Difference between revisions of "Dedekind domain"

m (I think an integral domain is usually defined to be commutative)
(Invertibility of ideals)
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* Dedekind domains have unique prime factorizations of [[ideal]]s (but not necessarily of elements).
 
* Dedekind domains have unique prime factorizations of [[ideal]]s (but not necessarily of elements).
* Ideals are invertible if we extend to [[fractional ideal]]s. Let <math>R</math> be a Dedekind domain with field of fractions <math>K</math>, and let <math>I</math> be any nonzero ideal of <math>R</math>. Then set <math>I^{-1}=\{a\in K\mid aI\subseteq R\}</math>. We call an ideal <math>I</math> '''invertible''' if <math>II^{-1}=R</math>. (Note that this is always a subset, but it is not always equal unless we are in a Dedekind domain.) 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.
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There are also various properties of [[homological algebra|homological]] importance that Dedekind domains satisfy.
 
There are also various properties of [[homological algebra|homological]] importance that Dedekind domains satisfy.
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==Invertibility of Ideals==
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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.
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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.
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We will need the following lemmas.
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'''Lemma 1:''' Every nonzero integral ideal <math>J</math> of <math>R</math> contains a product of [[prime ideal]]s (counting <math>R</math> as the empty product).
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''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>
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'''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>.
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''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>.
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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>
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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
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<cmath>(\gamma\beta)I = \gamma(\beta I) \subseteq \gamma II^{-1} = \gamma A \subseteq R,</cmath>
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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>).
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Indeed, the map <math>f:x\mapsto \gamma x</math> is an <math>R</math>-linear map from <math>I^{-1}\to I^{-1}</math>. As <math>R</math> is noetherian, <math>I^{-1}</math> must be a finitely generated <math>R</math>-module. Indeed, for some nonzero <math>r\in R</math>, <math>rI^{-1}</math> must be an integral ideal of <math>R</math>, which is finitely generated by the definition of noetherian rings. But if <math>rI^{-1} = Ry_1+\cdots + Ry_m</math> then <math>I^{-1} = R(y_1/r)+\cdots+R(y_m/r)</math>, so <math>I^{-1}</math> is finitely generated as well. Now take <math>I^{-1} = Rx_1+\cdots +Rx_m</math>, and let <math>M_f</math> be the [[matrix]] representation of <math>f</math> with respect to <math>x_1,\ldots,x_m</math>. Then <math>M_f</math> is an <math>m\times m</math> matrix with coefficients in <math>R</math> and we have:
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<cmath>M_f
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\left(
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\begin{array}{c}
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x_1\\
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x_2\\
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\vdots\\
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x_m
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\end{array}
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\right)
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=
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\gamma
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\left(
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\begin{array}{c}
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x_1\\
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x_2\\
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\vdots\\
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x_m
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\end{array}
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\right),
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</cmath>
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and so <math>\gamma</math> is an [[eigenvalue]] of <math>M_f</math>. But then <math>\gamma</math> is a root of the [[characteristic polynomial]], <math>g(t) = |I_m(t)-M|</math> of <math>M_f</math>. But as <math>M_f</math> has all of its entries in <math>R</math>, <math>g(x)</math> is a monic polynomial in <math>R[x]</math>. Thus as <math>R</math> is integrally closed in <math>K</math>, <math>\gamma\in R</math>.
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This is a contradiction, and so we must have <math>A = II^{-1} = R</math>,  and so <math>I</math> is indeed invertible. <math>\blacksquare</math>
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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.
  
 
[[Category:Definition]]
 
[[Category:Definition]]
 
[[Category:Ring theory]]
 
[[Category:Ring theory]]
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[[Category:Algebraic number theory]]

Revision as of 00:58, 29 November 2010

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

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 that all fractional ideals of $R$ are invertible.

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

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