weaker upper bound of class number of O_k
Clash Royale CLAN TAG#URR8PPP
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![statement[![2]](https://i.stack.imgur.com/xL22l.png)
![statement and the proof[![1]](https://i.stack.imgur.com/RBSXm.png)
can anybody explain how the proof of this corollary is concluded?
i understood the proof right before the underlined sentence but still do not get how the argument before that leads to the underlined sentence. thank you. (uploaded 2 photos)
number-theory algebraic-geometry arithmetic ideal-class-group
edited Aug 30 at 18:36
awllower
9,93642471
asked Aug 30 at 8:37
Kento
161
add a comment |Â
up vote
1
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can anybody explain how the proof of this corollary is concluded?
i understood the proof right before the underlined sentence but still do not get how the argument before that leads to the underlined sentence. thank you. (uploaded 2 photos)
number-theory algebraic-geometry arithmetic ideal-class-group
edited Aug 30 at 18:36
awllower
9,93642471
asked Aug 30 at 8:37
Kento
161
2
I think the idea is as follows. We have $j = q_1^a_1 cdots q_r^a_r$. An ideal of order $j$ is going to factor as a product of prime ideals above $q_1$ through $q_s$. In total, you expect at most $a_1 + cdots + a_r$ such prime ideals. Moreover, how many primes are there above a given $q_i$? At most $n$, because $K$ has degree $n$ over $mathbbQ$. Thus in the worst-case scenario you have make $sum_i a_i$ choices out of $n$ possibilities.
– Sofie Verbeek
Aug 30 at 13:59
thank you very much. couldnt notice that the ideal I is factored into a product of prime ideals above q_i s.
– Kento
Sep 1 at 1:01
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
can anybody explain how the proof of this corollary is concluded?
i understood the proof right before the underlined sentence but still do not get how the argument before that leads to the underlined sentence. thank you. (uploaded 2 photos)
number-theory algebraic-geometry arithmetic ideal-class-group
edited Aug 30 at 18:36
awllower
9,93642471
asked Aug 30 at 8:37
Kento
161
can anybody explain how the proof of this corollary is concluded?
i understood the proof right before the underlined sentence but still do not get how the argument before that leads to the underlined sentence. thank you. (uploaded 2 photos)
number-theory algebraic-geometry arithmetic ideal-class-group
number-theory algebraic-geometry arithmetic ideal-class-group
edited Aug 30 at 18:36
awllower
9,93642471
asked Aug 30 at 8:37
Kento
161
edited Aug 30 at 18:36
awllower
9,93642471
asked Aug 30 at 8:37
Kento
161
edited Aug 30 at 18:36
awllower
9,93642471
edited Aug 30 at 18:36
awllower
9,93642471
edited Aug 30 at 18:36
awllower
9,93642471
9,93642471
asked Aug 30 at 8:37
Kento
161
asked Aug 30 at 8:37
Kento
161
asked Aug 30 at 8:37
Kento
161
161
2
I think the idea is as follows. We have $j = q_1^a_1 cdots q_r^a_r$. An ideal of order $j$ is going to factor as a product of prime ideals above $q_1$ through $q_s$. In total, you expect at most $a_1 + cdots + a_r$ such prime ideals. Moreover, how many primes are there above a given $q_i$? At most $n$, because $K$ has degree $n$ over $mathbbQ$. Thus in the worst-case scenario you have make $sum_i a_i$ choices out of $n$ possibilities.
– Sofie Verbeek
Aug 30 at 13:59
thank you very much. couldnt notice that the ideal I is factored into a product of prime ideals above q_i s.
– Kento
Sep 1 at 1:01
add a comment |Â
2
I think the idea is as follows. We have $j = q_1^a_1 cdots q_r^a_r$. An ideal of order $j$ is going to factor as a product of prime ideals above $q_1$ through $q_s$. In total, you expect at most $a_1 + cdots + a_r$ such prime ideals. Moreover, how many primes are there above a given $q_i$? At most $n$, because $K$ has degree $n$ over $mathbbQ$. Thus in the worst-case scenario you have make $sum_i a_i$ choices out of $n$ possibilities.
– Sofie Verbeek
Aug 30 at 13:59
thank you very much. couldnt notice that the ideal I is factored into a product of prime ideals above q_i s.
– Kento
Sep 1 at 1:01
2
2
I think the idea is as follows. We have $j = q_1^a_1 cdots q_r^a_r$. An ideal of order $j$ is going to factor as a product of prime ideals above $q_1$ through $q_s$. In total, you expect at most $a_1 + cdots + a_r$ such prime ideals. Moreover, how many primes are there above a given $q_i$? At most $n$, because $K$ has degree $n$ over $mathbbQ$. Thus in the worst-case scenario you have make $sum_i a_i$ choices out of $n$ possibilities.
– Sofie Verbeek
Aug 30 at 13:59
I think the idea is as follows. We have $j = q_1^a_1 cdots q_r^a_r$. An ideal of order $j$ is going to factor as a product of prime ideals above $q_1$ through $q_s$. In total, you expect at most $a_1 + cdots + a_r$ such prime ideals. Moreover, how many primes are there above a given $q_i$? At most $n$, because $K$ has degree $n$ over $mathbbQ$. Thus in the worst-case scenario you have make $sum_i a_i$ choices out of $n$ possibilities.
– Sofie Verbeek
Aug 30 at 13:59
thank you very much. couldnt notice that the ideal I is factored into a product of prime ideals above q_i s.
– Kento
Sep 1 at 1:01
thank you very much. couldnt notice that the ideal I is factored into a product of prime ideals above q_i s.
– Kento
Sep 1 at 1:01
add a comment |Â
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2
I think the idea is as follows. We have $j = q_1^a_1 cdots q_r^a_r$. An ideal of order $j$ is going to factor as a product of prime ideals above $q_1$ through $q_s$. In total, you expect at most $a_1 + cdots + a_r$ such prime ideals. Moreover, how many primes are there above a given $q_i$? At most $n$, because $K$ has degree $n$ over $mathbbQ$. Thus in the worst-case scenario you have make $sum_i a_i$ choices out of $n$ possibilities.
– Sofie Verbeek
Aug 30 at 13:59
thank you very much. couldnt notice that the ideal I is factored into a product of prime ideals above q_i s.
– Kento
Sep 1 at 1:01