# Difference between revisions of "Dedekind domain"

(remove nonexistent category) |
m (→Invertibility of Ideals) |
||

Line 27: | Line 27: | ||

''Proof:'' Assume that this is not the case. Let <math>\mathcal S</math> be the collection of integral ideals of <math>R</math> not containing a product of prime ideals, so <math>\mathcal S</math> is nonempty and <math>R\not\in \mathcal S</math>. As <math>R</math> is noetherian, <math>\mathcal S</math> must have a maximal element, say <math>M</math>. Clearly <math>M</math> cannot be prime (otherwise it would contain itself), so there must be <math>x,y\in R</math> with <math>xy\in M</math> but <math>x,y\not\in M</math>. But then <math>M\subsetneq M+(x),M+(y)</math>, and so <math>M+(x)</math> and <math>M+(y)</math> contain products of prime ideals. But then <math>(M+(x))(M+(y)) = M+(xy)\subseteq M</math> also contains a product of prime ideals, contradicting the choice of <math>M</math>. <math>\square</math> | ''Proof:'' Assume that this is not the case. Let <math>\mathcal S</math> be the collection of integral ideals of <math>R</math> not containing a product of prime ideals, so <math>\mathcal S</math> is nonempty and <math>R\not\in \mathcal S</math>. As <math>R</math> is noetherian, <math>\mathcal S</math> must have a maximal element, say <math>M</math>. Clearly <math>M</math> cannot be prime (otherwise it would contain itself), so there must be <math>x,y\in R</math> with <math>xy\in M</math> but <math>x,y\not\in M</math>. But then <math>M\subsetneq M+(x),M+(y)</math>, and so <math>M+(x)</math> and <math>M+(y)</math> contain products of prime ideals. But then <math>(M+(x))(M+(y)) = M+(xy)\subseteq M</math> also contains a product of prime ideals, contradicting the choice of <math>M</math>. <math>\square</math> | ||

− | '''Lemma 2:''' For any proper integral ideal <math>J</math>, there is some <math>\gamma\in K\ | + | '''Lemma 2:''' For any proper integral ideal <math>J</math>, there is some <math>\gamma\in K\setminus R</math> for which <math>\gamma J\subseteq R</math>. |

− | ''Proof:'' Take any nonzero <math>a\in J</math>. By Lemma 1, <math>(a)</math> contains a product of prime ideals, say <math>(a)\supseteq P_1P_2\cdots P_n</math> with <math>n</math> minimal (i.e. <math>J</math> does not contain a product of <math>n-1</math> prime ideals). As <math>R\not\subseteq (a)</math>, <math>n\ge 1</math>. As <math>J</math> is a proper ideal, it must be contained in some maximal ideal, <math>P</math>. Since maximal ideals are prime in commutative rings, <math>P</math> is prime. But now <math>P_1P_2\cdots P_n\subseteq P</math>. Thus as <math>P</math> is prime, <math>P_i\subseteq P</math> for some <math>i</math> (if <math>P</math> is prime and <math>A,B</math> are ideals with <math>AB\subseteq P</math> then either <math>A\subseteq P</math> or <math>B\subseteq P</math>). But as <math>R</math> is a Dedekind domain, <math>P_i</math> must be maximal, so <math>P = P_i</math>. Now assume WLOG that <math>i = n</math>. By the minimality of <math>n</math>, <math>(P_1\cdots P_{n-1})\not\subseteq (a)</math>. Take any <math>b\in P_1\cdots P_{n-1}\ | + | ''Proof:'' Take any nonzero <math>a\in J</math>. By Lemma 1, <math>(a)</math> contains a product of prime ideals, say <math>(a)\supseteq P_1P_2\cdots P_n</math> with <math>n</math> minimal (i.e. <math>J</math> does not contain a product of <math>n-1</math> prime ideals). As <math>R\not\subseteq (a)</math>, <math>n\ge 1</math>. As <math>J</math> is a proper ideal, it must be contained in some maximal ideal, <math>P</math>. Since maximal ideals are prime in commutative rings, <math>P</math> is prime. But now <math>P_1P_2\cdots P_n\subseteq P</math>. Thus as <math>P</math> is prime, <math>P_i\subseteq P</math> for some <math>i</math> (if <math>P</math> is prime and <math>A,B</math> are ideals with <math>AB\subseteq P</math> then either <math>A\subseteq P</math> or <math>B\subseteq P</math>). But as <math>R</math> is a Dedekind domain, <math>P_i</math> must be maximal, so <math>P = P_i</math>. Now assume WLOG that <math>i = n</math>. By the minimality of <math>n</math>, <math>(P_1\cdots P_{n-1})\not\subseteq (a)</math>. Take any <math>b\in P_1\cdots P_{n-1}\setminus (a)</math> let <math>\gamma = b/a\in K</math>. We claim that this is the desired <math>\gamma</math>. |

First if <math>\gamma\in R</math> then <math>b = \gamma a\in (a)</math>, a contradiction, so <math>\gamma\not\in R</math>. Now for any <math>x\in J</math>, <math>bx\in P_1\cdots P_{n-1}J\subseteq P_1\cdots P_n\subseteq (a)</math>, and so <math>bx = ar</math> for some <math>r\in R</math>. But now <math>\gamma x = \frac{bx}{a} = r\in R</math>, and so <math>\gamma J\subseteq R</math>, as required. <math>\square</math> | First if <math>\gamma\in R</math> then <math>b = \gamma a\in (a)</math>, a contradiction, so <math>\gamma\not\in R</math>. Now for any <math>x\in J</math>, <math>bx\in P_1\cdots P_{n-1}J\subseteq P_1\cdots P_n\subseteq (a)</math>, and so <math>bx = ar</math> for some <math>r\in R</math>. But now <math>\gamma x = \frac{bx}{a} = r\in R</math>, and so <math>\gamma J\subseteq R</math>, as required. <math>\square</math> | ||

− | Now we return to the main proof. Assume that <math>A\ne R</math>. Then by Lemma 2, there is some <math>\gamma\in K\ | + | Now we return to the main proof. Assume that <math>A\ne R</math>. Then by Lemma 2, there is some <math>\gamma\in K\setminus R</math> for which <math>\gamma A\subseteq R</math>. By the definition of <math>I^{-1}</math>, for any <math>\beta\in I^{-1}</math> we have |

<cmath>(\gamma\beta)I = \gamma(\beta I) \subseteq \gamma II^{-1} = \gamma A \subseteq R,</cmath> | <cmath>(\gamma\beta)I = \gamma(\beta I) \subseteq \gamma II^{-1} = \gamma A \subseteq R,</cmath> | ||

and so <math>\gamma\beta\in I^{-1}</math>. It follows that <math>\gamma I^{-1}\subseteq I^{-1}</math>. We claim that this implies <math>\gamma\in R</math> (contradicting the choice of <math>\gamma</math>). | and so <math>\gamma\beta\in I^{-1}</math>. It follows that <math>\gamma I^{-1}\subseteq I^{-1}</math>. We claim that this implies <math>\gamma\in R</math> (contradicting the choice of <math>\gamma</math>). |

## Revision as of 16:30, 14 October 2017

A **Dedekind domain** is a integral domain satisfying the following properties:

- is a noetherian ring.
- Every prime ideal of is a maximal ideal.
- is integrally closed in its field of fractions.

Dedekind domains are very important in abstract algebra and number theory. For example, the ring of integers of any number field is a Dedekind domain.

There are several very nice properties of Dedekind domains:

- Dedekind domains have unique prime factorizations of ideals (but not necessarily of elements).

There are also various properties of homological importance that Dedekind domains satisfy.

## Invertibility of Ideals

Let be a Dedekind domain with field of fractions , and let be any nonzero fractional ideal of . We call **invertible** if there is a fractional ideal such that . We shall show the following

**Theorem:** All fractional ideals of are invertible.

*Proof:* Given any nonzero fractional ideal of define . is clearly an -module. Moreover, for any nonzero (such an alpha clearly exists, if for then ) we have by the definition of , and so must be a fractional ideal of . It follows that is a fractional ideal of as well, let . By definition, , and so is an integral ideal. We claim that in fact , and so is invertible.

We will need the following lemmas.

**Lemma 1:** Every nonzero integral ideal of contains a product of prime ideals (counting as the empty product).

*Proof:* Assume that this is not the case. Let be the collection of integral ideals of not containing a product of prime ideals, so is nonempty and . As is noetherian, must have a maximal element, say . Clearly cannot be prime (otherwise it would contain itself), so there must be with but . But then , and so and contain products of prime ideals. But then also contains a product of prime ideals, contradicting the choice of .

**Lemma 2:** For any proper integral ideal , there is some for which .

*Proof:* Take any nonzero . By Lemma 1, contains a product of prime ideals, say with minimal (i.e. does not contain a product of prime ideals). As , . As is a proper ideal, it must be contained in some maximal ideal, . Since maximal ideals are prime in commutative rings, is prime. But now . Thus as is prime, for some (if is prime and are ideals with then either or ). But as is a Dedekind domain, must be maximal, so . Now assume WLOG that . By the minimality of , . Take any let . We claim that this is the desired .

First if then , a contradiction, so . Now for any , , and so for some . But now , and so , as required.

Now we return to the main proof. Assume that . Then by Lemma 2, there is some for which . By the definition of , for any we have and so . It follows that . We claim that this implies (contradicting the choice of ).

Indeed, the map is an -linear map from . As is noetherian, must be a finitely generated -module. Indeed, for some nonzero , must be an integral ideal of , which is finitely generated by the definition of noetherian rings. But if then , so is finitely generated as well. Now take , and let be the matrix representation of with respect to . Then is an matrix with coefficients in and we have: and so is an eigenvalue of . But then is a root of the characteristic polynomial, of . But as has all of its entries in , is a monic polynomial in . Thus as is integrally closed in , .

This is a contradiction, and so we must have , and so is indeed invertible.

In fact, the converse is true as well: if all nonzero ideals are invertible, then is a Dedekind domain. This is sometimes used as a definition of Dedekind domains.

This result has a number of important applications.

First, as we clearly have for all fractional ideals , the set of fractional ideals, , of becomes an abelian group under multiplication. The identity element is and the existence of inverses is guaranteed by the above result. The set of principal ideal fractional ideals, forms a subgroup of this group. The quotient group is is then the ideal class group of , a object of great importance in algebraic number theory.

We may define divisibility of ideals in the "obvious" way. Namely for fractional ideals we say that iff there is some *integral* ideal for which . The above result gives a useful characterization of divisibility.

**Theorem:** For fractional ideals of , iff .

*Proof:* First, if , then there is some for which . But this gives . Conversely assume that . Consider the fractional ideal . We have . Thus is a fractional ideal and . It follows that is an integral ideal of . But now and so .

As an application of this, we get that and for all ideals and .