The arithmetic analogy between and is not perfect, as is not generally a unique factorization domain. For instance, note
in displays two essentially distinct ways to write as the product of irreducible elements. Nevertheless, one can recover in a different sort of factorization: every ideal may be uniquely written as the product of prime ideals. Any integral domain in which this is possible is called a Dedekind domain. We will see by the end of this post that working in the world of ideals rather than numbers does not preclude us from extracting useful number-theoretic information.
Our goal for this post is to prove some of the important results on Dedekind domains (in particular, that is a Dedekind domain), and it will not hurt to work in greater generality than the number field case. In particular, the setup is as follows: let be a Noetherian integral domain, and let denote its field of fractions.
4.1. “Modular arithmetic”
From the module-theoretic perspective, ideals of are just finitely generated (as is Noetherian) -submodules of . This point of view is useful because it allows us to naturally generalize the notion of an ideal to include larger subsets of the entire fraction field . In particular, a fractional ideal is a finitely generated -submodule of . Since ideals in are also fractional ideals of , for clarity we refer to ordinary ideals in asintegral ideals. We denote by the collection of fractional ideals in .
Given , , recall one can define their product and sum
which are themselves fractional ideals. This allows one to introduce a notion of invertibility: an ideal is invertible if there exists such that .
We will have to do some hard work now to demonstrate the following proposition.
Proposition 4.1 Let be a Noetherian integral domain. Then, the following are equivalent:
- is integrally closed (if satisfies a monic polynomial in , then ) and every prime ideal in is maximal.
- Every integral ideal in can be written uniquely as the product of prime ideals.
- Every fractional ideal of is invertible. In particular, is an abelian group under the module product.
If satisfies any of these conditions, one christens a Dedekind domain.
Before descending into the land of commutative algebra, let us note the corollary:
Corollary 4.2 is a Dedekind domain.
Proof: In the previous post, we showed is Noetherian and that every prime ideal in is maximal. Suppose satisfies a monic polynomial , given by
Since is a finitely generated -module, it follows from the fundamental theorem on modules over a principal ideal domain that
is finitely generated as a -module. Since is a -linear combination of , , , , it follows
is also finitely generated as a -module. Applying the fundamental theorem one last time, we have
is itself a finitely generated -module. Since , we conclude is an algebraic integer in , i.e. .
4.2. Proof of Proposition 4.1
As with many arguments in commutative algebra, we piece together the proof from a series of fairly sophisticated lemmas. We shall first show the direction in Prop. 4.1, so assume is Noetherian, integrally closed, and every prime ideal in is maximal.
Lemma 4.3 Assume . Let be an integral ideal of . Then, there exist prime ideals , , such that
Proof 1: (General). Denote by the collection of integral ideals which fail to satisfy the given condition. Suppose is nonempty. Since is Noetherian, an appeal to Zorn’s lemma admits a maximal element . Of course cannot be prime, so there exist , such that but , . There are the proper inclusions and , so each of and contains a product of prime ideals. However, , which is a contradiction.
Proof 2: (-specific). For the axiom of choice truthers among us, one can show this result in the specific case without invoking Zorn. Use that every quotient ring is finite, and choose an ideal for which is minimal. Then argument proceeds exactly as in the first proof.
For any , define the fractional ideal
By construction, . In fact, is the only candidate for the inverse of a fractional ideal: one can easily show if is invertible, then (exercise!).
Lemma 4.4 Assume . Let be a prime ideal of . Then, .
Proof: The argument proceeds in three steps.
Step 1: (). Choose a nonzero for which
such that is as small as possible. By the prime avoidance lemma, one of the (without loss of generality, assume ), is contained in . Then since is maximal. By the minimality of , one knows
so there exists such that . In particular, . However, by the choice of ,
Hence, . It follows .
Step 2: ( for any nonzero ideal ). Let , , be a basis for as an -module. Suppose and choose . Then there is a linear map given by , with matrix . Then is an eigenvalue of , so that . But the left-hand side is a polynomial in . By integral closure of , it follows . Thus, , contradicting step 1.
Step 3: (). By step 2, we see , so the inclusion is proper. Since is an -submodule of , it is an integral ideal. But is a maximal ideal, so as desired.
Using these lemmas, it is actually not so difficult to finally furnish the proof of . We demonstrate existence of the prime ideal factorization, and leave uniqueness as an (easy) exercise.
Lemma 4.5 Let be an integral domain and an integral ideal. If can be factored as the product of invertible prime ideals, then this factorization is unique.
Proof 1 of : (General). Let denote the collection of integral ideals that do not permit a prime ideal factorization. Suppose is nonempty. As earlier, Zorn’s Lemma gives a maximal element . Since every ideal is contained in a maximal ideal, there is some prime ideal such that . Then, one may factor , where
is an integral ideal. Since is maximal in , it follows has a prime ideal factorization. But this gives a prime ideal factorization, which is a contradiction.
Proof 2 of :(-specific). We shall prove by induction (on ) that if an integral ideal contains a product of nonzero prime ideals, then is a product of prime ideals. By Lemma 4.3, this is sufficient to prove . For , if for some prime ideal , then by maximality. Suppose now that . Since is finite, there is a maximal ideal containing . Then the prime avoidance lemma implies, without loss of generality, that . Multiply through the inclusion by :
Induction gives a prime ideal factorization for , and hence one for .
Next on the agenda: we must show .
Lemma 4.6 Choose any nonzero . If for fractional ideals , then each is invertible.
Proof: Simply set .
Lemma 4.7 Assume . Let be a nonzero prime ideal. If is invertible, then is maximal.
Proof: Choose any . Suppose that . Then we may factorize
where each , is a prime ideal. Let . Then in , we have the prime ideal factorizations
so Lemma 4.6 implies each , is invertible. Equating these factorizations as
we see Lemma 4.5 tells that and each appears exactly twice in the collection . Hence is the multiset . This gives the following chain:
Thus, if , then for some and . Since and by choice, we must have . This yields another chain:
since is invertible. But by assumption.
Corollary 4.8 Assume . Let be a nonzero prime ideal. Then, is invertible.
Proof: Choose any . We may factorize , where, by Lemma 4.6, each is an invertible prime ideal. In fact, each is also maximal, due to Lemma 4.7. The prime avoidance lemma then implies without loss of generality , so is itself invertible.
The meat of this portion is now done with, and we can give the proof of .
Proof: Let be a fractional ideal. Since is finitely generated, there exists such that . Factoring into prime ideals, we may write . Then is the inverse of .
Finally, it remains only to prove , which is (comparatively) not very tough at all.
Proof: Let be a prime ideal in , and suppose is a maximal ideal containing . Then , so the prime avoidance lemma implies either or . Suppose the former is true. Then, multiplying through by gives
which is impossible. Hence is maximal.
Suppose satisfies a monic polynomial of degree , whose coefficients are in . Consider the fractional ideal
Then , so multiplying through by proves . We conclude is integrally closed.
Let us conclude this section with a very brief exercise, which illustrates two important properties of Dedekind domains.
Lemma 4.9 Let be a Dedekind domain.
- “To contain is to divide”: the ideals , satisfy iff there exists an integral ideal such that .
- Comaximal equals relatively prime: the ideals and have no common prime ideal factors iff .
Proof: Easy exercise.
4.3. Introducing the ideal class group
We have strayed a bit far from number theory with all of these maximality arguments. Let us return by introducing the ideal class group of a number field , whose relation to number theory is beautiful and important.
Recall denotes the collection of fractional ideals of , which gains the structure of an abelian group under the module product. There is a subgroup which consists precisely of the principal fractional ideals, those generated by exactly one element in . The class group of is then defined as the quotient
Equivalently, one may place on the integral ideals of a relation given by iff there exist , such that
The equivalence classes thus obtained gain a natural group structure under ideal multiplication, and one easily sees this is isomorphic to the structure on . In the next post, we will prove the following extremely important proposition.
Proposition 4.10 For a number field , the class group is finite.
For now, let us take this fact as given. The order of is called the class number of , and denoted . Observe that iff is a principal ideal domain. Thus, in some sense, the class number measures the extent to which a ring of integers fails to have (or succeeds in having) unique factorization for elements.
Consequently, computation of the class number is significant in number theory, as it detects how well one can treat ideals as elements. This interpretation is scalable: finiteness of the class group implies for any integral ideal , the ideal is principal. One can use this fact to great effect in the solutions to certain Diophantine equations:
Proposition 4.11 Suppose is square-free, with . Let , and assume furthermore that . Then,
- If there exists for which , then the Mordell equation has the unique solutions .
- If there exists no such , then the Mordell equation has no solutions.
The proof of this proposition is actually already well within our grasp (given temporary suspension of disbelief regarding the finiteness of ). Recall that when , the ring of integers of is exactly .
Lemma 4.12 Assume . Suppose there are , such that for some ideal . Furthermore, suppose and are comaximal ideals. Then and for some , and units , .
Proof: Since and are comaximal, they share no prime ideal factors. Therefore, and for some ideals , with . The class of principal ideals is the identity in . By finite group theory, it follows that the order of must divide both and . As it was assumed , it follows is principal. Similarly, is a principal ideal. Writing and , the result drops out.
Proof of Prop. 4.11: Assume there exists a solution to Mordell’s equation . Due to the choice of , modulo considerations tell that is necessarily odd.
Additionally, we find . Otherwise, suppose there is a prime dividing both and . Then also divides , so divides , which was assumed square-free.
Now, factor the equation in and pass to ideals:
Suppose and are not comaximal, i.e. have a common prime ideal factor . Then , and hence . But also contains , thus by primality contains . Since is odd and , also , so there are integers and such that . This is a contradiction.
Therefore, and are relatively prime. By Lemma 4.12, there exists and a unit such that . The only units in are (exercise!). Setting and assuming without loss of generality, it follows
Equating coefficients gives and accordingly. A little more computation proves and as desired.
As lovely as this little characterization is, really applying it requires one to know the class number of . And in general, calculating is quite challenging. There do exist explicit formulas and algorithms designed to compute in the case where is a quadratic field, which we will discuss in the subsequent posts.