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). | ||
− | + | ||
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. | ||
+ | |||
+ | ==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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | '''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). | ||
+ | |||
+ | ''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>. | ||
+ | |||
+ | ''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>. | ||
+ | |||
+ | 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 | ||
+ | <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>). | ||
+ | |||
+ | 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: | ||
+ | <cmath>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), | ||
+ | </cmath> | ||
+ | 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>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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]] | ||
+ | [[Category:Algebraic number theory]] |
Revision as of 23:58, 28 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 that all fractional ideals of
are invertible.
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. Unknown error_msg) 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. Unknown error_msg) 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. Unknown error_msg) 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.