some $I\_N$. 
Then $(r) \subset I\_N$ since $I\_N$ is an ideal, so 
$$
I = (r) \subset I\_N \subset I, 
$$
which means
$$
I\_N = I.
$$
But this is impossible because the inclusion $I\_{N+1} \subset I$ 
becomes $I\_{N+1} \subset I\_N$ and we were assuming 
$I\_N$ was a proper subset of $I\_{N+1}$. 
Because of this contradiction, nonzero nonunits in a PID without 
an irreducible factorization do not exist. QED


Lang's argument only uses the axiom of choice in a countable way, as above, but the way he pulls it in makes the application of Zorn's lemma feel a lot more fussy. 


