Theorem 3.2.1. Fundamental Theorem of Arithmetic.
Every integer \(n
\gt 1\) is either prime or can be factored as a product of primes. And such a factorization is unique up to a rearrangement of the factors.
Section 3.2 Factoring in integral domains
Let’s review some ideas surrounding the concept of factorization in rings to remind ourselves of how certain terminology became relevant. Let’s begin with factorization in the integers \(\Z\text{.}\) The Fundamental theorem of arithmetic is often phrased as
Remark 3.2.2.
We simply take the statement of the fundamental theorem in this context at face value, but as soon as we begin to poke at the edges, all sorts of questions come up.
- So why is there this restriction to integers greater than one? Well, ok, we are smart enough to avoid 0 and 1, and perhaps even \(-1,\) but what’s the matter with the integers \(n \lt -1\text{?}\)
- Then there is the word prime which in the integers plays two roles. A prime in \(\Z\) acts as both an irreducible element and a prime element, but how exactly are those roles evidenced?
- Moreover, a prime in \(\Z\) seems to be slightly more restrictive in its meaning than in general (an integer \(p\ge 2\) whose only positive divisors are 1 and itself). Why those restrictions?
There are many properties that make the integers special, but the facts that the unit group consists of only two elements, \(\pm
1\text{,}\) and that irreducibles and primes are the same, strongly influence how the Fundamental Theorem is stated when compared to a statement describing unique factorization in a more general integral domain.
Let’s start with an arbitrary integral domain \(R,\) and consider what we might mean by unique factorization. Certainly we must begin with some statement which says we have factored the given element and it can’t be factored anymore. We should exclude trying to factor 0 or units in the ring, and we shouldn’t fuss about the order of factors nor associates. For example, we don’t want to distinguish factorizations like
\begin{equation*}
6 = 2\cdot 3 = 3\cdot 2\cdot 1 = (-2)\cdot (-3)\text{.}
\end{equation*}
But as we said in \(\Z,\) this is easy to control since there are only two units.
In \(\Q[x]\text{,}\) we would like to say that \(f(x) = 2x\) cannot be factored anymore (is irreducible) even though
\begin{equation*}
2x = 2\cdot x = (2/3)\cdot (3x) = \cdots\text{.}
\end{equation*}
On the other hand, in \(\Z[x],\) the polynomial \(2x\) is not irreducible since \(f(x) = 2x = 2\cdot x\) and neither 2 nor \(x\) are units in \(\Z[x].\)
So factoring into irreducibles seems like a pretty natural notion, though we still have to deal with associates.
Example 3.2.3.
So let’s take a nonzero, non-unit element \(a\) in an integral domain \(R\) and try to factor it into irreducibles. What works and what might go wrong?
Solution.
Factoring is a simple process. We ask if the given (nonzero, non-unit) element \(a\) is irreducible. If it is, we are done. However if it is not, that means there is a nontrivial factorization, that is as a product
\begin{equation*}
a = a_1b_1
\end{equation*}
where neither factor is a unit. Now we iterate.
If both \(a_1, b_1\) are irreducibles, we are done, otherwise assume \(a_1\) is not. Then \(a_1\) has a nontrivial factorization
\begin{equation*}
a_1 = a_2 b_2
\end{equation*}
where neither factor is a unit. And we iterate. So what’s the problem? There does not seem to be any mechanism to force this procedure to terminate.
In the example above, we see that it is natural to try to factor in any integral domain, but what is lacking is a mechanism to terminate the procedure. While not often discussed in a first algebra course, this leads quite naturally to the notion of a Noetherian ring, but before we define it, let’s make the transition from elements to ideals seem natural.
We are talking about the factorization of elements in a ring \(R\text{.}\) We say that the element \(b\) divides the element \(a\) (written \(b\mid a\)) if there exists a \(c \in R\) so that \(a=bc,\) in other words if \(b\) is a factor of \(a.\) Now one issue is that if \(b\mid a\text{,}\) then so does \(bu\) for any unit \(u\) since
\begin{equation*}
a = bc = (bu)(u^{-1}c)\text{.}
\end{equation*}
Notice that \(b\mid a\) if and only if the principal ideals \((a) \subseteq (b)\text{,}\) and in an integral domain, \((b)
=(b')\) if and only if \(b\) and \(b'\) are associates. So principal ideals capture the notion of factoring and of associates all in one concept.
Returning to our example of factoring, if \(a\) factored non-trivially as \(a = a_1b_1\text{,}\) and \(a_1\) was not irreducible, then \(a_1 = a_2b_2\) with neither \(a_2, b_2\) units. In terms of ideals we would have
\begin{equation*}
(a) \subsetneq (a_1) \subsetneq (a_2) \subsetneq \cdots\text{.}
\end{equation*}
This means that until at some point in the factorization an irreducible appeared in the factorization, this ascending chain of ideals would go on indefinitely. Addressing this issue is one of the conditions which define a Noetherian ring.
Theorem 3.2.4.
Let \(R\) be a commutative ring with identity. The following three conditions are equivalent and define what it means for the ring to be Noetherian.
- \(R\) satisfies the ascending chain condition (ACC), meaning that given any ascending chain of ideals\begin{equation*} I_1 \subseteq I_2 \subseteq I_3 \subseteq \cdots \end{equation*}in \(R,\) there exists an index \(r\ge 1\) so that\begin{equation*} I_r = I_{r+1} = I_{r+2} = \cdots. \end{equation*}
- Every ideal \(I \subseteq R\) is finitely generated.
- Every non-empty collection of ideals has a maximal element.
You certainly know of Noetherian integral domains since every PID is automatically Noetherian since every ideal is generated by a single element. The main takeaway here is the following result.
Theorem 3.2.5.
Let \(R\) be a Noetherian integral domain. Then every nonzero, non-unit in \(R\) can be factored as a finite product of irreducibles.
Proof.
From the example and discussion above, the Noetherian condition tells us that as we begin to fact a nonzero, non-unit an irreducible must eventually appear as one of its factors.
So if \(a = \pi_1 a_1\) where \(\pi_1\) is irreducible, we ask if \(a_1\) is a unit. If it is, we are done. Otherwise \(a_1 = \pi_2 a_2\) with \(\pi_2\) irreducible. Again if \(a_2\) is a unit we are done. Otherwise \(a_3 = \pi_3 a_3.\) So again at this stage we are building another ascending chain of ideals:
\begin{equation*}
(a) \subsetneq (a_1) \subsetneq (a_2) \subsetneq
\cdots
\end{equation*}
which must terminate, but terminating means at the last stage \(a_{r-1} = \pi_r a_r\) with \(\pi_r\) irreducible and \(a_r\) a unit, which terminates the factorization.
Example 3.2.6.
The ring \(\Z[\sqrt{-5}] = \{ a+ b\sqrt{-5} \mid a,b \in
\Z\}\) is a Noetherian integral domain, so every nonzero, non-unit factors as a finite product of irreducibles, just like in \(\Z.\) But where \(\Z\) enjoys unique factorization, \(\Z[\sqrt{-5}]\) does not. So primes in \(\Z\) are somehow different than irreducibles in \(\Z[\sqrt{-5}].\) We investigate how.
Solution.
That the ring \(R=\Z[\sqrt{-5}]\) is Noetherian follows from somewhat more advanced knowledge: Since \(\Z\) is a PID, it is Noetherian, and so by the Hilbert basis theorem, so is the polynomial ring \(\Z[x].\) Now an easier exercise is that that homomorphic image of a Noetherian ring is Noetherian, and our ring \(R\) is the homomorphic image of \(\Z[x]\) under the evaluation map induced by \(x\mapsto \sqrt{-5}.\) The ring is an integral domain since it is a subring of the field \(\C.\)
Now to come to the more relevant part. We can write down a sentence like
\begin{equation*}
6 = 2\cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})
\end{equation*}
and claim this shows the ring does not have unique factorization, but that is just an assertion without much proof.
What does it mean to say this is a counterexample to unique factorization? Well we would have to know that \(2, 3, 1\pm
\sqrt{-5}\) are all irreducible and not associates.
Both of these questions can be answered using the norm map \(N:\Z[\sqrt{-5}] \to \Z\) given by
\begin{equation*}
N(a+b\sqrt{-5}) = (a+b \sqrt{-5})(a-b\sqrt{-5}) = a^2 +
5b^2.
\end{equation*}
It is easy to see that since \(N(\alpha) =
\alpha\overline{\alpha}\text{,}\) the product of the element and its complex conjugate, that \(N(\alpha\beta) =
N(\alpha)N(\beta)\) for any \(\alpha, \beta \in R = \Z[\sqrt{-5}].\)
The first lemma to prove is that \(\alpha\) is a unit in \(\Z[\sqrt{-5}]\) if and only if \(N(\alpha) = \pm 1\) (actually \(+1\) in our case since \(a^2+5b^2 \ge 0\) for all \(a,b\)). It follows that the units of \(\Z[\sqrt{-5}] =
\{\pm 1\}\text{,}\) so it is at least clear that none of the factors in the two factorizations of 6 are associates.
To show that \(2, 3, 1\pm
\sqrt{-5}\) are all irreducible is done case by case. We show that 2 is irreducible in \(\Z[\sqrt{-5}].\) Suppose not. Then
\begin{equation*}
2 = \alpha \beta
\end{equation*}
where neither \(\alpha, \beta\) are units, i.e., have norm 1. But since the norm is multiplicative,
\begin{equation*}
4 = N(2) = N(\alpha)N(\beta)\text{,}
\end{equation*}
where now this equation is an equation in \(\Z\) where we have unique factorization. So the only possibilities are that \(N(\alpha) = 1, 2\text{,}\) or 4. We cannot have \(N(\alpha)=1\) since that means that \(\alpha\) is a unit, nor can be have \(N(\alpha) =4\) since that forces \(\beta\) to be a unit. So our only chance for a nontrivial factorization is if \(N(\alpha)=N(\beta) = 2.\) But since \(a^2 + 5b^2=2\) has no solutions for \(a,b\in \Z,\) we are forced to conclude that one of \(\alpha\) or \(\beta\) is a unit, meaning that \(2\) is irreducible in \(\Z[\sqrt{-5}]\text{.}\) The other cases are analogous.
The failure of unique factorization in the ring \(\Z[\sqrt{-5}]\) is a consequence of the fact that not all irreducibles are primes. Having shown that they are irreducible, it is easy to show the elements are not prime. Recall, we have
\begin{equation*}
6 = 2\cdot 3 =
(1+\sqrt{-5})(1-\sqrt{-5})
\end{equation*}
so that means that \(2,3,1\pm\sqrt{-5}\) all divide 6, so in particular
\begin{equation*}
2
\mid (1+\sqrt{-5})(1-\sqrt{-5}).
\end{equation*}
We need to show that \(2\) does not divide either \(1+\sqrt{-5}\) nor its conjugate. But an easy lemma is that in an integral domain, if \(\pi_1 \mid \pi_2\) where the \(\pi_i\) are irreducible, then they must be associates. Since we know the units of the ring are only \(\pm 1,\) it is clear they are not associate. So these 4 elements are irreducible, but not prime in \(\Z[\sqrt{-5}].\)
The theorem we can now prove is
Theorem 3.2.7.
Let \(R\) be a Noetherian integral domain in which every irreducible element is a prime element. Then \(R\) is a unique factorization domain, meaning that every nonzero, non-unit in \(R\) has a factorization into a finite product of irreducibles which is unique in the sense that if
\begin{equation*}
a = \pi_1\cdots \pi_r = \pi'_1 \cdots \pi'_s
\end{equation*}
are two factorizations of \(a\) into irreducibles, then \(r=s\text{,}\) and (after a possible reordering) \(\pi_j\) and \(\pi'_j\) are associates for all \(j=1, \dots, r.\)
(sketch).
Only the uniqueness statement remains to be shown, and the proof here is exactly as it is in \(\Z,\) so we simply sketch the argument.
Proceed by induction on \(r.\) If \(r=1,\) then by the definition of an irreducible, it can have no nontrivial factorizations which makes \(s=1\) and \(\pi_1 = \pi'_1.\) For \(r\ge 2\text{,}\) we know that \(\pi_1 \mid a\text{,}\) so
\begin{equation*}
\pi_1 \mid \pi'_1\cdots \pi'_s\text{.}
\end{equation*}
Since the irreducible \(\pi_1\) is prime, it must divide one of the factors on the right hand side, so without loss of generality (since products commute), we can say \(\pi_1 \mid \pi'_1\text{.}\) Again, both being irreducibles, they must be associate, so we may write
\begin{equation*}
\pi_1 \cdots \pi_r = \pi'_1 \cdots \pi'_s = \pi_1 \pi''_2\pi'_3\cdots
\pi'_s,
\end{equation*}
where \(\pi''_2\) is an associate of \(\pi'_2.\)
Since we are in an integral domain, the cancellation law holds yielding
\begin{equation*}
\pi_2 \cdots \pi_r = \pi''_2\pi'_3\cdots
\pi'_s,
\end{equation*}
and the argument finishes by induction.
The problem of course is how do we know if we are in a ring which has the property that every irreducible is prime? You may recall you spent time learning about Euclidean domains, PIDs, and UFDs and the relationship between them. What you leveraged to produce examples is that Euclidean domains are all PIDs, and that all PIDs are UFDs, but identifying a PID which was not Euclidean or a UFD which is not a PID was not immediately obvious. Let’s first collect a few more tools and then dive into those issues.
