Mixing
A step in the proof of a version of Kummer's theorem
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In a book I am reading (Cox, Primes), one of the exercises asks to prove the following version of Kummer's Theorem regarding factorization of primes in an extension:
Suppose $L/K$ is Galois, $L=K(alpha)$, $alphainmathcal{O}_L$, $f(x)$ the monic minimal polynomial of $alpha$ over $K$ (thus $f(x)inmathcal{O}_K[x]$). Suppose $mathfrak{p}$ is prime in $mathcal{O}_K$ and that $f(x)$ is separable modulo $mathfrak{p}$, say
$$f(x)equiv f_1(x)cdots f_g(x)mod{mathfrak{p}}$$
where the $f_i$ are distinct and irreducible. Then all of the $f_i$ have the same degree, $mathfrak{p}$ is unramified in $L$, and the inertial degree of each factor is the degree of the $f_i$.
(There is actually more detail, but this is all I need for my question.)
Following the order in the exercise, one first shows that if $mathfrak{P}$ is an ideal over $mathfrak{p}$, then $f_i(alpha)inmathfrak{P}$ for some $i$ (assume $i=1$). This is pretty straightforward.
The next step is: If $sigmain D_{mathfrak{P}}$, the decomposition group of $mathfrak{P}$, then $f_1(sigma(alpha)) = sigma(f_1(alpha))inmathfrak{P}$. Separability then implies that $text{deg}(f_1(x))ge |D_{mathfrak{P}}|$.
I don't see how to prove the degree statement. Presumably the idea is to show that the $sigma(alpha)$ are incongruent modulo $mathfrak{P}$, thus providing at least $|D_{mathfrak{P}}|$ roots of $f_1$ in $mathcal{O}_L/mathfrak{P}$. But as the answer below points out, this is in general not true (and is in fact equivalent to what we are eventually trying to prove, that $mathfrak{p}$ is unramified in $L$).
algebraic-number-theory
asked Jan 11 at 16:34
rogerlrogerl
17.8k22747
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add a comment |
$begingroup$
In a book I am reading (Cox, Primes), one of the exercises asks to prove the following version of Kummer's Theorem regarding factorization of primes in an extension:
Suppose $L/K$ is Galois, $L=K(alpha)$, $alphainmathcal{O}_L$, $f(x)$ the monic minimal polynomial of $alpha$ over $K$ (thus $f(x)inmathcal{O}_K[x]$). Suppose $mathfrak{p}$ is prime in $mathcal{O}_K$ and that $f(x)$ is separable modulo $mathfrak{p}$, say
$$f(x)equiv f_1(x)cdots f_g(x)mod{mathfrak{p}}$$
where the $f_i$ are distinct and irreducible. Then all of the $f_i$ have the same degree, $mathfrak{p}$ is unramified in $L$, and the inertial degree of each factor is the degree of the $f_i$.
(There is actually more detail, but this is all I need for my question.)
Following the order in the exercise, one first shows that if $mathfrak{P}$ is an ideal over $mathfrak{p}$, then $f_i(alpha)inmathfrak{P}$ for some $i$ (assume $i=1$). This is pretty straightforward.
The next step is: If $sigmain D_{mathfrak{P}}$, the decomposition group of $mathfrak{P}$, then $f_1(sigma(alpha)) = sigma(f_1(alpha))inmathfrak{P}$. Separability then implies that $text{deg}(f_1(x))ge |D_{mathfrak{P}}|$.
I don't see how to prove the degree statement. Presumably the idea is to show that the $sigma(alpha)$ are incongruent modulo $mathfrak{P}$, thus providing at least $|D_{mathfrak{P}}|$ roots of $f_1$ in $mathcal{O}_L/mathfrak{P}$. But as the answer below points out, this is in general not true (and is in fact equivalent to what we are eventually trying to prove, that $mathfrak{p}$ is unramified in $L$).
algebraic-number-theory
asked Jan 11 at 16:34
rogerlrogerl
17.8k22747
$endgroup$
add a comment |
$begingroup$
In a book I am reading (Cox, Primes), one of the exercises asks to prove the following version of Kummer's Theorem regarding factorization of primes in an extension:
Suppose $L/K$ is Galois, $L=K(alpha)$, $alphainmathcal{O}_L$, $f(x)$ the monic minimal polynomial of $alpha$ over $K$ (thus $f(x)inmathcal{O}_K[x]$). Suppose $mathfrak{p}$ is prime in $mathcal{O}_K$ and that $f(x)$ is separable modulo $mathfrak{p}$, say
$$f(x)equiv f_1(x)cdots f_g(x)mod{mathfrak{p}}$$
where the $f_i$ are distinct and irreducible. Then all of the $f_i$ have the same degree, $mathfrak{p}$ is unramified in $L$, and the inertial degree of each factor is the degree of the $f_i$.
(There is actually more detail, but this is all I need for my question.)
Following the order in the exercise, one first shows that if $mathfrak{P}$ is an ideal over $mathfrak{p}$, then $f_i(alpha)inmathfrak{P}$ for some $i$ (assume $i=1$). This is pretty straightforward.
The next step is: If $sigmain D_{mathfrak{P}}$, the decomposition group of $mathfrak{P}$, then $f_1(sigma(alpha)) = sigma(f_1(alpha))inmathfrak{P}$. Separability then implies that $text{deg}(f_1(x))ge |D_{mathfrak{P}}|$.
I don't see how to prove the degree statement. Presumably the idea is to show that the $sigma(alpha)$ are incongruent modulo $mathfrak{P}$, thus providing at least $|D_{mathfrak{P}}|$ roots of $f_1$ in $mathcal{O}_L/mathfrak{P}$. But as the answer below points out, this is in general not true (and is in fact equivalent to what we are eventually trying to prove, that $mathfrak{p}$ is unramified in $L$).
algebraic-number-theory
asked Jan 11 at 16:34
rogerlrogerl
17.8k22747
$endgroup$
In a book I am reading (Cox, Primes), one of the exercises asks to prove the following version of Kummer's Theorem regarding factorization of primes in an extension:
Suppose $L/K$ is Galois, $L=K(alpha)$, $alphainmathcal{O}_L$, $f(x)$ the monic minimal polynomial of $alpha$ over $K$ (thus $f(x)inmathcal{O}_K[x]$). Suppose $mathfrak{p}$ is prime in $mathcal{O}_K$ and that $f(x)$ is separable modulo $mathfrak{p}$, say
$$f(x)equiv f_1(x)cdots f_g(x)mod{mathfrak{p}}$$
where the $f_i$ are distinct and irreducible. Then all of the $f_i$ have the same degree, $mathfrak{p}$ is unramified in $L$, and the inertial degree of each factor is the degree of the $f_i$.
(There is actually more detail, but this is all I need for my question.)
Following the order in the exercise, one first shows that if $mathfrak{P}$ is an ideal over $mathfrak{p}$, then $f_i(alpha)inmathfrak{P}$ for some $i$ (assume $i=1$). This is pretty straightforward.
The next step is: If $sigmain D_{mathfrak{P}}$, the decomposition group of $mathfrak{P}$, then $f_1(sigma(alpha)) = sigma(f_1(alpha))inmathfrak{P}$. Separability then implies that $text{deg}(f_1(x))ge |D_{mathfrak{P}}|$.
I don't see how to prove the degree statement. Presumably the idea is to show that the $sigma(alpha)$ are incongruent modulo $mathfrak{P}$, thus providing at least $|D_{mathfrak{P}}|$ roots of $f_1$ in $mathcal{O}_L/mathfrak{P}$. But as the answer below points out, this is in general not true (and is in fact equivalent to what we are eventually trying to prove, that $mathfrak{p}$ is unramified in $L$).
algebraic-number-theory
algebraic-number-theory
asked Jan 11 at 16:34
rogerlrogerl
17.8k22747
asked Jan 11 at 16:34
rogerlrogerl
17.8k22747
edited Jan 12 at 14:39
rogerl
asked Jan 11 at 16:34
rogerlrogerl
17.8k22747
asked Jan 11 at 16:34
rogerlrogerl
17.8k22747
asked Jan 11 at 16:34
rogerlrogerl
17.8k22747
17.8k22747
add a comment |
add a comment |
1 Answer
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No two $sigma(alpha)$ are equal does NOT imply that no two are congruent modulo $mathfrak{P}$ (in fact this discrepancy is precisely measured by the inertia group).
answered Jan 12 at 0:29
PigPig
251113
$endgroup$
$begingroup$
Sorry, I meant no two $f_1(sigma(alpha))$ are congruent modulo $mathfrak{P}$, since $f_1(sigma(alpha)) = sigma(f_1(alpha))$ and $sigmain D_{mathfrak{P}}$. Edited.
$endgroup$
– rogerl
Jan 12 at 1:13
$begingroup$
Hey - sorry for late reply. In your particular situation, $sigma(a)$'s cannot be congruent modulo $mathfrak{P}$, because such mod $mathfrak{P}$-equality gives rise to double roots of $f(x) mod mathfrak{P}$, contradicting separability.
$endgroup$
– Pig
Jan 14 at 1:40
$begingroup$
Sorry to be so dim, but why does having double roots mod $mathfrak{P}$ contradict separability (which relates to having multiple roots mod $mathfrak{p}$)?
$endgroup$
– rogerl
Jan 14 at 15:21
$begingroup$
(Not dim at all) - separability is independent of field extension (since it's about $f(x)$ and $f'(x)$ being relatively prime), so it's the same to ask for separability in $O_K/mathfrak{p}$ vs $O_L/mathfrak{P}$
$endgroup$
– Pig
Jan 14 at 15:24
$begingroup$
Right. Thanks. I'd like to accept this answer, but you should probably rework it to contain the actual answer, which is currently in the comments!
$endgroup$
– rogerl
Jan 14 at 15:47
add a comment |
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$begingroup$
No two $sigma(alpha)$ are equal does NOT imply that no two are congruent modulo $mathfrak{P}$ (in fact this discrepancy is precisely measured by the inertia group).
answered Jan 12 at 0:29
PigPig
251113
$endgroup$
$begingroup$
Sorry, I meant no two $f_1(sigma(alpha))$ are congruent modulo $mathfrak{P}$, since $f_1(sigma(alpha)) = sigma(f_1(alpha))$ and $sigmain D_{mathfrak{P}}$. Edited.
$endgroup$
– rogerl
Jan 12 at 1:13
$begingroup$
Hey - sorry for late reply. In your particular situation, $sigma(a)$'s cannot be congruent modulo $mathfrak{P}$, because such mod $mathfrak{P}$-equality gives rise to double roots of $f(x) mod mathfrak{P}$, contradicting separability.
$endgroup$
– Pig
Jan 14 at 1:40
$begingroup$
Sorry to be so dim, but why does having double roots mod $mathfrak{P}$ contradict separability (which relates to having multiple roots mod $mathfrak{p}$)?
$endgroup$
– rogerl
Jan 14 at 15:21
$begingroup$
(Not dim at all) - separability is independent of field extension (since it's about $f(x)$ and $f'(x)$ being relatively prime), so it's the same to ask for separability in $O_K/mathfrak{p}$ vs $O_L/mathfrak{P}$
$endgroup$
– Pig
Jan 14 at 15:24
$begingroup$
Right. Thanks. I'd like to accept this answer, but you should probably rework it to contain the actual answer, which is currently in the comments!
$endgroup$
– rogerl
Jan 14 at 15:47
add a comment |
$begingroup$
No two $sigma(alpha)$ are equal does NOT imply that no two are congruent modulo $mathfrak{P}$ (in fact this discrepancy is precisely measured by the inertia group).
answered Jan 12 at 0:29
PigPig
251113
$endgroup$
$begingroup$
Sorry, I meant no two $f_1(sigma(alpha))$ are congruent modulo $mathfrak{P}$, since $f_1(sigma(alpha)) = sigma(f_1(alpha))$ and $sigmain D_{mathfrak{P}}$. Edited.
$endgroup$
– rogerl
Jan 12 at 1:13
$begingroup$
Hey - sorry for late reply. In your particular situation, $sigma(a)$'s cannot be congruent modulo $mathfrak{P}$, because such mod $mathfrak{P}$-equality gives rise to double roots of $f(x) mod mathfrak{P}$, contradicting separability.
$endgroup$
– Pig
Jan 14 at 1:40
$begingroup$
Sorry to be so dim, but why does having double roots mod $mathfrak{P}$ contradict separability (which relates to having multiple roots mod $mathfrak{p}$)?
$endgroup$
– rogerl
Jan 14 at 15:21
$begingroup$
(Not dim at all) - separability is independent of field extension (since it's about $f(x)$ and $f'(x)$ being relatively prime), so it's the same to ask for separability in $O_K/mathfrak{p}$ vs $O_L/mathfrak{P}$
$endgroup$
– Pig
Jan 14 at 15:24
$begingroup$
Right. Thanks. I'd like to accept this answer, but you should probably rework it to contain the actual answer, which is currently in the comments!
$endgroup$
– rogerl
Jan 14 at 15:47
add a comment |
$begingroup$
No two $sigma(alpha)$ are equal does NOT imply that no two are congruent modulo $mathfrak{P}$ (in fact this discrepancy is precisely measured by the inertia group).
answered Jan 12 at 0:29
PigPig
251113
$endgroup$
No two $sigma(alpha)$ are equal does NOT imply that no two are congruent modulo $mathfrak{P}$ (in fact this discrepancy is precisely measured by the inertia group).
answered Jan 12 at 0:29
PigPig
251113
answered Jan 12 at 0:29
PigPig
251113
answered Jan 12 at 0:29
PigPig
251113
answered Jan 12 at 0:29
PigPig
251113
251113
$begingroup$
Sorry, I meant no two $f_1(sigma(alpha))$ are congruent modulo $mathfrak{P}$, since $f_1(sigma(alpha)) = sigma(f_1(alpha))$ and $sigmain D_{mathfrak{P}}$. Edited.
$endgroup$
– rogerl
Jan 12 at 1:13
$begingroup$
Hey - sorry for late reply. In your particular situation, $sigma(a)$'s cannot be congruent modulo $mathfrak{P}$, because such mod $mathfrak{P}$-equality gives rise to double roots of $f(x) mod mathfrak{P}$, contradicting separability.
$endgroup$
– Pig
Jan 14 at 1:40
$begingroup$
Sorry to be so dim, but why does having double roots mod $mathfrak{P}$ contradict separability (which relates to having multiple roots mod $mathfrak{p}$)?
$endgroup$
– rogerl
Jan 14 at 15:21
$begingroup$
(Not dim at all) - separability is independent of field extension (since it's about $f(x)$ and $f'(x)$ being relatively prime), so it's the same to ask for separability in $O_K/mathfrak{p}$ vs $O_L/mathfrak{P}$
$endgroup$
– Pig
Jan 14 at 15:24
$begingroup$
Right. Thanks. I'd like to accept this answer, but you should probably rework it to contain the actual answer, which is currently in the comments!
$endgroup$
– rogerl
Jan 14 at 15:47
add a comment |
$begingroup$
Sorry, I meant no two $f_1(sigma(alpha))$ are congruent modulo $mathfrak{P}$, since $f_1(sigma(alpha)) = sigma(f_1(alpha))$ and $sigmain D_{mathfrak{P}}$. Edited.
$endgroup$
– rogerl
Jan 12 at 1:13
$begingroup$
Hey - sorry for late reply. In your particular situation, $sigma(a)$'s cannot be congruent modulo $mathfrak{P}$, because such mod $mathfrak{P}$-equality gives rise to double roots of $f(x) mod mathfrak{P}$, contradicting separability.
$endgroup$
– Pig
Jan 14 at 1:40
$begingroup$
Sorry to be so dim, but why does having double roots mod $mathfrak{P}$ contradict separability (which relates to having multiple roots mod $mathfrak{p}$)?
$endgroup$
– rogerl
Jan 14 at 15:21
$begingroup$
(Not dim at all) - separability is independent of field extension (since it's about $f(x)$ and $f'(x)$ being relatively prime), so it's the same to ask for separability in $O_K/mathfrak{p}$ vs $O_L/mathfrak{P}$
$endgroup$
– Pig
Jan 14 at 15:24
$begingroup$
Right. Thanks. I'd like to accept this answer, but you should probably rework it to contain the actual answer, which is currently in the comments!
$endgroup$
– rogerl
Jan 14 at 15:47
$begingroup$
Sorry, I meant no two $f_1(sigma(alpha))$ are congruent modulo $mathfrak{P}$, since $f_1(sigma(alpha)) = sigma(f_1(alpha))$ and $sigmain D_{mathfrak{P}}$. Edited.
$endgroup$
– rogerl
Jan 12 at 1:13
$begingroup$
Sorry, I meant no two $f_1(sigma(alpha))$ are congruent modulo $mathfrak{P}$, since $f_1(sigma(alpha)) = sigma(f_1(alpha))$ and $sigmain D_{mathfrak{P}}$. Edited.
$endgroup$
– rogerl
Jan 12 at 1:13
$begingroup$
Hey - sorry for late reply. In your particular situation, $sigma(a)$'s cannot be congruent modulo $mathfrak{P}$, because such mod $mathfrak{P}$-equality gives rise to double roots of $f(x) mod mathfrak{P}$, contradicting separability.
$endgroup$
– Pig
Jan 14 at 1:40
$begingroup$
Hey - sorry for late reply. In your particular situation, $sigma(a)$'s cannot be congruent modulo $mathfrak{P}$, because such mod $mathfrak{P}$-equality gives rise to double roots of $f(x) mod mathfrak{P}$, contradicting separability.
$endgroup$
– Pig
Jan 14 at 1:40
$begingroup$
Sorry to be so dim, but why does having double roots mod $mathfrak{P}$ contradict separability (which relates to having multiple roots mod $mathfrak{p}$)?
$endgroup$
– rogerl
Jan 14 at 15:21
$begingroup$
Sorry to be so dim, but why does having double roots mod $mathfrak{P}$ contradict separability (which relates to having multiple roots mod $mathfrak{p}$)?
$endgroup$
– rogerl
Jan 14 at 15:21
$begingroup$
(Not dim at all) - separability is independent of field extension (since it's about $f(x)$ and $f'(x)$ being relatively prime), so it's the same to ask for separability in $O_K/mathfrak{p}$ vs $O_L/mathfrak{P}$
$endgroup$
– Pig
Jan 14 at 15:24
$begingroup$
(Not dim at all) - separability is independent of field extension (since it's about $f(x)$ and $f'(x)$ being relatively prime), so it's the same to ask for separability in $O_K/mathfrak{p}$ vs $O_L/mathfrak{P}$
$endgroup$
– Pig
Jan 14 at 15:24
$begingroup$
Right. Thanks. I'd like to accept this answer, but you should probably rework it to contain the actual answer, which is currently in the comments!
$endgroup$
– rogerl
Jan 14 at 15:47
$begingroup$
Right. Thanks. I'd like to accept this answer, but you should probably rework it to contain the actual answer, which is currently in the comments!
$endgroup$
– rogerl
Jan 14 at 15:47
add a comment |
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