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?
MathOverflow is a question and answer site for professional mathematicians. It only takes a minute to sign up.
Sign up to join this communityIt 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?
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.
This is an extended comment on Keith's answer:
Another way to think about Keith's argument (or the more general argument for the existence of minimal primes in Noetherian rings) is that one constructs a tree as follows: beginning with an ideal $I$, if it not prime then we may find $J_1, J_2$ properly containing $I$ such that $J_1 J_2 \subset I$, and we make a "factorization tree" joining $I$ to $J_1$ and $J_2$, and then iterate this procedure.
Noetherianness says that each branch of the tree has finite length, and since each edge has finite valence, the whole tree is finite, hence we eventually must reach prime ideals, and we find a finite collection of prime ideals $P_i$ whose product is contained in $I$.
The fact about trees being used can be phrased (in the contrapositive) as: an infinite tree with every vertex of finite valence has an infinite branch. (This has a name, which I forget; it is also what underlies the proof of Bolzano--Weierstrass.) Now the proof of this fact uses DC (as far as I can tell), and it's probably well-known to logicians exactly whether or not it is equivalent to DC (or, at any rate, exactly how strong it is).
The question then becomes whether we can model any tree (whose internal vertices have valence 3) as a factorization tree (in a PID, say, although the possibility of allowing more general Noetherian rings seems reasonable, at least if it helps to give a positive answer).
Added: The result on trees is called Koenig's lemma. (Thanks to Artie Prendergrast-Smith for this.)
I should also note that my original movitation for formulating the above argument was similar to that of Keith: the usual proof of the existence of minimal primes in Noetherian rings uses Zorn's lemma and has a non-constructive feel, whereas the above reformulation feels more concrete.