Doubt regarding proof that $\mathbb{Z}, \mathbb{Z}[x]$ are unique factorization domains

The usual proof for unique factorization in $\mathbb{Z}$ proceeds via the concept of GCD (greatest common divisor) of two integers leading to the fundamental property of primes in $\mathbb{Z}$:

Theorem: If $p$ is prime and $a, b \in \mathbb{Z}$ such that $p\mid ab$ then either $p\mid a$ or $p\mid b$.

And this is the result which guarantees uniqueness of factorization. Note that the same procedure does not apply in $\mathbb{Z}[x]$ as we don't have a concept of GCD here. Thus for example we can't say that $2, x$ have GCD $1$ because we can't find polynomials $p(x), q(x) \in \mathbb{Z}[x]$ such that $$1 = 2p(x) + xq(x)$$ (contradiction arises when we put $x = 0$ in above equation).

The problem is eliminated by considering polynomials in $\mathbb{Q}[x]$ and then we have the GCD available here so that $\mathbb{Q}[x]$ is a unique factorization domain. Next we use the fact that $\mathbb{Z}[x]\subset\mathbb{Q}[x]$ to factorize elements of $\mathbb{Z}[x]$ as product of polynomials in $\mathbb{Q}[x]$ and then use Gauss lemma to prove that the factorization can also be done using polynomials in $\mathbb{Z}[x]$ only.

Thus it appears that existence of GCD is not necessary to guarantee existence of unique factorization. Does that mean we can prove unique factorization in $\mathbb{Z}$ via some other means rather than the approach I outlined in the beginning?

Also I would like to know if the property of prime numbers (mentioned in theorem at the beginning) is always a consequence of the existence of GCD in a more generally setting of integral domains?

Update: From the comments and in particular the wiki link given by Hand Lundmark it is clear that the ideas of the usual ring of integers have been generalized in many ways to give rise to the famous chain of class inclusions (see wiki link on GCD Domains) and this question is perhaps a very naive attempt to understand that all (or some of) those inclusions are proper.

Further Update: I was a bit hesitant about asking question related to a seemingly trivial matter (namely unique factorization in integers and polynomials with integer coefficients), but the way it has been received here is so much more than what I expected. MSE never ceases to amaze me (and perhaps other users too)! Thanks to all those who answered/commented/chatted. I have now got a lot of food for thought (and study).

• Your "property of primes" is the definition of a prime element in a general integral domain. The generalisation of "only divisible by itself and $1$" is called "irreducible" instead. They do not always coincide. – Arthur Apr 15 '17 at 7:07
• I think you are right. We can use the Jordan Holder Theorem to prove the fundamental theorem of arithmetic. See www-home.math.uwo.ca/~srankin/courses/302/2005/… – Tensor_Product Apr 15 '17 at 7:10
• @arthur: Is it true that in domains with GCD available "primes" are same as "irreducibles"? – Paramanand Singh Apr 15 '17 at 7:45
• In $\mathbb{Z}[X]$ you do have GCDs (the definition of which involves only divisibility); see Wikipedia (GCD domain), which says that if $R$ is a GCD domain, then so is $R[X_1,\dots,X_n]$. What your argument shows is that $\mathbb{Z}[X]$ is not a Bézout domain, but that's a different thing. – Hans Lundmark Apr 15 '17 at 8:30
• And yes, in GCD domains irreducibles are automatically prime. – Hans Lundmark Apr 15 '17 at 8:32

Thus it appears that existence of GCD is not necessary to guarantee existence of unique factorization.

Effectively there is not necessary the existence of GCD in order to prove the fundamental theorem of arithmetic. It's enough a weaker condition, namely we only need the four number lemma, see e.g., this answer. This lemma in a more abstract setting give rise to two special classes of domains known as Schreier domains and pre-Schreier domains (or Riesz domains).

Egreg wrote:

It turns out that this is one of the keys for a domain to have unique factorization. Indeed, a domain $R$ has unique factorization if and only if

1. $R$ has the ascending chain condition on principal ideals;

2. every irreducible element in $R$ is prime.

The first condition ensures existence of a factorization into a product of irreducible elements; the second condition ensures uniqueness.

Domains satisfying the first condition are called ACCP domains, whereas the domains satisfying the second condition are called AP-domains (atom $\implies$ prime), where atom=irreducible.

An important result is the following:

Theorem: If $R$ is an ACCP domain, then $R[x]$ is also an ACPP domain.

Proof: This is theorem 17a) in Pete L. Clark's notes on factorization in integral domains.

I'm not aware of any characterization of AP domains, but what I know is that every GCD domain is an AP-domain, even more: both Schreier and pre-Schreier domains are AP-domains and more exactly we have the following chain:

$$\text{GCD-domain} \implies \text{Schreier domain} \implies \text{pre-Schreier domain} \implies \text{AP-domain}.$$

Therefore we have the following characterization of unique factorization domains:

$$\text{UFD} \iff \text{ACCP domain}+\text{Schreier domain}\; (^*)$$

On the other hand, it turn out the being a Schreier domain behaves well under polynomial ring extensions, namely we have the following:

Theorem (Cohn): If $R$ is a Schreier domain, then so is $R[x]$.

Proof: This is theorem 2.7 in the paper Bezout rings and their subrings.

So we can apply (*) to the particular cases of $\Bbb Z$ and $\Bbb Z[x]$ in this way:

i) For $\Bbb Z$:

1. We can use induction (or the well-ordering principle) to ensure the existence of a prime factorization. Note that $\Bbb Z$ satisfying the well-ordering principle is equivalent to $\Bbb Z$ being an ACPP domain.
2. We can use the four number lemma to guarantee the uniqueness.

ii) For $\Bbb Z[x]$:

1. Since $\Bbb Z$ satisfies the ACCP condition, then $\Bbb Z[x]$ also satisfies the ACCP condition, so this will give us the existence of the irreducible factorization.

2. Since $\Bbb Z$ is a Schreier domain, then $\Bbb Z[x]$ is also a Schrier domain, so this will guarantee the uniqueness.

However, there is a little more to say. The classical proof of: $$D\; \text{UFD} \implies D[x]\; \text{UFD}\; (^{**})$$ uses, as you've noted, Gauss' lemma: the product of two primitive polynomials is primitive, but the above result isn't necessary because we can proof (**) without it. For instance, you can check theorem 27 of Pete L. Clark's notes cited lines above.

Returning to Gauss' lemma, this result gives rise to the class of domains known as GL-domains, and Anderson and Zafrullah proved that they are between the class of pre-Schreier and AP-domains. So in conclusion we have the chain:

$$\text{GCD-domain} \implies \text{Schreier domain} \implies \text{pre-Schreier domain}$$ $$\implies \text{GL-domain} \implies \text{AP-domain}.$$

For more info related to Schreier, pre-Schreier and GL-domains you can check these papers:

1. M. Zafrullah, On a property of pre-Schreier domains, Communications in Algebra, 15 (1987), 1895-1920.
2. S. McAdam and D. Rush, Schreier Rings, Bulletin of the London Mathematical Society, 10 (1978), 77-80.
3. J. Arnold and P. Sheldon, Integral domains that satisfy Gauss's lemma, The Michigan Mathematical Journal, 22 (1975), 39-51.
4. D. Anderson and M. Zafrullah The Schreier property and Gauss' lemma, Bollettino U. MI, 8 (2007), 43-62.
• A similar discussion (but not in so much detail) is also available in the chat I had with Hans Lundmark (see chat.stackexchange.com/rooms/57148/…). Thanks a lot (and +1) for your detailed answer. – Paramanand Singh Apr 15 '17 at 16:35
• @ParamanandSingh I fixed some mistakes and added a new reference. – Xam Apr 15 '17 at 22:30

There are many similarities between passing from $\mathbb{Z}$ to $\mathbb{Z}[x]$ and from $k[x]$ to $k[x,y]$ ($k$ a field), because both $\mathbb{Z}$ and $k[x]$ are principal ideal domains.

One can see it by “unifying” Gauss’ lemma.

Theorem. (General Gauss’ lemma) If $R$ is a unique factorization domain with quotient field $Q$ and $f(x)\in R[x]$ is a primitive polynomial which factors in $Q[x]$, then $f$ also factors in $R[x]$.

A nonzero polynomial is primitive if a greatest common divisor of its coefficients is $1$.

But what's a greatest common divisor? In general an element $d$ of a domain is a greatest common divisor (GCD) of $a$ and $b$ if

• $d$ divides both $a$ and $b$;
• for all $c$, if $c$ divides both $a$ and $b$, then $c$ divides $d$.

Of course a greatest common divisor, if it exists, is only determined up to multiplication by invertible elements.

In a unique factorization domain (UFD) a GCD exists for every pair of elements: just take the product of all common irreducible divisors with the minimum exponent (irreducible elements differing in multiplication by an invertible should be identified). This is the empty product, that is, $1$ if the two elements have no common irreducible factors.

Note also that in general domains one has to distinguish between prime and irreducible elements. A nonzero and noninvertible element $p$ is

• prime if, for all $a,b$, if $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$;

• irreducible if, for all $a,b$, if $p=ab$, then $a$ is invertible or $b$ is invertible.

It's easy to show that a prime element is also irreducible. On the other hand, an irreducible element may not be prime.

It turns out that this is one of the keys for a domain to have unique factorization. Indeed, a domain $R$ has unique factorization if and only if

1. $R$ has the ascending chain condition on principal ideals;
2. every irreducible element in $R$ is prime.

The first condition ensures existence of a factorization into a product of irreducible elements; the second condition ensures uniqueness.

The general Gauss’ lemma, which is proved essentially the same way as it is proved for $R=\mathbb{Z}$ (and $Q=\mathbb{Q}$), is the key step in proving that if $R$ is a UFD, then also $R[x]$ is a UFD.

On the other hand, the Bézout property of the GCD need not hold for a UFD. It holds when $R$ is a principal ideal domain (PID), but as soon as we pass to $R[x]$, the Bézout property does not hold, unless $R$ is a field. Indeed, the example is the same as you used: take $r\in R$ a nonzero and noninvertible element (which exists if $R$ is not a field); then $1$ is easily seen to be a greatest common divisor of $r$ and $x$ in $R[x]$ according to the above definition, but there do not exist $f,g\in R[x]$ such that $$1=rf(x)+xg(x)$$ (you have $r=2$ in $R=\mathbb{Z}$).

You can use $\mathbb{R}[x,y]$ to get a geometric interpretation of why the GCD cannot satisfy the Bézout property. Consider two irreducible distinct conics, say the circle defined by the polynomial $x^2+y^2-1$ and the parabola defined by $y-x^2$. A GCD of these polynomials is $1$, but there cannot exist polynomials $f(x)$ and $g(x)$ so that $$1=(x^2+y^2-1)f(x)+(y-x^2)g(x)$$ for the simple reason that the relation cannot hold at the two points where the circle and the parabola intersect.

• Thanks for putting all the discussion in comments into a well framed answer. +1 Will wait for some more time before accept. – Paramanand Singh Apr 15 '17 at 10:39
• Great answer. +1 – Xam Apr 15 '17 at 15:24