Mixing
Standard argument for making the class group of a number field trivial
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Let $K$ be a number field and let $R=mathcal{O}_K$ be its ring of integers. Let $a_1, ldots, a_h$ be ideals of $R$ generating the class group $Cl(K)$. Let $S$ denote the set of valuations corresponding to prime ideals dividing $a_1cdots a_h$. Let
$$R_S = {a in K mid v(a)geq 0, v text{ non-Archimedean valuation on } K, v notin S }$$
Silverman (p.213 Arithmetic of Elliptic Curves) claims that $R_S$ has trivial class group. I’m sure this is a standard and simple fact, but I’m having trouble proving it.
We have an inclusion $R to R_S$ and a map $Cl(R) to Cl(R_S)$. Let $I$ be an ideal of $R_S$, we want to show it is principal. We have $I cap R equiv prod a_i^{e_i}$ in the class group so there exists principal ideal $P,P’$ of $R$ such that
$$P(Icap R) = P’ prod a_i^{e_i}$$
$$P(Icap R)R_S = P’ prod a_i^{e_i}R_S$$
Got stuck. Somehow the $a_iR_S$ become principal in $R_S$. I know
$$R=bigcaplimits_mathfrak{p} R_mathfrak{p}$$
$$R_S=bigcaplimits_{mathfrak{p} notin S} R_mathfrak{p}$$
Where $mathfrak{p}$ ranges over all maximal ideals of $R$
algebraic-number-theory
asked Sep 28 at 13:23
usr0192
1,150411
add a comment |
up vote
3
down vote
favorite
Let $K$ be a number field and let $R=mathcal{O}_K$ be its ring of integers. Let $a_1, ldots, a_h$ be ideals of $R$ generating the class group $Cl(K)$. Let $S$ denote the set of valuations corresponding to prime ideals dividing $a_1cdots a_h$. Let
$$R_S = {a in K mid v(a)geq 0, v text{ non-Archimedean valuation on } K, v notin S }$$
Silverman (p.213 Arithmetic of Elliptic Curves) claims that $R_S$ has trivial class group. I’m sure this is a standard and simple fact, but I’m having trouble proving it.
We have an inclusion $R to R_S$ and a map $Cl(R) to Cl(R_S)$. Let $I$ be an ideal of $R_S$, we want to show it is principal. We have $I cap R equiv prod a_i^{e_i}$ in the class group so there exists principal ideal $P,P’$ of $R$ such that
$$P(Icap R) = P’ prod a_i^{e_i}$$
$$P(Icap R)R_S = P’ prod a_i^{e_i}R_S$$
Got stuck. Somehow the $a_iR_S$ become principal in $R_S$. I know
$$R=bigcaplimits_mathfrak{p} R_mathfrak{p}$$
$$R_S=bigcaplimits_{mathfrak{p} notin S} R_mathfrak{p}$$
Where $mathfrak{p}$ ranges over all maximal ideals of $R$
algebraic-number-theory
asked Sep 28 at 13:23
usr0192
1,150411
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $K$ be a number field and let $R=mathcal{O}_K$ be its ring of integers. Let $a_1, ldots, a_h$ be ideals of $R$ generating the class group $Cl(K)$. Let $S$ denote the set of valuations corresponding to prime ideals dividing $a_1cdots a_h$. Let
$$R_S = {a in K mid v(a)geq 0, v text{ non-Archimedean valuation on } K, v notin S }$$
Silverman (p.213 Arithmetic of Elliptic Curves) claims that $R_S$ has trivial class group. I’m sure this is a standard and simple fact, but I’m having trouble proving it.
We have an inclusion $R to R_S$ and a map $Cl(R) to Cl(R_S)$. Let $I$ be an ideal of $R_S$, we want to show it is principal. We have $I cap R equiv prod a_i^{e_i}$ in the class group so there exists principal ideal $P,P’$ of $R$ such that
$$P(Icap R) = P’ prod a_i^{e_i}$$
$$P(Icap R)R_S = P’ prod a_i^{e_i}R_S$$
Got stuck. Somehow the $a_iR_S$ become principal in $R_S$. I know
$$R=bigcaplimits_mathfrak{p} R_mathfrak{p}$$
$$R_S=bigcaplimits_{mathfrak{p} notin S} R_mathfrak{p}$$
Where $mathfrak{p}$ ranges over all maximal ideals of $R$
algebraic-number-theory
asked Sep 28 at 13:23
usr0192
1,150411
Let $K$ be a number field and let $R=mathcal{O}_K$ be its ring of integers. Let $a_1, ldots, a_h$ be ideals of $R$ generating the class group $Cl(K)$. Let $S$ denote the set of valuations corresponding to prime ideals dividing $a_1cdots a_h$. Let
$$R_S = {a in K mid v(a)geq 0, v text{ non-Archimedean valuation on } K, v notin S }$$
Silverman (p.213 Arithmetic of Elliptic Curves) claims that $R_S$ has trivial class group. I’m sure this is a standard and simple fact, but I’m having trouble proving it.
We have an inclusion $R to R_S$ and a map $Cl(R) to Cl(R_S)$. Let $I$ be an ideal of $R_S$, we want to show it is principal. We have $I cap R equiv prod a_i^{e_i}$ in the class group so there exists principal ideal $P,P’$ of $R$ such that
$$P(Icap R) = P’ prod a_i^{e_i}$$
$$P(Icap R)R_S = P’ prod a_i^{e_i}R_S$$
Got stuck. Somehow the $a_iR_S$ become principal in $R_S$. I know
$$R=bigcaplimits_mathfrak{p} R_mathfrak{p}$$
$$R_S=bigcaplimits_{mathfrak{p} notin S} R_mathfrak{p}$$
Where $mathfrak{p}$ ranges over all maximal ideals of $R$
algebraic-number-theory
algebraic-number-theory
asked Sep 28 at 13:23
usr0192
1,150411
asked Sep 28 at 13:23
usr0192
1,150411
asked Sep 28 at 13:23
usr0192
1,150411
asked Sep 28 at 13:23
usr0192
1,150411
asked Sep 28 at 13:23
usr0192
1,150411
1,150411
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$
newcommand{O}{mathcal O}
newcommand{Cl}{mathrm{Cl}}
newcommand{p}{mathfrak{p}}
$
Let $K$ be a number field, and $a_1, ..., a_h$ be ideals of $O_K$ which are representatives of the ideal classes of $Cl(O_K)$ (in particular, $h$ is the class number of $K$).
Let $S$ be a finite set of places of $O_K$, containing all the prime factors of all $a_i$, as well as all infinite places.
Claim : The ring $O_{K, S}$ is a principal ideal domain.
Proof :
By proposition 23.2 b) and d), the ring $O_{K, S}$ is a Dedekind domain, and every ideal of $O_{K,S}$ is an extended ideal, i.e. of the form $JO_{K,S}$ for some ideal $J subset O_K$.
By definition of the ideals $a_i$, there is an element $x in K^{times}$ and an index $i$ such that
$$J = x a_i.$$
The ideal $a_i^h = y O_K$ is principal (where $y in O_K$), since the class group is annihilated by $h$. One sees that $y in a_i cap O_{K,S}^{times}$ (because $a_i^h subset a_i$ and the only primes dividing $y$ are in $S$), which implies that
$a_i O_{K,S} = (1)$
and finally
$$J O_{K,S} = x O_{K,S} subset O_{K,S}$$
is a principal ideal, as claimed.
Remarks:
1)
Instead of using proposition 23.2 above, one can first show that $O_{K,S}$ is equal to a localization $Sigma^{-1}O_K$, namely for the multiplicative set $Sigma$ of elements $a in O_K setminus {0}$ such that the only prime factors of $a O_K$ belong to $S$, that is
$$Sigma = bigcap_{p not in S} (O_K setminus p) = O_K setminus bigcup_{p notin S} p.$$
(There seems to be a typo on page 41 of this book. The "multiplicative" set they are considering doesn't even contain $1$).
Notice that we want to show the equality
$$left( bigcap_{p not in S} (O_K setminus p) right)^{-1} O_K = bigcap_{p not in S} ((O_K setminus p)^{-1} O_K),$$
which is not obvious.
$$ $$
A) We clearly have $Sigma^{-1} O_K subseteq O_{K,S}$. The reverse inclusion is not obvious: it uses the fact that the class group of $O_K$ is torsion (because finite) — see here. In fact the argument is similar as above: pick $x in O_{K,S}$. For every $p in S$ such that $v_{p}(x) < 0$, write $p^h = b_{p} O_K$ as an principal ideal. Then
$b := prodlimits_{v_{p}(x) < 0} b_{p}^{-v_{p}(x)}$ is an element of $O_K$ only divisible by primes in $S$, so that $b in Sigma$.
Finally, $bx in O_K$, so we conclude that $x in Sigma^{-1} O_K$ as desired.
[Here is a wrong proof that $O_{K,S} subset Sigma^{-1} O_K$ : if $x in O_{K,S}$ is non-zero, then fix $b_p in p^{-v_p(x)}$ for every prime $p in S$ such that $v_p(x)<0$. Then $b = prod_p b_p in Sigma$ and $xb in O_K$, so that $x in Sigma^{-1} O_K$. The problem is that $b$ may have prime factors outside ${ p mid a_i }$, so that $b in Sigma$ is not true].
See also here, or here about relation between overrings and localizations. Moreover, remark 2.1 here might be of interest (but there is no proof there).
Finally, more general statements are given in theorem 2 here, mostly proven in Fossum, The Divisor Class Group of a Krull Domain.
$$ $$
B) Once you know that $O_{K,S} cong Sigma^{-1}O_K$, you can apply proposition 3.11.1) in Atiyah, MacDonald, Introduction to Commutative Algebra, to conclude that any ideal of $O_{K,S}$ is indeed an extended ideal $J O_{K,S}$, where $J$ is an ideal of $O_K$. (Moreover, any localization of a Dedekind domain is still Dedekind, see here).
$$ $$
C) You can also look at prop. 4.19 here, but notice that they define $O_{K,S}$ as a localization.
$$ $$
2) Here we are considering the multiplicative set
$Sigma = O_K setminus bigcuplimits_{p in mathrm{Spec}(O_K) setminus S} p.$
If instead you consider $Sigma' = O_K setminus p$ for a single fixed prime $p$ of $O_K$, you also get that $(Sigma')^{-1}O_K = (O_K)_{p}$ is a PID : it is even a DVR.
This is a result similar to the claim above, but which is in fact quite unrelated, since $Sigma$ deals with the cofinite set $mathrm{Spec}(O_K) setminus S$ of primes, while $Sigma'$ deals with the finite set ${p}$...
answered yesterday
Watson
15.7k92869
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1 Answer
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1 Answer
1
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active
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active
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up vote
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$
newcommand{O}{mathcal O}
newcommand{Cl}{mathrm{Cl}}
newcommand{p}{mathfrak{p}}
$
Let $K$ be a number field, and $a_1, ..., a_h$ be ideals of $O_K$ which are representatives of the ideal classes of $Cl(O_K)$ (in particular, $h$ is the class number of $K$).
Let $S$ be a finite set of places of $O_K$, containing all the prime factors of all $a_i$, as well as all infinite places.
Claim : The ring $O_{K, S}$ is a principal ideal domain.
Proof :
By proposition 23.2 b) and d), the ring $O_{K, S}$ is a Dedekind domain, and every ideal of $O_{K,S}$ is an extended ideal, i.e. of the form $JO_{K,S}$ for some ideal $J subset O_K$.
By definition of the ideals $a_i$, there is an element $x in K^{times}$ and an index $i$ such that
$$J = x a_i.$$
The ideal $a_i^h = y O_K$ is principal (where $y in O_K$), since the class group is annihilated by $h$. One sees that $y in a_i cap O_{K,S}^{times}$ (because $a_i^h subset a_i$ and the only primes dividing $y$ are in $S$), which implies that
$a_i O_{K,S} = (1)$
and finally
$$J O_{K,S} = x O_{K,S} subset O_{K,S}$$
is a principal ideal, as claimed.
Remarks:
1)
Instead of using proposition 23.2 above, one can first show that $O_{K,S}$ is equal to a localization $Sigma^{-1}O_K$, namely for the multiplicative set $Sigma$ of elements $a in O_K setminus {0}$ such that the only prime factors of $a O_K$ belong to $S$, that is
$$Sigma = bigcap_{p not in S} (O_K setminus p) = O_K setminus bigcup_{p notin S} p.$$
(There seems to be a typo on page 41 of this book. The "multiplicative" set they are considering doesn't even contain $1$).
Notice that we want to show the equality
$$left( bigcap_{p not in S} (O_K setminus p) right)^{-1} O_K = bigcap_{p not in S} ((O_K setminus p)^{-1} O_K),$$
which is not obvious.
$$ $$
A) We clearly have $Sigma^{-1} O_K subseteq O_{K,S}$. The reverse inclusion is not obvious: it uses the fact that the class group of $O_K$ is torsion (because finite) — see here. In fact the argument is similar as above: pick $x in O_{K,S}$. For every $p in S$ such that $v_{p}(x) < 0$, write $p^h = b_{p} O_K$ as an principal ideal. Then
$b := prodlimits_{v_{p}(x) < 0} b_{p}^{-v_{p}(x)}$ is an element of $O_K$ only divisible by primes in $S$, so that $b in Sigma$.
Finally, $bx in O_K$, so we conclude that $x in Sigma^{-1} O_K$ as desired.
[Here is a wrong proof that $O_{K,S} subset Sigma^{-1} O_K$ : if $x in O_{K,S}$ is non-zero, then fix $b_p in p^{-v_p(x)}$ for every prime $p in S$ such that $v_p(x)<0$. Then $b = prod_p b_p in Sigma$ and $xb in O_K$, so that $x in Sigma^{-1} O_K$. The problem is that $b$ may have prime factors outside ${ p mid a_i }$, so that $b in Sigma$ is not true].
See also here, or here about relation between overrings and localizations. Moreover, remark 2.1 here might be of interest (but there is no proof there).
Finally, more general statements are given in theorem 2 here, mostly proven in Fossum, The Divisor Class Group of a Krull Domain.
$$ $$
B) Once you know that $O_{K,S} cong Sigma^{-1}O_K$, you can apply proposition 3.11.1) in Atiyah, MacDonald, Introduction to Commutative Algebra, to conclude that any ideal of $O_{K,S}$ is indeed an extended ideal $J O_{K,S}$, where $J$ is an ideal of $O_K$. (Moreover, any localization of a Dedekind domain is still Dedekind, see here).
$$ $$
C) You can also look at prop. 4.19 here, but notice that they define $O_{K,S}$ as a localization.
$$ $$
2) Here we are considering the multiplicative set
$Sigma = O_K setminus bigcuplimits_{p in mathrm{Spec}(O_K) setminus S} p.$
If instead you consider $Sigma' = O_K setminus p$ for a single fixed prime $p$ of $O_K$, you also get that $(Sigma')^{-1}O_K = (O_K)_{p}$ is a PID : it is even a DVR.
This is a result similar to the claim above, but which is in fact quite unrelated, since $Sigma$ deals with the cofinite set $mathrm{Spec}(O_K) setminus S$ of primes, while $Sigma'$ deals with the finite set ${p}$...
answered yesterday
Watson
15.7k92869
add a comment |
up vote
1
down vote
$
newcommand{O}{mathcal O}
newcommand{Cl}{mathrm{Cl}}
newcommand{p}{mathfrak{p}}
$
Let $K$ be a number field, and $a_1, ..., a_h$ be ideals of $O_K$ which are representatives of the ideal classes of $Cl(O_K)$ (in particular, $h$ is the class number of $K$).
Let $S$ be a finite set of places of $O_K$, containing all the prime factors of all $a_i$, as well as all infinite places.
Claim : The ring $O_{K, S}$ is a principal ideal domain.
Proof :
By proposition 23.2 b) and d), the ring $O_{K, S}$ is a Dedekind domain, and every ideal of $O_{K,S}$ is an extended ideal, i.e. of the form $JO_{K,S}$ for some ideal $J subset O_K$.
By definition of the ideals $a_i$, there is an element $x in K^{times}$ and an index $i$ such that
$$J = x a_i.$$
The ideal $a_i^h = y O_K$ is principal (where $y in O_K$), since the class group is annihilated by $h$. One sees that $y in a_i cap O_{K,S}^{times}$ (because $a_i^h subset a_i$ and the only primes dividing $y$ are in $S$), which implies that
$a_i O_{K,S} = (1)$
and finally
$$J O_{K,S} = x O_{K,S} subset O_{K,S}$$
is a principal ideal, as claimed.
Remarks:
1)
Instead of using proposition 23.2 above, one can first show that $O_{K,S}$ is equal to a localization $Sigma^{-1}O_K$, namely for the multiplicative set $Sigma$ of elements $a in O_K setminus {0}$ such that the only prime factors of $a O_K$ belong to $S$, that is
$$Sigma = bigcap_{p not in S} (O_K setminus p) = O_K setminus bigcup_{p notin S} p.$$
(There seems to be a typo on page 41 of this book. The "multiplicative" set they are considering doesn't even contain $1$).
Notice that we want to show the equality
$$left( bigcap_{p not in S} (O_K setminus p) right)^{-1} O_K = bigcap_{p not in S} ((O_K setminus p)^{-1} O_K),$$
which is not obvious.
$$ $$
A) We clearly have $Sigma^{-1} O_K subseteq O_{K,S}$. The reverse inclusion is not obvious: it uses the fact that the class group of $O_K$ is torsion (because finite) — see here. In fact the argument is similar as above: pick $x in O_{K,S}$. For every $p in S$ such that $v_{p}(x) < 0$, write $p^h = b_{p} O_K$ as an principal ideal. Then
$b := prodlimits_{v_{p}(x) < 0} b_{p}^{-v_{p}(x)}$ is an element of $O_K$ only divisible by primes in $S$, so that $b in Sigma$.
Finally, $bx in O_K$, so we conclude that $x in Sigma^{-1} O_K$ as desired.
[Here is a wrong proof that $O_{K,S} subset Sigma^{-1} O_K$ : if $x in O_{K,S}$ is non-zero, then fix $b_p in p^{-v_p(x)}$ for every prime $p in S$ such that $v_p(x)<0$. Then $b = prod_p b_p in Sigma$ and $xb in O_K$, so that $x in Sigma^{-1} O_K$. The problem is that $b$ may have prime factors outside ${ p mid a_i }$, so that $b in Sigma$ is not true].
See also here, or here about relation between overrings and localizations. Moreover, remark 2.1 here might be of interest (but there is no proof there).
Finally, more general statements are given in theorem 2 here, mostly proven in Fossum, The Divisor Class Group of a Krull Domain.
$$ $$
B) Once you know that $O_{K,S} cong Sigma^{-1}O_K$, you can apply proposition 3.11.1) in Atiyah, MacDonald, Introduction to Commutative Algebra, to conclude that any ideal of $O_{K,S}$ is indeed an extended ideal $J O_{K,S}$, where $J$ is an ideal of $O_K$. (Moreover, any localization of a Dedekind domain is still Dedekind, see here).
$$ $$
C) You can also look at prop. 4.19 here, but notice that they define $O_{K,S}$ as a localization.
$$ $$
2) Here we are considering the multiplicative set
$Sigma = O_K setminus bigcuplimits_{p in mathrm{Spec}(O_K) setminus S} p.$
If instead you consider $Sigma' = O_K setminus p$ for a single fixed prime $p$ of $O_K$, you also get that $(Sigma')^{-1}O_K = (O_K)_{p}$ is a PID : it is even a DVR.
This is a result similar to the claim above, but which is in fact quite unrelated, since $Sigma$ deals with the cofinite set $mathrm{Spec}(O_K) setminus S$ of primes, while $Sigma'$ deals with the finite set ${p}$...
answered yesterday
Watson
15.7k92869
add a comment |
up vote
1
down vote
up vote
1
down vote
$
newcommand{O}{mathcal O}
newcommand{Cl}{mathrm{Cl}}
newcommand{p}{mathfrak{p}}
$
Let $K$ be a number field, and $a_1, ..., a_h$ be ideals of $O_K$ which are representatives of the ideal classes of $Cl(O_K)$ (in particular, $h$ is the class number of $K$).
Let $S$ be a finite set of places of $O_K$, containing all the prime factors of all $a_i$, as well as all infinite places.
Claim : The ring $O_{K, S}$ is a principal ideal domain.
Proof :
By proposition 23.2 b) and d), the ring $O_{K, S}$ is a Dedekind domain, and every ideal of $O_{K,S}$ is an extended ideal, i.e. of the form $JO_{K,S}$ for some ideal $J subset O_K$.
By definition of the ideals $a_i$, there is an element $x in K^{times}$ and an index $i$ such that
$$J = x a_i.$$
The ideal $a_i^h = y O_K$ is principal (where $y in O_K$), since the class group is annihilated by $h$. One sees that $y in a_i cap O_{K,S}^{times}$ (because $a_i^h subset a_i$ and the only primes dividing $y$ are in $S$), which implies that
$a_i O_{K,S} = (1)$
and finally
$$J O_{K,S} = x O_{K,S} subset O_{K,S}$$
is a principal ideal, as claimed.
Remarks:
1)
Instead of using proposition 23.2 above, one can first show that $O_{K,S}$ is equal to a localization $Sigma^{-1}O_K$, namely for the multiplicative set $Sigma$ of elements $a in O_K setminus {0}$ such that the only prime factors of $a O_K$ belong to $S$, that is
$$Sigma = bigcap_{p not in S} (O_K setminus p) = O_K setminus bigcup_{p notin S} p.$$
(There seems to be a typo on page 41 of this book. The "multiplicative" set they are considering doesn't even contain $1$).
Notice that we want to show the equality
$$left( bigcap_{p not in S} (O_K setminus p) right)^{-1} O_K = bigcap_{p not in S} ((O_K setminus p)^{-1} O_K),$$
which is not obvious.
$$ $$
A) We clearly have $Sigma^{-1} O_K subseteq O_{K,S}$. The reverse inclusion is not obvious: it uses the fact that the class group of $O_K$ is torsion (because finite) — see here. In fact the argument is similar as above: pick $x in O_{K,S}$. For every $p in S$ such that $v_{p}(x) < 0$, write $p^h = b_{p} O_K$ as an principal ideal. Then
$b := prodlimits_{v_{p}(x) < 0} b_{p}^{-v_{p}(x)}$ is an element of $O_K$ only divisible by primes in $S$, so that $b in Sigma$.
Finally, $bx in O_K$, so we conclude that $x in Sigma^{-1} O_K$ as desired.
[Here is a wrong proof that $O_{K,S} subset Sigma^{-1} O_K$ : if $x in O_{K,S}$ is non-zero, then fix $b_p in p^{-v_p(x)}$ for every prime $p in S$ such that $v_p(x)<0$. Then $b = prod_p b_p in Sigma$ and $xb in O_K$, so that $x in Sigma^{-1} O_K$. The problem is that $b$ may have prime factors outside ${ p mid a_i }$, so that $b in Sigma$ is not true].
See also here, or here about relation between overrings and localizations. Moreover, remark 2.1 here might be of interest (but there is no proof there).
Finally, more general statements are given in theorem 2 here, mostly proven in Fossum, The Divisor Class Group of a Krull Domain.
$$ $$
B) Once you know that $O_{K,S} cong Sigma^{-1}O_K$, you can apply proposition 3.11.1) in Atiyah, MacDonald, Introduction to Commutative Algebra, to conclude that any ideal of $O_{K,S}$ is indeed an extended ideal $J O_{K,S}$, where $J$ is an ideal of $O_K$. (Moreover, any localization of a Dedekind domain is still Dedekind, see here).
$$ $$
C) You can also look at prop. 4.19 here, but notice that they define $O_{K,S}$ as a localization.
$$ $$
2) Here we are considering the multiplicative set
$Sigma = O_K setminus bigcuplimits_{p in mathrm{Spec}(O_K) setminus S} p.$
If instead you consider $Sigma' = O_K setminus p$ for a single fixed prime $p$ of $O_K$, you also get that $(Sigma')^{-1}O_K = (O_K)_{p}$ is a PID : it is even a DVR.
This is a result similar to the claim above, but which is in fact quite unrelated, since $Sigma$ deals with the cofinite set $mathrm{Spec}(O_K) setminus S$ of primes, while $Sigma'$ deals with the finite set ${p}$...
answered yesterday
Watson
15.7k92869
$
newcommand{O}{mathcal O}
newcommand{Cl}{mathrm{Cl}}
newcommand{p}{mathfrak{p}}
$
Let $K$ be a number field, and $a_1, ..., a_h$ be ideals of $O_K$ which are representatives of the ideal classes of $Cl(O_K)$ (in particular, $h$ is the class number of $K$).
Let $S$ be a finite set of places of $O_K$, containing all the prime factors of all $a_i$, as well as all infinite places.
Claim : The ring $O_{K, S}$ is a principal ideal domain.
Proof :
By proposition 23.2 b) and d), the ring $O_{K, S}$ is a Dedekind domain, and every ideal of $O_{K,S}$ is an extended ideal, i.e. of the form $JO_{K,S}$ for some ideal $J subset O_K$.
By definition of the ideals $a_i$, there is an element $x in K^{times}$ and an index $i$ such that
$$J = x a_i.$$
The ideal $a_i^h = y O_K$ is principal (where $y in O_K$), since the class group is annihilated by $h$. One sees that $y in a_i cap O_{K,S}^{times}$ (because $a_i^h subset a_i$ and the only primes dividing $y$ are in $S$), which implies that
$a_i O_{K,S} = (1)$
and finally
$$J O_{K,S} = x O_{K,S} subset O_{K,S}$$
is a principal ideal, as claimed.
Remarks:
1)
Instead of using proposition 23.2 above, one can first show that $O_{K,S}$ is equal to a localization $Sigma^{-1}O_K$, namely for the multiplicative set $Sigma$ of elements $a in O_K setminus {0}$ such that the only prime factors of $a O_K$ belong to $S$, that is
$$Sigma = bigcap_{p not in S} (O_K setminus p) = O_K setminus bigcup_{p notin S} p.$$
(There seems to be a typo on page 41 of this book. The "multiplicative" set they are considering doesn't even contain $1$).
Notice that we want to show the equality
$$left( bigcap_{p not in S} (O_K setminus p) right)^{-1} O_K = bigcap_{p not in S} ((O_K setminus p)^{-1} O_K),$$
which is not obvious.
$$ $$
A) We clearly have $Sigma^{-1} O_K subseteq O_{K,S}$. The reverse inclusion is not obvious: it uses the fact that the class group of $O_K$ is torsion (because finite) — see here. In fact the argument is similar as above: pick $x in O_{K,S}$. For every $p in S$ such that $v_{p}(x) < 0$, write $p^h = b_{p} O_K$ as an principal ideal. Then
$b := prodlimits_{v_{p}(x) < 0} b_{p}^{-v_{p}(x)}$ is an element of $O_K$ only divisible by primes in $S$, so that $b in Sigma$.
Finally, $bx in O_K$, so we conclude that $x in Sigma^{-1} O_K$ as desired.
[Here is a wrong proof that $O_{K,S} subset Sigma^{-1} O_K$ : if $x in O_{K,S}$ is non-zero, then fix $b_p in p^{-v_p(x)}$ for every prime $p in S$ such that $v_p(x)<0$. Then $b = prod_p b_p in Sigma$ and $xb in O_K$, so that $x in Sigma^{-1} O_K$. The problem is that $b$ may have prime factors outside ${ p mid a_i }$, so that $b in Sigma$ is not true].
See also here, or here about relation between overrings and localizations. Moreover, remark 2.1 here might be of interest (but there is no proof there).
Finally, more general statements are given in theorem 2 here, mostly proven in Fossum, The Divisor Class Group of a Krull Domain.
$$ $$
B) Once you know that $O_{K,S} cong Sigma^{-1}O_K$, you can apply proposition 3.11.1) in Atiyah, MacDonald, Introduction to Commutative Algebra, to conclude that any ideal of $O_{K,S}$ is indeed an extended ideal $J O_{K,S}$, where $J$ is an ideal of $O_K$. (Moreover, any localization of a Dedekind domain is still Dedekind, see here).
$$ $$
C) You can also look at prop. 4.19 here, but notice that they define $O_{K,S}$ as a localization.
$$ $$
2) Here we are considering the multiplicative set
$Sigma = O_K setminus bigcuplimits_{p in mathrm{Spec}(O_K) setminus S} p.$
If instead you consider $Sigma' = O_K setminus p$ for a single fixed prime $p$ of $O_K$, you also get that $(Sigma')^{-1}O_K = (O_K)_{p}$ is a PID : it is even a DVR.
This is a result similar to the claim above, but which is in fact quite unrelated, since $Sigma$ deals with the cofinite set $mathrm{Spec}(O_K) setminus S$ of primes, while $Sigma'$ deals with the finite set ${p}$...
answered yesterday
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answered yesterday
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answered yesterday
Watson
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answered yesterday
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