Primes lying above a prime in a Galois extension
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Assuming $L$ is Galois over $K$, and that $O_L$ and $O_K$ are their respective rings of integers. Let $p$ be a prime ideal of $O_K$, is there a classification of the prime ideals laying above $p$ in $O_L$?
I understand that if $O_L$ is generated by a unique element over $O_K$, i.e. $O_L=O_K[theta]$, then there exists such a description, using the irreducible prime factors of $theta$'s integral dependence equation, after "dividing by $p$". However, as there is no particular reason for this to happen in general, I am curious as to whether there exists a more general classification.
field-theory galois-theory algebraic-number-theory galois-extensions
asked Dec 27 '18 at 20:01
kindasortakindasorta
9810
$endgroup$
add a comment |
$begingroup$
Assuming $L$ is Galois over $K$, and that $O_L$ and $O_K$ are their respective rings of integers. Let $p$ be a prime ideal of $O_K$, is there a classification of the prime ideals laying above $p$ in $O_L$?
I understand that if $O_L$ is generated by a unique element over $O_K$, i.e. $O_L=O_K[theta]$, then there exists such a description, using the irreducible prime factors of $theta$'s integral dependence equation, after "dividing by $p$". However, as there is no particular reason for this to happen in general, I am curious as to whether there exists a more general classification.
field-theory galois-theory algebraic-number-theory galois-extensions
asked Dec 27 '18 at 20:01
kindasortakindasorta
9810
$endgroup$
1
$begingroup$
The classification you are describing still holds even if $O_K[theta] subsetneq O_L$, provided that $p$ doesn't divide the conductor of $O_K[theta]$ inside $O_L$ (this should be sufficient to consider primes that don't divide the discriminant of $f$, see here) – this misses only finitely many primes. See Neukirch's book on ANT, I.8.3, or also math.stanford.edu/~conrad/154Page/handouts/dedekindcrit.pdf.
$endgroup$
– Watson
Dec 27 '18 at 20:51
add a comment |
$begingroup$
Assuming $L$ is Galois over $K$, and that $O_L$ and $O_K$ are their respective rings of integers. Let $p$ be a prime ideal of $O_K$, is there a classification of the prime ideals laying above $p$ in $O_L$?
I understand that if $O_L$ is generated by a unique element over $O_K$, i.e. $O_L=O_K[theta]$, then there exists such a description, using the irreducible prime factors of $theta$'s integral dependence equation, after "dividing by $p$". However, as there is no particular reason for this to happen in general, I am curious as to whether there exists a more general classification.
field-theory galois-theory algebraic-number-theory galois-extensions
asked Dec 27 '18 at 20:01
kindasortakindasorta
9810
$endgroup$
Assuming $L$ is Galois over $K$, and that $O_L$ and $O_K$ are their respective rings of integers. Let $p$ be a prime ideal of $O_K$, is there a classification of the prime ideals laying above $p$ in $O_L$?
I understand that if $O_L$ is generated by a unique element over $O_K$, i.e. $O_L=O_K[theta]$, then there exists such a description, using the irreducible prime factors of $theta$'s integral dependence equation, after "dividing by $p$". However, as there is no particular reason for this to happen in general, I am curious as to whether there exists a more general classification.
field-theory galois-theory algebraic-number-theory galois-extensions
field-theory galois-theory algebraic-number-theory galois-extensions
asked Dec 27 '18 at 20:01
kindasortakindasorta
9810
asked Dec 27 '18 at 20:01
kindasortakindasorta
9810
edited Dec 27 '18 at 20:28
kindasorta
asked Dec 27 '18 at 20:01
kindasortakindasorta
9810
asked Dec 27 '18 at 20:01
kindasortakindasorta
9810
asked Dec 27 '18 at 20:01
kindasortakindasorta
9810
9810
1
$begingroup$
The classification you are describing still holds even if $O_K[theta] subsetneq O_L$, provided that $p$ doesn't divide the conductor of $O_K[theta]$ inside $O_L$ (this should be sufficient to consider primes that don't divide the discriminant of $f$, see here) – this misses only finitely many primes. See Neukirch's book on ANT, I.8.3, or also math.stanford.edu/~conrad/154Page/handouts/dedekindcrit.pdf.
$endgroup$
– Watson
Dec 27 '18 at 20:51
add a comment |
1
$begingroup$
The classification you are describing still holds even if $O_K[theta] subsetneq O_L$, provided that $p$ doesn't divide the conductor of $O_K[theta]$ inside $O_L$ (this should be sufficient to consider primes that don't divide the discriminant of $f$, see here) – this misses only finitely many primes. See Neukirch's book on ANT, I.8.3, or also math.stanford.edu/~conrad/154Page/handouts/dedekindcrit.pdf.
$endgroup$
– Watson
Dec 27 '18 at 20:51
1
1
$begingroup$
The classification you are describing still holds even if $O_K[theta] subsetneq O_L$, provided that $p$ doesn't divide the conductor of $O_K[theta]$ inside $O_L$ (this should be sufficient to consider primes that don't divide the discriminant of $f$, see here) – this misses only finitely many primes. See Neukirch's book on ANT, I.8.3, or also math.stanford.edu/~conrad/154Page/handouts/dedekindcrit.pdf.
$endgroup$
– Watson
Dec 27 '18 at 20:51
$begingroup$
The classification you are describing still holds even if $O_K[theta] subsetneq O_L$, provided that $p$ doesn't divide the conductor of $O_K[theta]$ inside $O_L$ (this should be sufficient to consider primes that don't divide the discriminant of $f$, see here) – this misses only finitely many primes. See Neukirch's book on ANT, I.8.3, or also math.stanford.edu/~conrad/154Page/handouts/dedekindcrit.pdf.
$endgroup$
– Watson
Dec 27 '18 at 20:51
add a comment |
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$begingroup$
The classification you are describing still holds even if $O_K[theta] subsetneq O_L$, provided that $p$ doesn't divide the conductor of $O_K[theta]$ inside $O_L$ (this should be sufficient to consider primes that don't divide the discriminant of $f$, see here) – this misses only finitely many primes. See Neukirch's book on ANT, I.8.3, or also math.stanford.edu/~conrad/154Page/handouts/dedekindcrit.pdf.
$endgroup$
– Watson
Dec 27 '18 at 20:51