Difference between revisions of "Dedekind domain"

Latest revision as of 15:36, 14 October 2017

A Dedekind domain is a integral domain $R$ satisfying the following properties:

$R$ is a noetherian ring.
Every prime ideal of $R$ is a maximal ideal.
$R$ 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 $R$ be a Dedekind domain with field of fractions $K$ , and let $I$ be any nonzero fractional ideal of $R$ . We call $I$ invertible if there is a fractional ideal $I^{-1}$ such that $II^{-1}=R$ . We shall show the following

Theorem: All fractional ideals of $R$ are invertible.

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

We will need the following lemmas.

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

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

Lemma 2: For any proper integral ideal $J$ , there is some $\gamma\in K\setminus R$ for which $\gamma J\subseteq R$ .

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

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

Now we return to the main proof. Assume that $A\ne R$ . Then by Lemma 2, there is some $\gamma\in K\setminus R$ for which $\gamma A\subseteq R$ . By the definition of $I^{-1}$ , for any $\beta\in I^{-1}$ we have $(\gamma\beta)I = \gamma(\beta I) \subseteq \gamma II^{-1} = \gamma A \subseteq R,$ and so $\gamma\beta\in I^{-1}$ . It follows that $\gamma I^{-1}\subseteq I^{-1}$ . We claim that this implies $\gamma\in R$ (contradicting the choice of $\gamma$ ).

Indeed, the map $f:x\mapsto \gamma x$ is an $R$ -linear map from $I^{-1}\to I^{-1}$ . As $R$ is noetherian, $I^{-1}$ must be a finitely generated $R$ -module. Indeed, for some nonzero $r\in R$ , $rI^{-1}$ must be an integral ideal of $R$ , which is finitely generated by the definition of noetherian rings. But if $rI^{-1} = Ry_1+\cdots + Ry_m$ then $I^{-1} = R(y_1/r)+\cdots+R(y_m/r)$ , so $I^{-1}$ is finitely generated as well. Now take $I^{-1} = Rx_1+\cdots +Rx_m$ , and let $M_f$ be the matrix representation of $f$ with respect to $x_1,\ldots,x_m$ . Then $M_f$ is an $m\times m$ matrix with coefficients in $R$ and we have: $M_f \left( \begin{array}{c} x_1\\ x_2\\ \vdots\\ x_m \end{array} \right) = \gamma \left( \begin{array}{c} x_1\\ x_2\\ \vdots\\ x_m \end{array} \right),$ and so $\gamma$ is an eigenvalue of $M_f$ . But then $\gamma$ is a root of the characteristic polynomial, $g(t) = |I_m(t)-M|$ of $M_f$ . But as $M_f$ has all of its entries in $R$ , $g(x)$ is a monic polynomial in $R[x]$ . Thus as $R$ is integrally closed in $K$ , $\gamma\in R$ .

This is a contradiction, and so we must have $A = II^{-1} = R$ , and so $I$ is indeed invertible. $\blacksquare$

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

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

Theorem: For fractional ideals $A,B$ of $R$ , $A|B$ iff $B\subseteq A$ .

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

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

@@ Line 15: / Line 15: @@
 ==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 that all fractional ideals of <math>R</math> are invertible.
+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.
@@ Line 25: / 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>
-'''Lemma 2:''' For any proper integral ideal <math>J</math>, there is some <math>\gamma\in K\sm R</math> for which <math>\gamma J\subseteq R</math>.
+'''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}\sm (a)</math> let <math>\gamma = b/a\in K</math>. We claim that this is the desired <math>\gamma</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}\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>
-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\sm 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
+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>
 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>).
@@ Line 61: / Line 63: @@
 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:Ring theory]]
-[[Category:Algebraic number theory]]

Page

Toolbox

Search

Difference between revisions of "Dedekind domain"

Latest revision as of 15:36, 14 October 2017

Invertibility of Ideals