One of the most natural questions one can ask about a quotient of a polynomial ring is whether it is a Unique Factorization Domain (UFD). While there is no universal method for answering this question, there are a number of beautiful techniques from commutative algebra and algebraic geometry that often make the problem surprisingly manageable.
We begin with some elementary techniques and standard results from ring theory and commutative algebra that are frequently useful for deciding whether a ring is (or is not) a UFD. Some of the basic facts we shall repeatedly use are:
- A(i) A UFD is a normal domain.
- A(ii) A Noetherian integral domain is a UFD if and only if every height one prime ideal is principal.
- A(iii) In a UFD, irreducible elements are the same as prime elements.
In practice, these results are often more useful for proving that a ring is not a UFD. Combined with various techniques for establishing irreducibility—which is typically much subtler than checking primeness—such as the use of norm maps, they become extremely effective tools for proving non-UFDness. We shall illustrate these ideas through a number of examples.
There are, however, several powerful methods for proving that a ring is a UFD as well. Some of the fundamental results that we shall use are:
- B(i) Every Euclidean domain is a PID, and every PID is a UFD.
- B(ii) Localizations of UFDs are again UFDs.
- B(iii) Nagata’s Criterion.
As the article progresses, we shall develop more sophisticated techniques, drawing on ideas from commutative algebra and algebraic geometry, to prove that certain quotients of polynomial rings are UFDs—A theorem that will be of interest to us is the following: if is a field, is a nonsingular conic, and is an affine patch of , then the absence of -rational points on implies that the coordinate ring is a UFD. This is extremely helpful for determining when such rings are UFDs—particularly over and, more generally, over number fields. Rather than assuming such results, we shall develop the necessary theory as we proceed. In fact, one of the pleasant surprises is that B(iii) Nagata’s Criterion will emerge naturally as a corollary of the results we prove.
With the exception of Nagata’s Criterion, we shall not prove the bulleted results above. They are standard facts that can be found in almost any textbook on commutative algebra. Our focus will instead be on how these results can be applied effectively, and on developing the additional machinery needed to prove less obvious and more interesting examples of unique factorization.
EXAMPLE 1
In this example we shall prove that the following ring is not a UFD. The proof is a simple application of A(i)
Proof. Showing is irreducible in is pretty easy and hence it is prime by A(iii) since is a UFD, thus the ring (call it ) is an integral domain. We show that a certain element in the quotient field of that is not in is integral over .
Let be the quotient field of , consider the monic equation , then clearly satisfies this monic equation since . It remains to show that , so suppose it did, then for some , thus in , which implies which is clearly not possible since has degree in , and thus and is integral over , thus from A(i) we conclude that is not a UFD!
EXAMPLE 2
In this example, we shall prove that the following ring is not a UFD. The proof is an application of A(iii) and demonstrates the usefulness of norm maps as a tool for proving the irreducibility of elements.
Proof. First we will check that the element is not a prime element in this ring (call it ). To show this we quotient out by and check that it’s not an integral domain.
which clearly has nonzero nilpotent in it, hence is not an integral domain, thus is not prime in .
Now we claim that is irreducible in , it’s not immediately clear to see why it must be irreducible since there are relations in , one way to do is to pull it back to and do the computation there but it’s ugly, we will instead use the ingenious norm argument.
Firstly observe that any element in has a unique representative of the form where (this is fairly easy to see). Now we define the following map.
The great thing about this map is that it is multiplicative, that is .
Reason (click on it)
Since all elements in have a unique representative of the form and none of them are the same in , we can view as a free -module of rank with basis . Now consider the following map
given by
where is an -module homomorphism which takes . Now since determinant is multiplicative, we get that is a multiplicative map.
Now why is it relevant in our case? You see that the map defined above is the same as the one that we have defined. To see this, abbreviate as respectively. Then the matrix defined by is
and the determinant is exactly what we have defined above, hence the above defined map is indeed multiplicative.
From this we immediately see that if is a unit then is a unit in and hence . From this we can characterize all the units of , so
From this characterization and the fact that it’s easy to see that is a constant and .
Remark (click on it)
Over this conclusion would not hold! So be careful with the conclusions, here it holds simply because after squaring, all the coefficients are nonnegative and as a result the leading terms don’t cancel off.
So the units in are precisely . Now suppose
where are non-units in , then via the multiplicative property of norm we get
but since are non-units, and both have at least degree , thus we arrive at a contradiction! So is irreducible.
We get is irreducible but not prime in , thus from A(iii) we conclude that
is not a UFD!
EXAMPLE 3
In this example we shall prove that the following ring is a UFD.
Proof. Notice that since in this ring, it motivates us to think of as the coordinates of the unit circle. Call the ring . We define the map
by sending
Notice first of all that the map is well defined since
then clearly the map is surjective as
Construct the inverse
by
It’s easy to verify that is the identity. Thus we get that
Now we already know that is a UFD, and from B(ii) we get that is a UFD, thus is a UFD!
Remark. For those who already know some basic algebraic geometry, the assertion that
should almost be immediate since over , the two conics given by and are affinely equivalent, i.e., they are isomorphic as affine varieties, and hence their coordinate rings are isomorphic.
EXAMPLE 4
In this example we shall prove that the following ring is a UFD. The proof is computational and we will use Nagata’s Criterion B(iii) along with some other commutative algebraic results. In fact this ring is a PID but not an ED. The ring we consider is the following
Proof. We shall characterize all the prime ideals in this ring (call it ). First we observe that where is a prime ideal in . To see this we do some computation. Assume WLOG . Then in the ring
we have
thus
and the discriminant of this quadratic equation is . Hence we see that
Thus we see that are prime ideals, even maximal.
Now in the next half of the proof we use certain results from commutative algebra. First observe that
So is a one-dimensional integral domain, thus every nonzero prime ideal is maximal, and therefore is a field. Now let
where are the residue classes of in , respectively. Then is a finitely generated -algebra which is also a field. Now we use the following result from commutative algebra.
Zariski’s Lemma. If is a finitely generated -algebra such that is also a field, then is a finite field extension.
So the above lemma tells us that since is a field which is a finitely generated -algebra, then is a finite field extension of , and thus
In particular we observe that is linearly dependent over , which implies
for some . Thus
and we proved earlier that is a prime ideal. Since has height , we immediately get
From this we can immediately conclude that is a PID, hence a UFD by B(i), from this result.
But we take a different path and prove that is a UFD via Nagata’s Criterion.
Let be the multiplicative set generated by all prime elements with . The localization has no nonzero prime ideals, and therefore is a field, hence a UFD. Now Nagata’s Criterion tells us that is a UFD.
Remark. In this example, we have seen that the fact that has only one nontrivial finite algebraic extension, namely , plays a crucial role in the proof. If the base field is replaced by , or more generally by a number field, the situation changes dramatically: these fields admit infinitely many finite algebraic extensions, and consequently none of the arguments above carry over directly. Later in this article, we shall study quotients of polynomial rings over and number fields, and develop criteria that determine when such rings are UFDs and when they are not.
EXAMPLE 5
In this example we prove that the following ring is a UFD using Nagata’s Criterion B(iii) and a very useful dimension argument technique. We show that the following ring is a UFD.
Proof. Checking that is irreducible, and hence prime by A(iii) since is a UFD, is pretty easy. Thus the ring (call it ) is an integral domain. We check that is a prime element in . To do this we quotient out by and see that the resulting ring is an integral domain.
Now it’s easy to check that is irreducible in and hence prime by A(iii) since is a UFD. Therefore the resulting quotient is an integral domain, and we conclude that is prime in .
Now to use Nagata’s Criterion we localize at . Let
so that
Then in we have
We now claim that
Indeed,
since clearly all belong to the right-hand side using , while the reverse inclusion is immediate. Hence
Now we claim that are algebraically independent over . Let
and let be the quotient field of . Then
and from one can clearly see that are all rational functions of . Thus
Now we use the following theorem from commutative algebra.
Theorem. If is a finitely generated -algebra which is an integral domain and is its quotient field, then
Applying the theorem gives
Thus has transcendence degree , which implies that is algebraically independent over . Consequently,
where are indeterminates, and hence is a UFD.
Now is a localization of a UFD, so by B(ii) it is again a UFD. Therefore, by Nagata’s Criterion B(iii), we conclude that is a UFD.
In the remark following Example 4, we saw that verifying whether a quotient of a polynomial ring is a UFD becomes significantly more subtle when the ground field is , or more generally, a number field. The elementary techniques developed so far are no longer sufficient, and a deeper understanding of the underlying commutative algebra and algebraic geometry becomes necessary. In this section, we shall develop the required background from scratch and build up the theory needed to tackle such problems. As a pleasant by-product of this development, we shall recover Nagata’s Criterion as a corollary of the more general results that we prove.
Weil divisor class group and the Excision sequence
Throughout this section all rings and schemes are assumed to be Noetherian. Our goal in this section is to define the divisor class group of a scheme. We begin with the notion of Weil divisors in complete generality, and then impose additional hypotheses that allow us to define principal divisors.
Definition. Let be a Noetherian integral scheme and be its function field. A Weil divisor on is a formal finite integral linear combination of integral closed subschemes of codimension one. Thus a Weil divisor has the form
where the sum ranges over the integral codimension one closed subschemes of , each , and all but finitely many of the coefficients are zero.
To define principal divisors we shall now assume that is regular in codimension one, which is defined below.
Recall that a point is said to have codimension one if its closure is an integral closed subscheme of codimension one (for an affine scheme it consists of all the height one prime ideals). We say that is regular in codimension one if, for every codimension one point , the local ring is a discrete valuation ring (DVR).
Definition. Let be a codimension one point. Since is a discrete valuation ring, it induces a discrete valuation
The integer is called the order of at the point .
Proposition. Let be an integral Noetherian scheme which is regular in codimension one, and let . Then for all but finitely many codimension one points .
Proof. Since is a rational function, there exists a dense open subset on which is a regular invertible function, that is,
Now let be a codimension one point. The germ of at is then a unit in the local ring . Since units in a discrete valuation ring have valuation zero, it follows that . Hence, if , then necessarily .
At this point the Noetherian hypothesis becomes essential. Since is Noetherian, every closed subset has only finitely many irreducible components. Thus the closed subset has only finitely many irreducible components, and therefore contains only finitely many codimension one irreducible components. Their generic points are precisely the codimension one points contained in . Consequently for only finitely many codimension one points .
Definition. Let . The principal divisor associated to , denoted by , is defined by
where denotes the integral codimension one closed subscheme whose generic point is .
Since discrete valuations satisfy
it follows immediately that
Hence the set of principal divisors forms a subgroup of .
Definition. The Weil divisor class group of , denoted by , is the quotient group
Excision/Localization Sequence. Let be a Noetherian integral scheme, and let be a closed subset each of whose irreducible components have codimension one. Suppose has irreducible components. Then we have the following exact sequence of groups:
The map
sends each irreducible component of to its divisor class , while the map
is simply the natural restriction map.
Although the exactness of this sequence at each stage is straightforward to verify, it will play a crucial role in many of the arguments that follow later on.
UFD Characterization via Cl(X) and Nagata’s Criterion
We now present a fundamental characterization of Unique Factorization Domains in terms of the Weil divisor class group introduced previously. As an immediate application, we shall also prove Nagata’s Criterion, which was used earlier in Example 4 and Example 5. Before proceeding with the proof, we briefly state the celebrated Algebraic Hartogs’ Lemma, which will play a key role in the argument. We shall state the lemma without proof.
Algebraic Hartogs’ Lemma. Let be a normal Noetherian integral domain. Then
Remark. This says that any element which belongs to the localization for every height one prime ideal of must already belong to . It is easy to see that this statement is immediate for one-dimensional rings, since in that case all nonzero prime ideals are maximal and we have
where ranges over the maximal ideals of . This is analogous to the elementary fact that if a rational number has denominator divisible by no prime number, then .
Proposition. Let be an integral domain. Then
Proof. Assume first that is a UFD. Then from A(i) we already know that is normal, so it remains to show that every divisor is principal.
Let be a divisor. Since is a UFD, we know from A(ii) that every height one prime ideal is principal, so let where is a prime element. Define
We claim that . Indeed, if are height one prime ideals generated by distinct prime elements, then .
Therefore,
Conversely, assume that is normal and . We show that every height one prime ideal is principal, and hence is a UFD by A(ii).
Let be a height one prime ideal. Since , there exists such that .
By definition, , so , and therefore .Now let . Then for every height one prime ideal ,
Indeed, if , then , while if , then and since . Hence
which implies
Since is normal, Algebraic Hartogs’ Lemma gives
Thus , so . Together with , we conclude that
Hence every height one prime ideal is principal, and therefore is a UFD by A(ii).
Nagata’s Criterion. Let be a Noetherian domain and let be a prime element. If
is a UFD, then is a UFD.
Proof. We first prove that is normal.
Since is a UFD, A(i) implies that is normal.
Suppose
where and . It suffices to prove that .
Since is normal, for some .
Assume
Without loss of generality, assume
Multiplying the integral equation by gives
Since is prime, a contradiction. Hence , proving that is normal.
Now apply the Excision Sequence:
Since is a UFD, the previous proposition implies Hence
Thus it suffices to show that represents the trivial divisor class.
We claim that Indeed, Moreover, if is any other height one prime ideal and then which implies
Since itself is a height one prime ideal, this forces a contradiction.
Hence so is principal in the divisor class group, and therefore
Since we have already proved that is normal, the proposition above yields that is a UFD.
Riemann–Roch and Bézout for Curves over Arbitrary Fields
The celebrated theorems of Riemann–Roch and Bézout are usually presented over algebraically closed fields in most introductory textbooks on algebraic geometry. In this article, however, we shall require these theorems over arbitrary fields. Since our applications involve only curves (by which we shall always mean integral schemes of dimension one), we restrict ourselves to stating these results in the setting of curves over arbitrary fields.
One important point to keep in mind is that some of the familiar notions require slight modifications in this more general setting. We begin by recalling the relevant definitions.
Throughout this section, whenever we refer to a closed point of a curve , we mean a maximal ideal of the scheme . If is a closed point, its residue field is defined by
where denotes the local ring of at , and is its unique maximal ideal.
Definition. Let be a Weil divisor
Then we define
As usual, for a divisor , we define
It is easy to verify that is a finite-dimensional vector space over . We denote its dimension by
and let denote a canonical divisor on .
We may now state the Riemann–Roch theorem.
Riemann–Roch. Let be a smooth projective geometrically integral curve over a field with genus , and let be a Weil divisor on . Then
Corollary (Riemann–Roch, genus ). If has genus , then
We now state Bézout’s theorem in the setting of curves over arbitrary fields. We use the same notions as before.
Bézout’s Theorem. Let and be two projective plane curves over with degrees and . Assume they share no common irreducible components. Then
where
- is the set of all closed points in the scheme-theoretic intersection of and .
- is the intersection multiplicity at , defined by
where are the local equations of and , respectively.
Worked out example to gain some familiarity
The UFD Property of Coordinate Rings of Conics over Arbitrary Fields
Theorem. Let be a field and let be a nonsingular conic. Let
Suppose that there is no -rational point on . Then
is a UFD.
Proof. We make use of all the results that we have seen before in this proof.
To start off, let
and let which is the line at infinity of . Let
Now consider the Excision Sequence
where is the number of irreducible components of .
By Bézout’s Theorem we get that We shall show that . Suppose
Now by Bézout’s Theorem we know that
This forces both which implies that and are -rational points on , contradicting our hypothesis.
Thus the Excision Sequence becomes
To show that is a UFD, it suffices by the Proposition proved earlier to show that
or equivalently, that the map
is surjective. Since is nonsingular, it already follows that is normal.
Since is the only closed point at infinity, it follows from Bézout’s Theorem that Let be a Weil divisor on with
Since is a nonsingular conic, it has genus . Therefore two divisors are linearly equivalent if and only if they have the same degree. Hence
It remains to show that there is no divisor on of degree . Suppose is a divisor of degree . Then by the genus case of Riemann–Roch we have
Therefore there exists a nonzero rational function such that Let Then is effective, and
Hence for some closed point , and which implies that is a -rational point on , contradicting our assumption.
Thus there are no divisors of odd degree on , and every divisor of even degree is linearly equivalent to for some
Therefore is surjective.
Finally, from the Excision Sequence
we conclude that
Since is normal, the Proposition proved earlier implies that is a UFD.
A careful refinement of the above proof leads to even stronger results, three of which we state below.
Theorem (Stronger versions).
- If has no -rational points, then is a UFD.
- If has -rational points at infinity, then is a UFD.
- If has -rational points but none at infinity, then is not a UFD.
Applications
Application 1
Let us revisit Example 4, where we proved that
is a UFD. This now follows almost immediately from the theorem above.
Let be the projective closure of the affine curve, namely
On the affine patch , this becomes
It is immediate that
has no solution in , and hence has no -rational point. Therefore, by the theorem above,
is a UFD.
For the counterpart, we observe that
does have a point at infinity; for example,
lies on the conic. Hence, from the stronger version of the theorem,
is also a UFD.
As an exercise, apply this theorem to Exercise 3.
Application 2
The true strength of the theorem becomes apparent when the ground field is , or more generally, a number field. In this setting, the arithmetic question of whether the conic admits a rational point completely governs the algebraic question of whether its coordinate ring is a UFD. It is in these cases that the theorem reveals its full power.
Consider the ring
Let be the projective closure of the conic, namely
Clearly this has no point in . Hence, by the theorem,
is a UFD.
Now consider the same ring over :
The projective conic
has the rational point
However, if , then
so
which has no solution in since
Thus has a rational point, but none at infinity. Therefore, by the stronger version of the theorem,
is not a UFD.
Exercises
- Show that
is not a UFD.
- Show that
is a UFD.
Bonus!
Here is an extra example illustrating the usefulness of the preceding theory for the coordinate ring of the elliptic curve
over the algebraically closed field .
We shall prove that
Proof. Let be the projective closure of the affine curve
inside .
Its homogeneous equation is
When , we obtain
so there is exactly one point at infinity. Let the corresponding affine open subset be denoted by .
The Excision Sequence therefore becomes
Since
we obtain the exact sequence
Hence
A deep theorem of algebraic geometry, whose proof ultimately relies on the Riemann–Roch Theorem, states that
Since is an infinite group, it follows that
Therefore, by the Proposition proved earlier,
is not a UFD.
Remark. There is nothing special about the above example. More generally, the affine coordinate ring of any elliptic curve is never a UFD. Indeed, quotienting the group
by the image of a finitely generated free abelian group still leaves a large nontrivial group. Consequently,
so the coordinate ring cannot be a UFD. This illustrates the broader principle that affine coordinate rings of positive genus curves almost always fail to be unique factorization domains.
Conclusion
In this article, we explored a variety of techniques for determining whether quotients of polynomial rings are unique factorization domains. We began with elementary tools from commutative algebra, which are often effective in proving that a ring is not a UFD, before developing the geometric machinery of the Weil divisor class group to obtain powerful criteria for proving that a ring is a UFD. Along the way, we recovered Nagata’s Criterion as a natural consequence of this theory.
The final examples on affine conics beautifully illustrate the philosophy behind these ideas: over number fields, the arithmetic question of whether a conic has a rational point completely determines whether its coordinate ring is a UFD. This striking interplay between arithmetic, algebraic geometry, and commutative algebra is one of the many reasons why the study of unique factorization remains such a rich and fascinating subject.
Signing off!
Comments
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