Neukirch's interpretation of unit group and class groupMotivation behind the definition of ideal class groupRay class field and ring class fieldGeneralized class group of $mathbb Q(sqrt-5)$Artin reciprocity theorem for Hilbert class fieldComputing ideal class group by other means than the Minkowski bound?Surjection of idèle class group onto class group of a number field $K$?Example of principal ideals in the Class GroupClass number 1 equal to UFDConfusion about the class groupThere exist an integral ideal prime to a given nonzero integral ideal
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Neukirch's interpretation of unit group and class group
Motivation behind the definition of ideal class groupRay class field and ring class fieldGeneralized class group of $mathbb Q(sqrt-5)$Artin reciprocity theorem for Hilbert class fieldComputing ideal class group by other means than the Minkowski bound?Surjection of idèle class group onto class group of a number field $K$?Example of principal ideals in the Class GroupClass number 1 equal to UFDConfusion about the class groupThere exist an integral ideal prime to a given nonzero integral ideal
$begingroup$
In page 22 from Neukirch's Algebraic Number Theory, he defines the class group $Cl_K$ of a number field $K$ to be the quotient of group of fractional ideals $J_K$ by the subgroup of principal ideals $P_K$.
He shows the following exact sequence:
$$1to mathcalO^*to K^*to J_Kto Cl_Kto 1$$
Then he makes the following comment: "[...] the class group measures the expansion that takes place when we pass from numbers to ideals, whereas the unit group measures the contraction in the same process".
Two questions: 1) What does he mean by "expansion" and "contraction"? 2) How is this exact sequence illuminating in any sense?
number-theory algebraic-number-theory ideal-class-group
$endgroup$
add a comment |
$begingroup$
In page 22 from Neukirch's Algebraic Number Theory, he defines the class group $Cl_K$ of a number field $K$ to be the quotient of group of fractional ideals $J_K$ by the subgroup of principal ideals $P_K$.
He shows the following exact sequence:
$$1to mathcalO^*to K^*to J_Kto Cl_Kto 1$$
Then he makes the following comment: "[...] the class group measures the expansion that takes place when we pass from numbers to ideals, whereas the unit group measures the contraction in the same process".
Two questions: 1) What does he mean by "expansion" and "contraction"? 2) How is this exact sequence illuminating in any sense?
number-theory algebraic-number-theory ideal-class-group
$endgroup$
add a comment |
$begingroup$
In page 22 from Neukirch's Algebraic Number Theory, he defines the class group $Cl_K$ of a number field $K$ to be the quotient of group of fractional ideals $J_K$ by the subgroup of principal ideals $P_K$.
He shows the following exact sequence:
$$1to mathcalO^*to K^*to J_Kto Cl_Kto 1$$
Then he makes the following comment: "[...] the class group measures the expansion that takes place when we pass from numbers to ideals, whereas the unit group measures the contraction in the same process".
Two questions: 1) What does he mean by "expansion" and "contraction"? 2) How is this exact sequence illuminating in any sense?
number-theory algebraic-number-theory ideal-class-group
$endgroup$
In page 22 from Neukirch's Algebraic Number Theory, he defines the class group $Cl_K$ of a number field $K$ to be the quotient of group of fractional ideals $J_K$ by the subgroup of principal ideals $P_K$.
He shows the following exact sequence:
$$1to mathcalO^*to K^*to J_Kto Cl_Kto 1$$
Then he makes the following comment: "[...] the class group measures the expansion that takes place when we pass from numbers to ideals, whereas the unit group measures the contraction in the same process".
Two questions: 1) What does he mean by "expansion" and "contraction"? 2) How is this exact sequence illuminating in any sense?
number-theory algebraic-number-theory ideal-class-group
number-theory algebraic-number-theory ideal-class-group
asked Mar 11 at 20:33
rmdmc89rmdmc89
2,2931923
2,2931923
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Every element of $K^times$ generates a principal fractional ideal, and two elements of $K^times$ generate the same principal fractional ideal if they differ by a unit of $mathcalO_K$, i.e. by an element of $mathcalO^times_K$. In this sense the unit group $mathcalO_K^times$ measures the proportion of numbers to principal fractional ideals. Here numbers is understood to mean elements of $K^times$.
This characterization also shows that the group $P_K$ of principal fractional ideals of $K$ is precisely the cokernel of the map
$$1 longrightarrow mathcalO_K^times longrightarrow K^times.tag1$$
That is to say, the group $P_K$ fits into the short exact sequence
$$1 longrightarrow mathcalO_K^times longrightarrow K^times longrightarrow P_K longrightarrow 1.tag2$$
Of course not all fractional ideals are principal in general. That is to say, going from numbers to ideals we also gain new ideals, apart from those generated by single numbers (i.e. principal ideals). The proportion of fractional ideals to principal fractional ideals is measured by the quotient $J_K/P_K$. This quotient is called the class group of $K$, denoted by $operatornameCl_K$.
This can be phrased briefly by saying that the class group $operatornameCl_K$ of $K$ is the cokernel of the map
$$1 longrightarrow P_K longrightarrow J_K.tag3$$
Then by definition $operatornameCl_K$ fits into the short exact sequence
$$1 longrightarrow P_K longrightarrow J_K longrightarrow operatornameCl_K longrightarrow 1.tag4$$
Putting the two short exact sequence $(2)$ and $(4)$ together shows that $operatornameCl_K$ fits into the exact sequence
$$1 longrightarrow mathcalO_K^times longrightarrow K^times longrightarrow J_K longrightarrow operatornameCl_K longrightarrow 1.$$
This shows that the group of units and the class group are the kernel and cokernel, respectively, of the map $K^times longrightarrow J_K$. The former measures the how many numbers that contract to the same ideal, the latter measures the proportion of ideals to ideals coming from numbers.
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Every element of $K^times$ generates a principal fractional ideal, and two elements of $K^times$ generate the same principal fractional ideal if they differ by a unit of $mathcalO_K$, i.e. by an element of $mathcalO^times_K$. In this sense the unit group $mathcalO_K^times$ measures the proportion of numbers to principal fractional ideals. Here numbers is understood to mean elements of $K^times$.
This characterization also shows that the group $P_K$ of principal fractional ideals of $K$ is precisely the cokernel of the map
$$1 longrightarrow mathcalO_K^times longrightarrow K^times.tag1$$
That is to say, the group $P_K$ fits into the short exact sequence
$$1 longrightarrow mathcalO_K^times longrightarrow K^times longrightarrow P_K longrightarrow 1.tag2$$
Of course not all fractional ideals are principal in general. That is to say, going from numbers to ideals we also gain new ideals, apart from those generated by single numbers (i.e. principal ideals). The proportion of fractional ideals to principal fractional ideals is measured by the quotient $J_K/P_K$. This quotient is called the class group of $K$, denoted by $operatornameCl_K$.
This can be phrased briefly by saying that the class group $operatornameCl_K$ of $K$ is the cokernel of the map
$$1 longrightarrow P_K longrightarrow J_K.tag3$$
Then by definition $operatornameCl_K$ fits into the short exact sequence
$$1 longrightarrow P_K longrightarrow J_K longrightarrow operatornameCl_K longrightarrow 1.tag4$$
Putting the two short exact sequence $(2)$ and $(4)$ together shows that $operatornameCl_K$ fits into the exact sequence
$$1 longrightarrow mathcalO_K^times longrightarrow K^times longrightarrow J_K longrightarrow operatornameCl_K longrightarrow 1.$$
This shows that the group of units and the class group are the kernel and cokernel, respectively, of the map $K^times longrightarrow J_K$. The former measures the how many numbers that contract to the same ideal, the latter measures the proportion of ideals to ideals coming from numbers.
$endgroup$
add a comment |
$begingroup$
Every element of $K^times$ generates a principal fractional ideal, and two elements of $K^times$ generate the same principal fractional ideal if they differ by a unit of $mathcalO_K$, i.e. by an element of $mathcalO^times_K$. In this sense the unit group $mathcalO_K^times$ measures the proportion of numbers to principal fractional ideals. Here numbers is understood to mean elements of $K^times$.
This characterization also shows that the group $P_K$ of principal fractional ideals of $K$ is precisely the cokernel of the map
$$1 longrightarrow mathcalO_K^times longrightarrow K^times.tag1$$
That is to say, the group $P_K$ fits into the short exact sequence
$$1 longrightarrow mathcalO_K^times longrightarrow K^times longrightarrow P_K longrightarrow 1.tag2$$
Of course not all fractional ideals are principal in general. That is to say, going from numbers to ideals we also gain new ideals, apart from those generated by single numbers (i.e. principal ideals). The proportion of fractional ideals to principal fractional ideals is measured by the quotient $J_K/P_K$. This quotient is called the class group of $K$, denoted by $operatornameCl_K$.
This can be phrased briefly by saying that the class group $operatornameCl_K$ of $K$ is the cokernel of the map
$$1 longrightarrow P_K longrightarrow J_K.tag3$$
Then by definition $operatornameCl_K$ fits into the short exact sequence
$$1 longrightarrow P_K longrightarrow J_K longrightarrow operatornameCl_K longrightarrow 1.tag4$$
Putting the two short exact sequence $(2)$ and $(4)$ together shows that $operatornameCl_K$ fits into the exact sequence
$$1 longrightarrow mathcalO_K^times longrightarrow K^times longrightarrow J_K longrightarrow operatornameCl_K longrightarrow 1.$$
This shows that the group of units and the class group are the kernel and cokernel, respectively, of the map $K^times longrightarrow J_K$. The former measures the how many numbers that contract to the same ideal, the latter measures the proportion of ideals to ideals coming from numbers.
$endgroup$
add a comment |
$begingroup$
Every element of $K^times$ generates a principal fractional ideal, and two elements of $K^times$ generate the same principal fractional ideal if they differ by a unit of $mathcalO_K$, i.e. by an element of $mathcalO^times_K$. In this sense the unit group $mathcalO_K^times$ measures the proportion of numbers to principal fractional ideals. Here numbers is understood to mean elements of $K^times$.
This characterization also shows that the group $P_K$ of principal fractional ideals of $K$ is precisely the cokernel of the map
$$1 longrightarrow mathcalO_K^times longrightarrow K^times.tag1$$
That is to say, the group $P_K$ fits into the short exact sequence
$$1 longrightarrow mathcalO_K^times longrightarrow K^times longrightarrow P_K longrightarrow 1.tag2$$
Of course not all fractional ideals are principal in general. That is to say, going from numbers to ideals we also gain new ideals, apart from those generated by single numbers (i.e. principal ideals). The proportion of fractional ideals to principal fractional ideals is measured by the quotient $J_K/P_K$. This quotient is called the class group of $K$, denoted by $operatornameCl_K$.
This can be phrased briefly by saying that the class group $operatornameCl_K$ of $K$ is the cokernel of the map
$$1 longrightarrow P_K longrightarrow J_K.tag3$$
Then by definition $operatornameCl_K$ fits into the short exact sequence
$$1 longrightarrow P_K longrightarrow J_K longrightarrow operatornameCl_K longrightarrow 1.tag4$$
Putting the two short exact sequence $(2)$ and $(4)$ together shows that $operatornameCl_K$ fits into the exact sequence
$$1 longrightarrow mathcalO_K^times longrightarrow K^times longrightarrow J_K longrightarrow operatornameCl_K longrightarrow 1.$$
This shows that the group of units and the class group are the kernel and cokernel, respectively, of the map $K^times longrightarrow J_K$. The former measures the how many numbers that contract to the same ideal, the latter measures the proportion of ideals to ideals coming from numbers.
$endgroup$
Every element of $K^times$ generates a principal fractional ideal, and two elements of $K^times$ generate the same principal fractional ideal if they differ by a unit of $mathcalO_K$, i.e. by an element of $mathcalO^times_K$. In this sense the unit group $mathcalO_K^times$ measures the proportion of numbers to principal fractional ideals. Here numbers is understood to mean elements of $K^times$.
This characterization also shows that the group $P_K$ of principal fractional ideals of $K$ is precisely the cokernel of the map
$$1 longrightarrow mathcalO_K^times longrightarrow K^times.tag1$$
That is to say, the group $P_K$ fits into the short exact sequence
$$1 longrightarrow mathcalO_K^times longrightarrow K^times longrightarrow P_K longrightarrow 1.tag2$$
Of course not all fractional ideals are principal in general. That is to say, going from numbers to ideals we also gain new ideals, apart from those generated by single numbers (i.e. principal ideals). The proportion of fractional ideals to principal fractional ideals is measured by the quotient $J_K/P_K$. This quotient is called the class group of $K$, denoted by $operatornameCl_K$.
This can be phrased briefly by saying that the class group $operatornameCl_K$ of $K$ is the cokernel of the map
$$1 longrightarrow P_K longrightarrow J_K.tag3$$
Then by definition $operatornameCl_K$ fits into the short exact sequence
$$1 longrightarrow P_K longrightarrow J_K longrightarrow operatornameCl_K longrightarrow 1.tag4$$
Putting the two short exact sequence $(2)$ and $(4)$ together shows that $operatornameCl_K$ fits into the exact sequence
$$1 longrightarrow mathcalO_K^times longrightarrow K^times longrightarrow J_K longrightarrow operatornameCl_K longrightarrow 1.$$
This shows that the group of units and the class group are the kernel and cokernel, respectively, of the map $K^times longrightarrow J_K$. The former measures the how many numbers that contract to the same ideal, the latter measures the proportion of ideals to ideals coming from numbers.
edited Mar 11 at 21:28
answered Mar 11 at 20:51
ServaesServaes
28.3k34099
28.3k34099
add a comment |
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