# Difference between revisions of "Dedekind domain"

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==Invertibility of Ideals== | ==Invertibility of Ideals== | ||

− | Let <math>R</math> be a Dedekind domain with field of fractions <math>K</math>, and let <math>I</math> be any nonzero [[fractional ideal]] of <math>R</math>. We call <math>I</math> '''invertible''' if there is a fractional ideal <math>I^{-1}</math> such that <math>II^{-1}=R</math>. We shall show | + | Let <math>R</math> be a Dedekind domain with field of fractions <math>K</math>, and let <math>I</math> be any nonzero [[fractional ideal]] of <math>R</math>. We call <math>I</math> '''invertible''' if there is a fractional ideal <math>I^{-1}</math> such that <math>II^{-1}=R</math>. We shall show the following |

− | Given any nonzero fractional ideal <math>I</math> of <math>R</math> define <math>I^{-1} = \{\beta\in K|\beta I\subseteq R\}</math>. <math>I^{-1}</math> is clearly an <math>R</math>-[[module]]. Moreover, for any nonzero <math>\alpha \in I\cap R</math> (such an alpha clearly exists, if <math>x/y\in I</math> for <math>x,y\in R</math> then <math>x\in I</math>) we have <math>\alpha I^{-1}\subseteq R</math> by the definition of <math>I^{-1}</math>, and so <math>I^{-1}</math> must be a fractional ideal of <math>R</math>. It follows that <math>II^{-1}</math> is a fractional ideal of <math>R</math> as well, let <math>II^{-1} = A</math>. By definition, <math>A\subseteq R</math>, and so <math>A</math> is an integral ideal. We claim that in fact <math>A = R</math>, and so <math>I</math> is invertible. | + | '''Theorem:''' All fractional ideals of <math>R</math> are invertible. |

+ | |||

+ | ''Proof:'' Given any nonzero fractional ideal <math>I</math> of <math>R</math> define <math>I^{-1} = \{\beta\in K|\beta I\subseteq R\}</math>. <math>I^{-1}</math> is clearly an <math>R</math>-[[module]]. Moreover, for any nonzero <math>\alpha \in I\cap R</math> (such an alpha clearly exists, if <math>x/y\in I</math> for <math>x,y\in R</math> then <math>x\in I</math>) we have <math>\alpha I^{-1}\subseteq R</math> by the definition of <math>I^{-1}</math>, and so <math>I^{-1}</math> must be a fractional ideal of <math>R</math>. It follows that <math>II^{-1}</math> is a fractional ideal of <math>R</math> as well, let <math>II^{-1} = A</math>. By definition, <math>A\subseteq R</math>, and so <math>A</math> is an integral ideal. We claim that in fact <math>A = R</math>, and so <math>I</math> is invertible. | ||

We will need the following lemmas. | We will need the following lemmas. | ||

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In fact, the converse is true as well: if all nonzero ideals are invertible, then <math>R</math> is a Dedekind domain. This is sometimes used as a definition of Dedekind domains. | In fact, the converse is true as well: if all nonzero ideals are invertible, then <math>R</math> 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 <math>RI = I</math> for all fractional ideals <math>I</math>, the set of fractional ideals, <math>G</math>, of <math>R</math> becomes an abelian group under multiplication. The identity element is <math>R</math> and the existence of inverses is guaranteed by the above result. The set of [[principal|principal ideal]] fractional ideals, <math>H</math> forms a subgroup of this group. The [[quotient group]] is <math>G/H</math> is then the [[ideal class group]] of <math>R</math>, a object of great importance in algebraic number theory. | ||

+ | |||

+ | We may define divisibility of ideals in the "obvious" way. Namely for fractional ideals <math>A,B</math> we say that <math>A|B</math> iff there is some ''integral'' ideal <math>C</math> for which <math>B = AC</math>. The above result gives a useful characterization of divisibility. | ||

+ | |||

+ | '''Theorem:''' For fractional ideals <math>A,B</math> of <math>R</math>, <math>A|B</math> iff <math>B\subseteq A</math>. | ||

+ | |||

+ | ''Proof:'' First, if <math>A|B</math>, then there is some <math>C\subseteq R</math> for which <math>B = AC</math>. But this gives <math>B = AC\subseteq AR = A</math>. Conversely assume that <math>B\subseteq A</math>. Consider the fractional ideal <math>A^{-1}</math>. We have <math>A^{-1}B\subseteq A^{-1}A = R</math>. Thus <math>A^{-1}B</math> is a fractional ideal and <math>A^{-1}B\subseteq R</math>. It follows that <math>A^{-1}B</math> is an integral ideal of <math>R</math>. But now <math>A(A^{-1}B) = AA^{-1}B = RB = B</math> and so <math>A|B</math>. <math>\blacksquare</math> | ||

+ | |||

+ | As an application of this, we get that <math>\gcd(I,J) = I+J</math> and <math>\text{lcm}(I,J) = I\cap J</math> for all ideals <math>I</math> and <math>J</math>. | ||

[[Category:Definition]] | [[Category:Definition]] | ||

[[Category:Ring theory]] | [[Category:Ring theory]] | ||

[[Category:Algebraic number theory]] | [[Category:Algebraic number theory]] |

## Revision as of 00:29, 29 November 2010

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 $\gamma\in K\sm R$ (Error compiling LaTeX. ! Undefined control sequence.) 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 $b\in P_1\cdots P_{n-1}\sm (a)$ (Error compiling LaTeX. ! Undefined control sequence.) 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 $\gamma\in K\sm R$ (Error compiling LaTeX. ! Undefined control sequence.) 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 .