- Unique Factorization of Ideals
- Quotient and Localization
- A Nonzero Prime Ideal is Maximal
- Fractional Ideal
- A Homomorphism on a Fractional Ideal
- Invertible Ideal
- Unique Factorization of Fractional Ideals
- Dedekind implies Noetherian
- Invertible Ideal and Localization
- Localization Produces a DVR
- Integrally Closed and One Prime Ideal
- Invertible iff Projective
- 11 Definitions
- Fraction Ring is Dedekind
- UFD iff PID
- Finitely Many Prime Ideals
- To Contain is to Divide
- Two Generators
- Class Group
- The Index of an Ideal
- Conditions for a Finite Class Group

A dedekind domain is a beautiful generalization of a unique factorization domain (ufd). As you recall, a nonzero nonunit in a ufd is uniquely a product of prime elements. In the integers, 30 is 2×3×5, and that's the end of it. A dedekind domain is similar, but now we're working with ideals. Every nonzero proper ideal is uniquely a product of prime ideals. The multiples of 30 are the multiples of 2 times the multiples of 3 times the multiples of 5, and that's the end of it.

Like a ufd, a dedekind domain is an integral domain. I might lift this restriction from time to time, but then some strange rings become dedekind. The integers mod m for instance, or F[x] mod x2, or the quotient of any traditional dedekind domain. So unless otherwise stated, a dedekind domain is also an integral domain.

A field has no proper nonzero ideals, hence it is dedekind by default.

Let's show that a pid is dedekind. Let R be a pid, and H a nonzero proper ideal. Since R is a pid, H = xR for some generator x. H is a product of ideals iff x is the product of their generators, or an associate of this product. Also, an ideal is prime iff its generator is prime. Therefore the unique factorization of x implies a unique factorization of H.

There are ufds that are not dedekind. We'll see this later on.

If P is a prime ideal, and S and T are proper ideals in R, and P*S = P*T = H, then write H as a unique product of prime ideals. This product necessarily includes P. It also includes the primes of S, or the primes of T. The primes of S are therefore equal to the primes of T, and S = T. If PS = PT, you can cancel P, like an integral domain.

If J is an ideal, and JS = JT, write J as a product of prime ideals and cancel them one at a time. This leaves S = T.

Let S H and J be ideals, with J properly contained in H. SJ lies in SH, and if SJ = SH then cancel S to show J = H. Therefore SJ is properly contained in SH. Strict inequality persists even if S = R, or H = R, or J = 0. If S is a nonzero ideal, multiplication by S respects containment, and proper containment.

Multiplication by S maps a chain of ideals (ascending, descending, or linearly ordered) to another chain of ideals.

Now for the converse. Let SJ lie properly in SH. If SJ = 0 then J = 0 (integral domain), which is a proper ideal in H, so set this case aside. If S = R then SJ = J and SH = H, so set this case aside.

J+H is an ideal that I will call H′. With SJ in SH, SH = SH′. Cancel S, and H = H′. Therefore J lies in H. Since SJ is properly contained in SH by assumption, J < H. Chains of ideals correspond under the action of S.

If H is a proper ideal in R, multiply by H, and H2 lies properly in H. Continue by induction, and the powers of H build an infinite descending chain of ideals. Since R is an integral domain, each ideal in this chain is nonzero, though their intersection could be 0. If a ring exhibits factorization of ideals, then so does its quotient or localization. Start with the quotient. Of course you have to mod out by a prime ideal Q, just to get an integral domain. Let J in R/Q pull back to H in R, having some prime factorization. Each prime factor of H contains H, else the product could not be all of H. Each prime factor contains Q, and corresponds to a prime in the quotient ring. The image of the product of ideals is the product of the images, hence J is the product of prime ideals. This product might not be unique, but J does factor into prime ideals.

Localization about P is similar. J pulls back to H, which factors, and product and localization commute, and prime ideals correspond, hence J is a product of primes in the fraction ring.

Assume P is a minimal nonzero prime ideal and localize about P. In the local ring S, P is maximal, and the only prime. Every nonzero ideal is a power of P. Ideals are linearly ordered, hence S is a valuation ring. The valuation group is infinite, hence the powers of P go on forever, building a descending chain of ideals. Let U be the intersection of this chain. If U is nonzero it is some power of P, yet it is smaller than every power of P, hence the chain intersects in 0. Since localization respects both product and intersection, the intersection of Pn in R, when localized, gives the intersection in S, which is 0. Therefore the powers of P intersect in 0.

Let P be a minimal prime containing the prime Q. These are called adjacent primes, since there is no prime properly in between. Move to the quotient ring R/Q, where the powers of P intersect in 0. Pull back, and the powers of P intersect in Q. This works even if R is not an integral domain - because R/Q is an integral domain.

Given P containing Q, not necessarily adjacent, select any x in P-Q, and let S be a minimal prime containing q and x. Then ratchet Q up to T, a maximal prime in S that misses x. The union of an ascending chain of such primes is prime, is contained in S, and misses x, so there is such a prime T. A prime properly between S and T would contain x or not contain x, and either is a contradiction. Therefore S and T are adjacent. As described above, the powers of S close in on T. The powers of P must contain T, or the powers of S would not contain T. Thus the powers of P contain Q. In general, the powers of P contain every prime that is contained in P. Replace P with the intersection of the powers of P, and the set of prime ideals contained therein is the same.

Let H be the intersection of the powers of P. If there is any prime inside P, 0 for instance, this prime is also in H. Let Q be a maximal prime ideal inside H. A prime S between Q and P would be between Q and H, hence P and Q are adjacent. The powers of P close in on Q, which is the same as H. The powers of a prime intersect in another prime.

At this point the chain is set, from a maximal ideal down to 0. The powers of any prime P intersect in the next prime down, call it Q, and all other primes in P are in Q. March down the chain of prime ideals, and if the chain is infinite then it intersects in a prime ideal. Prime ideals correspond to ordinals, with the maximal ideal at 0. The ring is a set, so the chain stops at 0. Given two prime ideals in a prime P, one contains the other, i.e. the earlier in the descending chain contains the latter. All the prime ideals inside P are linearly ordered, in fact well ordered.

If R is not an integral domain then the chain of primes stops at the minimal prime ideal below P, whatever that is. Suppose somewhere in a dedekind domain, one prime contains another, contains another. This implies a chain of length at least 3. Let P Q and V be the top three primes in the chain. Take the quotient ring by V, then localize about P. The resulting ring, call it S, has primes P, Q, and 0. Within R, the powers of P, times Q, build a chain between Q and Q2. The powers of P, times Q2, build a chain between Q2 and Q3. Continue this process, and the ideals generated by P and Q are well ordered by the exponent on Q followed by the exponent on P. Push this forward to S, where every ideal is some power of Q times some power of P. All the ideals are linearly ordered, and S is a valuation ring.

If a is in P - P2, then a has the least positive valuation. As such, a generates the maximal ideal P, and P is principal. The valuation group is torsion free, hence the powers of P march on forever, building a descending chain of ideals. The intersection is the corresponding ideal Q.

Find b in Q - QP, and b has the least valuation of anything in the ideal Q, whence b generates Q. Within the valuation group, b is a cluster point from below. Divide b by a in the valuation group, and a is a cluster point from below. However, there are no ideals between P and S. This is the long sought contradiction. A chain of length 3 cannot exist. Every nonzero prime in a dedekind domain is maximal. (If R is not an integral domain, every prime ideal is maximal or minimal, with at most one minimal prime under each maximal prime.)

It is easy, then, to find a ufd that is not dedekind. If K is a field, K[x,y], the polynomials in x and y with coefficients in K, is a ufd. x generates a prime ideal, with quotient K[y], which is an integral domain. Yet x lives in the larger maximal ideal generated by x and y. Let R be an integral domain with fraction field F. A ufd for instance, or a dedekind domain. Since R has no zero divisors, R embeds in F via R/1.

Let H be an R module contained in F, and assume, for some d in R, d*H lies entirely in R. In other words H has a common denominator d, and when you multiply through by d you get an ideal in R. In this case H is called a fractional ideal.

If R is the integers and F is the rationals,
the integers and half integers form a **Z** module of dimension 2
that is also a fractional ideal with common denominator 2.
In contrast,
H could be the set of rational numbers whose denominators are powers of 2.
This is a **Z** module that is not a fractional ideal.

If H is contained in R it is a fractional ideal by default, hence every ideal is a fractional ideal.

If H is finitely generated, let d be the product of the denominators of the generators. d drives the generators, and all of H, into R. Therefore every finitely generated R module contained in F is a fractional ideal.

Let H1 and H2 be fractional ideals with denominators d1 and d2 respectively. Let H3 be all finite sums of products of elements from H1 cross H2, and show that H3 is an R module. If H1 and H2 are contained in R then H3 is the product of the two ideals. The denominator d1d2 drives H3 into R. Thus the product of fractional ideals is another fractional ideal.

Multiplying fractional ideals is both associative and commutative. Also, multiplication by R leaves any fractional ideal unchanged. The fractional ideals form a commutative monoid. Let R be an integral domain with fraction field F. Let w be a module homomorphism from a fractional ideal H into F. There is a symmetry here: a*w(b) = b*w(a). This assumes a and b are both in H, so that w can be applied to a or to b. The relationship holds if a and b come from R, whence aw(b) = w(ab) = bw(a). That's the definition of an R module homomorphism. But the relationship extends to all of F.

Write a as an/ad (numerator over denominator), and b as bn/bd. Follow along as we do some algebra. The second step multiplies top and bottom by something in R, and this works only because the denominators are not zero divisors.

aw(b) =

adbd aw(b) over adbd =

bdanw(b) over adbd =

bdw(anb) over adbd =

w(anbn) over adbd

The last expression is symmetric in a and b, so the same algebra applies to bw(a), giving aw(b) = bw(a). Let H1 and H2 be R modules contained in the fraction field F, and assume H1*H2 = R. In this case we say H1 and H2 are invertible, and each is the inverse of the other. Note that 0 is never an invertible module, because H*0 can never produce 1.

If the inverse exists, it is unique. Assume H1H2 = R, and let H2′ be the set of fractions x in F such that x*H2 lies in R. Show that H2′ is an R module containing H1.

At this point H2 has an inverse H1, and a possibly larger inverse H2′. Write the following equation.

H2′ = R*H2′ = (H1*H2)*H2′ = H1*R = H1

The inverse of H2 is unique, assuming H2 is invertible, and the inverse is precisely the set of elements that drive H2 into R. If H2 already lives in R, its inverse contains all of R.

Take any numerator n in H2 and note that n*H1 lies in R. Thus H1 is a fractional ideal. Reverse this to show H2 is a fractional ideal. If a submodule of F is invertible it is automatically a fractional ideal.

By definition, an invertible ideal is an invertible fractional ideal; it need not be an ideal in R. If you mean an ideal in R that happens to be invertible, you must say so.

If H2 is a fractional ideal that is not invertible, H2′′ might not equal H2.
Let H2 be the ideal in K[x,y] generated by x and y.
Here H2′ is the base ring K[x,y],
and H2′′ is the same ring,
which is larger than our ideal.
This is a round about way to prove an ideal is *not* invertible.

Let H2 be principal, with generator g. Invert g in F and let this generate H1. Now H1H2 = R, and the inverse H1 is principal. In other words, an invertible ideal is principal iff its inverse is principal, and their generators are inverses in F.

Whenever H2 is invertible, one can multiply an equation through by H1 to cancel H2, like an integral domain. The invertible ideals form a group within the monoid of all fractional ideals. Within the rational numbers, a fraction like 18/35 (which happens to be in lowest terms) is uniquely a product of primes, although some of the exponents may be negative. In this case 18/35 = 2 × 32 × 5-1 × 7-1. This result generalizes to the fractions of any ufd. Once reduced, the numerator and denominator have no primes in common, each has a unique factorization, and these can be combined to give a prime factorization for the fraction.

Extend this result to dedekind domains. Every fractional ideal is a unique product of prime ideals, where some of the exponents may be negative.

The product of finitely many fractional ideals is invertible iff each component is invertible. If each fractional ideal is invertible, multiply their inverses together to get the inverse of the product. Conversely, let J * (H1 * H2 * H3 * … * Hn) = R. Each factor Hi is multiplied by something else to get R, hence each Hi is invertible.

Let P be a prime ideal and let x be a nonzero element in P. Write x*R as a product of prime ideals Q1Q2Q3… If P contains a product of ideals it contains at least one of them. Suppose P contains Q1. One prime ideal cannot properly contain another, hence P = Q1. Thus P is one of the primes in the factorization of x*R. Since x*R is invertible, so is P.

An immediate corollary is that all nontrivial proper ideals are invertible, as they are products of prime ideals, which are invertible.

Now, let H be a fractional ideal with denominator d. Thus d*H is an ideal in R, with a unique decomposition. The ideal generated by d also has a unique decomposition. Since every prime ideal is invertible, divide through by the ideals of d*R to find a factorization of H. Some other factorization, when multiplied by the prime ideals in d*R, would produce a different factorization for d*H, which is a contradiction. Therefore H is uniquely represented as a product of prime ideals or their inverses. Every nontrivial fractional ideal is invertible, and the fractional ideals form a group. Let H be an invertible ideal with inverse H′. We know that H times H′ generates 1. Write the following equation, with generators a and b drawn from H and H′ respectively.

∑(i=1 to n) aibi = 1

Let c be an arbitrary element of H. Multiply by 1, and c is the sum of aibic. Now the elements of H′, when multiplied by anything in H, wind up in R. Each factor cbi is in R. Therefore c is spanned by the elements a1a2a3 … an, as an R module. Since c was arbitrary, H is finitely generated.

In the last section we showed every ideal in a dedekind domain R is invertible. Since invertible ideals are finitely generated, all the ideals of R are finitely generated, and R is noetherian. Let R be an integral domain with fraction field F, and let S be a multiplicatively closed set in R. S inverse of R, denoted R/S, is a ring somewhere between R and F. Conveniently, R/S has the same fraction field F.

Let H be an ideal in R that is invertible, with H′ as inverse. Recall that H embeds in, and generates, the ideal H/S inside the ring R/S. Simply apply all the denominators in S to H and to R. The embedding is implemented via H → H/1.

In the last section we showed H′ is finitely generated. Assign it the generators x1 x2 x3 … xn. Remember, these generators live in F, and might not be contained in R.

These generators span a finitely generated R/S module in F, which is a fractional ideal in the ring R/S.

Let a/b be any element of H/S and multiply by one of the generators, say x1. Since ax1 is in R, the result lies in the ring R/S. This holds for all generators times all elements in H/S, and all sums thereof, hence H/S times H′/S is contained in R/S.

H times H′ spans 1 in R. Therefore H/S times H′/S spans 1 in R/S, and H/S is invertible. An ideal that is invertible remains invertible in the fraction ring. A circle of implications starts with a dedekind domain. This is noetherian and locally a dvr. This characterization in turn leads to another, that all ideals are invertible. This in turn implies all prime ideals are invertible, which implies a dedekind domain, and we're back where we started. Thus there are several equivalent definitions for a dedekind domain. Let's get started.

We already showed the ideals of a dedekind domain are finitely generated, whence it is noetherian.
We only need show each localization is a dvr.
Let R be dedekind and let P be one of its prime/maximal ideals.
Let S be the localization of R about P.
In other words, S consists of fractions with numerators in R,
and denominators not contained in P.
Let Q be the localization of P in S.
Localization kills off all prime ideals except P → Q.
Prime factorization persists into S,
thus every ideal is a power of Q.
Ideals are linearly ordered,
hence S is a valuation ring.
Furthermore, the descending chain of ideals corresponds to the positive integers, the powers of Q.
This gives S a valuation group of **Z**,
hence S is
a dvr.
Remember that a dvr is a pid, which is dedekind, hence S is dedekind.

Next let R be noetherian and locally a valuation ring. Since R is noetherian, the localization of R about P is also noetherian. (An ascending chain of ideals in the localization would pull back to an ascending chain of ideals in R.) A noetherian valuation ring is a dvr. Thus every localization about a prime ideal produces a dvr and a pid.

Let H be a nonzero proper ideal of R, and let H′ be the elements of F that drive H into R. Recall that H′ is an R submodule of F. If H is invertible, then H′ is indeed the inverse of H.

Suppose H is not invertible. Thus H*H′ is a proper ideal of R. Embed HH′ in a maximal ideal P. Of course a maximal ideal is prime, which is why I'm calling it P.

Localize about P. In other words, bring in denominators from S, where S = R-P. Now R/S is a dvr, and a pid. Its maximal ideal is P/S, which I will call Q.

Remember that HH′ lies in P, and since H′ contains 1, H also lies in P. Since H lies in P, H/S lies in Q, and is principal. Let x generate H/S, where x is an element of H.

At this point we need the fact that R is noetherian. Let y1 y2 y3 … yn generate H in R. Bring in a denominator of 1, and these generators embed in Q. For each generator yi, write the following.

yi/1 = x * ai/si (x generates the ideal)

yi/x = ai/si

Let t be the product over si, an element in R-P. Within the fraction field F, consider t/x times yi. This is the same as t times yi/x. Since yi/x is the same as ai/si, multiplying by t gives something in R. Thus t/x times yi is in R. This holds for each yi, so t/x drives H into R, and t/x is a member of H′.

Multiply t/x times x. This is part of HH′. Therefore t lies in P. Yet we know t is in R-P. This is a contradiction, hence HH′ is all of R, and H is invertible. Every ideal of R is invertible.

If ideals are invertible then the prime ideals are invertible, so assume the prime ideals of R are invertible and move forward.

An invertible ideal is finitely generated, and if the prime ideals are finitely generated, then R is noetherian.

Finally complete the circle; R is noetherian and its maximal ideals are invertible. Recall the definition of a dedekind domain. Every ideal is a unique product of prime ideals. This has always been our assumption, but now we have to prove it. That will make all these definitions equivalent.

Let H be any proper, nonzero ideal of R. Pull out the factors of H, one by one, where each factor is a maximal ideal of R. Multiply H by M′, since each maximal ideal M is invertible. That's how we "pull out" the factor of M. Repeat this process until you are left with a maximal ideal. Thus H is a product of maximal ideals. Before we explore this process in detail, let's prove the factorization is unique.

Let H be the product of maximal ideals in two different ways. If the maximal ideal M is common to both factorizations, multiply through by M′. This effectively cancels M from both sides. Keep doing this, until the ideals in the first factorization are completely different from the ideals in the second.

Let the factors on the right be the maximal ideals that we pulled out of H. The factors on the left are other maximal ideals, or even other prime ideals that are not maximal. Let P be an ideal in this factorization. Now the product of the ideals on the right gives an ideal that is contained in P. Since P is prime, one of the ideals on the right is contained in P. Those ideals are maximal, hence P equals one of the maximal ideals on the right. This is a contradiction, hence the factorization of H into prime ideals is unique. We only need show that a factorization exists.

It is enough to find one factor M of H, and prove that HM′ is larger than H. Pull a maximal ideal out of H and the resulting ideal is larger. If this can always be done, then pull out one factor after another, and multiply H by M inverse at each step. This builds an ascending chain of ideals. Since R is noetherian the chain cannot continue forever. At some point H itself becomes a maximal ideal, and the process terminates. This establishes a factorization for H.

Let H be a nonzero ideal of R that is not maximal. Embed H in a maximal ideal M, and let M′ be the inverse of M. Note that M′ includes all of R, plus some fractions not found in R.

Let J = M′H. Since H is contained in M, M′H is contained in R. This makes J an ideal.

If J = R, write H = HR = HM′M = JM = RM = M. We already said H is properly contained in M, so this is a contradiction. Therefore J is a proper ideal of R.

Since M′ contains 1, M′H contains H, and J is a proper ideal that contains H. If J is larger than H we are done. That's all we need to finish the proof, and R becomes a dedekind domain. Suppose J = H and derive a contradiction.

Since M′H = H, multiply through by M, giving H = MH. Multiplying H by the maximal ideal that contains it doesn't change H at all.

Localize R about M, and recall that localization and product commute. Thus MH still equals H. Since R is noetherian, its localization is noetherian, and H is finitely generated. Apply nakiama's lemma, and H = 0. This is a contradiction, hence R is dedekind.

Here is a summary of equivalent conditions for an integral domain R.

- R is dedekind
- R is noetherian and locally a dvr
- The ideals of R are invertible
- The prime ideals of R are invertible
- R is noetherian and the maximal ideals are invertible

This section looks at the converse.
If R is local and integrally closed, with one prime ideal,
it is a valuation ring.
And since R is noetherian,
the valuation ring becomes a pid, and a dvr with valuation group **Z**.
So the two characterizations of a noetherian ring are equivalent:
valuation ring ⇔ integrally closed and one prime ideal.

This is a theorem about valuation rings, and you'd expect to find it under valuation rings, but the proof involves invertible fractional ideals, which is why it is here. In fact, valuation rings and dedekind domains are interrelated, and it's hard to talk about one without the other.

Let R be integrally closed with one prime ideal M. I'm calling it M because R has to have a maximal ideal, which is prime, so that is the one prime ideal.

Let F be the fraction field of R, and let H be a nonzero fractional ideal of R, and let S be the set of elements in F that drive H into itself. In other words, x ∈ S iff x*H ⊆H. Verify that S is a fractional ideal in F.

Since 1*H = H, S contains 1, and S contains R.

Since H is a fractional ideal, let dH lie in R. Here d is the common denominator of H. Let y be an element of H, and let x be an element of S. Since xy is in H, dxy is in R. Put another way, dy times x lies in R, for every x in S. Multiply dy by the denominator of y and call the result z. Now z drives S into R. Therefore S is a fractional ideal.

Multiplication by z implements an R module isomorphism, hence S is isomorphic to some ideal I in R. Remember that S contains R, thus zR is a submodule of I, or if you prefer, an ideal inside I. Select an x in S and do the same for the submodule in S generated by 1 and x. This becomes an ideal inside I, generated by z and zx. A larger submodule, generated by 1, x, and x2, becomes a larger ideal. Continue this process through the powers of x. If each xn is new, not spanned by the lower powers of x, then the result is an infinite ascending chain of ideals in R. Since R is noetherian, this is impossible. The submodules of S cannot ascend forever, because their isomorphic images in I cannot ascend forever. Therefore some power of x is spanned, as an R module, by the lower powers of x. This makes x integral over R, and since R is integrally closed, x is already in R. Therefore S = R. The elements of F that drive H into itself, for any fractional ideal H, belong to R, and that's the end of it.

Next, let J be an ideal, and consider J′, the elements of F that drive J into R. Of course J′ includes R. It may or may not include more than R.

Let S be the set of ideals J in R such that J′ properly contains R. The "inverse" of J includes fractions outside of R. If x is a nonzero element of M, the principal ideal generated by x belongs to S. The inverse of this ideal is generated by 1/x, which lies outside of R. All the proper principal ideals are contained in S.

Start with the ideal generated by x, and find a larger ideal in S (not necessarily principal), then a larger one, then a larger one, and so on, until the process stops, which it must, since R is noetherian. Let J be the maximal ideal in S containing x. Note that J is not all of R, since R is not in S at all. We want to show J is prime. Since there is but one prime ideal, that will prove J = M.

Let a*b lie in J, with a and b not in J, and let c be a fraction in J′ that is outside of R. Let a and J generate a larger ideal K. Note that cb drives K into R. Since J is maximal, the ideal K is not in S. That means K′ includes nothing outside of R, and cb is in R. Therefore c drives the ideal L, generated by b and J, into R. L is larger than J, and part of the set S, which is a contradiction. This makes J a prime ideal, and since there is but one prime ideal, J = M. The maximal ideal M belongs to S, and M′ includes fractions outside of R.

Let's use these facts to show M is invertible. MM′ is an ideal of R. If M is not invertible than this product becomes a proper ideal. R is a local ring, so MM′ is contained in M. We showed that precisely R drives an ideal into itself, so R drives M into M, and M′ is contained in R. Yet we just showed that M′ properly contains R. This is a contradiction, so MM′ = R, and M is an invertible ideal.

Since M is invertible every prime ideal is invertible, and that makes R a dedekind domain, as per the previous section. Every ideal is a power of M, ideals are linearly ordered, and R is a valuation ring.

That completes the proof.
When R is noetherian,
declaring it a valuation ring is the same as saying it is integrally closed with one prime ideal.
In either case it is a pid, and a dvr with valuation group **Z**.
An ideal is invertible iff it is a projective R module.
If you don't know what a projective module is,
don't worry about it.
This theorem isn't critical for an understanding of dedekind domains;
it's more of an interesting side note.
So you can skip ahead to the next section if you like.

Let H be an invertible ideal, (could be a fractional ideal), hence it is finitely generated. Let Y be a free R module whose rank is equal to the number of generators in H. If the generators of H are a1 through an, then a module homomorphism maps the basis e1 through en of Y onto these generators. The first coordinate e1 maps to a1, the second coordinate e2 maps to a2, and so on.

Let b1 through bn be elements of H inverse (some of them may be 0), such that the sum of aibi equals 1. Then build a map from H into Y as follows. For c in H, map c to the sum of cbiei. (Since cbi is in R, this map is well defined.) Verify that this map is indeed a module homomorphism.

Compose the two homomorphisms, from H into Y, and back onto H. The result is the sum of caibi, which is c. The sequence is split exact, H is a summand of Y, and H is projective.

Now for the converse. Let Y be a free module with a (possibly infinite) basis e1 e2 …, and let Y map onto a projective module H, that happens to be a fractional ideal in the ring R. I'll call this a projective fractional ideal.

To get things started, map Y onto H by mapping each generator onto a different element of H. Surely fewer generators would suffice, but this mapping works. Map e1 onto a1, e2 onto a2, and so on.

The sequence is split exact, so a reverse homomorphism v carries H back into Y. If the rank of Y is infinite, v maps H into the direct sum, rather than the direct product. That's the definition of a free R module.

Compose v with the projection from Y onto ej to give a homomorphism gj(H) into R. This is basically the jth component of v(H) in Y.

Let x be any one of the generators ai in H. Let wj = gj(x). These are the pieces of v(x) in Y.

Let c be any element of H. By the symmetry of module homomorphisms on H, described in an earlier section, cwj = xgj(c).

In the fraction field F, c times (wj/x) = gj(c), which is an element of R. This holds for every c in H. Therefore wj/x drives H into R, and lies in H inverse. With luck, the set wj/x, as x ranges over the elements of H and j ranges over the nonzero components of v(x), will generate H inverse. We only need show these generators span 1.

Select any nonzero c in H and let S be the finite set of indices for which gj(c) is nonzero. Thus v(c) in Y is the finite sum of gj(c)*ej, for all j ∈ S.

Remember that gj(c) is cwj/x. Make this substitution and get the following sum.

v(c) = ∑(j ∈ S) cejwj/x

Apply the ring homomorphism forward from Y into H. This is where we use split exact. Anything in R, times e[j], is multiplied by a[j], and v(c) becomes c. This gives the following.

c = ∑(j ∈ S) cajwj/x

Cancel c on both sides, and a linear combination of the generators wj/x is equal to 1. The submodule spanned by these generators is indeed the inverse of H. Therefore, a fractional ideal is invertible iff it is projective. The following conditions on an integral domain R, with a fraction field F, are equivalent.

- R is dedekind. That is, every nonzero proper ideal in R is uniquely a product of finitely many prime ideals.
- Every nonzero proper fractional ideal in R is uniquely a product of finitely many prime ideals or their inverses.
- R is noetherian and every nonzero maximal ideal is invertible.
- Every nonzero prime ideal in R is invertible.
- Every nonzero ideal in R is invertible.
- Every nonzero fractional ideal of R is invertible.
- Every ideal in R is projective.
- Every fractional ideal of R is projective.
- R is noetherian and integrally closed, and every nonzero prime ideal is maximal.
- R is noetherian, and each localization RP is a discrete valuation ring.
- R is noetherian, and every nonzero prime ideal is maximal, and every nonzero primary ideal is a prime power.

An earlier theorem takes care of 1 → 2, and 2 → 1 is obvious.

An earlier theorem takes care of: 1 → 10 → 5 → 4 → 3 → 1, so these are all equivalent.

Since you can always flip the exponents on the prime factors of a fractional ideal, 2 → 6. And 6 → 5 is trivial, so 6 is brought into the fold.

Invertible is the same as projective, so 8 ⇔ 6 and 7 ⇔ 5.

Let's show that 5 implies 9. Every ideal is invertible, and finitely generated, hence R is noetherian. Prime ideals are maximal, so we only need show R is integrally closed. If the fraction x is integral over R, then R[x] is a finitely generated R module inside F. Thus R[x] is a fractional ideal, and within a dedekind domain, every fractional ideal is invertible.

Consider R[x]*R[x], the product of the two fractional ideals, and remember that R[x] contains 1. The product is simply R[x]. If R[x]′ is the inverse of R[x], then write the following.

R[x] = R*R[x] = R[x]′*R[x]*R[x] = R[x]′*R[x] = R

This shows x lies in R, and R is integrally closed.

Show 9 → 10, and that will make 9 equivalent to the others. Let P be a prime ideal and let RP be the localization of R about P. An ascending chain in RP pulls back to an ascending chain in R, therefore RP is noetherian. The property of being integrally closed also survives localization, so RP is integrally closed.

Prime and maximal ideals are synonymous, so there are no prime ideals (other than 0) properly contained in P. Apply prime correspondence, and there are no prime ideals of the ring RP between 0 and the maximal ideal PP. In other words, RP has one nonzero prime ideal. This satisfies the conditions of an earlier theorem, hence every RP is a valuation ring, and a dvr. That brings 9 into the fold.

Condition 11 deals with primary ideals and laskerian rings. If you are not familiar with this topic, you can skip ahead to the next section.

Let R be a noetherian integral domain where all nonzero prime ideals are maximal. If R were integrally closed it would be dedekind by condition 9. Instead, we are given condition 11: every nonzero primary ideal is a prime power.

Since R is noetherian it is laskerian. Let J be a nonzero proper ideal and write it as the finite intersection of primary ideals. Since all primes are maximal, the intersection equals the product. Each primary ideal is a prime power, hence J is a finite product of prime ideals.

Suppose the representation is not unique. Some other product yields J, hence some other intersection yields J. By first uniqueness, the prime ideals used to build J have to be the same. Since these primes are maximal they are all isolated in σ(J). Apply second uniqueness, and the primary ideals are fixed. The representation of J is unique, and R is dedekind.

Now for 1 → 11. Let H be a P primary ideal in a dedekind domain R. H is a finite product of prime ideals. The radical of this product is the radical of the intersection. This in turn is the intersection of the radicals. As a given, rad(H) = P. This must equal the intersection of all the primes that divide H. Thus P lies in each prime factor of H, and since P is maximal, H is a power of P. A P primary ideal is a power of P.

That's condition 11, but as a bonus, let's prove the converse. Let H be a power of P. Since P is maximal, and H is a power of P, H is P primary. Therefore H is P primary iff H is a power of P. If R is dedekind and S is a multiplicatively closed set, R/S is dedekind. Let H/S be a proper ideal in R/S. Since H is invertible in R, H/S is invertible in R/S. All ideals are invertible, and by condition 5, R/S is dedekind. The prime ideals of R/S correspond to the primes of R that miss S, thus R/S is a smaller dedekind domain. If R is a dedekind domain, the following conditions are equivalent.

- R is a ufd.
- All maximal ideals are principal.
- All fractional ideals are principal.
- R is a pid.
- Every finitely generated torsion free R module is free.

Note that 3 → 4 → 1 is straightforward.

To show 1 → 2, select a minimal set of generators for a maximal ideal M. Let x be one of these generators, and write x as a product of irreducible elements. Since M is a prime ideal it contains at least one of these factors. Replace x with the prime factor that belongs to M. Do this for every generator in the set.

In a ufd, irreducible and prime are synonymous, and prime elements generate prime ideals. If M includes the generators x and y, then the prime ideals generated by x and y belong to M. However, there are no prime ideals properly contained in M, so x and y both generate M. We don't need both of them. If the set of generators is minimal, it consists of one generator, and M is principal. That completes 1 → 2.

Prove 2 → 3, and the first 4 criteria become equivalent. Assume the maximal ideals are principal. Each fractional ideal is a product of maximal ideals and their inverses, and the inverse of a principal ideal is principal. Thus all fractional ideals are principal, and 1 through 4 are equivalent.

A finitely generated module over a pid is well characterized. In particular, a torsion free module becomes free, and that completes 4 → 5.

For the converse, assume finitely generated torsion free modules are free. Let H be any ideal in R. Remember that R is noetherian, and an integral domain. H is finitely generated and torsion free, hence it is free. Suppose H has rank > 1. A free R module of rank 2 now embeds in R, which is a free R module of rank 1. This is a contradiction, hence H is principal and R is a pid. Let R be a dedekind domain with one prime ideal P. All ideals are powers of P, and are linearly ordered by containment. This makes R a valuation ring. The valuation group is isomorphic to the integers, as dictated by the powers of P. This makes R a dvr, and a pid.

Now let R be a dedekind domain with finitely many prime ideals. Since prime ideals do not contain each other, ideals are not linearly ordered, and R is not a valuation ring, or a dvr. However, R is still a pid.

By criteria 2 above, it is enough to show each prime ideal is principal.

Let H be the sum of any two prime ideals. If this does not span 1 then drive H up to a maximal ideal M, which contains our two prime ideals. M contains two different maximal ideals, hence one of them is not maximal after all. This is a contradiction. Prime ideals are pairwise coprime, and if J is the product of these prime ideals, the chinese remainder theorem tells us R is isomorphic to the direct product of fields R/Pi for each prime ideal Pi.

Generalizing the above, let H be the sum of two prime powers. If this does not span 1 then a maximal ideal M contains both prime powers, and since M is prime, M contains both prime ideals. Once again M contains two maximal ideals, a contradiction. If J is the product of prime powers, R/J is isomorphic to the direct product of the quotient rings R mod these prime powers.

Let P be one of the aforementioned prime ideals and let c lie in P-P2. Apply the chinese remainder theorem to P2 and all other prime ideals. Let g = c mod P2, and 1 mod all other primes. If g lies in some other prime Q then g becomes 0 in the field R/Q, which is a contradiction. Thus g, and the ideal generated by g, does not lie in Q or any power of Q. If Q is part of the unique factorization of g*R, then g would lie in Q. Thus g generates a power of P.

If g were a unit it would be invertible mod P, yet it is 0 mod P. Thus g*R is a positive power of P, at least P1. If g*R is P2 or higher, than g becomes 0 mod P2. Yet g = c mod P2, which is nonzero. Therefore g*R = P, and P is principal. This holds for every prime ideal P, and R is a pid. If H and J are ideals, H contains J iff H divides into J. The reverse implication is clear; if GH = J then J is contained in G and in H. So let's prove the forward direction.

Let P be any prime ideal dividing H. Localize about P, creating a dvr. If S is the complement of P, the new ring is denoted RP, or R/S, as you prefer.

Let Q correspond to P. Since product and localization commute, the image of Pn is Qn. Let U be the preimage of Qn, which is at least Pn. Factor U into prime ideals in R, and inject these prime ideals into R/S. Primes other than P become the entire ring, hence the image of U is Qj, where Pj divides U. Since the image is Qn, Pn divides U. Therefore U is at most Pn, and could be smaller if other prime ideals are involved. Yet U contains Pn, hence U = Pn. Therefore Pn is a saturated ideal.

Inject H into R/S, giving an ideal H/S. Once again, H/S is the product of the images of the prime powers that build H. Most of these extensions are the entire ring R/S. The only extension that survives is Qn, which equals H/S. Therefore H = Pn time some other prime ideals outside of P.

J is contained in H, thus J/S lies in H/S. J/S is Qn or a higher power of Q. By the above, J has at least n powers of P in its factorization. This holds for every P dividing H, thus H divides J. To contain is to divide.

As a corollary, finitely many ideals contain a given ideal J. These are the ideals that have the primes of J in their factorization, to the same exponents or less. The number of factors equals the product of the successors of the exponents on the primes of J. This may remind you of the corresponding function in number theory.

The intersection of two ideals G and H is the largest ideal J that is contained in both G and H, which is the largest ideal that is divisible by both G and H. In other words, J = lcm(G,H). To find Pe in J, find Pm in the factorization of G and Pn in the factorization of H, and set e to the larger of m and n.

The sum of two ideals G and H is the smallest ideal J that contains both G and H, which is the smallest ideal that divides both G and H. In other words, J = gcd(G,H). To find Pe in J, find Pm in the factorization of G and Pn in the factorization of H, and set e to the smaller of m and n.

This can be generalized to the sum or intersection of a finite set of ideals.

Suppose all the powers of an ideal H contain an ideal J. The powers of H divide J, and if H is divisible by P, arbitrarily high powers of P divide J. Yet the factorization of J presents a specific power of P. This is a contradiction, hence the powers of H collectively contain only 0.

In an earlier section I mentioned that the quotient of a dedekind domain is dedekind, with the caveat that it might not be an integral domain. In fact it is a pid.

Let H be a proper nonzero ideal in a dedekind domain R, and let S be the quotient ring R/H. The ideals containing H become the ideals in S. These are the ideals that divide H, as described above. S has finitely many ideals, and is noetherian and artinian.

The prime ideals of R that participate in the factorization of H correspond to the prime ideals of S. The factorization of an ideal in S is determined by the factorization of its preimage in R, and is unique. S is dedekind.

There are finitely manyh primes in S, so apply the proof in the previous section, and S becomes a pid. Every ideal in a dedekind domain can be generated using one or two generators.

Given any ideal H, take a nonzero element x in H, and consider the image of H in the quotient ring R/{x}. As shown in the previous section, the quotient ring is a pid. The image of H is principal, generated by y. In other words, y generates all the cosets of {x} in H. Therefore x and y generate H. Every dedekind domain R has a class group that indicates how far away R is from being a pid. The class group of a pid is trivial.

Since multiplication of ideals is associative, the ideals of R form a monoid under multiplication. Bring in the fractional ideals to round out the group. Since multiplication in R is commutative, the group is abelian.

The group is free, with maximal ideals acting as generators.
In fact it is isomorphic to **Z**n, where n is the number of maximal ideals.
If n is an infinite cardinal, we're talking about the direct sum, not the direct product.
That's because each ideal is a finite product of prime ideals.

The product of principal fractional ideals is principal,
and the inverse of a principal fractional ideal is principal.
Therefore the principal fractional ideals form a subgroup inside the group of fractional ideals.
The subgroup of a free group is free,
so this principal subgroup also looks like parallel copies of **Z**,
though these generators need not correspond with the generators of the larger group.

The quotient, ideals mod the principal ideals, gives the class group. The class number is the size of the class group. As mentioned earlier, a pid has a trivial class group, with class number equal to 1. On the other hand, an ideal that is not principal becomes a nontrivial element in the class group.

Two ideals, or fractional ideals, represent the same element in the class group if their quotient is principal. If H and J are ideals in R, and u and v are elements of R, write H/J = {v/u}, or {u}H = {v}J.

Every coset in the class group is represented by some fractional ideal, so multiply through by the common denominator, and find another cosrep that is an ideal in R. The ideals represent the entire class group. If H is not principal, there is some other ideal J representing the inverse of H, whereupon HJ is principal.

Let a(R) be a ring automorphism on R.
Conjugation is a typical example.
Extend this to the fraction field, and then to fractional ideals.
Finally, extend the automorphism to the group of ideals under multiplication.
Since principal ideals map to principal ideals, the automorphism extends to the class group.
The index of a subgroup H in a group G is the number of cosets of H in G.
It's no different with rings.
The index of an ideal H in a dedekind domain R is the cardinality of R/H.
Let the index of the 0 ideal be 0.
This is quite nice in **Z**, where the index is the absolute value of the generator.

If H contains J, represent the elements of R/J as cosets of J in H cross cosets of H in R. Thus the index of J is the index of H times the index of J in H.

The index of a product of ideals is the product of their indices. This is true if any ideal is 0, and it carries along if an ideal is all of R, so assume ideals are nonzero and proper. Demonstrate the relationship for H*M, where M is a maximal ideal. If this holds then the index of GH is the product of the indexes of every maximal ideal (including multiplicities) that contributes to GH. The same number can be realized by considering the prime factors of G in aggregate, then the prime factors of H. Thus the index of GH equals the index of G times the index of H.

Let R/M be the field K. Tensor H with R/M, as R modules, giving H/MH, which is a K vector space. Since H cannot equal MH, the dimension is positive. If the dimension is greater than 1, let x generate a one dimensional subspace inside the larger K vector space. Remember that x ultimately comes from H. Let W be the ideal spanned by MH and x. Now W is an ideal properly between H and MH, which is impossible. Therefore H/MH has dimension 1, and is isomorphic to K.

The index of MH is |K| times the index of H, where |K| is the index of M. That completes the proof.

When each R/M has a finite index, each ideal has a finite index.

If an ideal H has index n, let the inverse of H have index 1/n. Do this for all ideals, and build a group homomorphism from the nonzero fractional ideals into the positive rational numbers. The index of a fractional ideal has no meaning in terms of cosets or cardinality, but it is often convenient. Assume there is a positive integer l such that, for any ideal H, there is a nonzero x in H with |x| ≤ l×|H|. Here |H| is the index of H, and |x| is the index of the ideal generated by x.

It is perhaps simpler to say there is an x in H such that the index of x in H ≤ l.

Let the ideal H represent the inverse of an arbitrary member of the class group. (The entire class group can be represented by ideals.) Select x as above, and let G be the ideal x/H. To contain is to divide, so G is indeed an ideal. Also, G and H are inverses in the class group. In other words, G represents the designated member of the class group.

Write GH = x and take indexes. The index of G is bounded by l.

Every element of the class group is represented by an ideal with index bounded by l. If the number of ideals with index ≤ l is finite, then the class group is finite.

Assume there are finitely many prime ideals with index ≤ l. Each of these, when raised to a certain power, attains an index that exceeds l. Exponents on primes are bounded, and the number of ideals with index ≤ l is finite, and the class group is finite.

Each ideal with index ≤ l is generated by primes with index ≤ l. Thus the primes with index ≤ l generate the class group.